THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA LOS ANGELES A TREATISE ON DYNAMICS OF A PARTICLE. EonDon: C. J. CLAY AND SONS, CAMBEIDGE UNIVERSITY PEESS WAEEHOUSE, AYE MAEIA LANE. (BHasgofoi: 263, AEGYLE STREET. Eetpjig: F. A. BBOCKHAUS. gorfe: THE MACMILLAN COMPANY. Bcmbag : E. SEYMOUR HALE. A TEEATISE ON DYNAMICS OF A PARTICLE WITH NUMEROUS EXAMPLES BY EDWARD JOHN ROUTH, Sc.D., LL.D., M.A., F.R.S., &c., HON. FELLOW OF PETERH00SE, CAMBRIDGE ; FELLOW OF THE UNIVERSITY OF LONDON. CAMBRIDGE AT THE UNIVERSITY PRESS 1898 [All rights reserved.] Cambrtbge : PRINTED BY J. AND C. F. CLAY, AT THE UNIVERSITY PRESS. OA Mathematical Sciences Library PEEFACE. many questions which necessarily excite our interest and curiosity are discussed in the dynamics of a particle that this subject has always been a favourite one with students. How, for example, is it that by observing the motion of a pendulum we can tell the time of the rotation of the earth, or knowing this, how is it that we can deduce the latitude of the place ? Why does our earth travel round the sun in an ellipse and what would be the path if the law of gravitation were different ? Would any other law give a closed orbit so that our planet might (if undisturbed) repeat the same path continually? Is there a resisting medium which is slowly but continually bringing our orbit nearer to the sun ? What would be the path of a particle in a system of two centres of force ? When a comet passes close to a planet does it carry with it in its new orbit some tokens to prove its identity ? Such problems as these (which are merely examples) excite our curiosity at the very beginning of the subject. When we study the replies we find new objects of interest. Beginning at the elementary resolutions of the forces we are led on from one generalization to another. We presently arrive at Lagrange's general method, by which when a single function (worthily called after his great name) has been found we can write down, in any kind of coordinates, all the equations of motion cleared of unknown reactions. A little further on we find Jacobi's method 850120 VI PREFACE. by which the whole solution of a dynamical problem can be made to depend on a single integral. The last word has not yet been said on these problems. The student finds as he proceeds much left to discover and many new questions to ask. When we extend our studies so as to include the planetary perturbations and to take account of the finite size of the bodies the mathematical difficulties are much increased. In the dynamics of a particle we confine ourselves to simpler problems and easier mathematics. As the subject of dynamics is usually read early in the mathematical course, the student cannot be expected to master all its difficulties at once. In this treatise the parts intended for a first reading are printed in large type and the student is advised to pass over the other parts until they are referred to later on. The same problem may be attacked on many sides and we therefore have several different ways of finding a solution. In what follows the most elementary method has in general been put first, other solutions being given later on. For the sake of simplicity they have also generally been treated first in two dimensions. In these ways the difficulties of dynamics are separated from those of pure geometry and it is hoped that both difficulties may thus be more easily overcome. Some of the examples have been fully worked out, on others hints have been given. Many of these have been selected from the Tripos and College papers in order that they may the better indicate the recent directions of dynamical thought. I cannot conclude without thanking Mr Dickson of Peterhouse. He has kindly assisted me in correcting most of the proofs and has given material aid by his verifications and suggestions. EDWARD J. ROUTH. PETERHOUSE, July, 1898. CONTENTS. CHAPTER I. ELEMENTARY CONSIDERATIONS. AKT8. 1 30. Velocity and acceleration .... 31 38. Cartesian, polar and intrinsic components 39 40. Eelative motion 4145. Angular velocity 4648. Units of space and time .... 49 62. Laws of motion 63 67. Units of mass and force .... 68 72. Vis viva and work 73 79. The two solutions of the equations of motion 8091. Impulsive forces and impacts . 92 93. Motion of the centre of gravity 94. Examples on impacts . . . . PAGES 110 1113 1415 1516 1617 1723 2326 2627 2834 3541 4142 4246 CHAPTER II. RECTILINEAR MOTION. 95 103. Solution of the equation. Ambiguities .... 47 53 104 117. A heavy particle in vacuo and in a resisting medium. Bough chords 5360 118 122. Linear equation and harmonic motion .... 60 64 Vlll ARTS. 123136. CONTENTS. Centre of force. Discontinuity of friction, of resistance and of central forces 137 142. Small oscillations. Magnification 143 147. Chords of quickest descent. Smooth. Bough 148 150. Infinitesimal impulses 151 153. Theory of dimensions PAGES 6473 7376 7779 8082 8283 CHAPTER III. MOTION OF PROJECTILES. 154 161. Parabolic motion .... 162 167. Resistance varies as the velocity 168 180. Other laws of resistance. Special cases 8492 9295 95102 CHAPTER IV. CONSTRAINED MOTION IN TWO DIMENSIONS. 181190. The two resolutions. Work function .... 103110 191192. Bough curves . . . 110112 193196. Zero pressure 112115 197 198. Moving curves of constraint 115 117 199 203. Time of describing an arc. Subject of integration infinite 117 120 204 212. Motion in a cycloid. Smooth; rough; and a resisting medium 120126 213220. Motion in a circle. Coaxial circles 2 4F 2 (F^6p + (F+6) 2 ~(^6) 2t c being the distance between the centres of the generating circles, and l> the radius of the moving circle. [Math. Tripos.] 15. Acceleration. This word is used to express the rate at which the velocity is increasing. It may be either uniform or variable. If a, point move in such a manner that the increments of velocity gained in any equal times are the same in direction and equal in magnitude, the acceleration is said to be uniform. The increment of velocity in each unit of time measures the magnitude of the acceleration. 16. First, let the point move in a straight line. Let v be the velocity at any time t ; after a unit of time has elapsed, let v +f be the velocity. After a second unit of time the velocity must be v + 2/, because equal increments are gained in equal times. Hence after t 1 units of time the velocity has increased by f(t to)- If v be the velocity at the time t, we have v = v +f(t - t ). The quantity /is the acceleration. 17. If the point does not move in a straight line the explanation is only slightly altered. Let Oy represent the direction in which the constant increments of velocity are given to the point, and let Ox be the direction of motion at the time t = t . Let u , v<> be the components of the velocity in the directions of the axes Ox and Oy respectively at the time t . After a unit of time has elapsed the component of velocity parallel to Oy is v +f, but ART. 21.] ACCELERATION. 7 that parallel to Ox is unchanged because no velocity has been added in that direction. After t t units of time, the component of velocity parallel to Oy is v +f(t - t \ while that parallel to Ox is still u . If u, v are the components of velocity at the time t, we have U = U , V=V +f(t-t ). The magnitude of the acceleration is /, and its direction is Oy. 18. When the increments of velocity in equal times are unequal in magnitude, or not the same in direction, the accelera- tion is said to be variable. To obtain a measure we follow the method adopted to measure variable velocity. Acceleration when uniform is measured by the velocity generated in any unit of time. When variable, the acceleration at any instant is measured by the velocity which would be generated in the next unit of time if the acceleration had remained constant in magnitude during that interval and faced in direction. 19. To find the equations of motion of a point moving in a straight line with a variable acceleration f. Let v and v + dv be the velocities at the times t and t + dt. Assuming the principles of the differential calculus, dv being the increment in the time dt, it follows by a simple proportion that dv/dt is the velocity which would be added in a unit of time, if the acceleration had remained constant. Hence, by Art. 16, /= dvjdt. The argument is usually put into a more elementary form. Let Sv be the velocity generated in the time dt. Let f l , f 2 be the greatest and least accelerations of the particle during the interval dt. Then since the actual rate at which the velocity is increasing is always less than the one and greater than the other, the velocity added is less than J\8t and greater than / 2 5. In the limit /i and / 2 coin- cide and we have /= dvjdt. 20. Let the geometrical position of the point at the time t be determined by its distance s from a fixed point in the path. Let v be the velocity, / the acceleration, then _ds f_=0 l -6, we see by a simple resolution, that the resolved part of the space accele- ration of Q in the direction PQ is The resolved part perpendicular to PQ is = Sj - R sin + S cos . In the same way the resolved parts along and perpendicular to OP are R + R! cos - &! sin 0, S + R 1 sin + S 1 cos . M 2 /a and sin . w 2 /a, where = (u/6 - uja) t. Ex. 3. A point P describes a circle of 1 foot radius in 1 hour, and a point Q i describes a concentric circle of 4 feet radius in 14 hours, both points move in the <5ounter-clockwise direction ; show that the line joining them rotates in the counter- clockwise direction for a period of 43^ minutes followed by a period of 21^ minutes in the clockwise direction. [Coll. Exam. 1896.] Ex. 4. A circular wire of radius a moves in its own plane without rotation so that its centre has a simple harmonic motion of amplitude a (Art. 32): a bead ' moves on the wire uniformly, completing a circuit in the period of the simple harmonic motion, and being in the line of the motion of the centre when the centre is in its mean position and is moving in the direction towards the bead ; prove that ART. 43.] ANGULAR VELOCITY. 15 the acceleration of the bead is towards the centre of the simple harmonic motion and that its path is an ellipse of eccentricity (f ^/5 - f )^. [Coll. Exam. 1897.] Ex. 5. A railway passenger seated in one corner of the carriage looks out of the windows at the further end and observes that a star near the horizon is traversing these windows in the direction of the train's motion and that it is obscured by the partition between the corner windows on his own side of the carriage and the middle window while the train is moving through the seventh part of a mile. Prove that the train is on a curve the concavity of which is directed towards the star and which, if it be circular, has a radius of nearly three miles, the breadth of the carriage being seven feet and the breadth of the partition four inches. [Math. Tripos, I860.] 41. Angular velocity and acceleration. A rigid body is said to be turning round an axis OA when each point is describing a circle whose plane is perpen- dicular to OA and whose centre lies in OA. Let be the angle which the plane containing any point P of the body and the axis OA makes with some plane fixed in space and passing through OA. The rate at which the angle

/dt and the angular acceleration by d 2 0/jdt = dfajdt. If Q be any point of the body, r its distance from the axis OA and u=djdt be the angular velocity, the point Q is moving perpendicularly to the plane QOA with a velocity equal to wr. If the rotation continue only for a time dt the axis OA (by rotation about which the motion in that time can be constructed) is called the instantaneous axis and u is the instantaneous angular velocity. 42. An angular velocity w about an axis is geometrically represented by a length OA proportional to w measured along the axis. The direction of the rotation is determined by the convention used in Statics to indicate the direction of rotation of a couple. If OA be the direction in which the length is measured the rotation when positive, should appear to be in some standard direction to a spectator placed with his feet at A and head at B. This standard direction is often taken to be the direction of rotation of the hands of a clock. 43. Parallelogram of angular velocities. If two instantaneous angular velocities of a body are represented in magnitude and direction by two lengths OA, OB, the diagonal OC of the parallelogram constructed on OA, OB as sides is the resultant instantaneous axis of rotation, and its length represents the magnitude of the resultant angular velocity. Let Q be any point which at the time t lies in the plane AOB; r lf T 2 , the distances of Q from OA, OB, p its distance from OC. Let u^=OA, u 2 =OB, = OC. The velocity of Q due to the two rotations Wj, w 2 is u^ + wfa, while that due to the single rotation is Op. To prove that these are equal it is sufficient to notice that if OA, OB represented forces and OC the resultant, the equality merely asserts that the sum of the moments of OA, OB about Q is equal to that of the resultant OC. 16 VELOCITY AND ACCELERATION. [CHAP. I. Let v be the velocity of Q, then v = Wjfj + w.,r t = ftp. If Q lie on 00, /> = 0, and therefore every point of 00 is at rest. Hence OO is the resultant axis of rotation. Also since = v/p the angular velocity about the axis is Q. 44. The theorem of the parallelogram of angular accelerations follows from that of angular velocities, just as the parallelogram of linear accelerations follows from that of linear velocities. 45. The rule for compounding angular velocities being the same as that used in Statics to compound forces, we may interpret the limiting case when the inter- section is at infinity as we do the corresponding case in Statics. It is however simpler to deduce the result independently. Let the body have instantaneous angular velocities u, u', about two parallel axes OA, O'B distant a from each other. The resultant velocity of any point Q in the plane of OA, O'B and distant y and y + a from them respectively is where s is the space described, and v the velocity at the time t. This equation may be integrated in two ways. Taking the time t as the independent variable, we have mv-mv = /F 1 dt + fF 2 dt + (2), where v is the velocity at the time t , and the limits of integration are t to t. The forces F Jt F 2) &c. may not act during the whole time, thus F! might act from ti to ^ + a, F a might act from t z to 2 + y3 and so on. In such cases the limits of each integral should be from the time of beginning to the time of ending of the force. For the sake of conveniently using the equation we notice (what really follows at once from the second law) that each force F adds to the moving mass a momentum equal to fFdt, where tlie integration extends over the time of action of the force. This is called the time-integral of the force. The equation (2) is called the equation of momentum. 75. Taking the space s as the independent variable, we have (3). ART. 76.] THE TWO SOLUTIONS. 29 It follows that the increase of the kinetic energy of the mass moved is equal to the sum of the works of the several forces. Each force F communicates to the moving mass an amount of kinetic energy equal to jFds where the integration extends over the space described while F acts on the mass. This is called the space- integral of the force. The equation (3) is called sometimes the equation of vis viva and sometimes the equation of energy. If the velocity of the mass is the same at any two times, the momentum added on by some of the forces must be equal to that removed by other forces. If again the velocity is the same in any two positions, the work added on by some of the forces must be equal to that sub- tracted by other forces. In this way we obtain two equations to find the one quantity v. If the forces F l} F z , &c. are constant both the space and time- integrals can be at once found. We therefore use either or both the equations (2) and (3). If the forces are functions of either t or s, only one of the integrations can be immediately effected. We use the equations (2) or (3) according as the forces depend on the time or on the position of the particle. 76. When the system contains more than one par- ticle, their mutual actions may have to be taken into consideration. Suppose, for example, that two particles P, P', whose masses are m, m', are constrained to slide on the straight lines Ox, Ox', and are acted on by the forces F, F' in these directions. Let these be connected by a string of given length which passes over a smooth pulley C. The two equations of energy are \m O 2 -v it takes the simpler form \ (m + m') v 2 =g sin a (m - m') (s - a). These equations give s and v in terms of the time t. We notice that if m>m', the particle P descends along the rod Ox and finally draws P' up to 0. Ex. 2. Two small rings of masses HI, m' are moving on a smooth circular wire which is fixed with its plane vertical. They are connected by a straight weightless inextensible string. Prove that, as long as the string remains tight, its tension is ; , where 2a is the angle which the string when tight subtends at the m + m centre and 6 is the inclination of the string to the horizon. [Pemb. Coll. 1897.} Equate the tangential accelerations of the two particles. 32 THE EQUATIONS OF MOTION. [CHAP. I. Ex. 3. A bucket of mass 37 Ibs. is raised from the bottom of a shaft of depth h feet by means of a light cord which is wound on a wheel of mass m Ibs. The wheel is driven by a constant force which is applied tangentially at its rim for a certain time and then ceases. Prove that if the bucket just comes to rest at the top of the shaft, t seconds after the beginning of the motion, the greatest rate of working in foot-poundals per second is 2 ~-r- ---- jr . The mass of the wheel may be considered to be condensed in its rim. [Coll. Ex. 1896.] Let the force F act on the rim for a time t'. This force communicates a momentum Ft' to the system, which (since the system comes to rest after a time t) is equal to that removed by gravity in the whole ascent, therefore Ft'=Mgt. If *' is the space ascended in the time t', the force F communicates a work Fs\ which is equal to that removed by gravity in the whole ascent h, therefore Fa'=Mgh. Since the mass moved is M + m and F - Mg is the acting force we have also the two equations (M+ m) v' = (F- Mg) t', (M + m)8' = ^(F- Mg) t' 2 where v' is the velocity at the time t' (Art. 25). These four equations determine F, t, v', *'. The rate of adding work to the system is Fv (Art. 72), and this is greatest when v is greatest, i.e. when v = v'. The result follows without difficulty. Ex. 4. A train of mass m runs from rest at one station to stop at the next at a distance 1. The full speed is V and the average speed is v. The resistance at the rails when the brake is not applied is uV/lg of the weight of the train and when the brake is applied it is u'Vjlg of the weight of the train. The pull of the engine has one constant value when the train is starting and another when it runs at full speed. Prove that the average rate at which the engine works in starting the train is mF 2 (u + U)ll, where i = ? - ^ - ^ . [Coll. Ex. 1895.] There are three stages of the journey. During the first the engine pulls with force F, the acceleration is F /m - uVfl, and the velocity increases from zero to V. During the second stage the velocity is uniform and equal to V, the pull F' of the engine just balancing the resistance. During the third the engine stops working, the brake is applied and the acceleration is - u'V/l. Using the formulae of Art. 25, and remembering that the sum of the spaces in the three stages is /. while the average velocity is I divided by the sum of the times, we deduce F. The average rate of working is the quotient "work by time," Art. 72; during the first stage Ex. 5. The cage of a coal-pit is lowered for the first third of the shaft with a constant acceleration, for the next third it descends with xmiform velocity, and then a constant retarding force just brings it to rest as it reaches the bottom of the shaft. If the time of descent is equal to that taken by a particle in falling four times the whole depth, prove tiiat the pressure of the man inside on the bottom of the cage was at the beginning 23/48ths of his weight. [Coll. Ex. 1897.] The initial acceleration / is found to be 25 2 /, hence =T-T'- mg. d The vertical acceleration of m' is equal to that of m plus that due to the relative motion. Belatively to m it begins to describe a circle of radius b with a velocity v, the relative vertical acceleration is therefore v 2 /b, see Art. 39. Hence Solving these equations the result follows at once. Ex. 10. In the system of pulleys in which the string, passing round each pulley, has one end attached to a fixed beam and the other to the pulley next above, there is no "power" and no "weight." The n moveable pulleys are all of equal weight, they are smooth, and can all be treated as particles in calculating their motions. The string is without mass. Prove that the acceleration of the lowest pulley is 30/(2 + l). [Coll. Ex. 1896.] R. D. 3 34 THE EQUATIONS OF MOTION. [CHAP. I. The equation of momentum for the rth pulley counting downwards is md?y r ldt* =mg-2T r + T^ , where T^ and T n+1 , being the power and weight, are zero. Also the velocity of each pulley is half that of the one just above. Multiplying these equations by 1, 2, 2 s ... 2 1 *" 1 beginning at the lowest and adding the results the tensions disappear. Ex. 11. In the system of pulleys in which each string is attached to the weight, there are two pulleys, the weight of the moveable pulley being w, the power P and the weight IF. Prove that the acceleration of W is g W ~ . [Coll. Ex. 1897.] yr + W + W Ex. 12. A prism with axis horizontal and whose section by a plane perpen- dicular to it is a regular polygon ABCD... of 4n sides is fixed with the uppermost face AB horizontal, and n equal particles are placed at the middle points of AB, BG, &c. These are connected by a continuous string which passes over smooth pulleys at the corners B, C, &c. Assuming that the faces are smooth, prove that the initial acceleration is j- (cot -- - 1 ) . [Coll. Ex. 1897.] tT\t \ 471 J Ex. 13. Two equal particles are connected by a string one point of which is fixed and the particles are describing circles of radii a and b about this point with the same angular velocity so that the string is always straight. The string is suddenly released, prove that the tensions of the two portions are altered in the ratios (a + 6) : 2a and (a + b) : 2b. [Coll. Ex. 1895.] Before the release the tensions are mv-fja and mv/jb, where t; 1 /a=v 2 /fe = w. After the release the relative space velocity is v=v 1 + v 2 . The acceleration of each particle being Tjm, the relative acceleration is 2T/m. Since the relative path of either is a circle of radius r=a + b, the relative acceleration is v 2 /r. Equating these, the tension is mi> 2 /2r. The result follows. Ex. 14. A cubical box slides down a rough inclined plane, whose coefficient of friction is /*, two sides of the base being horizontal. If the box contain sufficient water just to cover the base of the vessel, prove that the volume of the water is \H times the internal volume of the vessel. [Coll. Ex. 1897.] The relative acceleration of a particle of water and the box must be perpendicular to the surface. 79. Linear and Angular Momentum. Let the momentum mv of any particle P of a system be represented in direction and magnitude (Art. 54) by a straight line PP'. Since velocities obey the parallelogram law, we may proceed as in Statics and replace the momentum PP' by three linear momenta at any assumed origin in the directions of the axes, and three couple momenta. Let the coordinates of the particle be x, y, z and the direction cosines be X, p., v. The three linear momenta being the resolved parts of mv are mv\, mvp, mw respectively. These are often called linear momenta. The three couple momenta are the moments of the momentum mv about the axes. We know by the corre- sponding theorem in Statics that these moments are x x y \ p V \ These are called the angular momenta about the axes. The linear momentum of a particle in any direction is the resolved part of the momentum in that direction. The angular momentum about a straight line is the moment of the momentum about that straight line. ART. 81.] IMPULSIVE FORCES. 35 Impulsive Forces. 80. Impulsive forces. In some cases the forces act only for a very short time, yet, being of great magnitude, produce perceptible effects. Let a force F act on a particle of mass m for a time T. Let v be the velocity at any time t less than T, and let V, F' be the velocities at the beginning and end of the interval T. We have T = , /. m at (V'-V)=\ T Fdt ........... (1). Jo Let the force F increase without limit while the duration T decreases without limit. The integral may have a finite limit, say P. The equation then becomes m(V'-V) = P ........................ (2). If v lt v 2 are the greatest and least velocities during the impact, the space described lies between v-fF and v 2 T, and both these are zero in the limit. The particle therefore has not had time to move, but its velocity has been changed from V to V. This sudden change of velocity is the distinguishing characteristic of an impulse. We may consider that a proper measure has been found for a force when from that measure we can deduce all the effects of the force. Since in the case of the limiting force the change of velocity is the only element to be determined we may measure such a force by the quantity P. When P is known, the change of velocity is given by (2). 81. An impulse or blow is the limit of a force whose magnitude is infinitely great and time of action infinitely small. A finite force F is measured by the momentum generated per unit of time. An impulse P is measured by the whole momentum generated during the whole time of action, that is, P=/Fdt. When the direction of the force F remains fixed in space during its time of action, the resolved part of P in any direction is also the limit of the resolved part of F. When the direction of F is not fixed in space, we resolve F into its components X, Y. The integrals of these, viz. X^jXdt, Y l =fYdt, are defined to be the components of the limiting impulse. 32 36 IMPULSIVE FORCES. [CHAP. I. Strictly speaking, there are no impulsive forces in nature, but there are some forces which are very great and which act only for a short time. The blow of a hammer is a force of this kind. Such forces should be treated as finite forces if the small displacements during the time of action cannot be neglected, and as impulses when these are imperceptible. 82. The general equations of impulsive motion follow from those of finite forces. If (u lt v^ are the Cartesian components of velocity we have, by Art. 73, where X, Y are the components of a finite force F. Let (u, y), (u, v) be the components of the velocity just before and just after the action of any impulse. Let X 1 =fXdt, Y 1 =fYdt be the components of the impulse, Art. 81. We then have by inte- gration, m(u' u) = Xi, m(v' v)=Y 1 . These equations may be summed up in the following working rule, / Res. Mom. \ _ f Res. Mom. \ _ /Resolved\ \after impulse/ \before impulse/ \ impulse /' 83. Elastic smooth bodies. When two spheres of any hard material impinge on each other they appear to separate almost immediately and a finite change of velocity is generated in each by the mutual action. Let the centres of gravity of the spheres be moving before impact in the same straight line with velocities u, v. After impact they will continue to move in the same straight line ; let u', v' be their velocities. Let ra, in' be the masses, R the action between them. The equations of motion are m(u'-u) = -R, m'(v'-v) = R ............ (1). These equations are not sufficient to determine the three quanti- ties u', v' and R. To obtain a third equation we must consider what takes place during the impact. Each of the balls is slightly compressed by the other, so that they are no longer perfect spheres. Each also in general tends to return to its original shape, so that there is a rebound. The ART. 83.] ELASTIC SMOOTH BODIES. 37 period of impact may therefore be divided into two parts. Firstly, the period of compression, during which the distance between the centres of gravity of the two bodies is diminishing and secondly, the period of restitution in which the distance is increasing. The first 'period terminates when the two centres of gravity have the same instantaneous velocity, the second when the bodies separate. The ratio of the magnitude of the action between the bodies during the period of restitution to that during compression is found to be different for bodies of different materials. If the bodies regain their original shapes very slowly the separation may take place before this occurs and then the action during restitution is less than that during compression. In some cases the force of restitution may be neglected, and the bodies are then said to be inelastic. In this case we have just after the impact u' = v. This gives mm , mu + m'v /e> . R = ,(u v\ :.u= (2). m + m ^ m + m If the force of restitution cannot be neglected, let R be the whole action between the balls, R the action up to the moment of greatest compression. The magnitude of R can be found by experiment. This may be done by observing the values of u' and v' and thus determining R by means of the equations (1). Such experiments were made in the first instance by Newton and led to the result that R/R is a constant ratio which depends on the materials of which the balls are made. Let this constant ratio be called 1 + e. The quantity e is never greater than unity; in the limiting case when e = 1 the bodies are said to be perfectly elastic. The Newtonian law R/R = 1 + e gives only a first approxi- mation to the motion, and is not to be regarded as strictly true under all circumstances. The value of e being supposed to be known the velocities after impact may be easily found. The action R must be first calcu- lated as if the bodies were inelastic, the value of R may then be deduced by multiplying by 1 + e. This gives , mu + m'v m'e . u ~~ m + m' m + m . mu + m'v V . 4- me , , 38 IMPULSIVE FORCES. [CHAP. I. The three equations comprised in (1) and (3) give the whole motion. Substituting from (3) in (1), we have (3). 84. We notice as a useful corollary that v' u' = e (v u) (4). The relative velocity after impact bears to the relative velocity before impact the ratio of e to 1. By the third law of motion the momentum gained by one ball is equal to that lost by the other ; the whole momentum being un- altered by the impact. Hence mu' + m'v' = mu + m'v (5). This result follows also by eliminating R between the equations (1). The equations (4) and (5) may be used to determine u', v', when the impulse R is not required. 85. When two perfectly elastic spheres of equal mass impinge on each other the bodies exchange velocities. In this case, by (3), R = m (u v) and the equations (1) then show that u' = v, v' = u. Conversely, we may show in the same way that if the spheres exchange velocities their masses are equal and the elasticity is perfect. 86. When a sphere impinges on a fixed plane, we regard the plane as an infinitely large mass. Putting m' infinite, we find R = mu (1 + e), u' = eu, v' = 0, the velocity of the sphere is therefore reversed in direction and its magnitude is multiplied by e. Ex. If the plane be in motion with a velocity V, prove that the velocity of the sphere after the rebound is - eu + V (1 +e). 87. If one sphere of mass m impinge directly on another of mass mf which is at rest and if m = m'e, the equation (3) gives R = mu. The impinging sphere therefore loses its whole mo- mentum and is reduced to rest. In the same way, let n spheres be placed in a row at rest and let their masses form a geometrical progression of ratio 1/e. ART. 89.] ELASTIC SMOOTH BODIES. 39 If any velocity is given to the first, it will strike the next in order and be reduced to rest. The second will strike the third and remain at rest and so on. Finally the last sphere will proceed onwards with the whole momentum communicated to the first. If the spheres are perfectly elastic, e = 1 and the same things happen when the masses are equal. If the spheres are placed close together, they are only in apparent contact ; and each impact will still be concluded before the next begins. Each ball transfers the momentum to the next in order and remains in apparent rest, the last ball moving onwards with the whole momentum communicated to the first. This may partly explain why, in some cases when blows have been given by the wind or sea to masses of masonry, the stones to leeward have been more disturbed than those exposed to the blows. 88. Ex. A series of perfectly elastic balls are arranged in the same straight line, one of them impinges directly on the next and so on ; prove that if their masses form a geometrical progression of which the common ratio is 2, their velocities after impact will form a geometrical progression of which the common ratio is 2/3. [Math. Tripos, I860.] 89. Two smooth homogeneous spheres A and B impinge obliquely on each other. To find the subse- quent motion. Let the common tangent plane at the point of contact be the plane of xy, and let the common normal be the axis of z. The spheres being smooth the mutual im- pulse acts along the axis of z. Let F,, F 2 be the velocities of the two spheres, before impact, F/, F 2 ' the velocities after. Let F 2 (l-e 2 ). The vis viva after impact is equal to the vis viva before only luhen e = l, that is, when the bodies are perfectly elastic. It is evident that w l cannot be equal to w 2 or e = 1. ART. 92.] MOTION OF A FREE SYSTEM. 41 91. Ex. I. Particles are projected from a given point A in all directions and obliquely impinge on a fixed plane of elasticity e. Prove that after reflexion the directions of motion diverge from a point B, where AB intersects the fixed plane at right angles in some point M, and BN=e . AM. Let AP be the path of a particle before impact, PQ that after. Let QP produced intersect the perpendicular AM produced in some point B. The com- ponent of velocity, u, along MP is unchanged by the impact, while that perpendicular, viz. v, becomes ev and is reversed in direction, /. tan QPx=evju=e tan APM. It immediately follows that MB = e .AM, so that every reflected path intersects the perpendicular from A in the same point. By using this theorem we can trace the course of a particle after successive reflexions from any number of fixed planes. To take a simple case, let it be required to find how a particle should be horizontally projected from a given point A on the floor, that after reflexion at two vertical walls Ox, Oy, it may pass through another given point A'. We draw a perpendicular AB to the first wall and take MB = eAM. A perpendicular is drawn from B to the second wall, and C is taken so that CN=e . BN. Then, since all the paths after the first and second reflexions pass through B and C respectively, the required path AQPA' is found by joining A' to C, Q, to B and P to A. Ex. 2. A particle of elasticity e is projected along a horizontal plane from the middle point of one of the sides of an isosceles right-angled triangle so as after reflexion at the hypothenuse and remaining side to return to the same point ; prove that the cotangents of the angles of reflexion are e + 1 and e + 2 respectively. [Math. Tripos, 1851.] 92. A free system of mutually attracting particles is in motion. Prove (1) that the centre of gravity moves in a straight line with uniform velocity, and (2) that the motion of the centre of gravity is not affected by any impacts between the particles. The mutual attraction between any two particles is measured by the momentum transferred from one to the other per unit of time ; the mutual impulse is measured by the whole mo- mentum transferred. In either case it follows by the third law of motion that the whole momentum of the two particles and the components in any directions, are unaltered by their mutual action. 42 IMPULSIVE FORCES. [CHAP. I. Let (a?,, T/J), (# 2 , t/ 2 ), &c. be the Cartesian coordinates and (u lt vj, 0*2, #2), &c. the components of velocity at any time t. Since x2,m = Sw#, 2/Sra = Sray, we have by differentiation w2ra = %mu, vm = 2mv. It has just been shown that the components 2ww, 2rav are unaltered by the mutual attraction or impact of any two particles. Hence the components of the velocity of the centre of gravity, viz. w, v, are constant throughout the motion. The path of the centre of gravity is therefore the straight line x = ut+ A, y = vt + B, and the velocity is the resultant of u, v, If all the particles were suddenly collected together at the centre of gravity, each particle having its momentum unaltered in direction and magnitude, the momentum of the collected mass would be the resultant of the transferred momenta. The equations u2,m = "Zmu, v^m = 2wv assert that the centre of gravity of the particles before collection moves exactly as the collected mass does. 93. The effect of the mutual action of two particles (whether attracting or impinging on each other) is to transfer a momentum from one to the other whose direction is the straight line joining the particles. Hence the moment of the momentum about any straight line is unaltered by the transference. The moment of the momentum of the whole system (that is, its angular mo- mentum, Art. 79), about any straight line is unaltered by the mutual actions of the particles. In a system of mutually attracting or impinging particles, the components of its linear momentum along, and the angular momenta about, any fixed straight lines are constant, except so far as they may be altered by the action of external forces. This is only the third law of motion more fully explained. 84. Examples*. Ex. 1. If a system of mutually attracting particles were suddenly to become rigidly connected together, determine the conditions that the rigid body should be at rest. The rigid body will possess the same momenta as the system but differently distributed. If the momenta of all the particles are in equilibrium, the rigid body has no component of momentum in any direction and no moment of momentum * Many of these examples are taken from the examination papers for the entrance and minor scholarships in the several colleges. AKT. 94.] EXAMPLES. 48 about any straight line. It is therefore at rest. By the rules of Statics the necessary and sufficient conditions for the equilibrium are (1) the whole linear momentum along each axis of coordinates is zero, (2) the angular momentum about each axis is zero. Ex. 2. Particles of equal mass travel round the sides of a closed skew polygon in the same direction, one starting from each corner and the velocity of each is proportional to the side along which it moves. Prove that their centre of gravity is at rest and that it coincides with the centre of gravity of the sides of the polygon supposing the masses of the sides to be equal. Prove also that if one particle be removed, the centre of gravity of the remaining particles describes a polygon whose sides are parallel and proportional to those of the original polygon. Since the sides exert no pressures on the particles the centre of gravity moves in a straight line with uniform velocity whatever the momenta of the particles may be. When, as in the problem, the momenta are parallel and proportional to the sides of a closed figure, the components Swut and ZTOV of Art. 92 are zero, and the centre of gravity is therefore at rest. The other parts of the question then follow at once. Ex. 3. An explosion occurs in a rigid body at rest, and the particles fly off in different directions. If in any subsequent positions they were suddenly connected together, prove that the rigid body thus formed would be at rest. Ex. 4. A number of particles originally in a straight line fall from rest, and rebound from a partially elastic horizontal plane. Prove that, at any time, the particles which have rebounded once lie in a parabola. [Coll. Ex. 1897.] Ex. 5. Two small spheres of equal mass can move inside a rough endless horizontal tube of length I. One sphere impinges with velocity v on the other at rest. If the friction of the tube produce a retardation / in either sphere and if after impact the spheres just meet again, prove that 2fl=v 2 e. [Coll. Ex. 1896.] Ex. 6. Four equal balls of the same material are projected simultaneously with equal velocities from the corners of a square towards its centre, and meet in the neighbourhood of the centre. Show that they return to the corners with velocities reduced in the ratio of the coefficient of restitution to unity. [Coll. Ex. 1892.] Ex. 7. Two equal spheres each of mass m are in contact on a smooth hori- zontal table, a third equal sphere of mass m' impinges symmetrically on them. Prove that this sphere is reduced to rest by the impact if 2m' = 3me, and find the loss of kinetic energy by the impact. [Coll. Ex. 1897.] Ex. 8. Two equal balls lie in contact on a table. A third equal ball impinges on them, its centre moving along a line nearly coinciding with a horizontal common tangent. Assuming that the periods of the two impacts do not overlap, prove that the ratio of the velocities which either ball will receive according as it is struck first or second is 4 : 3 - e, where e is the coefficient of restitution. [Math. Tripos, 1893.] Ex. 9. A heavy particle tied to a string of length I is projected horizontally with a velocity V from the point 'to which it is attached. Show that the energy lost by the impulse is a minimum when V i = lg/^/B: see Arts. 27, 90. [Coll. Ex. 1896.] 44 IMPULSIVE FORCES. [CHAP. I. /<.> . 10. A particle of mass m lies at the middle point C of a straight tube AB of mass M and length 2a, both of whose ends are closed. It is shot along the tube with velocity V. Prove that it will pass the middle point of the tube in the same direction after a time r ( 1 + - ) , e being the coefficient of restitution between the V\ ej particle and either end of the tube ; and that in this time the tube will have moved forward a distance -rr ( 1 + - ) . [Coll. Ex. 1895.1 M+m\ e) The particle traverses the length CA = a in a time a/V and after impact has a relative velocity eV. It therefore traverses the length AB = 2a in a time 2a/eV, and after impact at B has a relative velocity e*V. It traverses the remaining length BC=a in the time ajeW. The whole time T is the sum of these three times. The particle is now at the same point C of the tube as before, the distance traversed by the tube is therefore equal to that traversed by the centre of gravity of the system. Since the initial velocities of the particle and tube are V and zero, the velocity of the centre of gravity is v = mVI(M + m). The distance traversed is therefore vT. Ex. 11. A particle is projected inside a straight tube of length 2a, closed at each end, which lies on a smooth horizontal table and whose mass is equal to that of the particle. Prove that, at the moment just before the fourth impact the tube has described a distance 15a, if the coefficient of restitution is , and find the proportion of kinetic energy which has disappeared. [Coll. Ex. 1895.] Ex. 12. A smooth particle of mass m is at rest in a rectangular box of mass M which is free to move down a smooth plane inclined at an angle a to the horizon, the lowest edge of the box being horizontal, and the particle at its middle point. Suddenly the box is started down the plane with velocity V. Prove that if the coefficient of restitution be unity, the particle will strike the top and bottom of the box after equal successive intervals of time; and that the spaces travelled by the box in the first and second of these intervals are as V 2 + gl sin a : ~ V 2 + 3gl sin a, where 21 is the length of the box. [Coll. Ex. 1896.] Ex. 13. A perfectly elastic ball is projected vertically with velocity v lt from a point in a rigid horizontal plane, and when its velocity is v 2 an equal ball is projected vertically from the same point also with velocity v 1 ; show, (1) that the time that elapses between successive impacts of the two balls is vjg, (2) that the heights at which they take place are alternately (3^ - t>.j) (Wj + V 2 )j8g and (3^ + v 2 ) fa - vjfig, (3) that the velocities of the balls at the impacts are equal and opposite and alternately \ fa - v z ) and ^(v^ + v^. [Math. Tripos, 1896.] Since the balls exchange velocities at each impact, we may suppose that they pass through each other, one ball following the other at an interval T = (v 1 - v 2 )/g. Ex. 14. A weight of mass m and a bucket of mass m' are connected by a light inelastic string which passes over a smooth pulley. These bodies are released from rest when a particle whose mass is p and coefficient of elasticity e falls with vertical velocity V upon the bucket. Prove that a second collision will occur between the particle and bucket after a time- e (m + m') Vjmg and find the condition that the bodies should then be in their initial positions. [Coll. Ex. 1895.] ART. 94.] EXAMPLES. 45 Ex. 15. A particle is projected from a point on the inner circumference of a circular hoop, free to move on a horizontal plane. Prove that if the particle return to the position of projection after two impacts, its original direction must make with the radius through the point an angle tan" 1 {e 3 j(l + e + e 2 )}^. [Coll. Ex. 1897.] Ex. 16. Two balls of masses M, m (centres A and B), are tied together by a string, and lie on a smooth table with the string straight. A ball of mass m' {centre C) moving on the table with velocity V parallel to the string strikes the ball of mass TO, so that the angle ABC is acute and equal to a. Prove that M starts . , , . Vmmf cos 2 a (1 + e) , . , . with a velocity ...... ^ - ' - jr. e being the coefficient ol restitution Mm' sin 2 a + m (M + m + m) between m and m'. [Coll. Ex. 1895.] Let U' be the velocity of TO' after impact in the direction CB, v^ the common velocity of M, m in the direction AB, v% the velocity of m perpendicular to AB then m'(U'- Fcosa)= -R. Since R cos a has to move both M and m, while R sin a affects m only, (M + m)v 1 ' = Rcos a, mv 2 ' = Rsina. At the moment of greatest compression, the velocities of TO', m along CB are equal U' = j' cos a + v.y sin a. These equations give R. Multiplying the result by 1 + e the second equation then gives Wj'. Ex. 17. Three particles A, B, C whose masses are m, m', m", connected by straight strings, are placed at rest on a smooth table, and the obtuse angle ABC is TT - a. If A receive a blow F parallel to CB prove that C will begin to move with a m'J<'co8 2 a velocity - - - - . m 2m + mm sin-' a Let T, T be the impulsive tensions of AB, BC. Since A, B must have equal velocities along BA (Fcosa-T)lm = (T- T' cos a)/m'. Since B, C have equal velocities along BC (Tcosa- r')/m' = Z"/m". These equations determine T and T', and the result required is T'jm". Ex. 18. Two smooth spheres whose coefficient of restitution is e are attached by inextensible strings to fixed points. One of them, whose mass is TO, describing a circle with velocity v, impinges upon the other whose mass is m' and which is at rest. If the line of centres makes an angle 6 with the string attached to TO and the strings at that instant cross each other at right angles, then m! begins to describe a circle with velocity mvamecose ( l + e ) [Coll Ex- 1896>] m cos 2 e + m' sin 2 Let A, B be the centres of m, m', and let the strings be attached to D, E. Let DA intersect EB in C. The force R on m acts along BA and makes an angle 6 with AD. Let v', w' be the velocities of m, m' along EC and CD. Then m(v'-v)= - At the moment of greatest compression, the velocities of m, m' along AB are equal, /. v'siu0=w'coa 0. This determines the value of R, and the required velocity is R (1 + e) cos 01m'. 46 IMPULSIVE FORCES. [CHAP. I. Ex. 19. A smooth inelastic sphere of radius r and mass m is suspended by a string above a horizontal table, and another smooth inelastic sphere of radius r 1 and mass m' is moving on the table; prove that the cotangent of the angle through which the direction of motion of the second sphere is deflected by a collision is m ' r ?J where a and b are the vertical and horizontal distances of mb {( r + r ')2_ a s_ ft 2}J the centre of the first sphere from the path of the second before impact. [Coll. Ex. 1892.] We notice that the vertical motion of one sphere is stopped by the reaction of the table, while that of the other is not stopped by the tension of the string. Ex. 20. Four equal particles are connected by three equal strings AB, BC, CD and lie on a horizontal plane with the strings taut in the form of half a regular hexagon. An impulse is applied at A in the direction DA. Prove that the initial tension of BC is one-fourteenth of the impulse. [Coll. Ex. 1897.] Ex. 21. If three inelastic particles, m lf m 2 , m 3 , moving with velocities v 1 ,v 2 , v s making angles a, /3, y, with each other, impinge and coalesce, prove that the loss of . 110 3 - 1212 energy is *-* - - ^= - . [Coll. Ex. 1896.] &2jTf\> Ex. 22. A shot whose mass is m penetrates a thickness s of a fixed plate of // m\ mass M, prove that, if M is free to move, the thickness penetrated is s I ( 1 + ) . [Coll. Ex. 1896.] The mass m strikes M with a velocity v and continues to move inwards until m and M have the same velocity v l = mv l(M+m). If F be the resistance regarded as constant, x and .x + e the spaces described by M and m, m (V - 1> 2 ) = - 2F (x + ff), Mvj* = 2Fx. Eliminating x, we find 2Fff=v 2 Mml(M+m). When M is infinite, 2Fs=v z m. The ratio tr/s follows. This problem may also be easily solved by considering the relative motion. Ex. 23. A smooth uniform hemisphere of mass M is sliding with velocity V on an inelastic horizontal plane with which its base is in contact; a sphere of smaller mass m is dropped vertically so as to strike the first on the side towards which it is moving, at an inclination of 45; prove that if the hemisphere be V 2 (2M - em)' 2 stopped dead, the sphere must have fallen through a height ^ 2 z where e is the coefficient of restitution between them. [Math. Tripos, 1887.] CHAPTER II. RECTILINEAR MOTION. Solution of the Equation of Motion. 95. LET us suppose that a particle of mass m is constrained to move in a straight line, which we may call the axis of x, under the action of forces whose component along x is F. Let F= mX. We have seen in the previous chapter that the equation of motion d?x F IS -T.Z = = .A. dt 2 m Properly this equation gives X when # is a known function of t y and therefore answers the question, given the motion, what is the force ? Usually we require the solution of the converse problem, given the accelerating force X (Art. 68), find the motion. To deter- mine this, we must regard the equation of motion as a differential equation and seek for its solution. 96. In the general case X may be a function of x and t and also of the velocity v of the particle. But the equation can only be solved in limited cases. We shall examine these solutions in turn. Let us suppose that X is a function of t only, say X =f(t). By integration we have where suffixes have been used to represent integrations with regard to t. 48 SOLUTION OF THE EQUATION OF MOTION. [CHAP. II. In this way x has been expressed as a function of t, leaving the constants A and B undetermined. As this value of x satisfies the differential equation, whatever values A and B may have, there is nothing in that equation to help us in finding these two constants. We must have recourse to some other data. These are the initial conditions of the motion. Let us suppose that the particle was projected at a time t = a, from a point determined by x = b with a velocity v = c. Then remembering that v = dx/dt, we have c=f,(a) + A, b=f,,(a) + Aa + B. Solving these, we find A and B. The motion is therefore given by * =/, (0 + {' -/, (a)} t + {b - ac + of, (a) -/ (a)}. 97. Let X be afivnction of x only, say X=f(x). dx dx d*x ., . dx //r> ^j =2/ ; (a) + 4 ..................... (2),. ............... (3). To determine the value of A and the sign of the radical we use the initial conditions. Let us suppose that when t = a, x = b, and v = c. We then have c*-2f,(b) = A ........................ (4), c={2f,(b) + A}* ........................ (5). If c is not zero, the radical must have the same sign as c, i.e. the radical is positive or negative according as the direction of the initial velocity makes x increase or decrease. If however c = 0, we notice that the particle will begin to move in the direction in which the force acts; the radical therefore follows the sign of the initial value of X. Since X is a function of x only, it is obvious that if the initial value of X is also zero, the particle is at rest in a position of equilibrium and that there will be no motion. We now have ........ (6). f J ART. 99.] FORCE, A FUNCTION OF THE SPACE. 49 Representing the left-hand side of this equation, after the in- tegration has been effected, by (x), we have (x) = t + B (7). To find B we recur again to the given initial conditions, viz. that x b when t = a, hence B = (b) a. 98. The equation (7) determines t when x is known, i.e. it gives the time at which the particle passes over any given point of the straight line along which it moves. If we require the position of the particle at any given time, we must solve the equation and express * = *(*) ( 8 >- The solution of this algebraical equation may lead to different values of x, thus we may have x = fa (t), as = fa (t), &c. We have yet to determine which of these represents the actual motion. We notice that since the equation (7) is satisfied by x = b, t = a, one at least of these values of a; must satisfy this condition. All the others must then be excluded as not agreeing with the given initial conditions. If more than one of these solutions could satisfy this condition, the equation obtained by putting t = a in (7), viz. (#) = a + B, must have equal roots. Hence (x) = when x=- b. Since (#) represents the left-hand side of (6) it immediately follows that 2/i (6) + -4 is infinite. But by (5) this cannot happen if the initial velocity c is finite. 98. Subject of integration infinite. Other points requiring attention arise when the integrals which occur are such that the subject of integration is infinite at some point B of the path. Since the forces in nature are necessarily finite this cannot happen in the integral (2), for if /, (a;) were infinite its differential coefficient, f(x) for any finite value of x, would also be infinite. In the integral (6) the subject of integration is infinite when the velocity is zero. We can use the integral (6) to find the time of transit from any point A to a point P as near as we please to B on the same side of B as A. If the result is infinite the particle never reaches B. If the time of arrival at B is finite we have to find the subsequent motion. As the particle approaches B the velocity is numerically decreasing and there- fore the accelerating force X has the opposite sign to the velocity. Supposing X not also to vanish at B, the particle after arriving at B must begin to retrace its steps. Considering B as a new initial position, the subsequent motion may be deduced from (3) by putting c=0. It X=Q also at B, the particle, as explained above, will remain there in equilibrium. R. D. 4 50 SOLUTION OF THE EQUATION OF MOTION. [CHAP. II. 1OO. Ex. 1. A particle moves in a straight line under a central force tending to the origin and equal to n 2 /.r 3 . Investigate the motion. iPr iP Wehave ^= -- 3 (1). The minus sign is introduced because the left-hand side represents the acceleration in the positive direction of x and the force acts towards the origin. We then find (2). Let us suppose that the particle starts from rest at a very great or infinite distance from the origin ; then when x is infinite, dxldt = 0. Hence ^4=0, and the equation becomes Since the particle begins to move towards the centre of force the velocity is initially negative. We therefore take the negative sign. Multiplying by .x and integrating, we find x 2 =B-2nt ....................................... (4). Initially when t=0, the particle is infinitely distant from the origin, i.e. x is infinite and therefore B is infinite. It follows that the particle does not get within a finite distance of the origin until after the lapse of an infinite time. If the initial conditions are slightly altered we may obtain a finite result. Let us suppose the particle to be initially projected at a distance x=b (b being positive) with a velocity njb towards the centre of force. Proceeding as before we find A =0, and as it is given that the initial velocity of the particle is negative, the radical has still the negative sign. We thus again arrive at the equation (4). Since x = b when t = 0, we find B = b 2 , and a=(6S_2nf)i .................................... (5). Since x is initially positive we must give the radical the positive sign. As t increases we see that x continually diminishes and when t=& 2 /2;i the particle arrives at the origin. Its velocity at that moment is found by putting x = in (3) and is easily seen to be infinite. Cases in which either the velocity or the force is infinite do not occur in nature. If we constrqct a central force by placing some attracting matter at the origin there would be an impact before the particle reached the origin and the whole motion would be changed. But as a matter of curiosity we may enquire what would be the subsequent motion if our equations held true for infinite velocities and forces. In this case the particle arrives at the origin with a negative velocity, we must therefore suppose that the radical in (2) does not change sign when the quantity passes through infinity at the origin. Hence since x now becomes negative, we must take the positive sign in (3) instead of the negative one hitherto used. This gives x 2 = B + 2nt, where B need not necessarily have the same value as before. To find B we notice that at the initial stage of this part of the motion, .r = and t = fc 2 /2n; we easily find that B= -b 2 . The motion after the particle has passed the origin is therefore given by x= - (2nt - brf. ART. 101.] FORCE, A FUNCTION OF THE VELOCITY. 51 H-2 Ex. 2. If x at n we have -^ = At n ~ 2 =A (-) " , where A = an(n-l). Let us at \a/ suppose that n>2. A particle is placed at rest at the origin. Show that if acted on by X=At n ~- n-2 the subsequent motion is given by x = at n , but if acted on by X=A (x/a) n the motion is given by x = 0. Ex. 3. A particle is projected from the origin with a velocity #p* under the action of an accelerating force X= -^fjf(p-x)^. Prove that the particle comes to rest in the position of equilibrium denned by x =p. 101. Let the acting force X be a function of the velocity only, say X =f(v). The equation of motion now takes the form dv _ f/ \ /i \ dt Integrating this, we have I dv -t+A (9\- lm- t+A (2) ' //(*) writing (f> (v) for the integral on the left-hand side, this becomes (v) = t + A (3). Supposing as before that the particle is initially projected at a time t = a, with a velocity c, we have A = (c) a. Two rules are given in the theory of differential equations for the solution of the equation (3). The first rule requires us to solve the equation for v and find v = ty(t), and as already ex- plained that solution is to be chosen which makes v = c when t = a. Remembering that v = dxjdt we then obtain x by inte- gration. If the equation (3) cannot be solved for v, we use the second rule. This requires us to recur to the form (1), eliminating dt by using the equation v = dx/dt, we have vdv vdv Thus after integration both x and t are expressed by (2) and (4) in terms of a subsidiary quantity v. We notice also that this subsidiary quantity has a dynamical meaning, viz. the velocity of the particle. 42 52 SOLUTION OF THE EQUATION OF MOTION. [CHAP. II. 1O2. Ex. 1. A particle is projected with a velocity V in a medium whose resistance is m> n , where n is a positive quantity. The equation of motion is then *--* .......................................... <" ^~ n (2). .- . " 1-n Measuring t from the moment of projection we have when t=0, v=V, hence yi-n A = = . We therefore find 1-n v i--V 1 -*=-(l-n)ict .............................. (3). If n < 1 the velocity decreases continually from its initial value V, and vanishes yin after a finite time, viz. t = - . The particle will then remain at rest, since {i n) K X=0. If n>l, writing (3) in the form we see that the velocity decreases continually and vanishes after an infinite time. If n = l, these equations take an indeterminate form. Beturning to the equa- tion (2) we have logt>= -Kt + A; .: v = Ve~ Kt ........................... (5). It follows that the velocity decreases continually and vanishes after an infinite time. In all these cases we can find the space described in any time t. Remembering that v = dxldt, we have from (3), *=/{ - (1 - n) Determining B from the condition that x=0 when t=0 we find 2-n -( < 2-n)KX={V 1 - n -(l-n)Kt} l - n -V*- n ..................... (6). We may also find the velocity after the particle has described any space ;c. We begin with dv V -=-=- KV n . dx :. v 1 ~ n dv=-Kdx; .: v 2 ~ n =V 2 - n -(2-n) KX .................. (7). Let us find the space described by the particle when v=0. yz-n yl-n If nl and <2, we have x=^ - . while t is infinite; the particle therefore (2 - n) K comes to rest after describing a finite space in an infinite time. If n > 2, we find that v vanishes when x is infinite and the particle describes an infinite space in an infinite time before it comes to rest. Ex. 2. If the resistance is KV, show that the particle comes to rest after describing the finite space V/K in an infinite time. ART. 104.] MOTION OF A HEAVY PARTICLE. 53 Ex. 3. If the resistance is KV Z , prove that the particle describes an infinite space in an infinite time before coming to rest. 1O3. Ex. 1. If X=(v).f(x) or X=(v)f(t), prove that the equation of motion can be solved by separating the variables. In the former case we use vdvjdx = X, in the latter dvjdt=X. Ex. 2. If X=f (x)v n + F(x) v 2 show that the equation of motion becomes linear by writing v 2 ~ n =y. Ex. 3. If X=f(v z lx) show that the equation of motion becomes homogeneous, and that the variables can be separated by writing v^ = xy. Motion of a heavy particle. 104. A heavy particle starting from rest slides down a rough straight line which is inclined to the vertical at an angle 6. It is required to find the motion. Let be the initial position of the particle, 0V the vertical, Q the particle at any time t. The accelerating force due to gravity is g cos 6. The pressure on the straight line being mg sin 6, the retarding force due to friction is fig sin 6, where ft is the coefficient of friction. The whole accelerating force is therefore fg (cos 6 fj, sin 9) = g sec e . cos (6 + e), where fj, = tan e. Writing OQ = s, the equation of motion is d?s dv ' COS ( + ^ ................ ^ )' We Integrating, we find v- = 2gs sec e cos (0 + e) + A. Since the particle starts from rest, v and s vanish together. therefore have A = 0, and v 2 = 2gs sec e cos(# + e) ..................... (2). To interpret this formula we make the angle VON = e and draw any straight line NVQ perpendicular to ON cutting the 54 MOTION OF A HEAVY PARTICLE. [CHAP. ii. vertical in V and the straight line along which the particle travels in Q. Then ON = s cos (0 + e). It follows that the velocity acquired in describing any chord OQ is independent of 6 and is equal to that acquired in describing 0V. If the chord OQ is taken on the same side of the vertical 0V as N, the angle 6 as above measured becomes negative. Since the friction varies as the pressure taken positively, it must now be represented by \ig sin 6. The theorem therefore only applies to the chords on the side of the vertical opposite to ON. If we make the figure turn round the vertical V, the straight line 0V will describe a right cone having 0V for its axis and |TT for the semi- vertical angle. The velocity acquired in descending any chord from rest at to the surface of this cone is equal to that acquired in descending 0V. 105. By integrating (1) twice with regard to t, and re- membering that both s and ds/dt vanish when t = 0, we find s = \g sec e cos (0 + e) t- (3). We may interpret this formula by a similar geometrical con- struction. Making as before the angle VON e, we see that, when t is constant, (3) represents the polar equation of a circle whose radius vector is s and whose centre C is situated on ON. We have therefore the following theorem. Describe any circle passing through and having its centre on ON, and let it cut the vertical through in some point V. The time of descent from rest at down any chord OQ of this circle is the same as that down V. The chord OQ must be on the side of V remote from the centre. In the same way if the circle is drawn above 0, we can show that the time of descent from rest at any point Q of the circle to is equal to the time down FO. ART. 108.] CHORDS DESCRIBED IN EQUAL TIMES. 55 1O6. When the straight line down which the particle slides is smooth ON coincides with the vertical. The cone in Art. 104 becomes a horizontal plane, and the circle in Art. 105 has OF for a diameter. We thus fall back on the well-known theorems (1) that the velocity acquired in descending from rest to a given hori- zontal plane is the same for all chords, (2) that the time of descending from rest at the highest point of a circle to the circle is the same for all chords. 107. If the motion take place in the air we must make allowance for its resistance. Supposing the resistance to vary as the velocity, the equation of motion is (1), where f= g sec e cos (0 + e). Remembering that v = ds/dt we find by integration a-> ............................ < 2 >- the constant being omitted because s and v vanish together. Transposing KS, the equation can be integrated again by following the ordinary rule for linear equations. We have %(***)**&*', =f (te Kt - - e Kt + c j . Noticing that s should vanish when t = 0, we have c = I/K. Hence, restoring the value of/, s= - 2 sec e cos (6 + e) [ict - 1 + e'**} (3). K When t is constant and (0 + e) is regarded as variable we see that (3) is again the equation of a circle having its centre on ON, The theorem of Art. 105 is therefore also true when the particle slides on a rough chord in a medium resisting as the velocity. The times of descent from rest at down all chords of the circle are equal. 1O8. There is another method of proof by which the solution of the diffe- rential equation is evaded. We notice that if we write s = <7cos (0 + e), the equation (1) of Art. 107 becomes dV do- _ =i;sece _ K _, from which the angle 6 has disappeared. The initial conditions now become ,. A particle is projected vertically upwards with a velocity V in a medium resisting as the square of the velocity. It is required to find the motion. During the ascending motion the resistance acts downwards and the equation of motion is dv dv v 2 5 - '-'!? where L is the limiting velocity. When the particle descends the resistance acts upwards, but since v 2 does not change sign with v, the equation of motion must be changed to dv dv v 2 where in both equations s and v are measured positively upwards. This discon- tinuity occurs whenever the power of v in the law of resistance is even. Following the second rule given in Art. 101 we express both s and t in terms of v. We have for the ascending motion Ldv .9 _,V *=-[ L J 2gs_ [2vdv _ . ~ ~ the constants being determined by the condition that v=V when t = and s = 0. The time T of ascent and the space h ascended are deduced by putting v=0. We thus find The time of ascent and the space ascended are less than in a vacuum, for both gravity and the resistance join in bringing the particle to rest. For the descending motion we have in the same way the constants being determined from the condition that when v Q, t=T, s^ ART. 116.] RESISTING MEDIUM. 59 The velocities at which the particle passes upwards and downwards through any given point of space are connected by a simple relation. Taking the given point as the point of projection upwards, let the two velocities be V and V. Putting s = in (5) we find 2gh . - =l Eliminating h between this equation and (3) we arrive at If 1 and the latter H > 2. It is remarkable that both these limits are finite, though the upward velocity of projection may be as great as we please. For the descending motion it is often convenient to measure s downwards from the highest point. We thus avoid using a negative velocity. Adopting this plan, the equation of motion is dv dv Putting v = xL as before, we find for the time and space necessary to acquire a velocity aL, g T' _ f* dx fjh' _ f a xdx ' These integrals can be found without difficulty when n is an integer by using the method of partial fractions, see Greenhill's Differential and Integral Calculus, Art. 190. Roberts' Integral Calculus, Art. 35. The result when n has its general integral value is too complicated to be reproduced here. 60 THE LINEAR DIFFERENTIAL EQUATION. [CHAP. II. 117. Ex. 1. A heavy particle is projected upwards with a velocity L in a medium resisting as the wth power of the velocity. Prove that the whole space (up and down) described when the velocity downwards is V is equal to LT when L is the limiting velocity and T is the time in which the particle falling from rest in the medium will acquire a velocity F 2 /L. Ex. 2. A particle is projected upwards with velocity L in a medium resisting as the cube of the velocity. Show that the whole time and space of the ascent 2ir L 2 are connected by the equation s + LT=.- . g The linear differential equation. 118. The Linear equation. The most important equation of motion which occurs in this part of dynamics is the linear equation with constant coefficients. The simplest form of this equation is where b and c are two constants. When 6 = the equation represents the motion of a particle acted on by a constant accelerating force equal to c, and the solution is obviously x = kct* + At + B ........................ (2). When b is not zero, we can simplify the equation by putting x = c/b + ................................. (3), we then have This can be solved without difficulty by the method already explained in Art. 97. But a simpler solution can be obtained by following the rules for solving equations with constant co- efficients given in books on differential equations. We assume as a possible solution % = Ae" .............................. (5). Substituting we find A (\ 2 + b) e** = 0. The equation is therefore satisfied if X = + V( b). If b is negative and equal to b', we have two real values of X, either of which give a solution. The equation is clearly satisfied by Be- t ^' ..................... (6), ART. 119.] HARMONIC OSCILLATION. 61 and this is the complete integral because it contains the two arbitrary constants A and B. If b is positive, X is imaginary; but remembering that an imaginary exponential is a trigonometrical expression, we replace the assumption (5) by %=Asm(\t + B) ........................ (7). Substituting we find A ( X 2 + &) sin (\t + B) = 0. The equation is therefore satisfied by X = + \/b. These two values of X give the same solution, the effect of changing the sign of X being merely that of changing the signs of the arbitrary constants A and B. The complete integral is therefore x = c/b + Asm(t^b + B) .................. (8). It may also be written in either of the forms x = c/b + A ' sin t^b + B' cos t V& ............... (9), .................. (10). 119. Harmonic Oscillation. The dynamical meaning of the linear equation is important. Consider first the case in which b is positive. Putting b = n 2 , we have (2). First, we notice that as t continually increases the value of x alternates between the limits c/w 2 + A. We therefore infer that the differential equation (1) represents an oscillatory motion and that the arc of oscillation is constant. The semi-arc of oscillation is A and its magnitude depends on the initial conditions. The semi-arc is called the amplitude of the oscillation. Secondly. The middle point of the arc is determined by x = c/n*, and this point is independent of the initial conditions. If the particle is placed at rest in the position defined by this value of x, the equation (1) shows that the accelerating force (viz. d^x/dP) is zero. The middle point of the arc of oscillation is therefore a position of equilibrium. Thirdly. When t is increased by 2ir/n, the values of x recur in the same order, but when increased by irjn they recur with opposite signs. The period of a complete oscillation is therefore 62 THE LINEAR DIFFERENTIAL EQUATION. [CHAP. II. 2?r/7i. This period is independent of the initial conditions. The quantity n is called the frequency of the oscillation. The time of a complete oscillation is the time occupied by the particle in describing twice the whole arc of oscillation starting from any point and returning finally to the same point again. When the period is independent of the length of the arc, the motion is sometimes called tautochronous. Fourthly. The constant B depends on the instant from which the time t is measured, thus if we write t + a for t, nothing is changed except that B is increased by na. Fifthly. Let # = #, dx/dt = v be the given values of x and v at the time t . Writing the equation (2) in the form (9) of Art. 118 and equating the values of x and dxjdt to X Q and v l} when t = t , we find the values of A' and B'. The solution there- fore becomes n / n Comparing this with the solution (2) we see that A sin B = C c/n 2 , A cos B = v /n. The semi-arc A of oscillation is therefore given by 120. Consider next the case in which b is negative. Writing b = n 2 , the differential equation and its solution become First, we notice that the motion is not oscillatory. Secondly. If A is not zero the particle travels in an infinite time to an infinite distance from the origin. If A = the particle after an infinite time arrives at the point determined by x = c/n 2 . Thirdly. The position of equilibrium is given by x = c/n*. Fourthly. The particle can change its direction of motion only once. This change occurs when ART. 121.] THE GENERAL CASE. 63 This gives 2nt = \og(B/A). This is imaginary if A and B have opposite signs, and gives only one real value of t if A and B have the same sign. The particle can change its direction only if this real value of t is subsequent to the beginning of the motion. Fifthly. If the values of x and v are respectively x and v at the time t = t , the value of oc at any time t is > 2 V n* nj 2 V n- 121. When the equation of motion is dx we take as the trial solution * = l + Ae " < 2 >- It is easily seen that this satisfies the differential equation if If a- -b is positive, the roots of the equation are real. Representing these by \! , \. 2 , the solution is where A l , A. 2 are two arbitrary constants. If a~-b is negative, say = -n 2 , the two roots are -an*J(-l). By an easy reduction the solution (4) becomes x = - + e~ I?, sin (nt + B) (5), 6 where B lt B. 2 are two arbitrary constants. If a a -6 = 0, the general solution is ' at .. ...(6). Considering the solution (5) as the more important of the three, we notice that the trigonometrical term vanishes whenever nt + B 2 is a multiple of TT, the particle therefore passes through the position defined by x=cjb at intervals each equal to v In. Since it necessarily passes through this point alternately in opposite directions, the interval between two consecutive passages in the same direction is 2ir/n. This is called the time of a complete oscillation. The point defined by x = cjb is evidently the position of equilibrium. To find the times at which the system comes momentarily to rest we put dxjdt^O. This gives tan (wH- 2 ) = n/a. The extent of the oscillations on each side of the position of equilibrium may be found by substituting the values of t given by this equation in the expression for x - cjb. Since these occur at a constant interval equal to ir\n we see that the amplitude of the oscillation continually decreases and the successive arcs on each side of the position of equilibrium form a geometrical progression whose common ratio is c ~ a "/ n . -r- cos nt + iix sin nt = \ (t) cos ntdt + B. 64 MOTION UNDER A CENTRE OF FORCE. [CHAP. II. 122. The following differential equations occur in dynamics. d?x (1) Solve jp+ a * = *() Multiplying by sin nt, both sides become perfect differentials, hence dx f - sin nt - nx cos nt = I (f> (t) sin ntdt + A. at j Multiplying by cos nt, both sides are again perfect differentials, dx ( sin nt = I (t) we use e nt and e~ as the two successive multipliers. (It (2) When (t) is trigonometrical another method can be used. Let the equation be M X -r-r + n*x = E sin (\t + F). at Assuming x=Msm(Xt + F) as a trial solution, we see at once that the equation is satisfied if M (-X 2 + n 2 ) = . Adding the solution found in Art. 118 we see that the complete integral is E x=Asin (nt + B) + _ ^sin (** + *') This method fails when X = ??. In this case we take x = Mt cos (\t + F) as a trial assumption. We find - 2Mn = E. The complete integral is therefore Ft - cosn in Motion under a centre of force. 123. Central force varying as the distance. A particle constrained to move on a smooth straight line C p A is acted on by a central force tending to a fixed point outside the straight line, whose magnitude varies as the distance of the particle from 0. Let OC=h be the perpendicular on the straight line AC. Let P be the particle, CP = x. The force on P being n 2 . OP, the component along PC is iPx. Supposing the straight line to be smooth and the motion to take place in vacuo, the equation of motion is This is the standard form discussed in Art. 119. The particle ART. 125.] ROUGH PATH. 65 therefore oscillates about C as the middle point of the arc, and the time of a complete oscillation is 2?r/n. To find the time of oscillation numerically the magnitude of the force must be known at some given distance from the centre 0. Suppose that the force is equal to gravity at a distance a, then n*a=g, and the time of a complete oscillation is 2?r >J(a/g). If g = 32'18, the distance a must be measured in feet and the formula gives the time in seconds. The extent of the arc of oscillation depends on the initial conditions. If the particle start from a point distant x from C with an initial velocity v measured positively from G, the whole subsequent motion is expressed by the fifth result of Art. 119. 124. Ex. Any two places on the surface of the earth are joined by a straight tunnel. A particle dropped from one falls towards the other under the sole attraction of the earth. Assuming that the resultant attraction tends to the centre and varies as the distance therefrom, prove that the particle will arrive at the second place after about 42 minutes, the radius of the earth being taken as 4000 miles. 125. Ex. Effect of friction. If the straight line in Art. 123 is sensibly rough, it is required to take account of the friction. Since the normal pressure on the straight line is equal to n z h and is therefore constant, the limiting friction is also constant. Let us represent this by /. The equation of motion is therefore We notice that the frictional accelerating force acts opposite to the direction of motion, so that the sign must be negative or positive according as the particle is A! D' G D * moving in the direction in which x is measured or the opposite. The equation therefore presents the discontinuity which so frequently occurs whenever friction has to be taken account of. Let the particle start from rest at A where CA=a. Initially the resolved attraction is ri*a and unless n 2 a is greater than the friction /, the particle will not move. Supposing this inequality to hold we write the equation in the form d 2 x The motion therefore from A towards C is the same as if the centre of force were displaced a distance CD =//?i 2 towards A. The particle comes to rest at a point A' on the other side of D where DA'=AD. On the return journey we take CD' also equal to fjn 2 and the particle moves as if D' were the centre of force. Thus the centre of force is alternately moved at each oscillation a constant distance, R. D. 5 66 MOTION UNDER A CENTRE OF FORCE. [CHAP. II. always opposite to the direction of motion. The friction reduces the extent of each successive semi-arc of oscillation by 2//n 2 . The particle comes finally to rest when the extent of the semi-arc is less than //n 2 . 126. Resistance of the air. If the motion take place in the air its resistance must be allowed for. As a sufficient illustra- tion of the general effects of this force, let us suppose that the resistance varies as the velocity. Excluding friction the equation of motion is then d?x dec Assuming n > K the solution is (Art. 121) x = Aer" sin (pt + B) ..................... (2), where _p 2 = w 2 K\ The constancy of the period of oscillation is therefore unaffected by the resistance of the medium, Art. 121. The time of oscillation is however longer than in a vacuum. The successive arcs on each side of the position of equilibrium decrease continually in geometrical progression and vanish only after an infinite time. In many cases the resistance of the medium is very slight compared with the other forces acting on the particle. The quantity K is then small, and we see that the period of any one oscillation differs from that in a vacuum by the squares of small quantities. In using the equation (2) we must however remember that when the position of the particle after a great many oscilla- tions is required we cannot regard pt as the same as nt ; for though p and n differ by a very small quantity, that difference is here multiplied by the time t. 127. By making observations on the lengths of the arcs of oscillation we may test the correctness of the assumed law of resistance. A convenient method of trying the experiment is to use the particle as a pendulum. It may be shown that when the oscillations are small the resolved action of gravity represents the force v&x while the resistance is 2/e dx/dt. The measurements show that the successive arcs do decrease in geometrical pro- gression when the arcs are small, but the decrease follows another law when not small. This, as Poisson remarks, is a justification of the statement that for small velocities the resistance varies nearly as the velocity. ART. 129.] DISCONTINUITY OF RESISTANCE. 67 The common ratio of the geometrical progression is e~ KW/ P. By measuring successive arcs the numerical value of K can be found. 128. Discontinuity of resistance. When the resistance varies as the velocity the analytical expression 2/cv changes sign with v. It therefore represents the retardation due to the re- sisting medium both in sign and magnitude. If the resistance varies as the square (or any even power) of the velocity, the analytical expression 2/cw 2 represents the retardation in magnitude only. Whenever the particle changes its direction of motion it will then be necessary to change the sign of K. Thus a dis- continuity is introduced into the equations similar to that which occurs when friction acts on the particle, Arts. 125 and 115. 129. Ex. 1. A particle oscillates in a straight line under the action of a central force tending to a fixed point C on the straight line and varying as Jf the distance therefrom. Supposing the motion to take place in a medium resisting as the square of the velocity, find the relation between any two successive arcs on each side of C. A' B' C I Supposing that the particle is moving in the negative direction (Art. 128) the equation of motion is vdvjdx = - n z x + KV Z . By Art. 103 this gives v*e KX C-\ ( x + ^~ ) e x . If x , x l be two successive K \ &K] / IN / 1 \ arcs, Xj being negative, we have [x 1 + ^-\e KXl (^o + ^) e ~ ** \ AKJ \ */ We notice that this relation is independent of the strength of the attractive force. To interpret this relation we trace the curve y = (x + ^\e x . If the particle start from rest at any place A it will come to rest again at A 1 where the ordinates of A and A' are equal. Taking CB = CA', the third point of rest is at a distance CB' from C on the side of C opposite to A', the ordinates of B, B' being equal, and so on. Thus if the particle start from rest at an infinite distance from C it will^irst come to rest at K, where C.K"=l/2/c numerically. The general character of the motion is that the successive arcs decrease rapidly at first, but afterwards become more and more nearly equal, the motion never ceasing. If CI is the abscissa of the point of inflexion, CI= CM= CK. Ex. 2. Prove that the times of describing all chords of a circle starting from rest at the same point A under the action of a centre of force situated on the diameter through A and varying as the distance are equal. The chords are to be regarded as smooth and the motion to be in a vacuum. 52 68 MOTION UNDER A CENTRE OF FORCE. [CHAP. II. Ex. 3. A heavy particle whose mass is m is suspended from a fixed point by an elastic string whose unstretched length is a. If the particle oscillate up and down in a vacuum, prove that the complete period of an oscillation is 2ir N /(?n/) > where E is Young's modulus. Ex. 4. A particle oscillates in a straight line in a medium whose resistance per unit of mass is K times the square of the velocity. There is a centre of force situated in the straight line whose attraction is /j. times the square of the distance from the centre of force. If a and 6 are the distances from the centre of force of two successive positions of instantaneous rest, and /x. is not zero, prove that 2 * a =l. [Art. 135.] 130. The inverse square of the distance. A particle, constrained to move in a straight line, is acted on by a central force tending to a fixed point external to the line and varying inversely as the square of the distance therefrom. It is required to find the motion. Let OC be a perpendicular on the straight line, OC=h. Let P be the particle, CP = x,OP = r. See fig. of Art. 123. Let the angle POC = (f>, then sin < = x/r. The accelerating force on P being ft/r 2 , the component along PC is found by multiplying by sin $ and is therefore fix/r 3 . The equation of motion is dx~ r 3 Since r 2 = A 2 + a?, we have rdr = xdx. Hence If the particle start with a velocity u at some point A distant a from 0, we have ...(2). If the particle is projected from C along GA with a velocity u greater than - v / (2/^/a), it is clear that the velocity v cannot vanish or change sign. The particle therefore will move continually away from the centre of force. 131. When the centre of force lies on the straight line of motion, the time occupied by the particle in travelling from the initial position A to any point P can be found without difficulty. We put x = b cos 2 6 , .: dx/dt = - 26 sin 6 cos 6 dO/dt. The equation of motion is ART. 133.] THE INVERSE SQUARE. 69 We notice that x begins at x = a with dxjdt initially positive ; x then increases until dxjdt = 0, i.e. until x = b. At this point the particle begins to return and dxjdt becomes negative. To represent these changes we make 6 begin at 6= -ft where cos/3= +/J(alb) because this makes dxjdt positive when x = a. We then make 9 increase through zero and finally become \ TT when the particle arrives at the centre of force. Thus the two times at which the particle passes through any point P are distinguished by the sign of 6. Since, according to this arrangement, continually increases with the time we give the positive sign to the radical in the expression for dOjdt. We then find after integration that the time from 0= -j3 to 6 is ^ (0 + sin20 + /3 + sin2/3). The time from rest at a distance x=a follows from the preceding or may be found independently. We have - dt the limits being x=a to x. Putting a; = acos 2 we easily find that the time t of moving from x = a to x is The time of arriving at the centre of force starting from rest at a distance a is found by putting 0= \ir. The result is 54/5-. a \f sufJi 132. Ex. 1. A particle falls from rest at a point A whose altitude above the surface of the earth is equal to the radius. Show that the velocity on arriving at the surface is equal to that acquired by a particle falling from rest through half that space under a constant force equal to g, where g represents gravity at the surface of the earth. Notice that if ^/r 2 is the attraction of the earth, a the radius, /t/a 2 =, where if> represents the angle POG. Inte- grating as before, we find 4u a = (cos - cos + f sin - f sin ) . = - sec e sin in ( ~2 ~ 6 )' V / h 2 where , e are the angles CO A, COM, so that /=tane. It is evident that v ( Ex. 2. If the force to vary as the inverse fourth power of the distance and the particle starting from rest at A come to rest again at B, prove that the angles CO A, COB are complements of each other when sin 2 (COA) = (4f- 2)/(/+l). Thus if /= a particle starting from rest at an infinite distance will just reach C. Ex. 3. A particle is constrained to move in a straight rough tube CA, and is acted on by a central repulsive force X/r, where r is the distance from the centre of force O and OCA is a right angle. The particle is projected from A away from C with a velocity v ; prove that if it come to rest at a point P, the angle COP is a value of & satisfying the equation fj.0 - log sec 6 = v 2 /2X, where /* is the coefficient of friction. [Coll. Ex. 1893.] 134. Ex. 1. The earth and moon being held at rest, find the least velocity V with which a particle must be projected from the moon to reach the earth. Let a be the radius of the earth, 6 = T 3 T a that of the moon, 60a their distance apart from centre to centre. Let E and ^ T E be the masses of the earth and moon. If x is the distance of the particle from the centre of the moon, the equation of motion is d?x E IE (1). dt 2 (60a-ar) 2 81 x 2 " This equation can be integrated by the rule of Art. 97. The constant of inte- gration can be found in terms of V by remembering that dx/dt= V when x = b. There is evidently a certain point between the earth and moon where the attractions of these bodies balance each other. By equating the right-hand side of (1) to zero, this point is easily seen to be at a distance 6a from the centre of the moon. If V is such that dx/dt vanishes when x = 6a, it follows that a velocity of projection ever so slightly greater than V will carry the particle to the earth. Remembering that E/a?=g and taking a to be 4000 miles, we find that V is approximately 1J miles per second. Ex. 2. If the earth and moon were placed at rest, they would fall towards each other under the influence of their mutual attractions. Supposing the initial distance to be equal to their present distance from each other show that they would meet after about four and a half days. Consider their relative motion. If E, M be the masses of the earth and moon, the attraction on the earth per unit of mass is M/r 2 . By Art. 39 we apply this, reversed in direction, as an acceleration to both bodies. The earth is thus reduced to rest, while the moon is acted on by the two accelerating forces Mjr 2 and E/r 2 . ART. 135.] DISCONTINUITY OF A CENTRAL FORCE. 71 The whole accelerating attraction on the moon causing the relative motion is therefore (E + M)jr 2 . We must also apply to each an initial velocity equal and opposite to that of the earth (Art. 10), but this, in our problem, is zero. The time is then found as in Art. 131. Ex. 3. Two mutually attracting spheres, each one foot in diameter, and the density of each the same as the mean density of the earth, are placed at rest in a vacuum, the distance between their surfaces being one quarter of an inch. Prove that they will meet in less than 250 seconds. This problem is due to Newton, its history is given in Todhunter's History of the Theory of Attractions, &c., Art. 725. Ex. 4. Two particles A, B, mutually attracting each other according to the Newtonian law, are placed at rest at a given distance a apart. The particle B is now constrained to move away from A along the straight line joining them with a uniform velocity u, show that A will catch B up if u 2 <2/u/a where p is the mass of B. Show also that the time will be (7r + 2 + sin 2/3) N /6 3 /2^ where cos 2 j8=a/6 and 2/t/6 = 2ju/a-w 2 . [Reduce B to rest, see Art. 131.] Ex. 5. A body of mass M is moving in a straight line with velocity U, and is followed at a distance r by a smaller body of mass m, moving in the same line with a smaller velocity u. The two bodies attract each other with a force varying as the inverse square of the distance and equal to K for two unit masses at unit distance. Prove that the smaller body will overtake the other after a time ( 1 ) * ( r ) * -7H J 1, ( {T + N/U-c^+cos-'w}, (K(M+m)) (1 + w) where K (M+m)(l - u) = (U-u) 2 r. [Math. Tripos, 1887.] 135. Discontinuity of a centre of force. A particle constrained to move on a smooth straight line is acted on by a force X tending to a point C situated on the line and varying as the nih power of the distance therefrom. It is required to find the motion. Let the particle P start from rest at A, GA = a, CP = x. The equation of motion is d?x dv M= V dx = - X (1) > .-. v 2 = -?- (a n+1 - # n+1 ) (2), the constant of integration being determined by the condition that v = when x= a. If n = 1 the integral takes a logarithmic form. If n is an odd integer this equation shows that the velocity is again zero at a point A' determined by x= a. The particle therefore oscillates on each side of C, the amplitudes on each side being equal. 72 MOTION UNDER A CENTRE OF FORCE. [CHAP. II. If n is an even integer, the expression for v vanishes for no real value of a; except a; = a. Since the particle must obviously oscillate on each side of C through equal arcs, it follows that the equation (2) cannot represent the dynamical facts of the problem. The reason is that the force X (as given in the question) varies as the nth power of the distance taken positively and always acts towards C. Now x is the distance of the particle from C taken with its proper sign. We must therefore write X = -px n or +n(-x) n .................. (3), according as the particle is on the positive or negative side of the origin G. These are identical if n is odd and in that case the equation (2) holds throughout the motion. If n is even, different equations of motion hold on each side of the origin. The particle arrives at C with a velocity v obtained by putting x = Q in (2). This is a finite velocity if n is positive. After passing C, the equation of motion (1) must be changed to vdv/dx = p ( x) n = fix* 1 ..................... (4), since n is even. We then find (5), the constant of integration being found by the condition that (2) and (5) must agree when x = 0. The equation (5) shows that v is again zero when #= a, so that the particle in its oscillations describes equal arcs on each side of C. After the particle has passed through G on its return journey the equation of motion resumes the form (1). The integration is the same as before, but the constant G must now be determined from the condition that the value of v at the origin is the same as that given by (5). The resulting value of v 2 is however the same as that given by (2), so that the motion on the positive side of the origin is always that represented by (2), and the motion on the negative side that represented by (5). 136. The time of travelling from A to C is given by dx n+1 To integrate this in terms of gamma functions we write x n+1 = a n+l or =a B+1 / ART. 137.] SMALL OSCILLATIONS. 73 according as n + 1 is positive or negative. We then have Ex. 1. A particle starts from rest at a distance a from a centre of force which attracts as the inverse cube of the distance. Show that the time of arriving at the centre is a 2 /^t. Ex. 2. A particle starts from rest at a distance a from a centre of force which attracts inversely as the distance. Prove that the time of arriving at the centre is Small Oscillations and Magnification. 137. Small Oscillations. A particle, constrained to describe a straight line, is under the action of a force tending to a point external to the straight line and varying as some given function of the distance from 0. It is required to discuss the motion when the arc of oscillation decreases without limit. Let OG be a perpendicular on the straight line, P the particle, OC = h, GP = x, OP = r. Let the accelerating force be rf(r). The equation of motion is therefore Since r z = h- + x*, we can expand xf(r) in powers of x. The equation then takes the form dticldP = Ajos + Aj? + ..................... (2), where A lt A 2 , &c. are known constants. Supposing the series to be convergent when x decreases without limit, we may ultimately omit all the terms after the first which does not vanish. Assum- ing x to be initially small we proceed to discuss the subsequent motion. When A 1 is not zero, the equation reduces to d?as/dP = A l as ........................... (3). The motion represented by this equation has been discussed in Art. 119. If A 1 is negative and equal to n 2 , the time of a complete oscillation is 2?r/ri. It appears therefore that when the arc of oscillation is continually diminished, the displacement and velocity of the particle are ultimately zero, but the limiting time 74 SMALL OSCILLATIONS AND MAGNIFICATION. [CHAP. II. is finite. This finite time is called the time of a small oscillation, and the equilibrium position is said to be stable. If A l is positive, we know by Art. 120 that the value of x contains a real exponential and that the motion is not oscillatory. As the displacement x does not remain small we cannot continue to reject the higher terms of the series (2) as compared with the first. The subsequent motion is not represented by equation (3). The equilibrium position is then unstable. If A l = 0, let the first power which does not vanish be the nth. The equation is then ultimately d?xldt*=A n x n (4). This equation has been discussed in Art. 135. If A n is negative the time of oscillation has been found in gamma functions, with a factor a~* (n ~ 1) , where a is the semi-arc of oscillation. The limiting time of oscillation is therefore infinite if n is positive and greater than unity. If A n is positive, the value of x becomes great and the higher powers of x cannot be neglected. 138. Ex. 1. If Saturn's ring were rigid and held at rest show that the position of Saturn placed at its centre would be one of unstable equilibrium for displacements in the plane of the ring. If the force between the ring and the planet were repulsion instead of attraction that position of Saturn would be stable and the time of a small oscillation would be 2ir v /(2a 3 /i), where a is the radius of the ring and M its mass. Show also that the time measured in seconds is 2ir x /(2a 3 /n6 2 <7) where n is the ratio of the mass of the ring to that of the earth, b the radius of the earth, and g is gravity at the surface of the earth, a and b being measured in feet. To prove this, we let x be the distance of Saturn S from the fixed centre C of the ring. Let P be a point on the ring, PCS = 0, SP=p. The attraction on S in M [ adO a cos 6 - x the direction CS is then seen to be F= . Substituting 2air J p 2 p p=a-xco&0, expanding in powers of xja and integrating, we find F=Mxj2a 3 . This force being positive, the equilibrium is unstable. Reversing its sign the time of a complete oscillation follows by Art. 123. The time in seconds is found by using the equation Ejb 2 =g, see Art. 134. Ex. 2. If the ring attract Saturn, show that the central position of the planet is stable for displacements perpendicular to the ring, and that the time of a small oscillation is 1w>J(a?lM). Ex. 3. A particle is in equilibrium under the influence of two centres A, B of repulsion each varying as the inverse nth power of the distance. Prove that the position of equilibrium is stable for displacements in the straight line AB and that the time of a small oscillation is 2ir,J(abln(a + b) F), where a, b are the distances of the particle from A and B, and F is the accelerating repulsion of either force on the particle in the position of equilibrium. ART. 139.] MAGNIFICATION. 75 139. Magnification. A particle, oscillating in a straight line under the action of a centre of force whose acceleration is rPsc, is also acted on by the two accelerating forces X = E cos \t, Y= F cos pi. It is required to find the motion. The equation of motion is d 2 x/dt 2 = -ri*x + E cos \t + F cos pt. The solution of this, by Art. 122, is x = A cos (nt + B) + E' cos \t + F' cos pi, E F where E' = -^, JVf-v w 2 A, 2 n z p? If the particle start from rest at a distance a from the origin when t = 0, we have A = a E' F' and 5 = 0. The motion of the particle is therefore compounded of three oscillations, one has the period ^irjn due to the central force, while the other two have the same periods, viz. 2?r/X and STT//*, as the forces X and Y. This example is important because it shows that the dynamical effects of oscillatory forces are not necessarily in proportion to their magnitudes, but depend also on their periods. Thus the ratio of E' to F' is a function of \ and p as well as of E and F. If the period of the force X is nearly equal to that of the oscillation caused by the central force, ri* X 2 is small, while, if no such near equality hold for the force Y, ri* p? is not small. It follows that if E and F are nearly equal, E' is much greater than F'. If also E and F were so small that the effect of Y on the motion of the particle were insensible, that of X might still be very great. The general result is, that of two forces X, Y, that one produces (cseteris paribus) the greatest oscillation whose period is most nearly equal to the period of the oscillation due to the central force. On the other hand we notice that a near equality between the periods of the forces X and Y has no dynamical significance. The reason is that these forces being explicit functions of the time do not modify each other, each producing its own effect. But the central force, viz. ri*oc, depends on the abscissa of the particle and this is more or less altered by the action of the forces X and Y. The solution shows that the alteration is considerable when 76 SMALL OSCILLATIONS AND MAGNIFICATION. [CHAP. II. the period of either X or Y is nearly equal to that due to the central force alone. If the period of X is exactly equal to that of the oscillation due to the central force the solution of the differential equation takes a different form. By reference to Art. 122 we see that ry x = a cos nt + =- sin nt + F' cos pt, so that the amplitude of the oscillation becomes very great as t increases. We may also notice that if \ is very great the terms which contain E' as a factor are very small. It follows that an oscillatory force whose period is very short produces very little effect on the motion of the particle. 140. As an example of these effects consider how great an oscillation can be generated in a heavy swing by a series of little pushes and pulls if properly timed. If we push when the swing is receding and pull when it is approaching us, the motion is continually increased and the amplitude of the oscillations becomes greater at each succeeding swing. Such a series of alternations of push and pull is practically an oscillatory force, such as X, whose period is exactly equal to that of the swing. If however the alternations of push and pull follow each other at an interval only nearly equal to that of the period of the swing, a tune will come when the effects are reversed. The push will be given when the swing is approach- ing us and the pull when the swing is receding. Thus, though a great oscillation of the swing is at first produced, that oscillation will be presently destroyed only to be again reproduced and so on continually. 141. Second approximations. In determining the small oscillations of a particle in Art. 137, it is explained that the terms containing x 2 , &c. are usually neglected. These terms are indeed very small in the differential equation, but we know from Art. 139 that their effects may in certain conditions be so magnified that they become perceptible in the value of x. It is therefore sometimes necessary to proceed to a second or a third approximation before we can find a value of x which represents the actual motion. Some examples of this will be given later on, but the reader will find the theory given at length in the Author's Rigid Dynamics, vol. ii. chap. vn. 142. Ex. A heavy particle P is suspended at rest from a point A by an elastic string whose initial and unstretched length is a. The point A at the time t=0 begins to oscillate up and down, so that its displacement (measured downwards) at the time t is c sin \t. Prove that the length of the string at, the time t is a + ^ (1 - cos nt) - J-^ sin nt + ^-^ sin \t. Discuss the interpretation of this result (1) when X is nearly equal to n, and (2) when X is very great. Notice that if cPxjdt 2 is to be the acceleration of P, x must be measured from a point fixed in space, say the initial position of A . ART. 143.] CHORDS OF QUICKEST DESCENT. 77 Chords of quickest descent. 143. To find the straight lines of quickest and slowest descent from rest at a given point to a given curve. The straight line is supposed to be smooth and the motion to be in vacuo. The solution of this problem depends on the theorem that the curve which possesses the property, that the times of descent Fig. 1. Fig. 2. down all radii vectores from rest at are equal, is a circle having for the highest or lowest point. See Art. 106. Describe a circle having its highest point at and touching the given curve in some point P. There are two cases, according as the circle touches the given curve on one side or the other. These are represented in figures (1) and (2). If OQ be any chord cutting the circle in R, the time down OP is equal to the time down OR and is therefore less than the time down OQ in fig. (1) and greater than that time in fig. (2). Thus OP is the chord of quickest or slowest descent according to the mode in which the circle of construction touches the given curve. If C is the centre of the circle, the angles CPO and COP are i ... . equal. Since CO is vertical the chord of quickest or slowest descent from rest at meets the given curve at a point P such that OP bisects the angle between the vertical and normal at P. If the position of a point P on a given curve is required such that the time of descent from P to a given point is a maximum or minimum, we follow the same construction except that is to be the lowest point of the circle of construction instead of the highest. The result is that the particle must start from a point P such that PO bisects the angle between the vertical and normal at P. 78 CHORDS OF QUICKEST DESCENT. [CHAP. II. 144. To find the chords of quickest and slowest descent from rest at one given curve to another given curve. Let PQ be the required chord. Then since the time down PQ is less than the time down any neighbouring chord drawn from P to the other curve, PQ must bisect the angle between the normal and vertical at Q. Similarly by fixing Q and varying P we see that PQ must bisect the angle between normal and vertical at P. The points P, Q are therefore such that they satisfy these two conditions, (1) the normals at P, Q are parallel, (2) the chord makes equal angles with each normal and the vertical. 145. To find the chord of quickest descent from rest in a medium whose resistance varies as the velocity we use the same construction, because the times of descent down all chords of a circle from rest at the highest point are equal. Art. 107. If the resistance vary as the square of the velocity the curve which possesses the property of equal times for the chords is not a circle ; see Art. 110, Ex. 4. The geometrical construction is therefore inapplicable. 146. If the chords of quickest or slowest descent are rough we slightly modify the rule. To find the rough chord of quickest descent from to a given curve we describe a circle to touch the given curve in some point P, but such that the diameter through O makes an angle with the vertical equal to the angle of friction, Art. 105. The result is that the required chord meets the curve at a point P such that OP makes equal angles with the normal at P and a straight line inclined to the vertical at the angle of friction. 147. Ex. 1. A point A and a straight line BC are given in the same vertical plane. Show how to draw (when possible) a straight line from A to BC, so that the time of descent from rest under gravity may be equal to a given time (. When there are two such lines, intersecting BC in P and Q, prove that the radius of the circle described about APQ is gt 2 . Ex. 2. Two parabolas are placed in the same vertical plane with their foci coin- cident, axes vertical and vertices downwards. Prove that the chord of quickest descent from the outer to the inner parabola passes through the focus and makes an angle equal to J ir with the axis. The normals at the extremities of the chords are parallel and the parabolas are similar. The chord therefore passes through the centre of similitude, i.e. the focus S. If PG be a normal, the second condition of Art. 144 shows that the tri- angle SP6 is equilateral, i.e. each angle is equal to TT. ART. 147.] EXAMPLES. 79 Ex. 3. Find the smooth chord along which a particle must travel starting from rest at some point on one given curve and ending at another given curve, so that the velocity acquired may be a max-min. The force acting on the particle tends to a fixed centre O and varies as some function of the distance from O. The result is that if P be either extremity of the required chord, either the force is zero at P or OP is a normal to the given curve at P. To prove this, let the central force be f'(r). We then find v- = '2f(r 1 ) -2/(r 2 ) where r lt r z are the distances of the extremities P, Q from 0. Fixing Q let us vary P along the arc (as in Art. 144), then p-* (3). It follows that the time of arriving at the centre of force O varies as the (1 - n)th power of the initial distance. If the central force vary as the distance, w=l and the time of arrival at O is the same for all initial distances; a theorem which has been proved in Art. 136 by integrating the equation of motion. If the central force vary according to the Newtonian law, n 2 and the square of the time varies as the cube of the initial distance, a result in accordance with one of Kepler's laws. The symbol A represents a number and as it has no dimensions its magnitude cannot be deduced from the theory of dimensions. 153. Ex. 1. A particle moves with an acceleration g, prove that the velocity acquired in describing a space s varies as *J(gs), and that the time varies as V( s /#)- Ex. 2. A particle starts from rest at a given distance from a centre of force whose attraction varies as the distance and moves in a medium whose resistance varies as the velocity. Prove that the time of arriving at the centre of force is independent of the initial distance. See Art. 126. Ex. 3. A particle P moves from rest under the action of a constant accelerat- ing force / and a centre of force whose attraction is /j. times the distance, both tending to the same point and the initial distance OP=a. Prove that t where t is the time of arrival at O. 62 CHAPTER III. MOTION OF PROJECTILES. Parabolic Motion. 154. General principle. The particle moves under the action of a force which, being fixed in direction and magnitude, is independent of the position of the particle. It follows that all the circumstances of the motion parallel to any fixed direction are independent of those of the motion parallel to any other direction. These circumstances may therefore be deduced from the formulae for rectilinear motion by taking account solely of the resolved initial velocity and the resolved force of gravity. 155. Cartesian axes. Let the particle be projected from a point with an initial velocity V in a direction making an angle a with the horizon. Let v be the velocity at any point P of the path; v x , v y its horizontal and vertical components. Consider the horizontal motion. Since the component of gravity in this direction is zero, the horizontal velocity is constant throughout the motion and is equal to V cos o. We therefore have x = V cos at, v x =V cos a (1). This gives an obvious and useful rule to find the time of describing any arc of the trajectory, viz. the time of transit is equal to the horizontal space divided by the horizontal velocity. Consider next the vertical motion. Since the component of gravity is g we infer from the formulas of rectilinear motion (Art. 25) that y = Vsmat-^gF, vj = V' 2 sin 2 a - 2gy (2). ART. 156.] PARABOLIC MOTION. 85 The Cartesian equation of the path is found by eliminating t between (1) and (2) ; we have y = x tan a \ga?l V 2 cos 2 a. (3). This is the well-known equation of a parabola. To find the greatest altitude of the particle. We consider only the descending motion; the particle starts downwards with a zero vertical velocity and arrives at the level of the original point of projection with a vertical velocity which, by the theory of rectilinear motion, is equal to that with which it was projected upwards. If h is the greatest altitude we have V 2 sin 2 a = 2gh. To find the time of /light. We again consider the vertical descending motion, disregarding the horizontal motion. If T be the time of ascent and descent, we have V sin a = ^gT. To find the range on a horizontal plane. We consider the horizontal motion ; the constant horizontal velocity is V cos a, and the time of flight has just been found. The range is there- fore V' 2 sin 2a/<7. The range is greatest for a given velocity when the direction of projection makes an angle of 45 with the horizon, and continually decreases as the angle increases to a right angle or decreases to zero. 156. When the motion with regard to an inclined plane passing through the point of projection is required, it is useful to take the axis of x along the line of greatest slope and the axis of y perpendicular to the inclined plane. If the direction of projection is not in the plane of xy, let F and W be the components of the velocity in and perpendicular to that plane. The motion perpendicular to the plane of xy is uniform and z = Wt. Turning our attention to the motion in the plane of xy, let 7 be the angle the direction of the velocity V makes with Ox and (3 the inclination of the plane to the horizon. The initial component velocities being Fcos . SP . rj. We deduce the following useful rule. The velocity at any point P of the path is that due to the distance of P from either the focus or directrix. Since the velocity at the highest point of the path is equal to the horizontal velocity, it follows that one quarter of the lotus rectum, i.e. AS, is equal to V 2 co$?a/2g. See Art. 155. We have also another formula to find the time of transit along any arc PQ. Let the vertical at either end, say Q, intersect the tangent at the other end in T, then the time of describing the arc PQ is the $ame as that of describing QT from rest under the action of gravity. It is also the same as that of describing PT with a uniform velocity equal to that at P. 158. Ex. 1. If three heavy particles be projected simultaneously from the same point in any directions with any velocities, prove that the plane passing through them will always remain parallel to itself. [Math. T. 1847.] If gravity did not act, the plane of the particles would be always parallel to a fixed plane. When gravity acts each particle is pulled through the same vertical space in the same time, hence the theorem remains true. ART. 159.] POINT TO POINT. 87 Ex. 2. Two tangents PE, QR are drawn to a parabolic trajectory, prove (1) that the velocities at P and Q are proportional to the lengths of those tangents, and (2) that the vertical through R divides the arc PQ into two parts which are described in equal times. Draw QT vertically to intersect the tangent PR in T. Then by the triangle of velocities, the sides RT, RQ, TQ represent in direction and magnitude the velocity at P, that at Q, and that added on by gravity during the time of transit. Since the diameter through R bisects the chord PQ, the results given above follow easily. Ex. 3. Two balls A, B equal in all respects are on the same horizontal line. The ball A is projected towards B with velocity v, while at the same instant B is let fall. Prove that the balls will impinge and that after impact, the coefficient of restitution being unity, A will fall vertically and Pi will describe a parabola of latus rectum 2v^g. [Coll. Ex. 1895.] The balls will impinge because the straight line joining their centres moves parallel to itself. At impact they exchange their horizontal velocities. Ex. 4. If v, v', v" are the velocities at three points P, Q, R of the path of a projectile, where the inclinations to the horizon are a, a - j8, a - 2j8, and if t, t' be the times of describing PQ, QR respectively, prove that v"t = vf, 1 + 4 = r- [Math. T. 1847.] V V V Besolve along and perpendicular to the middle tangent. Ex. 5. Three heavy particles P, Q, R are projected at equal intervals of time from the same point to describe the same parabola. Prove that the locus of the intersection of the tangents at P, R is a parabola. Prove also (1) that at any time t after the projection of Q, the tangent at Q is parallel to PR, (2) that each of these lines is parallel to the straight line joining to the position of Q at the time 2(. 159. To project a particle from a yiven point P with a given velocity V so that it shall pass through another given point Q. The velocity at P being known the common directrix HK of all parabolic paths from P to Q is constructed by drawing a horizontal at an altitude V*/2g above P. With centres P, Q and radii PH, QK we describe two circles intersecting in 8 and 8'. Then 8, 8' are the foci of the parabolic trajectories which could be de- scribed from P to Q. There are therefore two parabolic paths. The two foci are at equal distances from the chord PQ, one lying on each side. The two directions of projection may be found by bisecting the angles HPS and HPS'. If y 1} y< 2 are the angles these directions of projection make with the chord PQ, and 88 PARABOLIC MOTION. [CHAP. lit. ft the angle PQ makes with the horizon, it easily follows that We notice that the three sides of the triangle PSQ are known, viz. PS=V*/2g, if y be the altitude of Q above P, Q8 = PS-y, and PQ is the known distance of Q from P. It is clear that when PQ is greater than the sum of the radii PH, QK, the two trajectories are imaginary. The greatest possible distance of Q from P in any given direction PQ is found by making the foci 8, S' coincide and lie on PQ. In this case PH+ QK = PQ. Drawing a horizontal line H'K' above HK so that HH'=PH, it immediately follows that QK' QP. The locus of Q is therefore a parabola whose focus is P and directrix H'K'. This new para- bola therefore touches HK at its vertex H. It is represented in the figure by the dotted line. Unless the point Q lie within the space enclosed by this parabola, it is impossible to project a particle from P with the given velocity V, so that it shall pass through Q. If the particle is to be projected from P with the least velocity which will enable it to reach Q, the direction of projection must bisect the angle HPQ and Y 2 = g (r + y), where r is the distance PQ. 16O. Ex. 1. A particle is projected from a point P with velocity V, so as to pass through a point Q whose coordinates referred to P as origin are x, y, the axis of y being vertical. Prove that the directions of projection are given by the quadratic 2J/2 2F 2 tan 2 a -- tana + lH --- .5- =0, gx gx* and that the two times of transit are the positive roots of g*t* - 4 ( F 2 - gy) t 2 + 4r 2 = 0. Prove that the product of the times of transit is independent of the initial velocity F and is equal to the square of the time occupied by a particle falling from rest vertically through a distance equal to PQ. Prove also that the polar equation of the bounding parabola is V 2 lgr=l + cos&, where the origin is at P and is the angle r makes with the vertical. See Arts. 154 and 155. Ex. 2. Prove that every parabolic trajectory meets the bounding parabola in a point whose abscissa is x = 2h cot o, and whose depth below the directrix of the trajectory is h cot 2 a, where h is the height of that directrix above the point of projection. If they meet, the curves must touch for otherwise it would be possible to find a trajectory which would pass through a point beyond the boundary. ART. 160.] EXAMPLES. 89 Ex. 3. The point P being fixed and Q having any position, the tangents at P, Q to one parabolic path from P to Q meet in T, those to the other in 2", the velocity at P being given. Prove that the locus of the middle point of TT' is the directrix of either parabola. Prove also that for either parabolic path, the velocities at P, Q are as PTto TQ, and for the two paths the times of transit from P to Q are as P2' to PT'. Ex. 4. A fort of vertical height /c stands on a plane hill-side which makes an angle a with the horizon. Prove that a gun which can fire with muzzle velocity V from the top of the fort commands a district whose shape is an ellipse of eccentricity sin a, and whose area is TTSCC a F 2 (F 2 sec 2 a + 2kg) /# 2 . [Coll. Ex. 1896.] The paraboloid whose focus is the top of the fort and whose directrix plane is at an altitude V 2 jg is the boundary of all places which the shot can reach, Art. 159. The paraboloid cuts the plane hill-side in an ellipse whose projection on a horizontal plane is a circle. The rest follows easily. Ex. 5. At a horizontal distance a from a gun there is a wall of height h which is greater than a-ga^/v 2 ; prove that if the shot be fired off with a velocity v in a vertical plane at right angles to that of the wall, there will be a distance on the other side of the wall commanded by the gun equal to - - ^ ^&. (*** - a2 # 2 ~ 2ft 2 a)^, g (a + h ) provided this expression is real. [Coll. Ex. 1893.] Ex. 6. A particle is projected with velocity V along a straight frictionless tube of length I, inclined at an angle a to the horizontal, and after leaving the tube it describes a parabolic trajectory : prove that its range on the horizontal plane through F' 2 f / 2gl \b] the point of projection is I cos cH cos a sin a -j 1 + 1 1 + ^r, 2 . j j- , where F' 2 = F 2 - 2gl sin a. [Coll. Ex. 1893.] Ex. 7. Two smooth planes are at right angles with their edge of intersection horizontal and are equally inclined to the horizon. Prove that a perfectly elastic particle projected horizontally in a direction perpendicular to the common edge from a point vertically above it will return to its original position after two rebounds. [Coll. Ex. 1896.] Ex. 8. Two parabolas have their axes vertical and vertices downwards and the focus of each curve is on the other. A particle, whose coefficient of restitution is unity, is projected so as to rebound from the curves at each focus in succession ; prove that it will after the second rebound pass through its point of projection and follow its original path again. [Coll. Ex. 1897.] Ex. 9. Two particles are projected from the same point at the same instant with velocities v, v', and in directions o, a'. Prove that the time which elapses between their transits through the other point which is common to both their paths is ? ^n(a -.0_ [Math . T> 184L] g v cos o + v' cos a Ex. 10. A man travelling round a circle of radius a at speed v throws a ball from his hand at height h above the ground with a relative velocity F so that it alights at the centre of the circle. Prove that the least possible value of F is given by V*=v 2 + g{J(a? + h?)-h}. [Coll. Ex. 1896.] If the man were stationary, the least value of F 2 is given in Art. 159. To find the relative velocity we add to this ( - v) 2 . 90 PARABOLIC MOTION. [CHAP. in. 161. Ex. 1. A particle is projected from a point A with a velocity V in a direction making an angle a with the horizon. After rebounding from a vertical wall, elasticity e, it hits the ground, elasticity e'. Find the condition that after the second rebound the particle may pass through A. Problems of this kind are solved by considering the motion in two directions separately and equating some element (usually the time) common to both motions. Consider first the horizontal motion ; the blow at C is vertical and does not affect the horizontal motion, but the blow at B must be taken account of. Let ON=h, and let t 1? t 2 be the times of transit along the arc AB and the broken arc BCA. Then h=Vcoaat 1 , and the horizontal velocity of the rebound at B being eFcosa, we have also h = eVcosat. The whole time is h Kcoset Consider next the vertical motion, the blow at B may now be neglected while that at C has to be allowed for. Let t, , t 4 be the times of transit along A BC and GA. If k= AN \vel\a\e - k = V sin a< 3 - gt 3 2 . One root of this quadratic is negative and the other is positive. The former indicates the time before leaving A at which the particle might have passed the level of the ground and is here inadmissible. We take the positive root. If V be the vertical velocity of arrival at C taken positively, V'-= F 2 sina + 2gk, k = e'V't i - %gtf. Both the values of < 4 thus found are positive, and give the times of transit from C to A according as the particle passes through A on the up or on the down journey. Taking both these values we see that the required condition is found by equating ( x + U to either of the values of t 3 + t 4 . Ex. 2. A ball is projected from a point A on the floor of a room, so as to rebound from the wall (elasticity e) and hit a given point B on the floor. Let the intersection of the floor and wall be the axis of y and let A be on the axis of x. If M, v, w be the components of velocity of projection and .T, y the given coordinates of B, prove that euy=eva + vx, and 2viv=gy. Ex. 3. A particle is projected from a given point O on an inclined plane in a direction making an angle y with the plane, the inclination of the plane being /3. Investigate the condition that the particle passes through O at the nth impact. We consider the motions parallel and perpendicular to the plane separately. The motion parallel to the plane is not affected by the impacts. If T represent the whole time of transit from O to O again, we have V cos y = ^g sin /3Z 1 . The motion perpendicular to the plane is affected by each impact. The particle starts with a velocity Vainy, hits the plane at A l with the same normal ART. 161.] EXAMPLES. 91 velocity after a time T 1 , where V sin 7 = \ g cos /31\ . The particle rebounds with a perpendicular velocity eVsiny and the time of transit from A l to A. 2 is found as before. The whole time of transit is therefore Tj + r 2 + &c. = {2 V sin yjg cos 8} (1 + e + . . . + e n ~ l ). Equating the two complete times, we have the condition cot 7 . cot B = (1 - )/(! - e), which we notice is independent of the velocity of projection. Let B lt B. 2 , &c. be the points at which the tangents to the path are parallel to the inclined plane. The time of transit from O to B l is obviously equal to \ T 1 , while that from B l to B 2 is ^ (T 1 + T 2 ), and so on. If C 1 be the point at which the tangent is perpendicular to the plane, the time from to C l is clearly equal to i T. Ex. 4. A ball whose elasticity is e is projected with a velocity V and rebounds from an inclined plane which passes through the point of projection. If R lt R 2 , R 3 be any three consecutive ranges on the inclined plane, prove that R 3 -(e + e 2 )R^ + e s R 1 = 0. [Math. T. 1842.] Ex, 5. At two points A, B of a parabolic path the directions of motion are at right angles. If D be the distance AB, 6 the inclination to the horizon, V the velocity at A or B, prove that V 2 =gD (1 sin 6). Ex. 6. A particle is projected from a point on a rough horizontal plane with a velocity equal to that which would be acquired in falling freely through a height h, and in a direction making an angle a with the plane. The particle is inelastic and the coefficients of both the frictions are taken equal to unity, prove that the range from the point of projection to where the particle comes to rest is equal to h (1 + sin 2o). [Coll. Ex. 1897.] The particle describes a parabola with a range 2h sin 2a. On arriving at the plane, there is an impulsive friction which reduces the horizontal velocity from ucosa to v' = v cos a - v sin a. After describing a space a', when v' 2 = 2gs', the particle is reduced to rest by the finite friction. The whole range is 2/jsin 2a + s'. Ex. 1. A perfectly elastic particle slides down a length I of a smooth fixed inclined plane, and strikes a smooth rigid horizontal plane passing through the foot of the inclined plane. Prove that the maximum range of the ensuing parabolic path, as the inclination of the inclined plane is varied, is 82/3^/3. [Coll. Ex. 1896.] Ex. 8. A smooth inclined plane of mass M, inclined to the horizon at an angle a, is free to move parallel to a vertical plane through the line of greatest slope. A particle, mass m, is projected from a point in the lowest edge, up the face of the plane with a velocity V making an angle ft with the line of greatest . V- sin 28 M + m sin 2 a slope. Prove that the range of the particle on the plane is r . . //sin a M+m [Coll. Ex. 1897.] Ex. 9. Two inclined planes intersect in a horizontal line, their inclinations to the horizon being a and /3; if a particle be projected at right angles to the former from a point in it so as to strike the other at right angles, the velocity of projection is sin 8 [2#a/ { sin a - sin 8 cos (a + B)} ] , a being the distance of the point of projection from the intersection of the planes. 92 RESISTANCE VARIES AS THE VELOCITY. [CHAP. III. Ex. 10. A heavy particle descends the outside of a circular arc whose plane is vertical. Prove that when it leaves the circle at some point Q to describe a para- bola the circle is the circle of curvature at Q of the parabola. Thence show that the chord of intersection QR of the circle and parabola and the tangent at Q make equal angles with the vertical. Prove also that the axis of the parabola divides the chord QR in the ratio 3:1. The first part follows from Art. 36. Since the pressure is zero at Q, v*]p, and therefore p, must be the same for the circle and the parabola. The rest follows from conies. Ex. 11. A particle projected horizontally from the lowest point A of a circle whose plane is vertical leaves the circle at C and after describing a portion of a parabola intersects the circle at D. If B is the highest point of the circle prove that the arc El) is three times the arc BC. [Despeyrous, Cours de Mec.] Ex. 12. A particle is projected so as to enter in the direction of its length a smooth straight tube of small bore fixed at an angle 45 to the horizon and to pass out at the other end of the tube; prove that the latera recta of its path before entering and after leaving the tube differ by ^/2 times the length of the tube. [Math. Tripos, 1887.] Ex. 13. A man standing on the edge of a cliff throws a stone with given velocity u at a given inclination in a plane perpendicular to the edge. After an interval r he throws from the same spot another stone, with given velocity v at an angle i IT + 6 with the line of discharge of the first stone and in the same plane. Find T so that the stones may strike each other ; and prove that the maximum value of T for different values of 6 is Zv^jgw, and occurs when sin0=v/M, ? being tf's vertical component. [Math. Tripos, 1886.] Ex. 14. A particle is projected from the highest point of a sphere of radius c so as to clear the sphere. Prove that the velocity of projection cannot be less than [Math. Tripos, 1893.] Resistance vanes as the velocity. 162. To determine the motion of a heavy particle when the resistance of the medium varies as the velocity. Let the particle be projected from any point with a velocity V in a direction inclined at an angle a to the horizon. The equations of motion are d*x _ doc d*y _ dy : . dx/dt + KX = V cos a, dyfdt + tey = gt+V sin a. Both these equations are of the linear form, multiplying by e? 1 and integrating, we find KX = Fcos a (1 e~ Kt ) Ky = -gt + (Vsmct + L)(l-e- ART. 164.] THE RESULTANT ACCELERATION. 93 where rcL = g, so that L is the limiting velocity, Art. 111. The horizontal and vertical velocities at any time t are dxjdt = V cos ae- Kt , dyfdt = -L + ( Fsin a + L) e~ Kt . . .(2). 163. From these equations we deduce the general character- istics of the motion. We notice that when t is in- finite KX = V cos a. There is therefore a vertical asymp- tote at a horizontal distance OH = Fcos a/ K from the ori- gin. Let the tangent at intersect the asymptote in T , then OT = V/K and V K . OT . Since any point P may be taken as the origin, it follows that the velocity at any point P is proportional to the length PT of the tangent at P cut off by the vertical asymptote. Tracing the curve backwards we make t = oo ; we then find that both x and y are infinitely great. Since the exponential is infinitely greater than t, both yjx and dyjdx have ultimately the same ratio. Representing this ratio by tan ft, we have tan/3 = tan a + L/V cos a (3). The curve has therefore an infinite branch, the tangent or asymp- tote to which makes an angle ft with the horizon, determined from the initial conditions by this equation. This asymptote is at an infinite distance from the origin. 164. Eliminating the exponentials from the values of x and y, Art. 162, we find y = x tan ft Lt (4), a linear equation which must hold throughout the motion. Drawing a straight line OB parallel to the oblique asymptote, this equation shows that the vertical distance of P from OB is PB = Lt, where L is the limiting velocity. The perpendicular distance of P from OB being Lt cos ft, the resolved velocity at P perpendicular to the oblique asymptote is constant. The resultant acceleration at P is therefore parallel to BO. 94 RESISTANCE VARIES AS THE VELOCITY. [CHAP. III. 165. General principle. Since the resistance varies as the velocity, the resolved resistance in any direction is proportional to the resolved velocity in the same direction. The general principle proved in Art. 154 for motion in a vacuum will therefore apply to the motion with this law of resistance. The circumstances of the motion parallel to any fixed straight line are independent of those in any other direction. 166. Let the particle be projected from a distant point E on the oblique branch with such a velocity that it describes the trajectory. Consider the oblique resolution of the motion in the direction of (1) the tangent or asymptote at E and (2) the vertical. In the former motion the particle is acted on only by the resistance, and the acceleration at any time is therefore - KM, where u is the oblique component of velocity parallel to the asymptote. In the latter motion the particle starting from rest is acted on by gravity as well as by the resistance and has thus acquired its limiting velocity L. This component is constant in direction and magnitude so that the acceleration is zero. Combining these two motions, we see that in any position P of the particle, the velocity v along the tangent PT is the resultant of the vertical limiting velocity L and a velocity u parallel to the oblique asymptote. If U and u be cor- responding velocities at any two points O and P of the trajectory, u=Ue~ "*, where t is the time of transit from O to P; Art. 102. We also notice that the resultant acceleration at P is equal to -KM. Taking a parallel OB to the oblique asymptote and the vertical as axes of reference we have = [7 (1-e -")/*, r,=Lt (1), where = OB, t) = BP. If we refer the motion to the tangent at O and the vertical as axes, we have ' = OA, -q' = AP. We find by considering the motions in these directions separately K?=V(l-e-* t ), KT>'=gt-L(l-e- Kt ) (2). 167. Ex. 1. Particles are projected from a given point at the same instant with equal velocities in different directions ; prove that the locus at any time is a sphere. Refer the motion of any particle to the tangent OA and the vertical as axes of , i\. Both |, T\ are evidently functions of t which are independent of a. The locus is therefore a sphere whose radius is and whose centre is at a depth TJ below O. Art. 166. Ex. 2. Particles are projected from a given point O at the same instant with different velocities in the same direction OA, prove that at any subsequent time their locus is a straight line parallel to OA. Art. 165. Ex. 3. If the axis of x is inclined at an angle i to the horizon and the direc- tion of projection make an angle y with x, prove that KX= - g sin it + (V cos y + L sin i) (1 - e~ Kt )) KIJ = - g cos i t + (V sin y + L cos t) (1 - e ~ "Of ART. 168.] THE THREE EQUATIONS. 95 It' M be the point at which the tangent is parallel to x, prove that the time t^ of reaching M and the coordinates of M are ,* F sin y l = l - = - < V sm 7, j , i / j COS / ^(F cos 7 + 1, sint)= LFf cos (1 + 7); the latter equation being also true for all points on the trajectory. Ex. 4. A projectile moves under gravity in a uniform medium whose resistance varies as the velocity. Prove that the hodograph of the trajectory is a straight line and that the velocity of the point on the hodograph is proportional to the horizontal velocity of the projectile. [Coll. Ex.] Resistance varies as the ?i th power of the velocity. 168. To find the motion of a heavy particle, when the resistance varies as the w th poiver of the velocity. Let i/r be the angle the tangent at any point P of the path makes with the horizontal, p the radius of curvature measured positively downwards so that p = ds/d-^r. Let v be the velocity, u the horizontal component. Following John Bernoulli, 1721, we resolve the motion normally and horizontally, we thus have v 2 du = g COS "Y, -j- = KV n COS -ur. p dt Since v cos ty = u and p = vdt/d-^, these become dt u du KU H oty g cos 2 ty ' dt (cos We obtain one integral by eliminating dt, du K u V l+1 ! lcn .(A). a (cos ^) n+1 ' where a is the angle the initial direction of motion makes with the horizon, and u the initial horizontal velocity. To effect this integration we put ^ = tan^, we then have, except n = 0, l._J-*_^/(l+n*<- 1 >cto ............... (B), u n u n g J v ^ ' the sign of the radical when n is even being such that the subject of integration is positive between the limits ty = a and i/r = | TT, i.e. p =p and p = oc . 96 RESISTANCE VARIES AS W th POWER OF VELOCITY. [CHAP. III. We can conveniently take either u or p as the independent variable, and thus we obtain the two sets of relations, du i r t , if du X = U*dp = I r-yr j gj K)u n - l (l + p-)t(n- 11 ~ I u*pdp = / , . ,, n i f f ^ i n,n 1/1 i ,v2\*(n i) The first follows from equation (A), the second and third from the obvious relations dx udt, dy = updt. The limits in all the integrals being p =p to p or u = U Q to u. In this manner all the circumstances of the motion can be expressed in terms of one independent variable which may be either p or u. It is evident that the integral (B) has considerable importance in this theory. Putting we see that when n = 2 or n 3, W z = ${p(I +frf + log (p + (1 +^ 2 )*)}, W a = ' We may also find a general formula of reduction, viz. When the resistance is a constant force, say Kg, n = 0, and the integral (B) takes the form 'M\ S _ /1+sin^y i a/ \1 sini/r/ ' where a is the velocity when the particle is moving horizontally. 169. The equations (C) have been applied to the calculation of the trajectories of shot in various ways*. When the angle of elevation is not more than 10 to 15, as in the case of direct fire, * Bashforth, Phil. Trans. 1868, Treatise on the motion of projectiles, 1873; supplement, 1881. Proceedings of R.A. Institution, 1871 and 1885. W. D. Niven, On the calculation of the trajectories of shot, Proceedings of the Royal Society, 1877. Ingall, Exterior Ballistics, 1885. An account of Siacci's method is given by Greenhill in the Proc. of the R. A. Institution, vol. xvn. See also Artillery, its progress and present position by E. W. Lloyd and A. G. Hadcock, 1893. Greenhill, On the motion of a projectile in a resisting medium, Proceedings of the R. A. Institution, vols. xi., xn., xiv., 1880 to 1886. AKT. 171.] THE LAW OF RESISTANCE. 97 we may regard the trajectory as so flat that we can reject the square of p. Taking u as the independent variable the integration can then be effected without difficulty. When the path is more inclined we can divide the whole path into subsidiary arcs for each of which p may be regarded as approximately constant though of a different value in each arc. If the arcs were small enough the initial value of p in each arc might be taken as the proper value for that arc. For longer arcs it becomes necessary to give p a mean value taken over the whole subsidiary arc. 17O. In artillery practice the values of the integrals (C) are commonly inferred from tables especially constructed for that purpose, different tables being used to find t, x and y. Opinions differ as to the best methods of constructing and using these tables. Bashforth represents the law of resistance by KV S where K is a function of the velocity whose values are deduced from experiment. These values for a shot of given cross section and weight and for air of given density are tabulated for every few feet of velocity. In effecting the integrations (C) the quantity K is regarded as constant and in a long arc a value suitable to a mean velocity over the arc has to be found. This difficulty having been overcome, the integrals (C) are tabulated for different values of K and between certain ranges of angle. In the Italian method a quantity allied to the velocity is taken as the indepen- dent variable. To enable the integrations to be effected the quantity p is taken as constant throughout the subsidiary arc. The integrals (C) are then determined either by the use of tables or by giving the index n the value suitable to the range of velocity in the trajectory. An account of the methods of constructing and using these various tables would take us too far from our present subject. We must refer the reader to special treatises on Artillery. 171. Law of resistance. Many attempts have been made to discover the law of resistance to the motion of projectiles. Passing over the earlier experiments of Robins and Hutton we may mention as the most important the long-continued series made by F. Bashforth with the help of his chronograph. By this instrument the times taken by the same projectile in passing over a succession of equal spaces can be measured with great accuracy. Other experiments have also been made on the con- tinent, for example by Mayevski in 1881. It appears from all these experiments that the resistance cannot be expressed by any one power of the velocity. The general result is that for low and high velocities the resistance varies as the square of the velocity, and for intervening velocities as the cube and even a higher power of the velocity. R. D. 7 98 RESISTANCE VARIES AS 71 th POWER OF VELOCITY. [CHAP. III. To be more particular, let v be the velocity measured in feet per second, d the diameter of the ogival headed shot in inches, w the weight in pounds. Then taking the resistance to be ^2 / y \n ft ( TQOO ) ' Bashforth's experiments show that v < 850 ?i = 2 13= 61-3 v> 850 < 1040 w = 3 ft = 74'4 v> 1040 < 1100 ?z = 6 ft = 79-2 v> 1100 < 1300 = 3 = 108'8 v> 1300 < 2780 w = 2 /3= 141-5. Mayevski's experiments led to similar results except that the highest power of n was n 5. The values of ft were also different because the shots were more pointed than in those of Bashforth. We may notice that though the resistance for low and high velocities follows the same general law, yet the value of the coefficient ft is much greater for the high than the low velocities. When the velocity of the shot approximates to that of the velocity of sound in air, we might expect a considerable change in the law of resistance and this is shown in the results given above. 172. To discuss the motion when the resistance varies as the square of the velocity. In this case we can obtain two first integrals of the equations of motion. Resolving normally and horizontally as before, we find v- du ds -=gco&\!/, -5-= -KV-COSIJ/= -KU ..................... (1). p (I! it! Dividing the latter equation by u and integrating logu=A-KS, .: u = u e~ KS ........................... (2), where w is the horizontal velocity at the point O of projection and is measured from 0. Besides this we have the integral (B) already obtained in the general case by eliminating dt from the equations of motion. Writing p= - ds/d\f/ and v cos $=u, in (1), we find as before dt u du KU 2 ^ * '' dif/~ ~0cosV dt~ coa\f/ 1 2* f df 2/c f .., , .-. ,= -- I ^-r= --- I VO- + F )dp ..................... (4), u 2 g J cos 3 f g J * where p = tan ^ as before, and the radical is to be taken positively. Integrating (5). ART. 174.] THE SQUARE OF THE VELOCITY. 99 Eliminating u between (2) and (5) we find = C .............. (6). s Ku o This is the intrinsic equation of the path. 173. To discuss the form of the curve it will be convenient to place the origin at the highest point so that initially p = 0. We then have ............... (7). When s increases to positive infinity we see from (2) and (7) that u tends to zero and p to minus infinity. Since by (3) or (C) gdt= -udp and gdx=gudt= -u 2 dp, it follows that both dt/dp and dxjdp are ultimately zero. We shall prove that while t becomes ultimately infinite, x tends to a finite limit. We therefore infer that the curve has a vertical asymptote at a finite distance on the positive side of the highest point. To prove this we refer to (5) and retaining only the highest powers of p, we see that 1/w 2 is of the order p 2 . Putting u = bjp when p is very great, we find gt = - \udp = - b log p, gx= - Jw 2 dp = Wjp. Taking these between the limits p = p to infinity where p l is any large finite quantity, the first gives the time the particle takes to travel from the position defined by p =p to that defined by p = - ao , and the second gives the corresponding horizontal space. We see that the first is infinite and the second finite. 174. Consider next the other extremity of the trajectory. When the arc is negatively very great, we see by (2) that u is positive and infinite. It also follows by (7) that p tends to a limit m given by the equation - fi r/KM 2 + m x /(l + m 2 )+log(m + V(l + a 2 )) = .................. (8). Since the left-hand side passes from a negative quantity to positive infinity as m varies from zero to infinity, it is clear that this equation has at least one positive root. If the equation could have two real roots, the differential coefficient of the left-hand side would vanish for some intervening value of m. But since the differential coefficient is 2 N /(l + m 2 ) this is impossible. It follows that the curve on the negative side of the highest point has an asymptote inclined at a finite angle to the horizon. We shall now prove that this asymptote is at a finite distance from the highest point. To prove this we examine the limiting value of the intercept of the tangent on the axis of y, viz. y-xp, whenp = 7n. Eemembering that gdx= -u 2 dp, dy=pdx, we have g (y - xp) = - jpu 2 dp +p$u?dp, the limits being p=0 to p. As we only wish to determine whether the limit is finite or not we shall integrate from p = m-% 1 to p=m, where t is some finite quantity as small as we please. The remaining parts of the integrals will be included in two finite constants M and N. Writing p=m - , we have 9 (y ~ xp) = J(m - ) u 2 d - (m - )Ju 2 d - M + pN, the limits being = : to 0. To find what function u is of when is small, we refer to equation (5). Remembering that U = l/w 2 since U=U Q when p=0, we _ 100 RESISTANCE VARIES AS 71 th POWER OF VELOCITY. [CHAP. III. ' write that equation in the form a i -/<*)/(->- U J K Expanding and remembering that d//rfp = 2 A /(l+p 2 ) we find after subtracting (8) I = where 1-, A' and B' are finite constants. Substituting we find by an easy integration 9 (y - *P) =-& 2 + & 2 ! log + &c., where the + 2 sin 3 log cot + sin 2$. [Coll. Ex.] 176. Ex. 1. When the resistance varies as the nth power of the velocity, prove that the curve has a vertical asymptote at a finite distance on the positive side of the highest point. We have v = u */(! +/> 2 ) where u is given by equation (B). Now, by the action of gravity, p continually decreases from one end of the trajectory to the other. After the projectile has passed the summit p becomes negatively great and (B) then gives u=Ljp, where L is the limiting velocity. We thus have v = L when p = . Sub- stituting u=Llp in (C) and integrating from p=p l to oo , where p^ is any large finite quantity, we find that t and y are infinite and x finite. Ex. 2. Prove that, when the resistance varies as the nth power of the velocity, n being > 2, the arc of the trajectory on the negative side of the highest point begins at a point at a finite distance from the origin. Prove also that the tangent at this point makes an angle tan" 1 m with the horizon given by where w and p are any contemporaneous values of u and p. See Art. 116. As in Art. 174, this equation has one positive root. ART. 177.] VARIOUS RESULTS. 101 In the extreme initial position of the particle the velocity is infinite. Since v= x /(l+j> 2 ) we must there have either u or p infinite. If p = a> , (B) gives u=L\p and this makes v finite. The equation giving m is therefore obtained by putting M=OO in (B). To determine the position of the particle when this occurs we express M in terms of p and use the equations (C). Let the initial position defined by P = Po De such that p = m - ^ where 1 is a finite quantity as small as we please. Substituting p=m- in (B) and using the equation given above to find m, we have u n =b n j where b is a constant. Substituting in (C) and integrating from =$! to we find that t, x, and y are finite when =0. Ex. 3. When the law of resistance is the wth power of the velocity, and u, u' are the horizontal velocities at any two points of the trajectory at which the 112 tangents make equal angles with the horizon, then + -7- = where a is the u n u n a n velocity at the highest point. Ex. 4. When the resistance is ic' + KV n , investigate the linear equation du~ n K r n i + 9 (l+p*f 9 where u is the horizontal velocity and p is the tangent of the inclination to the horizon. Thence show that the determination of t, x, y may be reduced to inte- gration. [Allegret, Bulletin de la Societe" Math. 1872.] Ex. 5. When the resistance is constant and equal to Kg, the highest point being the origin and the velocity being a, prove that the horizontal velocity M at any point of the path is M = a(tan0)" where 20=\l/ + %ir. Thence deduce from the integrals (C), Art. 168, the values of t, x, y in terms of tan 0. If K< or =1, the subsequent path has a vertical asymptote which is at the finite distance x = 2ica z lg (4/c 2 - 1) if K> , but is at an infinite distance if K<|. If K>! the particle arrives at a point C at which the tangent is vertical in the finite time K.a\g (/c 2 - 1), the coordinates of G being %Ka 2 /g (4* 2 - 1) and - a 2 /4# (/c 2 - 1). On the negative side of the origin, the curve begins with a vertical asymptote which is infinitely distant and the time of describing the arc is infinite. 177. When the resistance varies as the cube of the velocity, the equation (B) of Art. 168 takes the form 1 K . -^= --(p-m)( u 3 g^ the origin being taken at the point at which the velocity is infinite and m being the corresponding value of p. To discuss the motion we substitute this value of u in the integrals (C). For the reduction of these integrals to elliptic forms we refer the reader to a paper by Greenhill in the Proceedings. of the Royal Artillery Institution, vol. xiv. 1886. Ex. Show that for the cubic law of resistance the velocity is a minimum at the point given by the negative root of the quadratic p z -m(m 2 + 3)p=l. Show also that when the direction of motion is perpendicular to the oblique asymptote, the horizontal velocity u is given by =m + where L is the limiting velocity. u m 102 RESISTANCE VARIES AS W th POWER OF VELOCITY. [CHAP. III. 178. Some formulae have also been given by the late Prof. Adams to determine the coordinates of a particle projected at any inclination to the horizon on the supposition that the resistance varies as the nth power of the velocity and that the path is not very curved. These were first published in the Proceedings of the Royal Society and proofs were given in Nature, vol. XLI., 1890. These appear to be long, but they admit of great abbreviation. 179. The equation of a trajectory being given in the form cos ^=/(pcos ^-), it is required to find the law of resistance. We notice that the equation can be written in this form, except when p cos \f/ is constant, for in that case p cos \f/ cannot be taken as the independent variable. This excepted curve is the catenary of equal strength. Resolving horizontally and tangentially, we have -j-(vcos^) = -.Rcos^, j- = -R-gsimp .................. (1). Eliminating dt R cos \f/ = (v cos \f/) (R + g sin \f/) ; ..................... (1). Remembering that the normal resolution gives v 2 /p=gcoe \{/, we have cos \f/ =f (v'jg). Substituting this value of cos^, the expression for the resistance R has been found. We may also write the expression in the form *"g=-*d -/)*<"/) ..................... (2), where f=f(v 2 lg) and the sign of the radical follows that of sin ^. 18O. Ex. 1. Find the law of resistance when the trajectory is a cycloid with the cusps pointing downwards. In this curve p=2acos^, .-. f=vl*J1ag. We then find that the resistance R= - 2g (1 - v*l2ag)%. Since the radical follows the sign of sin \f/, R accelerates the particle on the ascending and retards it on the descending branch. Since v = cos \(/ N /2ads (1), where < is the angle the direction of F makes with the tangent to the path, see Art. 70. That the work of the two forces X, Y is equal to that of their resultant is proved in Statics. It is also seen to be true by resolving the forces along the tangent; we then have y dx v dy _ /7ds + C is called sometimes the force function and sometimes the work function. The definite integral UU is the work done by the force F as the particle moves from the position (x , y ) to the position (x, y). Here U represents the same function of x , y that U is of x, y. 186. Work of a central force. Let the central force F be regarded as repulsive in the standard case. Let it tend from the centre S and be equal to f(r) where r is the distance of the particle from S. Then since dr/ds is the cosine of the angle the distance r makes with the displacement ds of the particle, the part of the work function due to F is fFdr. The integration is to be taken from the initial position A to the final position B of the particle. When the force under consideration is gravity the centre S is regarded as being infinitely distant. We then replace dr by + dy, the upper or lower sign being taken according as y is measured downwards or upwards. Supposing the weight of the particle to be mg and that y is measured downwards, the work of the weight is This rule is usually read thus, the work done by gravity is the weight midtiplied by the vertical space descended. It should be ART. 188.] THE WORK FUNCTION. 107 noticed that the work is independent of the horizontal displace- ment. See Art. 70. 187. Work of an elastic string. The case in which the particle is attached to a fixed point S by an elastic string differs from that of a central force tending to the same point in a certain discontinuity. If I be the unstretched length, r the actual length and E Young's modulus, the tension T is given by Hooke's law r I T = E . when the string is tight, i.e. when r > I, but the tension L is zero when the string is slack, i.e. r < I. Let the work be required when the string is stretched from a length I-L to h, and let T lt T 2 be the tensions at these lengths. If both j and 1 2 are greater than I, the work is /:<- The work done by the tension is therefore equal to minus the arithmetic mean of the tensions multiplied by the extension. The work done by the force which stretches the string in opposition to the tension is the same taken with the positive sign. This rule is of considerable use when the length of the string undergoes many changes during the motion, being sometimes greater than the unstretched length and sometimes less. It is important to notice that the rule, as given above, holds in all these cases provided the string is tight in the initial and final states. If the string is slack in either terminal state, we may still use the same rule provided we suppose the string to have its natural or unstretched length in that terminal state. 188. The equation of vis viva holds also when the particle is free from constraint and is acted on by any conservative system of forces. For, whatever curve the particle may describe, we may suppose it to be constrained, like a bead on an imaginary wire, to describe that path. The pressure is then zero throughout the motion, but, what more immediately concerns us here, is that the equation (6) of vis viva continues to hold under these circumstances. 108 CONSTRAINED MOTION. ^GHAP. IV. 189. The whole area or space taken into consideration when the forces are expressed in terms of the coordinates is called the field of force. Such a field is usually defined by expressing the force function (when there is one) as a function of the co- ordinates. It follows from the principle of vis viva that when a single particle moves in a field defined by a force function the kinetic energy of the particle in any and every position differs from the value of the force function at that point by a constant. The constant is independent of the direction of motion, so that two particles of equal mass projected from the same point with equal velocities but in different directions will always have equal velocities whenever they pass over a given point of the field. 19O. Examples. Ex. 1. A particle is projected from a given point on a smooth curve and is acted on by no forces. Prove (1) that the velocity is constant and (2) that the pressure varies as the curvature. Ex. 2. A heavy particle P describes a curve and in any position a normal PQ is drawn outwards, so that PQ is equal to half the radius of curvature at P. Prove that the velocity v and the pressure R on the particle measured inwards are given by v 2 = 2gz, Rp = 2mgz', where z, z' are the depths of P and Q below a certain horizontal straight line, which may be called the level of no velocity. Prove also that the particle leaves the curve when Q crosses the level of no velocity. Supposing that the axis of y in the standard figure of Art. 181 is drawn upwards, the two fundamental equations for a heavy particle are Jmt; 8 - $mV= -mg(y- 1/ ), mv 2 /p = - mg cos \f/ + R. If we draw a horizontal straight line at an altitude y lt such that #J/i= we see that The results to be proved follow immediately. If the particle is constrained to remain on the curve merely by the pressure R it will leave the curve when R changes sign. But this is what happens when Q crosses the level of no velocity. /'.> . 3. A particle is swung round a fixed point at the end of a string in a vertical plane. Prove that the sum of the tensions of the string when the particle is at opposite ends of a diameter is the same for all diameters. [Coll. Exam. 1896.] Ex. 4. A heavy particle, constrained to describe an ellipse whose plane is vertical and major axis inclined at an angle o to the horizon, is projected from the upper extremity A of the major axis with a velocity v . Find the velocity v l with which it passes the upper extremity B of the minor axis and the pressure at that point. ART. 190.] EXAMPLES. 109 Since the altitude of B above A is the difference between the projections of CA and GB on the vertical, the equation of vis viva gives Jm (v^ - v 2 ) = - mg (b cos a - a sin a). This gives two equal values of v 1 with opposite signs. One or the other is to be taken according as the particle is pro- jected from A upwards or downwards. If the values of v 1 are imaginary the particle will not reach B. The pressure Ej at B is found by re- solving the forces along BG inwards. We have jm;-, 2 -i =mg cos a + jRj, Pi where /> 1 = a a /6. Let us suppose that in addition to its weight the particle is acted on by a centre of force at the focus S such that the attraction at a distance r is /j.r n . The equa- tion of vis viva would then have on the right-hand side the additional term - J/xr n dr, the limits being the initial and the final values of r, i.e. r=a (1 + e) and r = a, Art. 186. The velocity v^ is then given by a n+l %m (Vi z -v 2 ) = - mg (b cos a - a sin a) - fj. ^ {1- (l + e) n+1 } ft "T~ J- - .' and the pressure is determined by mv 1 2 & - 1 = mg cos a 4- /j.a n .- + R 1 . Pi Let us next attach the particle to the centre C by an elastic string whose natural length is Z. The effect of this is to add another term to each equation. If I < b and < a the string is stretched throughout and the term to be added to the equation of vis viva is -\ (Tg + TJ (b -a) where T and T r are the tensions at A and B, see Art. 187. In our case T =E(a-l)/l and Tj = (6-l)/L If however l>b and b and >a the string is slack throughout and no term is to be added. The equation of pressure will also have an additional term on the right-hand side. This term is T lt where T l has the same value as in the equation of vis viva. In this way the velocity of the particle and the pressure at any point may be found with ease no matter how complicated the forces may be. Ex. 5. A small ring without weight can slide freely on a smooth wire bent into the form of an ellipse. An elastic string whose natural length is I also passes through the ring and has one end attached to the focus S and the other to the centre C. The ring being projected from the extremity A of the major axis, prove that the velocity v l , and the pressure R 1 at the extremity B of the minor axis are given by m (V - V) = (T 1 + T ) (a + ae- b), Stf.*,*?,!**, PI where T ft = E (2a + ae - 1)11, T l = E (a + b - l)jl provided the string is stretched at the beginning and end of the transit. 110 CONSTRAINED MOTION. [CHAP. IV. Ex. 6. A heavy bead is initially at the extremity of the horizontal diameter of a uniform heavy smooth circular wire whose plane is vertical. The system falls from rest through a space equal to the radius. The circular wire is then suddenly fixed in space. Find the subsequent motion of the bead, and determine if it ever comes finally to rest. Find also the pressure on the wire for any possible position of the particle. Ex. 7. A particle, constrained to describe a circular wire, is acted on by a central force tending to a point on the circumference and varying inversely as the fifth power of the distance, prove that the pressure is constant. Ex. 8. A particle is constrained to describe an equiangular spiral and is acted on by a central force tending to the pole whose acceleration is /*r n . The particle being projected with a velocity v at a distance a from the pole, prove that the velocity and pressure are given by 2 v 2 - v,?= - - (r n+l - a n+1 ). R f , 2/j. ..A am a n + 3 - = ( t> 2 + ^ a n+1 ) ? fir" sm a. m \ n+1 ) r n+1 If n=-3 and v =*J[*.la, the pressure R0. The spiral is therefore a free path when the force varies as the inverse cube of the distance, and since any point may be regarded as the point of projection, the velocity at every point is given by v = J fj.fr. Ex. 9. A particle is constrained to move in an ellipse along which it is pro- jected, and the straight line joining the foci attracts according to the Newtonian law. Prove that the resultant attraction varies inversely as the normal and that the velocity is constant. Ex. 10. A particle of unit mass moves in a smooth circular tube of radius a, under the action of a centre of force which repels as the inverse square of the distance. If the centre of force be midway between the centre of the circle and the circumference, and the particle be projected from the end of the diameter through the centre of force remote from that point, with a velocity whose square is 4/n (\/3 - l)/3a, the particle will oscillate through an arc 27ro/3 on either side of the point of projection. [Coll. Ex. 1897.] Ex. 11. A particle is constrained to describe a lemniscate and is under the action of two central forces tending to the foci and varying inversely as the cube of the distance. Supposing the forces to be equal at equal distances from the foci, prove that the pressure at any point P varies as the distance of P from the centre of the curve. Ex. 12. A particle slides down a smooth curve in a vertical plane. If the pressure on the curve is always X times the weight of the particle, prove that the differential equation to the curve is y + c=a (dxjds - X)* 2 . [Math. Tripos, 1863.] 191. Rough Curve. When the particle slides on a rough curve the friction acts opposite to the direction of motion and its magnitude is /* times the normal pressure taken positively. The ART. 192.] ROUGH CURVES. Ill equations of motion are by Art. 181 fit) mv -T- = X cos ty + Fsin fy pR, = X sin i/r + F cos ^ + R. P It is important to determine the signs of the terms containing R before proceeding with the solution. The initial value of the velocity being known the second equation determines the initial direction of R. Taking R to act positively in the direction thus found, it will continue to be positive during the subsequent motion until it vanishes. The initial direction of the velocity being known, the friction p,R must be made to act in the first equation opposite to that direction. If the particle start from rest the friction jj,R must be made to act opposite to the direction of the tangential force. The sign of /-i will then continue un- changed until either the pressure R or the velocity v vanishes and becomes reversed in direction. To solve the equations of motion we in general eliminate R. Remembering that when s and ty increase together p = ds/dty, we obtain an equation of the form By using the geometrical properties of the curve we express P in terms of i/r. The equation being linear, we then have The value of v being found, the value of R follows from either of the equations of motion. 192. Examples. Ex. 1. A particle is projected with a velocity F along a rough horizontal circle in a medium whose resistance varies as the square of the velocity. Prove that ~-~-8t v-Ve'fc - ypl, ' , where v is the velocity after a time t, s the arc described, and /3 is a constant. Ex. 2. A small bead of unit mass is constrained to move along a rough wire, bent into the form of an equiangular spiral of angle o, in a medium whose resistance is t> 2 cos a/c and is under the action of no other forces. If the coefficient of friction is cot a, prove that the time of travelling from a distance c to a distance b from the pole is e* (b - c)/F cos a where ci=b- c, and F is the velocity at the first of these points and is directed from the pole. 112 CONSTRAINED MOTION. [CHAP. IV. Ex. 8. A heavy particle moves on a rough cycloid placed with its convexity upwards and vertex uppermost. The particle is started with an indefinitely small velocity at the point at which the tangent makes with the horizon an angle e equal to the angle of limiting friction. Prove that the velocity at a point at which the tangent makes an angle with the horizon is 2 v //y sin (- e) and that the particle will leave the curve at the point at which the velocity is *J2ag (cos e - sin %e). [Coll. Ex. 1889.] Ex. 4. A particle is projected horizontally with velocity V along the inside of a rough vertical circle from the lowest point, prove that if it complete the circuit it will return to the lowest point with a velocity t; given by l). [Coll. Ex. 1887.] 193. Condition that a constrained motion is also free. It has already been pointed out that the required condition is that the pressure R must be zero throughout the motion, see Art. 190, Ex. 8. In this way we easily obtain several useful cases of free motion. If T and N be the tangential and normal components of the accelerating force estimated positively in the directions in which the arc s and the radius of curvature p are measured, we may prove that the condition JR = leads to the result 2T=-r- (pN). This is obtained by eliminating t> 2 between the normal and tangential resolutions in Art. 181 and differentiating the result. This form of the criterion though necessarily true is not sufficient to make R = 0. As no notice is taken in it of the initial velocity, it is generally less convenient than the simple rule that R=0. 194. Examples. Ex. 1. A particle is constrained to describe a smooth circle under the action of two centres of force tending to fixed points S, S' on the same diameter, the accelerating forces being njr 6 and /t'/ 7 "' 5 where r, r' are the distances of the particle from the centres of force. If S and S' are inverse points, prove that the pressure can be made zero by giving ju'/M and the velocity of projection suitable values. Let a be the radius; 6, b' the distances of S, S' from the centre C. Since the points are inverse bb'=a?. If P be the particle the triangles SPC, S'PC are similar and r'/r=a/&. The fundamental resolutions give ' - = ^. cos SPC + -,. cos S'PC+ - . a r 5 r 5 m From these we easily obtain E 1 ART. 195.] MOTION ALSO FREE. 113 In order that R = we have two conditions '/ " f* I J ' I / o.. 4 "*" b 'i ' Since r'jr=alb, the first condition shows that the tangential accelerations due to the two forces are equal at all points of the circle. Since any point may he regarded as the point of projection the second condition gives the velocity at all points of the orbit. Since v is zero at an infinite distance, this formula shows that the velocity at any point of the orbit is the same as if the particle were con- ducted from rest at an infinite distance to that point ; Art. 181. If the two centres of force are indefinitely near to each other the resultant attraction at any point P at a finite distance from them is the same as that of a single centre of force of double the intensity of either. Hence we arrive at Newton's theorem that a circle can be described freely under a single centre of force whose acceleration varies as the inverse fifth power, the centre of force being on the cir- cumference. When the particle comes indefinitely close to the two centres of force, they cannot be considered as one centre. The particle passes between the two centres with an infinite velocity. The two centres of force attract the particle in opposite directions with forces jt/(a - ft) 5 and v.'l(b' - a) 5 , both being infinite. The resultant force tending to the centre of the circle is therefore /*/a (o - b)* which is also infinite. This last force gives the initial curvature to the subsequent path. Ex. 2. A particle describes a catenary under the action of a force parallel to the ordinate. Show that if the pressure is zero, both the force and the velocity vary as the ordinate. Ex. 3. Show that a particle can describe a parabola under a repulsive force in the focus varying as the distance and another force parallel to the axis always three times the magnitude of the former. Prove also that if two equal particles describe the same parabola under the action of these forces, their directions of motion will always intersect on a fixed confocal parabola. [Coll. Ex.] Ex. 4. If a curve be described under the action of a force P tending to the pole and a normal force N, prove that 195. Does the particle leave the curve ? If the particle is a small ring which slides on the curve it is obvious that it cannot separate from the curve. In this case the pressure R may have any sign. If the particle slide on one side of the curve the pressure on the particle must tend towards that side on which the particle moves. The pressure R must therefore have the sign which suits this direction and must keep that sign throughout the motion. When therefore the analytical expression for R given by the normal resolution (Art. 184) changes sign the particle separates from the curve. R. D. 8 114 CONSTRAINED MOTION. [CHAP. IV. Since the forces in nature cannot be infinite the points at which R can change sign are found by putting R = Q in the normal resolution. Let mf be the resultant force, and let its direction make an angle < with the normal. Then mv 2 , = mfcos(f> + R. The possible points of separation are therefore given by v 2 =fp cos <. Now 2p cos < is the chord of curvature in the direction of the force mf. Representing one-fourth of this chord by c, the equation becomes v 2 = 2/c. Hence the particle can leave the curve only at a point such that the velocity is that due to one-fourth the chord of curvature in the direction of the resultant force. Art. 25. 196. Examples. Ex. 1. A heavy particle is suspended from a fixed point C by a string of length a. A horizontal velocity v is suddenly communicated to the particle so that it begins to describe a vertical circle. It is required to determine whether the particle will oscillate or the string become slack. The equation of vis viva shows that the velocity v at an altitude y above the lowest point of the circle is given by v*=v *-2gy ....................................... (1). The tension R is given by (2). If the particle oscillate the velocity is zero at the extremities of the arc of oscillation. It follows from (1) that the altitude of this point above the lowest point is v 2 /2<7. If the string becomes slack the tension vanishes at the point of separation. It follows from (2) that this occurs at an altitude (v z + ag)/3g above the lowest point. These points cannot be real points unless their altitudes are less than the diameter. We also notice that the altitude of the first of these points is greater or less than that of the second according as u 2 is greater or less than 2ag. If v 2 >5ag neither point is real. The particle must describe the whole circle and the string does not become slack. If v *<2ag the velocity vanishes at an altitude less than that at which the tension vanishes. The particle therefore oscillates and the string does not become slack. If v 2 <5ag but >2ag the string becomes slack before the velocity vanishes. The particle therefore leaves the circle and describes a parabola freely in space. If the particle, instead of being suspended by a string, were constrained to move like a bead on a vertical smooth circle of radius a the particle could not separate from the circle. It therefore oscillates or describes the whole circle according as v 2 < or >4a<7. ART. 197.] MOVING CURVES. 115 Ex. 2. A bead can slide on a horizontal circle of radius a and is acted on only by the tension of an elastic string, the natural length of which is a, fixed to a point in the plane of the circle at a distance 2a from its centre ; find the condition that the bead may just go round. Prove that in this case the pressures at the extremities of the diameter through the fixed point will be twice and four times the weight of the bead if that weight be such as to stretch the string to double its natural length. [Math. Tripos, I860.] Ex. 3. A heavy particle is allowed to slide down a smooth vertical circle of radius 27 from rest at the highest point. Show that on leaving the circle it moves in a parabola whose latus rectum is 16a. [Coll. Ex. 1895.] Ex. 4. A particle moves on the outside of a smooth elliptic cylinder whose axis is horizontal. The major axis of the principal elliptic section is vertical and the eccentricity of the section is e. If the particle start from rest on the highest generator, and move in a vertical plane, it will leave the cylinder at a point whose eccentric angle is , where e 2 cos 3 = 3 cos

. [Coll. Ex. 1892.] \ffj Jo |8-e"+2eoa# Ex. 6. A particle, projected inside a smooth circular tube, moves under an attractive force varying inversely as the square of the distance from a point within the rim of the tube and in its plane. Prove that the pressure cannot vanish at any point if the particle is performing complete revolutions. [Coll. Ex. 1897.] 197. Moving curves of constraint. To find the equations of motion of a particle constrained to slide on a curve moving in its own plane. Let be any point of the plane of the curve which it will be convenient to take as origin. Let / be the acceleration of this point, then the motion relative to will be unchanged if we apply to every point of the curve and to the particle an accelera- tion equal and opposite to that of 0. If we also apply to every point an initial velocity equal and opposite to that of 0, we may regard as a fixed point. The point is then said to have been reduced to rest. We shall now take as the origin of the polar coordinates r, 0, where 6 is measured from a straight line fixed relatively to the curve. Let o> be the angular velocity of Og referred to a straight line Ox fixed in space. Let be the angle the radius 82 116 CONSTRAINED MOTION. [CHAP. IV. vector r makes with the tangent. The equations of motion are rfV (dO \ s P , R . j- rj- + co = -- A -- sin dt* \dt J m m id --Jl r dt where P, Q are the components of the impressed forces, and/ x ,/ 2 those of/ These equations may be written in the forms R d0 \ Q f n r jl + = -/I + COS dt J m m r j d*r fd0\* P , -j- r -r, = -- d& \dtj m sm m 1 d (^ de \ Q f do > , ( R o "\ - -j- r 2 -j- = -/ 2 - -j- r + --- 2wv cos 6, r dt\ dt) m J dt \m J since rdd/dt = v sin 0, c?r/cft = ?; cos . These are the equations of motion we would have obtained if we had supposed the curve to be fixed in space and the particle to be acted on (in addition to the impressed forces) by three fictitious forces. The introduction of these forces is said to reduce the curve to rest. These forces are, (1) the force F l = mf by which the origin is reduced to rest ; (2) the force F^ = mw*r acting on the particle along the radius vector from the origin ; (3) F s = mr -^- acting perpendicularly to the radius vector in the direction tending to increase 0. We also observe that the expression R Zmwv takes the place of the pressure of the curve on the particle. Here v represents the velocity relatively to the curve. The velocity in space is the resultant of v and the velocity of the point of the curve occupied by the particle. By resolving the impressed and the fictitious forces along the tangent we obtain an equation free from the reaction, and from this the velocity v of the particle relatively to the curve may be found. This equation is dv IT . ,.dr ( d 2 (3r 2 -4ar)/2a, where a is the radius of the circle and m the mass of the bead. To reduce the circle to rest we apply the fictitious accelerating force F 2 = uPr. Hence %v*=$w-r 2 + C. Since the bead is initially at rest in space, it has a velocity relatively to the curve v= - u . 2a when r=2a. Hence C = and v= - wr through- out the motion. To find the pressure, we have v 2 r R = -drr . H --- 2uv. a 2a m Substituting for v its value, this gives the result. Ex. 2. A bead is at rest on an equiangular spiral of angle a at a distance a from the pole. The spiral begins to turn round its pole with an angular velocity w. Prove that the bead comes to a position of relative rest when r=acosa, and that the pressure is then mw 2 a sin 2o. Prove also that when the bead is again at its original distance from the pole, the pressure is ?w 2 a sin a (3 + sin 2 a). 199. Time of describing an arc. A heavy particle is in stable equilibrium at the lowest point A of a smooth fixed curve. Find the time of a small oscillation. Let < be the angle the normal at any point P near A makes \vith the vertical, s the arc AP, p the radius of curvature at A. Then is ultimately equal to s/p. The equation of motion is d*s . when sin < is expanded in powers of s. If the arc of oscillation is sufficiently small we may reject all the terms after the first powers of s. The time of a complete oscillation is therefore 2?r \lp[g. The time of oscillation is therefore the same as if the constraining curve were replaced by the circle of curvature at A. When it is necessary to take account of the small quantities of the order s 2 , it is more convenient to replace the equation of motion by its first integral, as in Art. 200. 118 CONSTRAINED MOTION. [CHAP. IV. /,'.-. 1. A particle P makes small oscillations about a position of stable equi- librium at the point A of a smooth curve under the attraction of a centre of force situated at a point C on the normal OAC to the curve, the magnitude of the force being/ (r) where r = CP. Prove that the time of oscillation is 2w \. ^r-J where (( a +p)n F=f(a), a=AC taken positively when C is on the convex side of the curve and /> = OA is the radius of curvature. Notice that the time is independent of the law of force but depends on its magnitude F at A. Ex. 2. A smooth wire revolves with constant angular velocity u about a fixed point in its plane and a bead is in relative equilibrium on the wire at an apse at distance a from the fixed point ; prove that, if slightly disturbed, the period of a small oscillation is A / ^ - , where p is the radius of curvature of the wire at u V a-p the apse and is less than a. [Coll. Ex. 1887.] Reduce the curve to rest, and use Art. 199. 200. Time of describing a finite arc. By using the equation of vis viva the determination of the time can be reduced to integration. The equation of vis viva is where U= (#, y) is a known function of the coordinates (x, y}. The constant G is known when the velocity is given at some point B whose coordinates are (h, k). We use the known equations of the curve to express any two of the variables x, y, s in terms of the third. Choosing s as this variable we have U = ty (s). Hence ds the integration being taken from one extremity of the arc de- scribed to the other. 2O1. Ex. 1. A heavy particle is projected from a point A of a vertical circle, centre 0, with such a velocity that it would come to rest at the highest point B. Prove that the time of transit from A to P is A /-log ~- where BOA = a r V 9 cotja BOP=0 and a is the radius. We notice that the time of arriving at the highest point is infinite. Ex. 2. Prove that the curve such that the time of descent of a heavy particle from rest at a given point A down any arc AP is equal to the time down the chord is a lemniscate. Taking A for origin and using polar coordinates, 6 being measured from the /0 ds t 7" 77 -^.= 2 . / -T . Differentiating DV^ 1 cos 0) V cos0 both sides and solving the differential equation we find that r^ .4 sin 20. The condition that the lower limit on the left-hand side is zero is found on trial to be ART. 202.] TIME IN AN ARC. 119 satisfied by this value of r. The required curve is therefore a lemniscate with the axis inclined at an angle of 45 to the vertical. J. A. Serret remarks that if the ratio of the times were k : 1, the differential equation would be ~]^ T \ tv WUIXi V if T" U. city This quadratic gives drjrd0=f(0), and the solution is reduced to integration. The history of this problem is given in the Bulletin de la Societe Mathgrnatiques, vol. xx. 1892. It was first solved by Euler in his Mecanique 1736 and afterwards by Fuss in the Memoires <&c. de Saint Petersbourg, 1824. Rispal gives a geometrical proof in Liouville, xn. 1847. Ex. 3. A particle is acted on by a centre of force varying as the distance. If the time of describing from rest an arc from a given point A is equal to the time of describing the chord, prove that the curve is a lemniscate. Ossian Bonnet, Liouville, vol. ix. Ex. 4. If the time of descent of a heavy particle from rest at a given point A down any arc AP bears to the time of descent down the chord a ratio equal to the ratio that the length of the arc bears to k times the length of the chord, prove s ~ Cy, where y is the vertical ordinate of P and C is a constant. 202. Subject of integration infinite. A difficulty some- times arises in finding the time of describing a finite arc AB if the velocity is zero at either limit. Let a particle be projected from a point A in such a manner that the velocity of arrival at B is zero. It is required to find the time of describing the arc AB. Let the points A, B be determined by s = a, s = b. Since the velocity at B is zero, we have C= ty(b). The time of describing the arc AB or BA is therefore given by the limits of integration being a, b. The subject of integration is infinite at the limit s = b, but the integral itself may be finite. If we write s = b + 2. If, as usually happens, Taylor's expansion holds true, we have n = 1. The time to or from a position of rest is then finite. 120 MOTION IN A CYCLOID. [CHAP. IV. If the point B is a position of equilibrium as well as of rest, we have dUjds = when ', 9 the angle the normal at any point P near B makes with the vertical. The equation of vis viva is then The time t of describing a small angle a is therefore given by 2p\* ~ The time of transit from A to B is therefore infinite unless the radius of curvature p at B is zero. 2O3. Examples. Ex. 1. A heavy particle is constrained to describe the curve x$ + y$=a$ , the axis of y being vertical. Show that the radius of curvature at every cusp is zero. Show also that a particle projected from the lowest cusp with a velocity (2gaft will arrive at the next cusp in a time which is three times that of falling freely from rest at the origin to the lowest cusp. [Despeyrous' problem.] Ex. 2. A small ring can slide freely on a smooth wire bent into the form of a cycloid. The axes of x and y being the tangent and normal at the vertex B, the force function is given by U=My m where m is positive and < 1. Prove that if the particle is projected from a point P whose ordinate is h with a velocity (2Mh m )^ the I-n time of arrival at B is t where Ml (1 - m) t = 2a* h * . Ex. 3. If the only force acting on the particle is gravity U=gy. If y = Ms* + ... prove that p = N* 2 """ + . . . where N~ l = Mn (n - 1) , provided n > 1. Hence n < 2 when p=0 and n=2 or is >2 when p is finite or infinite at the position of equilibrium. , , cPy dx (. fdy\>)* Use the theorem FT* **7 > & - 1 ? I > r ds" ds ( \dsj) Motion in a cycloid. 204. A heavy particle is constrained to move in a smooth fixed cycloid whose plane is vertical and vertex downwards. It is required to find the motion. Let A, A' be the cusps, the vertex, OQD a circle equal to the generating circle placed with its diameter on the axis OD, ART. 204.] FUNDAMENTAL PROPERTIES. 121 C its centre. Let PQN be a perpendicular on the axis drawn from any point P on the cycloid. The following geometrical properties of the cycloid are given in treatises on the differential calculus. (1) The tangent at P is parallel to the chord OQ and the arc OP is twice the chord OQ. (2) The radius of curvature at P is parallel to the chord QD and is equal to twice that chord. (3) The distance PQ is equal to the circular arc OQ. Let the angle QDO = , and let a be the radius of the gene- rating circle. The tangential and normal resolutions at P give (Art. 181) - ' a tf R - = q cos dH p m } .(1). The first equation shows at once that the motion is oscillatory, Art. 118. The time of a complete oscillation is 4-Tr A /- and is V 9 independent of the arc described. Let t be measured from the instant at which the particle P passes the vertex, let c be the semi-arc OB of oscillation. The first equation gives s = c sin A / -7 W 4 It follows that if two particles oscillate in the same or in equal cycloids both starting from the vertex, the two arcs described in equal times are in a constant ratio, viz. that of the complete arcs. If therefore the circumstances of the motion of a particle oscillating from cusp to cusp are known, those of a particle oscillating in any smaller arc can be immediately deduced. 122 MOTION IN A CYCLOID. [CHAP. IV. 205. If I is the depth below the cusp of the extremity B of the arc of oscillation, we have by the principle of vis viva v 2 = 2g (2a - b - ON). It follows at once from the geometrical properties of the curve that 2mqb R = zmq cos . P The first term is twice the resolved weight of the particle along the normal at P ; the second is the centrifugal force of a particle moving uniformly with the velocity due to the depth below the cusp of the extremity B of the arc of oscillation. 2O6. Examples. Ex. 1. A particle oscillates in a complete cycloid from cusp to cusp. Prove the following properties. (1) The velocity v at any point P is equal to the resolved part of the velocity V at the vertex along the tangent at P, i.e. v = Fcos . (2) The time of describing an arc OP is proportional to the angle ODQ, i.e. 1 V (3) The particle moves as if it were rigidly attached to the generating circle, that circle being supposed to roll with a uniform angular velocity on the base A A'. This follows from the last result because dldt is constant. (4) The centrifugal force at any point P is equal to the resolved part of the weight along the normal at P, and the pressure is twice either of these. Ex. 2. A heavy particle starts from rest at a point A of a cycloid, prove that the time T of transit from any point P to any point Q is given by where p, q, I are the depths of P, Q and the vertex below the level of A, and a is the radius of the generating circle. Ex. 3. A particle slides down a smooth cycloid starting from rest at the cusp. Prove that the whole acceleration at any instant is in magnitude equal to g and that its direction is towards the centre of the generating circle. [Coll. Ex.] The required acceleration is equivalent to the resultant of g and R/m; the result follows at once from the triangle of accelerations. Ex. 4. A smooth cycloid is placed with its axis AB inclined to the vertical, and its convexity upwards; a particle begins to slide down the arc from A, and leaves the curve at P ; the perpendicular from P on AB cuts at Q the circle on AB as diameter, and QR is a diameter of this circle ; prove that PR is horizontal. [Math. T. 1888.] 207. When a pendulum is removed from one place to another the number, n, of oscillations in any given time (such as a day) is altered by the change in the force of gravity and the alteration ART. 209.] THE CONVERSE PROBLEM. 123 of the length I of the pendulum due to a change of temperature. Since the number of oscillations in a given time varies inversely as the time of a single oscillation, we have n 2 = Gg/l where C is some constant. Taking the logarithmic differential, we find 9 Sn _ Sg _ SI n~ g I ' This formula is a very convenient first approximation to the value of Sn. 208. Ex. 1. Prove that a seconds pendulum brought to the summit of a mountain x miles high loses about 22x seconds per day if the attraction of the mountain can be neglected. If the mountain is of the form of table-land, the loss is only five-eighths of the above amount. The length of the pendulum is supposed to be unaltered. By Dr Young's rule the attraction at the top of table-land is g ( 1 - . - - \ nearly where a is the radius of the earth. Ex. 2. A railway train is running smoothly along a curve at the rate of 60 miles per hour, and a pendulum which would ordinarily oscillate seconds is observed to oscillate 121 times in two minutes. Show that the radius of the curve is approximately a quarter of a mile. [Coll. Ex. 1895.] Ex. 3. If the moon be in the zenith, prove that a seconds pendulum would be losing at the rate of ^ 7 th of a second per day. The moon attracts the earth as well as the pendulum and its disturbing effect is measured by the difference of its attractions at the centre of the earth and at the 2M /a\ 3 pendulum. This is --- I - 1 g where M=^E is the mass, and r=60a is the distance of the moon. 209. Ex. 1. A heavy particle oscillates on a smooth fixed curve, and the periods of oscillation in all arcs are the same. Prove that the curve is a cycloid. Let the axis of y be measured vertically upwards from the lowest point of the curve and let y = h be the initial value of y. Let the equation of the curve be s=f(y), where s is the arc measured from the lowest point. Since v^=2g (ft -y} the time t of reaching the lowest point is given by j5t=/X,! y)< fr. Put y hz, then Since the time t is to be the same for all values of ft, we have dtldh=0. Hence This equation requires that the second factor under the integral sign should be zero. If this were not true we could, by taking ft small enough, make that factor keep the same sign, while ftz varies from ftz = to hz = ft. Every term of the integral would then have the same sign and the sum could not be zero. Hence h-f'(hz) is 124 MOTION IN A CYCLOID. [CHAP. IV. independent of h, and therefore/' (hz) = M(hz)~^ where Mis a constant independent of h and z. We thus find by an easy integration that the &rcf(y) 2My^. This is the equation of a cycloid having the line joining the cusps horizontal. /.'r . 2. A body of mass M can slide on a perfectly smooth horizontal plane and has attached to it a thin tube in the vertical plane containing the centre of gravity. The form of the tube is such that the periods of the oscillations of a particle of mass m placed in it are the same for all arcs. Prove that the form of the tube may be derived from a cycloid by elongating the ordinates perpendicular to the axis in the ratio ^(M+m)l,JM. This problem is due to Clairaut; Mm. de VAcad., Paris, 1742. 210. Resisting medium. If the particle oscillate on a smooth cycloid in a medium resisting as the velocity, the tangential equation of motion becomes d*s ds *-* where w 2 = #/4a. This problem has been discussed in Arts. 121 and 126. The interval between two successive passages through the lowest point is always the same and the successive arcs of descent and ascent are in geometrical progression. If the resistance vary as the square of the velocity, the motion is discussed in Art. 129. 211. Tautochronous curves. When a particle oscillates on a given smooth curve either in a vacuum or in a medium whose resistance varies as the velocity, we know that the oscilla- tion is tautochronous about the position of equilibrium if the tangential force F=m z s where s is the length of the arc measured from the position of equilibrium and m is a constant, Art. 118. If therefore any rectifiable curve is given a proper force to produce a tautochronous motion can at once be assigned. A catenary is a tautochronous curve for a force acting along the ordinate equal to m z y because the resolved part along the tangent is obviously vn?s. The equiangular spiral is tautochronous for a central force pr tending to the pole, because the resolved part along the tangent being ra 2 s where m 2 = ft cos 2 a, the time of arrival at the pole is the same for all arcs. In the same way the epicycloid and hypocycloid are tauto- chronous curves for a central force tending from or to the centre ART. 212.] TAUTOCHRONOUS CURVES. 125 of the fixed circle and varying as the distance, because since r* = As 2 + B, the resolved part along the tangent, viz. firdrjds, varies as s. In all these cases the time of arrival at the position of equilibrium is the least positive root of tann = n/tc (Art. 121), where 2/cv is the resistance and n 2 + K? = ra 2 . The whole time from one position of momentary rest to the next is irfn. The properties of tautochronous curves are more fully discussed in the author's Rigid Dynamics. A historical summary is also there given. 212. Rough cycloid. A particle slides from rest on a rough cycloid placed with its aods vertical in a medium whose resistance varies as the velocity. Prove that the motion is tauto- chronous. The descending motion is given by cLv v^ -j- = fj,R g sin ZKV, - = R g cos ......... (1), (.it O where v is really negative. Eliminating R fa - *tf + 2*v + -3 sin (d> - e) = 0, dt p cos e where tan e = ytt. This may be written -r- (e u v) + 2ic (e u v) + $ e u sin (6 - e) = 0, dt cos e provided -j- = /j, - , that is u = /i0. Put e u ds = dw\ dL p . dw g . . , . ~ Now w = je~^ 4>a cos (j>d = 4,a cos e e~** sin ( e). The equation therefore reduces to d * W I ;; dw I gW =0 dt 2 dt 4a cos 2 e This is the linear equation, Art. 121. We infer that at what- ever point of the cycloid the particle is placed at rest, it arrives at the point E determined by w = 0, that is < = e, in the same time. Such a motion is called tautochronous. The point E is clearly an extreme position of equilibrium in which the limiting friction just balances gravity. 126 MOTION IN A CIRCLE. [CHAP. iv. The time of arrival at E is given by the least positive root of the equation tan nt = n/tc where n? + K- = <7/4a cos 2 e. The whole time from one position of momentary rest to the next is TT/U. So long as the particle is moving in the same direction the constant p, retains the same sign. The motion is therefore given by e -M* s i n (< _ e ) _ Ae- Kt sin (nt + B). When the particle arrives at the next position of rest, it will begin to return or will remain there at rest according as the value of at that point is greater or less than the angle of friction. Motion in a circle. 213. A heavy particle is constrained to move in a fixed circle luhose plane is vertical. It is required to find the time of describing an arc. Let G be the centre, A and B the lowest and highest points of the circle, a its radius. Let P be the position of the particle at any time t } the angle GBP. Let the particle be projected from the lowest point with a velocity F. The equation of vis viva gives f 2a -j9-J F 2 = Iga (1 cos 20). Let us put F 2 = 2gh, so that the velocity of projection is that due to a height h ; we also put h = 2a . /e 2 . If tc> 1, the velocity at the lowest point is more than sufficient to carry the particle to the highest point of the circle, the particle therefore goes continually round the circle in the same direction. If K < 1 the velocity at the lowest point is insufficient to carry the particle round the circle, the particle therefore oscillates. If K = 1 the particle arrives at the highest point with a velocity zero, but only after an infinite time has elapsed, Art. 201. Substituting for F 2 in the equation of vis viva, we have ART. 214.] THE ELLIPTIC INTEGRAL. 127 If t be the time of describing the arc AP which subtends an angle 20 at the centre, we have * d< t> , A/0* 2 - sin 2 f) - where one radical is positive and the other has the same sign as d/dt. If x=l, the integral is a known form. We have <"* , \ (A \ iT+-k (4), when < = ^TT, t is infinite so that the particle takes an infinite time to reach the highest point. If K> 1, we write the integral in the form g g If* d^ 1 - This elliptic integral* gives the time of describing the arc which subtends an angle < at the highest point of the circle. The time of arriving at the highest point is found by writing JTT for the upper limit. 214. When K < 1, we put K = sin a. We see from (2) that sin cannot exceed K and that the velocity is zero when sin < = K; the particle therefore oscillates on each side of the lowest point through an arc AD or AE which subtends an angle a at the highest point. Let sin = K sin ty, so that ty varies from zero to \TT. We then find after an easy substitution in (3) lg_ f * Jo V(l-* 2 sin 2 i/r)" This elliptic integral determines the time of describing an angle where and ty are related by the equation sin = K sin yfr. We can construct the angle ty geometrically. Describe a circle with centre C to touch ED, and let BP intersect this circle in Q ; then the angle BQC = ty. For another construction we draw a chord A'P' equal to the chord AP, then the angle CB'P' = -f. * The reader is referred to Prof. Greenhill's Treatise on the applications of elliptic functions. He begins with the problem of the simple circular pendulum as being the best introduction to the theory of these functions. 128 MOTION IN A CIRCLE. [CHAP. IV. In obtaining (6) we supposed the sign of cos ty to be the same as that of the radical in (3) and therefore the same as that of d/dt. Since cos < is positive, it then follows from (6) that d-^r/dt is positive. The point Q therefore travels round the circle, being the lower or upper intersection of BP with the circle accord- ing as P is moving from A to D or from D to A. 215. Series for the time of oscillation. We may approxi- mate very closely to the time of a complete vibration by using a series. If T be this time, the formula (6) gives \T when the upper limit is \ir. We have by the binomial theorem By a theorem in the integral calculus ^ (sin Tlr^d-b = l ' 3 ' 5 " ^ n ~ *) - f o a ^ ; " 2.4.6... 2n ' 2 It immediately follows that where K = sin a and a is the angle subtended at the highest point of the circle by the half-arc of oscillation. It is also useful to notice that K, is the ratio of the chord of the half-arc to the diameter of the circle. The first term of this series represents the time of an infinitely small oscillation. The other terms are regarded as small correc- tions to this time, and are sometimes called the "reduction to infinitely small arcs." The second term is usually a sufficient correction. Thus suppose the arc of oscillation on each side of the vertical to subtend an angle of 36 at the point of suspension, then a = 18 and K = -fa. The second term is only about ^th and the third th of the first. 216. Relation between continuous and oscillatory motions. Comparing the formulae (5) and (6) we see that the integrals are the same except that the moduli K and I/K are reciprocals. This leads to a theorem by which we connect a motion all round the circle with an oscillatory motion. Let two particles P, P' be projected from the lowest points A, A', of two circles of radii a, a', and let these be acted on by unequal gravitational forces g and g'. Let the velocities of projection F, V be such that the moduli are reciprocals. Then K being less than unity, we have F 2 =4a0/c 2 , F' 2 =4ay/x s . It then follows from ART. 218.] TWO KINDS OF MOTION. 129 what precedes that the particle P' travels round the circle and P oscillates in a semi- arc equal to AD, where the angle DBA = a and K=sin o. Let P, P' be the positions of the particles when the angles ABP=, A'B'P'ty, where sin = /c sin ^. If t, t' be the times of describing the arcs AP, A'P' we have / g t- V It follows therefore that \f > t '- K \/ * points P, P' therefore corre- spond to each other in the two motions, and it is easy to see that they are geometrically connected by the relation chord AP _ tut _ chord AD chord A'P' ~ a' ~ diam. A'B' ' It is obviously convenient that the particles should occupy corresponding points at the same instant of time. We therefore choose the constants a', g', so that t=t'. We then have g'ja'=K 2 gla. The equations of motion take the forms a d$ la'*? df ..I- -?- = lcCOB\b, ./ T ~ = COS, V g dt V 9 at where the coefficients on the left hand are equal. If we make the radii equal we can suppose both particles to describe the same circle. We then have a=a, g = V'=-V, A'P'=-.AP. 217. Ex. 1. If the circle described by P' has AM for its diameter, prove that P, P' move so as to be always on the same horizontal line, the gravitational forces being g and #/c 4 respectively. Ex. 2. If the circles are equal and the arc PP' is bisected by a point Q, prove that Q moves on the circle as if it were a third heavy particle acted on by a gravi- tational force g"=gK. The velocity of Q at A (and at all points) is equal to the mean of the velocities of P and P'. Prove also that Q goes half round while P' goes all round. Sang, Edinburgh Trans. 1865, vol. 24. These results follow at once from Art. 216. 218. Relations between two oscillatory motions. The investigation of these relations is properly a part of the theory of elliptic integrals, but the following theorem will serve as an example. K. D. 9 130 MOTION IN A CIRCLE. [CHAP. IV. If T, T' be the periods of oscillation corresponding to two semi-arcs which subtend angles a, a' at the highest point of the circle and so related that sin o = (tan a') 2 , then will T= T' (cos J a') 2 . The half arc of oscillation being denned by sin o = K, the time t of describing the angle is given by /fl /> where sin

+ \j/, so that is the angle the arc A'Q in the figure of Art. 213 subtends at B'. Eliminating we find tan^= ^ .. We shall now change the independent variable from ty to 0. The simplest (though not the shortest) method of effecting this is to find d\f/ by differentiation and sin 2 \f/ by trigonometry both in terms of 0. The substitution is then obvious and we have* /> d$ _ JJ_ [6 d8 J o ' = X sin 0. Let two particles P, P' oscillate in the circle APB through arcs AD, AD' which subtend angles a, a' at the highest point B, then the last equation shows that the times t, t', of describing corresponding angles , ', are connected by the relation To compare the changes of the values of these corresponding angles we refer to the figure of Art. 213. As P moves from A to D and back to A, Q travels round the semicircle A'QB', 20 increases from to IT, and 0' increases from to a'. Thus the oscillation from A to D and back to A corresponds to the oscillation A to D' only, i.e. a complete oscillation of P corresponds to half a complete oscillation of /". If T, T' be the times of a complete oscillation of P, F, we have therefore T=T'I(1 + K ). The two angles o, a' are connected by the relation 2 JK 1 cos a' sina =X = --^-; .'. JK= : - j- . 1 + K sin a Since *<1 and a' <^ir we take the lower sign in the value of JK. Hence sin a = (tan \ a') 2 . It follows also that t = 2t' (cos a') 2 . Ex. If a, , a.,, ... be a series of angles connected by the relation and if 7', be the time of a complete revolution in an arc subtending 4a, at the point of suspension, prove that Tj^sec^ttj.sec^a.j... to oo ) 2 . 27r /v /(a/#). [sang.] a 19. Co-axial Circles. Two heavy particles, constrained to describe the same vertical circle, are projected from any two points with velocities due to their depths below the same horizontal line. It is required to prove that the straight line joining the particles always touches a co-axial circle. * Cayley's Elliptic Functions, Art. 243. ART. 219.] CO-AXIAL CIRCLES. 131 Let Oy be the radical axis of two co-axial circles whose centres are G, C'. Let a tangent at any point T of one circle intersect the other in two points P, Q. Let PM, QN be perpendiculars on the radical axis. By a known property of co-axial circles the tangents PT, QT drawn from points on the outer circle satisfy the relations * PT 2 = 2 . CC' . PM, Qr~=2.CC'. QN. In the time dt let the tangent move into the position P'TQ'. Then since the elementary arcs QQ', PP', make equal angles with the chord P'Q', the triangles QTQ', FTP' are similar : hence arc QQ'/arc PP' = QT/PT. It follows from these two geometrical theorems that (vel. of Q) 2 /(vel. o(P) 2 =QNIPM. If then the point P move with a velocity equal to (2g . PM)*, the point Q must move with a velocity equal to (2' < /> Jo V(l-* 2 sinV) J V(i- where sin ^' = 2 sin \f/ cos ^ (1 - /c 2 sin 2 ^)* (1 - /c 2 sin 4 ^)~ l . [Cayley's Elliptic Functions, Art. 249.] CHAPTER V. MOTION IN ONE PLANE. Moving Axes. 221. THE components of velocity and acceleration along the axes of coordinates, the tangent and normal to the path and in some other directions have been already considered in Chapter I. The solution of the more difficult problems in dynamics requires however that we should have at our command a greater power of resolution than is given by these. We shall now investigate the general components for any moving axes in one plane. 222. To avoid the continual repetition of the same argument, we shall use the term vector to represent the subject under con- sideration, whether it be a velocity or an acceleration. Let us understand by a vector any quantity which has direction as well as magnitude, and which obeys the parallelogram law. Thus the radius vector of a point P is a vector and its resolved parts along the axes are the coordinates x and y. Again the velocity of P is a vector, and its resolved parts along the axes are dx/dt and dyjdt. The acceleration of P is also a vector and the resolved parts are d 2 x/dtf and d*y/dt 2 . Lastly if R be any vector whose direction makes an angle ty with the axis of x, its components along the axes, supposed to be rectangular, are R cos -fr and R sin i/r. 223. Fundamental theorem. A vector R having been resolved in the directions of two rectangular axes 0%, Or) which turn round a fixed origin in a given manner, it is required to find the rates at which these components are increasing with the time. Let P be the position of the moving point at any time t. Draw a straight line PQ to represent the instantaneous direction ART. 223.] FUNDAMENTAL THEOREM. 135 and magnitude of the vector R. Let u, v be the resolved parts of the vector in the directions of the axes Of, Or,. After a time dt, the point P will occupy a position P', the vector R will become R + dR and may be represented by the straight line P'Q'. The axes Of, Of] will turn round through a small angle d and will take the positions Of, Orf. The resolved parts of R + dR along these new axes will be u + du, and v + dv. At the time t the component of the vector in the direction Of is u. At the time t + dt the component in the same direction (i.e. in the direction Of not Of) is (u + du) cos d(f> (v + dv) sin d. The rate of increase of u in the direction Of is found by sub- tracting the component at the time t from that at the time t + dt and dividing by dt. If we represent the rate of increase in the direction Of by u lt we have _ [(u + du) cos d (v + dv) sin d] u Ul ~ ~~dT When we reject the squares of small quantities according to the rules of the differential calculus, we write unity for cos d and d for sin d. We therefore have _du d Ul ~Tt~ v dt' In the same way if the rate of increase in the direction Orj be Vi, we have _ (u + du) sin d + (v + dv) cos d v Vl ~ ~dT d dv ~ U dt + dt' 136 MOVING AXES. [CHAP. V. 234. This theorem is of great importance and particular attention should be given to the meaning of the letters. The rate of increase of u in the direction of the moving axis O is . Its rate of increase in the direction of an axis jixed in space which is coincident with the position of at the time t and which is left behind when 0| moves into some other position 0' is -v~ . It is the latter rate of increase not the former which i* required in dynamics. To make this point clear let us suppose that u represents the component velocity of a point P. Then _ /component along 0'\ /component along 0\ \ at time t + dt ) \ time t ) ' -' al ng S\ - ( Comp ' along \ timet + dt ) \ time t When it is necessary to distinguish between these two we may call the first the relative rate and the second the space rate of increase of the vector. 225. There is another method of establishing the fundamental theorem which is very generally used and which puts the argument into a more algebraic form. Let the moving axis Og make an angle with an arbitrary direction Ox fixed in space. Then if U be the component of the vector along Ox, U=u cos v sin ; dU fdu dd>\ I d/dt is not zero. By definition dU/dt = u l , and therefore _du d ltl ~dt~ V ~dt' Again let Ox coincide with Orj and let it be left behind when Orj moves to OTJ'. Since is the angle 0% makes with Ox measured from Ox round positively in the direction 77, the instantaneous value of <> is ^TT though as before it is increasing at the rate d/dt. By definition dU/dt is now v lt and hence dv ART. 227.] COMPONENTS OF ACCELERATION. 137 226. Ex. 1. To deduce the components of velocity and acceleration along and perpendicular to the radius vector, Art. 35. We take the arbitrary axis of to coincide with the radius vector, then =6. Begarding =r, 17 = as the components of the vector r, the space components of the velocity are _d dO _dr _dij dd _ d9 ^dt'^di'di' '~~dt + *di~ r di' Taking the velocity as a second vector, the components are u = drjdt, v rdO/dt, and the space components of the acceleration are _du_ dj)_d*r_ /de\- l ~~di V dt~dt 2 ~\dt) ' _dv dt dt ~ r dt V dt t Ex. 2. To deduce the components of acceleration along the tangent and normal, Art. 36. Taking the axis of parallel to the tangent, we have 0=^. Let the velocity be the vector, then u represents the velocity and v = 0. The components of accelera- tion are therefore _ du dif/ _ du _ dv d\f/ _ d\j/ dt dt dt dt dt dt 227. To find the components of velocity and acceleration with regard to moving axes. Let the position of the moving point P be given by its co- ordinates (, r)) with regard to two rectangular axes 0, Orj which turn round a fixed origin with an angular velocity d(f>/dt. Let (u, v) be the components of the velocity of P parallel to the instantaneous positions of 0|, Orj. Let (X, Y) be the components of the acceleration of P. The relations between (, 77), (u, v), (X,Y) follow at once from the general theorem. We have d dd) dri ., dd) , ,. ^ u = ~ 77-^, v -j7 + ?jT (-A-); dt dt dt * dt du dd) Tr dv dd) Y - f ^_ n\ '_ Y ^ \ i/ '_ / l-c \ dt dt' dt* dt' Substituting for u, v in the latter expressions their values given by the former, we have ~V ^ = d?~ Y _^rj_ (d\* l^ffa^l (C ^' ~ dt? \dt ) g dt\ dt)] If the origin is also in motion, these equations require some modification. Let p, q be the components of the space velocity 1 d / dd)\ I - -j- \rf -, } t) dt \ dt J I 138 MOVING AXES. [CHAP. V. of the origin in the directions of the axes. Let u, v continue to represent the components of the space velocity of the point P. To find u, v we add to the expressions (A) for the relative component velocities the component velocities of 0, Art. 10. We thus have d% d$ drj ^dd) These equations give the motion of P referred to a system of moving axes having any fixed origin but always remaining parallel to the original moving axes. With these values of u, v, the accelerations X, Y will continue to be expressed by the formulae (B). 228. We may deduce the expressions (C) for the accelerations X, T in terms of the coordinates , 77 from the theory of relative motion, explained in Art. 10. The motion of P in space is made up of the velocity relative to M together with that of M in space ; see fig. of Art. 223. Now OM is the radius vector of M, and the component velocities in the directions OM, MP are ' and <', while the accelerations in the same directions are r-ft* and ~(Pf) where accents represent differentiations with regard to the time. Again regarding M as fixed, MP is the radius vector of P, hence the component velocities of P along MP and parallel to MO (not OM ) are 77' and rj', while the accelerations in the same directions are 77" r)- and - j- (?7 2 <'). Adding together these components, ij \4/v we obviously obtain the values of u, v ; X, Y already given in Art. 227. 329. Relative and actual path. When the motion of a point is referred to moving axes 0%, Or) it is necessary to distinguish between the path in space and the path relative to the moving axes. Suppose a sheet of paper to be attached to the moving axes and to turn round the fixed point with them. The point P traces out on this sheet the relative path which is not the same as that traced out on a sheet fixed in space. The coordinates of P in the relative motion are (, 17) and the displacements parallel to these axes are d and di). The direction of the tangent of the relative path and the radius of curvature of that path are therefore found by the ordinary rules of the differential calculus. The coordinates of P in the path in space are also(, rj), but the displacements have just been proved to be d^-ijd^ and dq + i-dQ. These must be used instead of dx and dij in the formula; of the differential calculus. ART. 231.] RELATIVE AND ACTUAL PATH. 139 Let us represent by accents the differential coefficients with regard to any independent variable t. The formula of the differential calculus giving the space motion of P referred to fixed axes may be adapted to moving axes by writing u, v for x', y' respectively, where M=f --n', = V + ?$'> and u lt v^for x", y" where Thus, if if/, x be the angles the tangents to the relative and actual paths make with Of, and p, B be the radii of curvature of these paths, we have tan^ = p , 2-i. = f V - ?/f ", i jj i- = UV' - VU' + (V? + t> 2 ) A'. p P* When we apply kinematical theorems to purely geometrical properties in which the idea of time is absent, we regard t as an auxiliary arbitrary quantity introduced to represent the independent variable. If we wish the arc s to be the independent variable, we write t s. The effect of these changes may be exhibited in a figure. Let P, P' be the positions in space of the moving point at the times t, t + dt, and Of, Of 'the positions of the axis of reference at the same times. If PM, P'N be perpendiculars on Of, Of, we have f (A). Let P'M', PH be perpendiculars on Of and P'M' respectively. The coordinates of P, P' referred to axes Of, Orj fixed in space for a time dt are (B). These values of MM', P'H follow at once from Art. 223, but they may be obtained by projecting the broken line ON, NP' on Of, Oij. If x De the angle the tangent PP' makes with Of and d' is the angular velocity of f , and 0' = s'/p. Deduce an expression for the radius of curvature of the space locus of P. 140 MOVING AXES. [CHAP. V. Ex. 2. A particle P is attached to the extremity of a string of length I which is being wound on to a fixed curve after the manner of an involute. Prove that the component accelerations of P along and perpendicular to the straight portion of the string are respectively where ' is the angular velocity of . Also '= -%lp. Ex. 3. Assuming the earth to be uniformly describing a circle of radius a about the sun with velocity U, and the sun to be moving in a straight line in the plane of the earth's orbit with a uniform velocity F, prove that the radius of . . . , curvature at any point of the earth s orbit in space is - -- 2 = : , where \j i (J T v sin vj is the angle the line joining the earth and sun makes with the direction of the sun's motion. [Coll. Ex. 1892.] Ex. 4. A fine string wound round a circle has a particle P attached to its extremity and the circle is constrained to turn round its centre in its own plane with a uniform angular velocity w. The particle is initially in contact with the circle and has a velocity V normal to the circle. If | be the length of string unwound at the time t, prove that 2 = Ex. 5. A particle P is attached by a rod PA without mass to the extremity of another rod AS, n times as long, which revolves about the other extremity B, the whole motion taking place in a horizontal plane. If be the inclination of the rods, u the angular velocity of AB at the time t, prove that ** + *? + (^cos + w 2 sin e] = 0. [Math. Tripos, I860.] (tit (tli \ (tt J 232. Oblique axes. The general method of finding the resolved velocities and accelerations of a point referred to moving axes may be extended to oblique axes. These extensions however are not of any great importance because oblique axes are seldom used in mechanics. Let 0, OTJ be any two axes which make angles 6, with an axis Ox fixed in space. These angles we shall suppose to be perfectly arbitrary so that the angle 0t) between the axes is not necessarily constant. See figure of Art. 223. Let PQ represent any vector ; u, v its components obtained by oblique resolu- tion according to the parallelogram law. Let MJ , v x , represent as before the rates of increase of the components of the vector in directions fixed in space but coin- cident with the positions of 0, Oij at the time t. Let us resolve the vector in a direction perpendicular to 0. The resolved parts of w x and v l are clearly zero and v 1 sin ( - 0). Since 0'=d0, i)07)' = d, the resolution gives [(u + du) sin d& + (v + dv) sin ( - 6 + d)] - [v sin ( - 6)] v l sm(-0)= ^ d6 dv . = u di + Tt sm( + By resolving in a direction perpendicular to Orj we obtain in the same way M! sin ( - 6) = - v -^ + - sin ( - 6) - u cos (0 - 0) . ART. 234.] HYPER-ACCELERATIONS. 141 If , 77 are the oblique coordinates of P, the space velocities u, v of P are similarly wsin (0-0)= -1, 2 + f t sin (0-0)- The advantage of resolving perpendicularly to 0% and Oi) is that only one of the components u lt v lt enters into the resolution. We thus obtain each indepen- dently of the other. If we resolve in the directions 0, Oij we obtain the values of Mj + v : cos ((p - 6) and v 1 + MJ cos (0 - 6) and, from these, MJ and v l can be obtained by solving the equations. These values of MJ , v 1 were first given by H. W. Watson in the Math. Tripos of 1861. 233. Hyper-accelerations. It is seldom that we use higher differential coefficients with regard to the time than the second, Art. 21. When these are required the general theorem on vectors (Art. 223) gives the components for differ- ential coefficients of any order. Let x, y be the coordinates of a moving point referred to fixed axes, then X n =d n x/dt n , Y n =d n yjdt n are the components of the space hyper-acceleration of the n th order, Art. 21. Let 0, Or) be any set of moving axes, the relations between the space components of two successive orders of acceleration are X ^ Xn - Y Y _^^n . V ^ dt n dt dt n dt The reader may consult a Note Sur les Principes de la Mdcanique by Abel Transon, Liouville's Journal, vol. x. 1845, for another mode of treatment. Ex. 1. A point moves along a curve with velocity u, prove that the components along the tangent and normal of the acceleration of the third order are respectively ^_H?. - ( dt z fp ' ds \ p Ex. 2. A point P moves along a curve with uniform velocity. Prove that tan d ^cot5' where d, d' are the angles the diameter of the parabola of closest contact and the direction of the hyper-acceleration make with the normal at P. Show also that the semi-latus rectum of this parabola is p cos 3 d. D'Alembert's Principle. 234. When a single particle moves under the action of given forces the equations of motion may in general be found by re- solving the forces in some convenient directions. In the case of a system of particles the mutual reactions must also be taken into the account; these are in general unknown and will have to be eliminated from the equations. It is important to be able to write down some of the results of this elimination without first forming the equations of motion of every particle. Various 142 D'ALEMBERT'S PRINCIPLE. [CHAP. v. methods have been given to effect this either completely or partially. When in a statical problem, we wish to avoid introducing into our equations the mutual reactions of two bodies, we treat the two as one system. We resolve and take moments for the two bodies as if they were one. We may adopt the same method in dynamics. 235. In applying this principle to dynamics, it will be found convenient to use the term effective force. This may be defined as follows. When a particle is moving as part of a system, it is acted on by the external forces and the reactions of the other particles. If we consider this particle to be separated from the system and all these forces removed there is some one force which, with the same initial conditions, would make it move in the same way as before. This force is called the effective force on the particle. It follows that the effective force is statically equivalent to the impressed forces which act on the particle and the reactions of the rest of the system, but is differently expressed. Let m be the mass of the particle, (a:, y) the Cartesian coordinates ; the components of the force which must act to produce any given motion have been proved to be md^a/df* and md^y/dt 2 , these then are the components of the effective force. In the same way if v be the velocity and 1/p the curvature of the path, the tangential and normal components of the effective force are mdv/dt and mtf/p. See Art. 68. 236. Considering any one particle of the system, we know that the resolved parts of the effective forces in any directions .are equal to the corresponding resolved parts of the impressed forces and the reactions. It immediately follows that the effective forces on each particle, if reversed, are. in equilibrium with the impressed forces and the reactions. But, by Newton's third law, the mutual reactions of any two particles are in equilibrium. Making then any selection of the particles of a system, the reversed effective forces of those particles are in equilibrium with the external forces which act on them, excluding their mutual reactions, but including the pressures (if any) of the remainder of the system. ART. 238.] RESOLUTIONS AND MOMENTS. 143 Some of the equations of motion may therefore be found (1) by equating the sum of the resolved parts of the effective forces in any convenient directions to the sum of the resolved parts of the external forces, (2) by equating the sum of the moments of the effective forces about any point to the sum of the moments of the external forces. The resolved parts and moments of the external forces may be written down by the rules of statics. The components of the effective forces in various directions have been found in the preceding articles. The moment about any point then follows by multiplying that component by the length of the perpendicular from 0, Art. 6. If (x-i, T/J), (# 2 , 3/2) & c - are the Cartesian coordinates of a system of mutually attracting particles whose masses are m 1} w 2 &c., and if these are acted on by the external accelerating forces (X lt Fj), (X 2> F 2 ) &c., the equations of resolution and moments are ^ fd 2 y\ where the 2 implies summation for all the particles. 237. So long as we confine our attention to resolutions and moments it is unnecessary to include the mutual actions of the particles under consideration. If however we use the principle of virtual velocities to express the conditions of equilibrium we must remember that the particles may not be rigidly connected together. Now the work of two equal and opposite forces F, - F, acting on two particles distant r from each other is proved in statics to be Fdr. It is obvious that this does not vanish unless the distance r is invariable. This point is impor- tant in using the principle of vis viva. The most convenient way of applying the principle of Virtual Velocities to Dynamical problems is to use Lagrange's equations. 238. When the selected system of particles is a rigid body, the mutual distances of the particles composing it are invariable. It is proved in statics that the position of such a body in space of two dimensions can be denned by three quantities usually called coordinates. For example, these might be the Cartesian coordinates of some point and the angle which some straight line fixed in the body makes with some straight line fixed in space. Three independent equations of motion, free from mutual 144 D'ALEMBERT'S PRINCIPLE. [CHAP. v. reactions, are therefore necessary and sufficient to determine the position of the system at any time t. These three are supplied by the two resolutions and the equation of moments above described. It is proved in statics that a system of forces can be reduced to a single force R acting at some convenient point and a couple G. The components of the force R are equal to the sums of the components of all the forces of the system, and the couple G is equal to the sum of their moments about 0. This is usually called Poinsot's method of compounding forces. We shall now apply this method to find the resultants of a system of effective forces. 239. A system of particles, rigidly connected, moves in space of two dimensions. The coordinates of the centre of gravity are (x, y), the angle which a straight line fixed in the body makes with a straight line fixed in space is and the whole mass is M. It is required to prove that the effective forces of the whole system d*x cfeu are equivalent to two effective forces M-J-, M -~ acting at the at at JZfL centre of gravity, and an effective couple Mk 2 -~ , where Mk 2 is a constant which depends on the form and structure of the body or system. Let m be the mass of any particle of the body, x = x + , y = y + r) be its coordinates. Then since 2ra/2m, 2m?y/2m are the coordinates of the centre of gravity referred to the centre of gravity as origin, it is clear that 2w = 0, Smi; = 0. The sum of the resolved parts of the effective forces parallel to the axis of x is d?x d^ d 2 - , -tfoc The resolved part parallel to the axis of y may be found in the same way. These two effective forces are the same as the effective forces of a particle whose mass is M placed at the centre of gravity and moving with that point in space. 240. To find the effective couple we take moments about the centre of gravity. Remembering that , 77 are the coordinates ART. 241.] THE EFFECTIVE COUPLE. 145 of the particle m when referred to the centre of gravity, the couple is Since ^m^=0, 2w7; = 0, the right-hand side reduces to the first term. Let p, 6 be the polar coordinates of the particle m, referred to the centre of gravity as origin, then gdvj tjd^ = pdd. The couple is therefore d /cfy dA d cfy dA d -if - 17 -77 ) = j-. dt dtj d T* -i - -77 j-. -r- dt v dt dtj dt\ dt We shall now introduce the condition that the particles are rigidly connected together. When this is the case the dd/dt of every particle is equal to d/dt, and the length of every p is constant during the motion. For, let a be the angle the radius vector p of any particle m makes with the straight line fixed in the body, then = + a. Though a may be different for every particle, yet its value does not change during the motion, hence doL/dt = 0, and dd/dt = d/dt. The effective couple is (2ra/> 2 ) - . at~ 241. The constant 2mp 2 is called the moment of inertia of the system about an axis drawn through the centre of gravity perpendicularly to the plane containing the particles. To find the moment of inertia of any system about any axis, we multiply the mass of every particle by the square of its distance from the aads and add the results together. When the particles are so close together that they form a continuous body, the sum is an integral. Thus for a circular area of radius a and density D, the area of any element is p dOdp ; hence the moment of inertia about an axis drawn through the centre perpendicular to its plane is 2mp*=ffDpdedp . p 2 = D [p<] . [0], where the square brackets imply that the quantity is to be taken between the limits of integration. These limits being p = to a, and = to 2?r, the moment of inertia about the centre is \Mo?. In the same way the moment of inertia of a rectangle whose sides are 2a and 26 about an axis drawn through the centre of gravity perpendicular to its plane is $M (a z + 6 s ). R. D. 10 146 D'ALEMBERT'S PRINCIPLE. [CHAP. v. The moment of inertia of a sphere of radius a about a diameter s The moment of inertia of a triangular area about any axis is the same as that of three particles each one-third of its mass placed at the middle points of the sides. 242. The moment of inertia is of special importance in rotational motions, for, in a certain sense, it measures the dynamical significance of the form and structure of the moving body. Thus all free bodies having equal moments of inertia rotate with equal angular accelerations when acted on by equal couples. The translational motion depends on the mass and the position of the centre of gravity, Arts. 92, 239. 243. Sufficiency of the equations. The equations of motion of a particle moving freely are ffix_ (t), y = ^(t); the equations may be written in the functional forms = YT^ where X, Y are known functions of (t) and \f/ (t). Representing the initial time by t = 0, we suppose that the four initial values 0(0), ^(0), *(T)-0(0), ^(r)-^(O) ........................ (2) are given. Putting t=0 in (1) we deduce the values of (2r), \f> (2r) ; again putting t = r we obtain (3r), ^(3r), and so on. Thus by a continual repetition of the process the values of (nr), i^(nr) and therefore of (t), ^ (t) can be found. That the solution of the two equations of motion of the second order leads to results which contain four arbitrary constants (to be determined by the initial conditions) is also proved in treatises on differential equations; see Forsyth's Differential Equations, Art. 173. 244. On general and particular integrals. The Cartesian equations of motion of a free particle are x"=X, y" = Y .................................... (1), where accents denote differential coefficients with regard to the time. These are usually solved by combining them together so as to obtain a perfect differential. We then have by integration F(x t y, x', y', t) = C ................................. (2), where C is a constant. When an integral is obtained in this manner there is nothing to limit the initial conditions. However the particle may be projected the ART. 246.] ON GENERAL INTEGRALS. 147 equation (2), after determining the proper value of C, must be true throughout the whole motion. Such an integral is called a general integral. An integral which is true only for special initial conditions is called a particular integral. 245. If any equation such as (2) be arbitrarily written down containing one arbitrary constant we may enquire what the, dynamical problem is of which that equation is a general integral. To answer this we differentiate (2) and substitute from (1). We then have dF , dF , dF v dF v dF . - j -x' + y' + - r -,X+,Y+ = Q (3). ax ay ax dy dt Since the state of motion at any time t may be taken as the arbitrary initial motion the quantities x, y, x', y' are really arbitrary. The forces X, Y must there- fore be such as to make (3) an identity. To determine X, Y we differentiate (3) partially with regard to any of the four letters x, y, x', y', treating the others as constants. Supposing that X, Y are intended to be functions of x, y only, they are constants when we differentiate partially with regard to x', y'. In this way we may obtain, by successive differen- tiations, several equations each containing X, Y in the first degree. If these equations lead to inconsistent values of X, Y we infer that the given equation cannot be a general integral. It may also happen that all the equations to find X, Y are identical, and in this case the forces X, Y are to a certain extent arbitrary. Bertrand has shown that this can happen only when the integral (2) has the form (4). This therefore, when X, Y are functions of x, y only, is the only general integral which can be common to several dynamical problems. Liouville's Journal, 1852. Ex. 1. If a;' 2 + ?/' 2 - 2/ (x, y) = C be taken as the general integral, prove that X=dfldx, Y=dfjdy. This is the equation of vis viva. Ex. 2. Prove that xy'x'y = C with the upper sign cannot be a general integral; but, with the lower sign, is a general integral when the resultant force tends to the origin. The Principle of Vis Viva. 246. To investigate the principle of vis viva for a system of particles. Besides the external forces which act on the several particles we must here take into account their mutual actions and re- actions. Let m be the mass of any one particle ; x, y its coordinates ; let X, Y be the components of all the forces which act on that 102 148 THE PRINCIPLE OF VIS VIVA. [CHAP. V. particle. The equations of motion of that particle are d-y Multiplying these by dec/at and dy/dt respectively and adding the results, we have Summing this for all the particles of the system, we have The right-hand side of this equation, after multiplication by dt, is the work done by the forces as the system makes a small dis- placement, Art. 185. Amongst the forces X, Y are included the unknown reactions on the several particles, but it is clear that we may omit from the right-hand side all the reactions which would disappear in the principle of work in statics. When the remaining forces are such that the work integral fr(Xdx + Ydy) = U+C ..................... (4), where U is a known function of the coordinates of the particles, these forces are said to form a conservative system. Art. 181. Representing by v the velocity of the particle m, the integral of (3) becomes &mv* = U+C ......................... (5). Let U be the same function of the initial coordinates that U is of the coordinates at the time t, and let v be the initial value of v. The equation of vis viva may also be written in the form = U - U ..................... (6). 247. The principle of vis viva is important for several reasons. (1) The principle is of general application. The forces in nature are such that there is a work function, and the unknown reactions, in general, disappear from the equation. (2) When there is only one way in which the system can move, that motion is determined by the principle. ART. 249.] THE FORCE FUNCTION. 149 (3) The principle gives a relation between the circumstances of the motion in any stated position of the system and those at the initial stage. When the intermediate motion is not required this is particularly important. 248. The force function. The equation of vis viva can be usefully em- ployed only when the integrations necessary to obtain the force function U can be effected. It is also important to notice beforehand what forces and reactions may be omitted in forming that equation. The acting forces may be classified thus, (1) the external forces which act on the particles, (2) the mutual actions of such of the particles as are rigidly connected together, (3) the mutual attractions of independent particles, (4) the pressures due to any fixed curve or surface on which some of the particles are constrained to move. The external forces are in general central forces tending to or from fixed points. It follows from Art. 186 that, when each force is some function of the distance from the fixed point, the contribution of each to the work function can be integrated. Let R be the mutual action between two particles whose instantaneous distance apart is r, and let R be measured positively when the action tends to increase r. It is proved in statics that the work of both the action and reaction is Edr. It follows from this that the reaction between any two particles which keep an invariable distance from each other throughout the motion disappears from the equation of vis viva, for in such a case dr=Q. If any two independent particles repel each other with a force R which is a known function of their distance r, the contribution of this force to the work function can be integrated. If two particles are connected together by a tight string, even if bent by passing over smooth pulleys, fixed or moveable, the work of the tension is - Tdl, where I is the whole length of the string. If the length of the string is invariable the work is zero. The action of an inextensible string may therefore be omitted in the equation of vis viva. If the string is extensible and the tension obeys Hooke's law, the corresponding work can be found by integrating - Tdl, see Art. 187. 249. If one of the particles is constrained to move on a smooth fixed curve ichose equation is f(x, J/) = 0, let R be the normal pressure. The work of R is R cos ds ; this is zero because , being the angle between the direction of R and the arc of the path, is ir. If however the curve is itself constrained to move, the angle is not necessarily a right angle and the work may not be zero. Since the equation of the moving curve will contain t, this is usually expressed by saying that the geometrical relations must not contain the time explicitly, if the reactions ore to disappear. If the curve or surface is rough, the friction acts along the tangent to the path, and the work is zero only when the particle in contact is not in motion. 150 THE PRINCIPLE OF VIS VIVA. [CHAP. V. 250. Energy. Selecting some geometrically possible ar- rangement of the particles as a standard position, the work done by the forces as the particles move or are moved from any other given arrangement to the standard position is called the potential energy in the given position. Let the standard position be called S ; let the system move from some given initial position A and at the time t let its position be P. It has already been proved (Arts. 69, 246) that Kin. En. at P - Kin. En. at A = work A to P. But Pot. En. at P = work P to S, Pot. En. at A = work A to S. .-. Kin. En. at P + Pot. En. at P = Kin. En. at A + Pot. En. at A. It follows therefore that the sum of the kinetic and potential energies is constant throughout the motion. This sum is called the energy of the system, and it has just been proved that the energy of the system is constant and equal to its initial value. This theorem is true whatever standard position may be chosen, but it will be found convenient to so choose this position that the system may finally arrive there. When this choice is made the potential energy represents the whole work which can be obtained from the forces as the system moves to its final position. 251. As a simple example, let a heavy particle fall from rest at the ceiling of a room to the floor ; the kinetic energy after falling a distance z is $mv z = mgz. Let us take the floor (i.e. z = h) as the standard position, because the particle cannot descend any lower; the potential energy at the depth z is mg (h z). The whole energy is therefore mgh, which is constant throughout the motion. At the ceiling the energy is wholly potential because the particle starts from rest ; on arriving at the floor the energy is wholly kinetic, all the available potential energy having been changed into kinetic energy. 252. Degrees of freedom. If a system contain n particles free to move in space of two dimensions, its position can only be defined by the use of the 2w coordinates of the particles. There are evidently just 2?i different ways in which the particles can be moved, all other displacements being compounded of these. ART. 253.] VIS VIVA OF A BODY. 151 The system is then said to have 2?i degrees of freedom. If some of the particles are constrained to move on K given curves, or more generally if there are K given relations between the 2w coordinates, only 2w K coordinates are necessary to fix the position of the system and there are then 2n K degrees of freedom. The degrees of freedom of a system may be defined to be the number of coordinates required to fix its position. 253. Vis viva of a rigid body. When some or all of the particles of a system are rigidly connected together a simple and useful expression for the vis viva can be found. Let (x, y) be the coordinates of the centre of gravity, the angle which a straight line fixed in the body makes with a straight line fixed in space, and M the mass. The vis viva is then +' . dtj ) \dtj where M& is the constant called the moment of inertia of the body about the centre of gravity, see Art. 241. To prove this, let x = x + , y = y + 17 be the coordinates of any particle m, then v fdx\" ,~ . fdx\* /- d\ dx ^ ( fdM * m U) - < 2m > U) + 2 1 2m i ) at + 2 r W } ' Since 2w = as in Art. 240 the middle term is zero. Hence This equation expresses the proposition that the whole vis viva of a moving system, whether rigid or not, is equal to that of a particle of mass M moving with the centre of gravity together with the vis viva of the motion relative to the centre of gravity. To introduce the condition that the system is rigid we change to polar coordinates by writing (d%f + (drj)* = (ds)- = (dp)* + (pd6}\ Remembering that d0/dt is now the same for all the particles and equal to dfyjdt (Art. 240) and that dp/dt is zero, we find j 152 THE PRINCIPLE OF VIS VIVA. [CHAP. V. 364. Examples. Ex. 1. An endless light string of length 21, on which are threaded beads of masses M and m, passes over two small smooth pegs A and 11 in the same horizontal line and at a distance apart a, one bead lying in each of the festoons into which the string is divided by the pegs. The lighter bead m is raised to the mid-point of All and then let go. Show that the beads will just meet if \. [Math. Tripos, 1897.] We notice that only two positions of the system are contemplated in the problem, viz. (1) the initial position in which the bead m lies in All, and (2) the position in which the beads are in contact. In both these cases the kinetic energy is zero. The principle of vis viva asserts that the change of kinetic energy is equal to the work. It immediately follows that the work done when the system passes from the first to the second position is zero. Let x be the depth below AB at which the beads meet. Then omitting the tension, Art. 248, we have mgx + M {x - V(P - al)} =0. We also have by geometry 4x* + a?=P. Eliminating x we obtain the result. The circumstances of the motion when the beads TO, M are at any depths y, 17 below AB may also be deduced from the principle. We have $(mv 2 + Mv'*) = mgy + Mg {r,-J(P-al)} ..................... (1). Since the sum of lengths joining m and M to A is I, we have the geometrical equation v/(ia2 + j,2) + V(ia 2 + 1 ,*)=J .............................. (2). Differentiating the second equation, we have - Joining this to (1) we have the values of v, v' when y and 17 have any values not inconsistent with (2). Ex. 2. A particle of mass TO has attached to it two equal weights by means of strings passing over pulleys in the same horizontal line and is initially at rest half way between them. Prove that if the distance between the pulleys be 2a, the velocity of m will be zero when it has fallen through a space -. ^ ^ . 4m - - uf [Coll. Exam.] Ex. 3. Two pails of weights W, w, are suspended at the ends of a rope which is coiled round the perfectly rough rim of a uniform circular disc of radius a supported in a vertical plane on a smooth horizontal axis, and the pails can descend into a well so that when one comes up the other goes down. If the pails be allowed to move freely under gravity, and, when the heavier has descended a distance b from rest, a drop of water be thrown oft from the highest point of the rim of the disc, prove that this drop will strike the ground at a horizontal distance x from the axis of the disc given by where W is the weight of the disc, and h is the vertical distance above the ground of the highest point of the rim of the disc. [Math. Tripos, 1897.] The equation of vis viva gives J/'& 2 w 2 + (3/+TO) t> 2 =2 (M- m)gb. ART. 254.] EXAMPLES. 153 The theory of parabolic motion gives x = vt, and h=$gt'*. Putting w = t/-/a and fc 2 =^a 2 , we obtain the required value of x. Ex. 4. Two small holes A, B are made in a smooth horizontal table, the distance apart being 2a. A particle of mass M rests on the table midway between A and B ; and a particle of mass m hangs beneath the table, suspended from M by two equal weightless and inextensible strings, passing through the two holes. The length of each string is a (1 + sec a). A blow J is applied to M in a direction perpendicular to AB; show that if J 2 > ZMmag tan a, M will oscillate to and fro through a distance 2a tan a. But if J 2 is less than this quantity and equal to 2Mmag (tan a - tan ), the distance through which M oscillates will be 2 { P (P + 2) }*, where p = sec a - sec j8. [Coll. Ex. 1895.] The effect of the blow J is to communicate an initial velocity V= JIM to the mass M, leaving m initially at rest. Ex. 5. Two particles M , m are connected by a string passing over a smooth pulley, the lesser mass m hangs vertically, and M rests on a plane inclined at an angle a to the vertical. M starts without initial velocity from the point of the inclined plane vertically under the pulley. Prove that M will oscillate through a distance 5 ^~ . - where h is the height of the pulley above the initial m 2 - M" cos 2 a position of M, m is greater than .Mcos a but less than M. [Coll. Ex. 1897.] Ex. 6. Two equal particles connected by a string are placed in a circular tube. In the circumference is a centre of force varying as the inverse distance. One particle is initially at rest at its greatest distance from the centre of force, prove that if v, v' be the velocities with which they pass through a point 90 from the centre of force, -*/M + e -t^M = i. [Coll. Exam.] Ex. 7. A thin spherical shell of mass M is driven out symmetrically by an internal explosion. Prove that if when the shell has a radius a the outward velocity of each particle be V, the fragments can never be collected by their mutual attraction unless V"*, without mass, has a heavy particle rigidly attached to it at a point distant c from its centre, and its inner surface is constrained to roll on the outer surface of a fixed circle of radius a (b being greater than a), under the action of a repelling force from the centre of the fixed circle equal to /* times the distance. Prove that the period of small oscillations of the 11 v o & + c /6- hoop will be 2r - ) a \ CM J 154 THE PRINCIPLE OF VIS VIVA. [CHAP. V. Prove that when c = b, all oscillations large or small have the same period; and prove further that in the general case the hoop may be started so that it will continue to roll with uniform angular velocity equal to { /*&/(& - a)}^. [Math. Tripos, 1886.] The following is a simple (but not necessarily the shortest) method of writing down the equation of vis viva in problems of this kind. Having selected some independent variable to fix the position of the system, say, the inclination of the straight line joining the centres C, of the two circles to the vertical, we find the coordinates x, y of the particle in terms of by projecting 0(7, CP on the vertical and horizontal. The vis viva, being the sum of m (dxjdt) 2 and m (dyjdt) 2 , follows immediately. Equating the half of this sum to the force function \mn. (70 2 +C we have an equation giving do/dt in terms of i>. It is then easily seen that, if the constant C be properly chosen, the value of dOjdt reduces to the constant given in the question. To find the small oscillations, we differentiate the equation of vis viva and reject the squares of 6. When c=a, the path of the particle is an epicycloid and the oscillations large or small are, by Art. 211, tautochronous. 255. Rotating field of force. When a particle moves in a field of force which rotates round the origin with a uniform angular velocity n, an integral of the equations of motion can be found which reduces to that of vis viva when n = 0. Let 0, Ot] be two rectangular axes which rotate with the field of force, and let X, Y be the component accelerating forces. We then have by Art. 227 ^^ 2fc ~y dt ' .(1). Multiplying these by dg/dt and drj/dt and adding, we find eft di? + ^Tt d 2 ~ n V dt ^ dt) ~ dt dt ' We introduce the condition that the field of force rotates by making X, Y such functions of 17 only that X = dU/di; and Y=dU/di). Then U is a function of r) only and not of *. The equation then becomes %(tf-' where U is a function of the moving coordinates , ij, z. When 'A = V+C .................................... (7), where ' is the angular velocity of the field supposed to be constant. In this form we may omit from U all the actions and reactions which disappear in the principle of virtual work. 267. Coriolis' theorem on relative vis viva. A system of particles is referred to moving axes 0, Oij. Supposing the system at any instant to become fixed to the moving axes, let us calculate what would tJien be the effective forces on the system. If we apply these as additional impressed forces on the system, but reversed in direction, we may use the equation of vis viva to determine the relative motion as if the axes were fixed in space. Let mj, m 2 , &c. be the masses of the particles; (X IJ Yj), (X 2 , F 2 ), &c. the components of the impressed forces. Let also p, q be the resolved velocities of the origin, then, including these as explained in Art. 227, the equations of motion of any representative particle m are where = The left-hand sides of these equations measure the components of the effective forces on the particle m, Art. 227. The corresponding components on an imaginary particle of the same mass m attached to the moving axes and momentarily coin- ciding with the real particle are found by treating , 77 as constants. These are du , dp These we represent by X , Y for the sake of brevity. Transposing these terms to the other sides of the equations of motion, we have (3). These equations may also be used to supply another proof of the theorem in Art. 197. Multiplying these respectively by d^dt, dijjdt and adding, we have, as in Art. 255, 8S*3$H-*2+*-'3- Summing this representative equation for all the particles and integrating If the axes rotate round a fixed origin with a uniform angular velocity, u is constant and p, q are zero. The equation of Coriolis then takes the simpler form (5), ART. 260.] CORIOLIS ON VIS VIVA. 157 where r is the distance of the particle m from the origin and v is its velocity rela- tively to tlie axes. For a single particle this is the same as Jacobi's integral. If the angular velocity u is not uniform and p, q not zero, the system of additional forces (X Q) Y ) is not conservative and the integration in (4) cannot be effected except in special cases. The equation is however still important, for the first step in the integration of the equations (1) must be to eliminate the unknown reactions, if any such exist. Now the equation (4) is free from all the reactions which would disappear in the principle of vertical work, and that equation therefore supplies us at once with one result at least of the elimination. For the purposes of this proposition the forces measured by A' , Y are called the forces of moving space. When the origin of coordinates is fixed, these take the simple form 9 du ,. du . X =-^-r, Tt , ?=--, + _ (6). This theorem is due to Coriolis ; see the Journal Poll/technique, 1831. 258. Laisant's theorem. Ex. A particle moves under the action of a force whose Cartesian components are X=v n -, Y=v n ~, where v is the velocity. Prove that the equation of vis viva is v 2 ~ n =(2-n) U+C. See the Bulletin de la Societe Mathematique, 1893, vol. xxi. Moments and Resolutions. 259. The equation of Moments. If P, Q are the com- ponents of the force on a single particle resolved along and transverse to the radius vector, it is clear that Qr is equal to the moment of the forces about the origin. Representing this moment by M, the transverse polar equation of motion becomes 260. When a system of mutually attracting particles moves under the action of external forces we have by adding together the transverse polar equations of each particle d f C .d0\ (2) If R be the attraction of ra t on ra 2 , the reaction of m z on m^ is R, and the sum of the moments of these two must disappear from the right-hand side. If then the external forces are such that their resultant passes through the origin, we have 'S.MQ, and therefore by integration = # (3), 158 MOMENTS AND RESOLUTIONS. [CHAP. V. where H is a constant. This equation expresses the proposition that when a system of mutually attracting particles moves under the action of external forces such that the sum of the moments about a fixed point is zero, the sum of the angular momenta of all the particles about that point is constant. For example, if any number of mutually attracting planets move under the influence of a fixed sun, the sum of their angular momenta is constant. See also Art. 93. Since xdy ydx = r*dd (Art. 7), the equation (3) of moments when written in Cartesian coordinates takes the form 261. Rigid system. When a system of particles is rigid it is useful to have an expression for the resultant angular mo- mentum about the origin. Let (x, y) be the coordinates of the centre of gravity, the angle a straight line fixed in the body makes with a straight line fixed in space, and M the mass. The angular momentum of the whole mass is then where Mk 2 is the moment of inertia about the centre of gravity. See Art. 241. To prove this, let (x, y) be the coordinates of the particle m, then x = x + %, y = y + i)> Remembering that 2ra = 0, 2m?; = as in Art. 239, we find by substitution that ^ / dy dx\ .^ \f-dy ~dx\ _, /..dtj dg\ 2m (x -f- - y -.-- = (Sm) - . I -f * (| -jf ^ ^ 1 . \ dt J dtj ' \ dt dt) V dt dtj Since dx/dt, dy/dt are the components of the velocity of the centre of gravity, the first term is the moment of the velocity of a particle of mass M placed at the centre of gravity and moving with it. The equation therefore asserts that the angular momentum about any point is equal to that of the whole mass collected at the centre of gravity together with the angular mo- mentum round the centre of gravity of the relative motion. To introduce the condition that the system is rigid we change to polar coordinates by writing %di) rjd^ = p*d6. The second 7/J term then becomes ^mp z --. Remembering that dd/dt is the ART. 263.] ANGULAR MOMENTUM OF A BODY. 159 same for every particle and equal to d(j>/dt (Art. 240), this term becomes --. at It follows that, when a rigid body is acted on by any forces whose moment about the origin is G, the equation of moments is d 262. Ex. 1. A particle moves in a field of force defined by the force function . Show how to find the coordinates r, 6 in terms of the time. The force transverse to the radius vector is Q=dU/rd6. The equation of moments therefore becomes I r 2 ) = 2 -~ . Multiplying by r 2 ddjdt, the inte- Clt \ Oft J T Uv gration can be effected and we find (1), where A is an arbitrary constant. This integral is equivalent to a result given by both Jacobi and Bertrand. The equation of vis viva is Eliminating d6/dt by the help of (1) we arrive at an equation giving dt/dr as a function of r. The determination of t in terms of r has thus been reduced to an integration. The relation between 6 and t may then be found from (1) by another integration. Ex. 2. A particle is placed at rest at the point x=0, r=a in a field defined by cfix U=m --3- . Show by writing down the equations of vis viva and moments that the path is a circle. 263. The equation of resolution. If a system of particles moves under the action of external forces, we have by resolving parallel to the axis of x, (Art. 236), 2m ,-- = 2mX, at' where X is the typical accelerating force on the particle m. In this equation we may omit the mutual attractions of the particles, for the action and reaction being equal and opposite, these dis- appear in the resolution. If any direction fixed in space exist such that the sum of the components of the impressed forces in that direction is zero, we 1GO MOMENTS AND RESOLUTIONS. [CHAP. V. can take the axis of x parallel to that direction. We then have 2wJf = 0, .' . 2m - r - = A , at where A is a constant. This result is the same as that already arrived at, and more fully stated, in Art. 92. 264. Summary of methods of integration. When the system of particles moves in a given field of force the equation of vis viva in general supplies one integral of the equations of motion. If the system has only one degree of freedom, this integral is sufficient to determine the motion. When another integral is required, there is no general method of proceeding. We usually search if there is any direction fixed in space in which the sum of the resolved parts of the forces is zero, or any fixed point about which the sum of the moments is zero. In either of these cases an additional integral is supplied by the methods of Arts. 263 and 260. The first case usually occurs when the acting force is gravity, the second when the force is central. When these methods fail we have recourse to some artifice suited to the problem. Suppose that we have some reason for believing that a particle describes a certain path, we constrain the particle by a smooth curve. If the pressure can be made zero by the proper initial conditions, the constraint may be removed and the particle will describe the path freely, Art. 193. 265. Examples. Ex. 1. Two particles, of masses 771, M, placed on a smooth table, are connected by a string of length a + b, which passes through a fine ring fixed at a point on the table. The particles are projected with velocities U and V perpendicularly to the portions of the string attached to them, and the initial lengths are respectively a and 6. Find the motion. Let (r, 0), (p, ) be the polar coordinates of m and M at the time t. By the principles of angular momentum and vis viva, we have We have also the geometrical equation r + p=a + b ....................................... (3). Eliminating p, 6, , we find . (M+m} _ + __ + _-- m ^ + MF- ............... (4). ART. 265.] EXAMPLES. 161 In this differential equation, the variables can be separated and thus t can be expressed in terms of r by an integral. The integration cannot be generally effected. If the system oscillate, the extreme positions are determined by putting dr[dt = 0. We thus have 'l^ 22 = " (5). Since the left-hand side is positive when r^O and r=a + b and vanishes when r=a there is a second positive root less than a + b. This second root may be proved to be greater or less than a according as mU 2 ja is greater or less than MF 2 /b. These values of r determine the extreme positions of the system. We notice that if F be very small, the second root is very nearly equal to a+ b. If F=0 the particle M arrives at the origin, but the appearance when r=a + b of the singular form 0/0 in the equation (5) is a warning that the motion changes its character in this case. In fact if the third term on the left-hand side of (4) is removed, the velocity of arrival at is finite instead of being infinitely great. To find the tension T of the string, we use the radial equation of motion for one of the particles. This gives dV_ (de\ 2 __T dt?~ r \di) ~ ~m' Differentiating (4) we find drjdt in terms of r and after some slight reductions rp F 2 & 2 \ b-r) 3 )' M+m \ r 3 (a + b The string therefore does not become slack. Ex. 2. Two particles whose masses are in the ratio 1 : 2 lie on a smooth horizontal table, and are connected by a string that passes through a small ring in the table : the string is stretched and the particles are equidistant from the ring : the lighter particle is then projected at right angles to its portion of the string. Prove that the other particle will strike the ring with half the initial velocity of the first particle. [Coll. Ex. 1896.] Ex. 3. One A of two particles of equal mass, without weight, and connected by an inelastic string moves in a straight groove. The other B is projected parallel to the groove, the string being stretched. Prove that the greatest tension is four times the least. [Coll. Ex.] Keduce A to rest, then B is acted on by T and T cos 0, the latter being parallel to the groove, where is the angle AB makes with the groove. The particle B now describes a circle, and the normal and tangential resolutions give the angular velocity and the tension. Ex. 4. Two particles m, M, are connected by a string, of length a + b, which passes through a hole in a smooth table ; M hangs vertically at a depth b below the hole, m is projected horizontally and perpendicularly to the string with velocity I" from a point on the table distant a from the hole. Prove that if M just rise to the table, mF 2 (2a& + ft 2 ) = 2Mgb (a + b) 2 . Prove also that if M oscillates, mF 2 + 2Mga > 3 (M 2 mV Va 2 )*. What is the motion if mV' 2 =Mga f > R.D. 11 162 MOMENTS AND RESOLUTIONS. [CHAP. V. Ex. 5. Two small spheres of masses m and 2m are fixed at the ends of a weightless rigid rod AB which is free to turn about its middle point 0; the heavier sphere rests on a horizontal table, the rod making an angle 30 with it. If a sphere of mass m falling vertically with velocity u strike the lighter sphere directly, prove that the impulse which the heavier sphere ultimately gives to the table in f mu(l + e), where e is the coefficient of restitution between the two spheres, the table being perfectly inelastic. [Coll. Ex. 1893.] At the first impact we take moments for the two particles in, 2m about O to avoid the reaction at O. We therefore have Smv'a = Ra cos o, m (u r - u) = - R where a =30. At the moment of greatest compression the velocity of approach of the centres is zero, /. u' = v'cosa, and R = $mu. Since the complete value of R is found by multiplying this by 1 + e, the velocity of either end of the rod after impact is , 4 - wcosa (1 -;- <). The balls in and 2m rotate with the rod round through some angle, and 2m finally hits the table with a velocity v'. Taking the same equation of moments as before R'a cos a= Smv'a, .: R'=$mu (1 + e). Ex. 6. One end of a string of length I is attached to a small ring of mass m which can slide freely on a smooth horizontal wire, and the other end supports a heavy particle of mass m'. If this particle be held displaced in the vertical plane containing the groove, the string being straight and then let go, prove that the path of m' is part of an ellipse whose semi-axes are l t lml(m + m'), the major axis being vertical. [Coll. Ex. 1896.] Only the horizontal resolution and the geometrical equation are required. Ex. 7. A rectangular block of wood of mass J/ is free to slide between two smooth horizontal planes, and in it is inserted a smooth tube in the shape of a quadrant of a circle of radius a, one of the bounding radii lying along the lower plane, and the other being vertical. A particle of mass m is shot into the tube horizontally with velocity V, rebounds from the lower plane, and leaves the tube again with a relative velocity V, prove that V' 2 = e*V*-2ga(l- e 2 ) (M + m)/M, where e is the coefficient of restitution for the lower plane. [Coll. Ex. 1895.] Ex. 8. If in the case of three equal particles the units are so chosen that the energy integral is (v-f + , 2 + v^) -- h -- 1- - -, where r 12 is the distance r ss r si r i2 r between the particles whose velocities are v l and r 2 , and if r is a positive constant, the greatest possible value of the angular momentum of the system about its centre of inertia is f v/(2r). [Math. Tripos, 1893.] Ex. 9. Two equal particles are initially at rest in two smooth tubes at right angles to each other. Prove that whatever be their positions and whatever their law of attraction, they will reach the intersection of the tubes together. [Coll. Ex.] Ex. 10. Three mutually attracting particles, of masses m lt ni 2 , n^, are placed at rest within three fixed smooth tubes Ox, Oy, Oz at right angles to each other. The attraction between any two, say m^, wi 2 , is |Aw 1 7n 2 r 3 ' c where r 3 is the distance. If the triangle joining the particles always remains similar to its initial form, prove that the initial distances satisfy the equations wig mar,*" 1 " ffjj + m 3 - OTJ HI, + m-i - ._, wjj -f m 2 - m s ART. 267.] EXAMPLES. 163 266. Double answers. Ex. A cube, of mass M, constrained to slide on a smooth horizontal table, has a fine tube ACB cut through it in the vertical plane through its centre of gravity, the extremities A, B being on the same horizontal line and the tangents at A, II horizontal. A particle, of mass m, is projected into the tube at A with velocity F, deduce analytically from the equations of linear momentum and vis viva that the velocity of emergence at B is also V. Let u, v be the velocities of the cube and particle at emergence. The principles referred to give Mu + mv = m V, Mu? + mv^ = m V 2 . These give two solutions, viz. (1) u=0, v = V, and (2) M = 2mF/S, w = (wi- where S = m + M. To interpret these we notice that there are two sets of initial conditions which give the same linear momentum and vis viva. These are determined by the values of M, v just written down. We have therefore really solved two problems and have thus obtained two results. To distinguish the solutions, we investigate the intermediate motion. Let P be any point in the tube and let p be the tangent of~the angle the tangent makes with the horizon. If u, v now represent the horizontal velocities at P, the same two principles give Mu + mv = mV, Mv? + m (u 2 + p 2 x' 2 ) = mV 2 , where x' = v-u is the relative velocity. These give F Now v=V initially whenp = 0, hence the radical must have the positive sign and must keep that sign until it vanishes. On emergence therefore, when p is again zero, r = F. The negative sign of the radical evidently gives the initial conditions of the other problem. 267. Bodies without mass. Ex. 1. A heavy bead is free to slide along a rod whose ends move without friction on a horizontal circle ; prove that when the mass of the rod is negligible compared with that of the bead, the bead will, when started, continue to slide along the rod with an acceleration varying inversely as the cube of its distance from the middle point. [Math. Tripos, 1887.] The reaction between the rod and the particle is zero because the rod has no mass. To prove this, let R be the reaction, M the mass of the rod, then, taking moments about the centre of the circle, we have MK?duldt = Rp, where w is the angular velocity of the rod. Hence JR = when M=0. The particle P, being not acted on by any horizontal force, describes a straight line in space with uniform velocity b. If x be the distance of P from the middle point (7 of the rod ; a, c, the perpendiculars from O on the path and on the rod, we have x- + c 3 = OP 2 = a 2 + 6 2 2 . This gives d^xjdt 2 = b 2 (a- - c 2 )/^ 3 . Ex. 2. A rigid wire without mass is formed into an arc of an equiangular spiral and carries a heavy particle fixed in the pole. If the convexity of the wire be placed in contact with a perfectly rough horizontal plane prove that the point of contact will move with a uniform acceleration equal to g cot o, where a is the angle of the spiral. [Math. Tripos, I860.] 112 164 MOMENTS AND RESOLUTIONS. [CHAP. V. 368. Equation of the path. Let P, (,> be the resolved accelerating forces acting on the particle respectively along and perpendicular to the radius vector. Let P be regarded as positive when acting tmoards the origin. The equations of motion are v.de/ ~ rdt\ dt. To find the path we eliminate t. The second equation, after multiplication by integration, as in Art. 262, becomes For the sake of brevity we represent the right-hand side by H*. Putting also u= 1/r, we find d8ldt=Hv?. We then have dr_ 1 du d6_ _Trdu dt~~u z dedt~~ d' . + U = KW ....................................... < 4) " This can be integrated when P is a function of u alone, a case which is considered in the chapter on central forces. It can also be integrated when P=u 2 F(0), the method of solution being that shown in Art. 122. When P=u 3 F(6) the equation is linear. If one solution of the differential equation is known, say u=$ (0), the general integral may be determined by substi- tuting = z (0). After integration we find z = A + B \[ (0)]~ 2 d6. 369. When P=w s F(0), Q = u 3 f (0), the differential equation of the path takes the linear form (5). The various cases in which this equation can be integrated are enumerated in treatises on Differential Equations. ART. 270.] EQUATION OF THE PATH. 165 By multiplying the equation by the proper factor we can make the left-hand side a perfect differential. Conversely choosing any factor, we can find the relation between P and Q that this may be the proper integrating factor. If we wish the relation between P, Q to be independent of the initial conditions, the terms containing It- as a factor must be made a perfect differential independently of the remaining terms. The coefficient of h 2 is -j^ + u and this is made a perfect d0" differential by either of the factors sin 6 or cos 0. The remaining terms must therefore also become a perfect differential by the same factor. The condition that L , + M + Nu is & perfect differential is N- -r- + - ^ = 0, and the integral is d0- d0 do aO T du ( , dL\ known to be L + I If- -= ) u. at) \ do J Multiplying equation (5) by sin 0, the product is a perfect differential if {2f (6) -F(0)}sine-{ sin Of (6) } + 2 { sin 0f(0) } = 0, which reduces at once to = -^ -^ + 3 cot Q -^ ................................ (6). u 3 dO u 3 u* The integral, since f (0) = Qju 3 , becomes (h* + 2 l-^d0\ (sin0^-co&0u\ - ^sin0tt=C .......... (7), V j u J \ d& J u where C is a constant. This is a linear equation of the first order and can be integrated a second time when Qju 3 is given as a function of 6. The determination of the path can therefore be reduced to integration when the relation (6) is satisfied. In the same way, if we multiply (5) by cos 0, we find that the product is a perfect differential if ^ = ^-.. - 3 tan0 ?, ................................. (8), w 3 dO M J U A and the integral is { h- + 2 [ ^ dO ) ( cos 8 - + sin 6u ) - ^. cos 0u= C' ...... (9), \ J u J \ d9 ) u which is linear and can be integrated a second time. Another case in which the integration of (3) can be effected may be deduced from Art. 262. The equation (3) is P [ If then -g=/(M) + 2 I 3 de, the integral is (10). 37O. Ex. 1. If P=u*F(0) and Q=Pt&n0, prove that M = ^isin0 is a par- ticular solution of the linear equation (5). Thence obtain the general integral by putting w=zsin0, where z is a function of which is determined by solving a linear equation of the first order. Ex. 2. A particle moves under the forces prove that an integral of its motion is -j^ sin - u cos 01 +/j. h (sin - sin 30) " + cos 30wl = C. ) (- dO J 166 SUPERPOSITION OF MOTIONS. [CHAP. V. Obtain also a similar integral if P=/tiu 3 cosn0 -In [Coll. Exam. 1892.] Ex. 3. If the Cartesian accelerating forces X, Y are unrestricted, prove that the differential equation of the path is where A is a constant depending on the initial conditions. Prove also that the determination of y as a function of x can be reduced to integration when both X, Y are functions of x only. Ex. 4. If X and F/y are functions of x only, the differential equation of the A Y path is linear. Prove that it can be integrated when Y=y -= , and that the first (tX integral is (A + 2 \Xdx) ^-Xy = C. v n\ ^\ Prove also that when = - H -- , the differential equation can be integrated y ax x and that the first integral is (A + 2 \Xdx) x ^ - (A + 2 \Xdx + xX) y = G. Ex. 5. Prove that the Cartesian equations of motion can be completely integrated when the force function satisfies d?U_d?U_ cPU dx z dj/ 2 ~ dxdy ' To prove this we notice that U= (y + ax) + f(y + a'x), where a, a' are the roots of a* -*=!. We then change the variables to =y + ax and ij=y + a'x. The new coordinates , 17 are also rectangular. The equations of motion become d 2 /d< 2 = <'(), V 2 . provided the velocity of projection at any point A is given by The same ellipse can also be described about a centre of force in the other focus H 2 whose law of attraction is i^Jr^ provided the velocity v 2 has the corresponding value. It immediately follows that the particle can describe the ellipse freely about both centres of force acting simultaneously, provided (1) the velocity v at any point A is given by 2 /2 1\ /2 1\ v =Ml I --- ) +M 2 ( --- ) . \1 / \2 / and (2) the direction of projection at A bisects externally the angle between the focal distances. According to this mode of proof both the centres of force should be attractive, for it is evident that an ellipse could not be freely described about a single centre of repulsive force situated in either focus. But the law of continuity shows that this limitation is unnecessary. Supposing ^ and /^ to have arbitrary positive values, it has been proved that the equations of motion of a particle moving freely under both centres of force become satisfied when this value of v z is substituted in them. The equations contain only the first powers of ^ and jj? (see Art. 271) and can be satisfied only by the vanishing of the coefficients of these quantities. They will therefore still be satisfied if we change the signs of either ^ or /^ . In the same way we may introduce other changes into the theorem, provided always we can obtain a dynamical interpretation of the result. 276. Ex. 1. Prove that a particle can describe an ellipse freely under the action of three centres of force ; one in each focus attracting as the inverse square and the third in the centre attracting as the direct distance. Find also the velocity of projection. Ex. 2. Particles of masses m } , n/., , &c. projected from the same point in the same direction with velocities MJ, w 2 , &c. under the action of given forces F lt F 2 , &c. describe the same curve. Show that a particle of mass M projected in the same direction with a velocity V under the simultaneous action of all the forces F,, F z , &c. will also describe the same curve, provided MV* = J^V + m,,^ 2 + . . . . Ossian Bonnet, Note iv. to Lagrange's Mecanique. Ex. 3. A bead is projected along a smooth elliptical wire under the action of two centres of force, one in each focus, and attracting inversely as the square of the distance. If TP, TQ be any two tangents to the ellipse, prove that the pressure when the bead is at P : pressure when the bead is at Q : : TQ 3 : TP 3 . Initial Tensions and radii of Curvature. 276. Particles, of given masses, are connected together by in- elastic rods or strings of given lengths and are projected in any given manner consistent with these constraints. It is required to find the initial values of the tensions and the radii of curvatures of the paths. ART. 277.] INITIAL TENSIONS. 1H9 The peculiarity of the problems on initial motion is that the velocities and directions of motion of all the particles are known. It will thus not be necessary to integrate the differential equations of motion, for the results of these integrations are given. Supposing that there are n particles, we shall require besides the 2n equations of motion a geometrical equation corresponding to each reaction. To show how the geometrical equations may be formed, let us suppose that two particles m 1} ra 2 are connected by a rod or straight string of length I. The component velocities of the two particles in the direction of the string being necessarily equal, their relative velocity is the difference of their component velocities perpendicular to the rod; let these be Pi, F 2 . If < be the angle the rod makes with some fixed straight line, the geometrical equation is I -3- = P - Civ 2 The simplest method of obtaining the relative equations of motion is perhaps to reduce m l to rest. To effect this we apply to both particles (1) an acceleration equal and opposite to that of m^, and (2) an initial velocity equal and opposite to that of m^. The path of m 2 being now a circle whose centre is at rrii and whose radius is I, the relative accelerations are those for a circular motion. (Art. 39.) Let X l} X 2 be the components along the rod of junction of all the forces and tensions which act on m^, w 2 respectively. We then have (Art. 35) --'-- 1 . a) dt I m% m^ " In this way we may form as many equations as there are re- actions. By solving these the initial values of the reactions become known. If the angular accelerations of the rods are also required, let YH Y 2 be the component forces perpendicular to the rod which act on Wj, m 2 . Then ig-^-S ........................ (2). dv m 2 m-i 277. To find the curvatures of the paths, we refer to the equa- tions of motion in space. The velocity and direction of motion of 170 INITIAL TENSIONS AND CURVATURE. [CHAP. V. each particle being known, we may conveniently use the tan- gential and normal resolutions. We thus have 2n equations of the form & \r dv w m- = N, m-f- = T (3), p (if where N, T are linear functions of the forces and tensions which act on the particle m. These reactions having been found by considering the relative motion, we substitute in (3). The first of these determines the radius of curvature p of the path of m, and the second the tan- gential acceleration, if that be required. When any one of the particles is constrained to describe a given curve, the initial pressure of that curve is one of the unknown reactions. This pressure will be determined by the normal resolu- tion of (3) since the radius of curvature of the path is the same as that of the constraining curve. 278. If some or all the particles start from rest, the equations of relative motion are simplified, for we then have (f> = where the accent denotes d/dt. Since however the direction of motion of a free particle at rest is not given, the tangential and normal resolu- tions are then inappropriate. We can however use the Cartesian or polar resolutions in space. Since 6' = 0, the polar resolutions reduce to r" and r0" which are very simple forms. We must however bear in mind that if we require to differentiate the equations of motion this simplification must not be introduced until all the differentiations have been effected, Art. 281. We may also use Lagrange's equations, when the curvatures and not the tensions are required. These modifications of the general method are more especially useful in Rigid Dynamics and are discussed in the first volume of the author's treatise on that subject. 379. Examples. Ex. 1. Particles are attached to a string at unequal distances, and placed in the form of an unclosed polygon on a smooth table. The particles are then set in motion without impacts and are acted on by any forces. It is required to find the initial tensions and curvatures. Let ABCD &c. be any consecutive particles, and let the tensions of AB, BC, &c. be T lt T 2 , &c. Let the given forces be F lt F 2 , &c. and let them act in directions making angles a, /3, Ac. with AB, BC, &c. Let Ijdfjdt, 1.4^/dt, &G. stand for the known difference of the velocities of the consecutive particles resolved perpen- dicular to the rod or string joining them. ART. 279.] EXAMPLES. 17 1 The particle B being reduced to rest, G is acted on by T 3 /z 3 along CD, T S /7M 3 along CB, T 2 /m 2 along CB, TJm,, parallel to AB. Besides these there are the impressed accelerating forces -F 3 /m 3 and -.F 2 /m,. Since G describes a circle relatively to B, we have for the particle C d - at 3 z z . 2 where A, B, C, &c. are the internal angles of the polygon. The second resolution may be omitted if the angular accelerations of the several portions of string are not required. An equation, corresponding to the first of these, can be written down for each of the ?i particles, beginning at either end, except the last. We thus form (n - 1) equations to find the (n - 1) tensions. To find the initial radius of curvature of the path in space of any particle C we resolve along the normal to the path. Let the directions of motion of the particles be A A', BB', &c. and let v lt v 2 , &c. be the velocities of the particles. Then ^1 = T 3 sin DCC' + T, sin BGG' - F s sin (DGC' - y). Ps If the particle m s is initially at rest, v s =0 and the last equation fails to deter- mine p 3 . The initial tensions may still be deduced from the first equation. The initial direction of motion of the particle coincides with the direction of the resultant force and is therefore known when the initial tensions have been found. The tangential acceleration is also known for the same reason. The determination of the radius of curvature requires further consideration. Ex. 2. Heavy particles, whose masses beginning at the lowest are ?n lf m 2 , &c., are placed with their connecting strings on a smooth curve in a vertical plane. Find the initial tensions. In this problem the arc between any two particles remains constant, so that the tangential accelerations of all the strings are equal. Let this common accelera- tion be /. Taking all the particles as one system, the tensions do not appear in the resulting equation, we have therefore (?! + m 2 + &c. )/= - 7j<7 sin \f/ 1 - m*g sin \(/. 2 - &o., where \fr lt ^ 2 , &c. are the angles the tangents at the particles make with the horizon. Considering the lowest particle, we have h/= ~ m \9 sin V^i + T \ 172 INITIAL TENSIONS AND CURVATURE. [CHAP. V. Considering the two lowest, (>w, + 2 )/= - m,<7 sin f j - m.,g sin ^.,+ T 2 , and so on. Thus all the tensions T l , T. 2 , &c. have been found. If any tension is negative, that string immediately becomes slack. We also notice that the initial tensions are independent of the velocities of the particles. To find the initial reactions, we use the normal resolutions. If v be the initial velocity of the particle m, we thns find = -mgoon \1/ + R. P Ex. 3. Three equal particles are connected by a string of length a + b so that one of them is at distances a, b from the other two. This one is held fixed and the others are describing circles about it with the same angular velocity so that the string is straight. Prove that if the particle that was held fixed is set free the tensions in the two parts of the string are altered in the ratios 2a + b : 3a and 2b + a : 36. [Coll. Ex. 1897.] Ex. 4. Three equal particles tied together by three equal threads are rotating about their centre of gravity. Prove that if one of the threads break, the curva- tures of the paths instantaneously become 3/5, 6/5, 3/5ths respectively, of their former common value. [Coll. Ex. 1892.] Ex. 5. Two particles are fastened at two adjacent points of a closed loop of string without weight which hangs in equilibrium over two smooth horizontal parallel rails. Prove that when the short piece of string between the particles is cut the product of the tensions before and after the cutting is equal to the product of the weights of the particles. [Coll. Ex. 1896.] Ex. 6. Two particles of equal weight are connected by a string of length I which becomes straight just when it is vertical. Immediately before this instant the upper particle is moving horizontally with velocity *Jgl, and the lower is moving vertically downwards with the same velocity. Prove that the radius of curvature of the curve which the upper particle begins to describe is -f?*J5l. [Coll. Ex. 1897.] Just after the impulse the upper particle begins to move in a direction inclined tan" 1 1/2 to the horizon. Ex. 1. Two equal particles A, B, are connected by a string of length I, the middle point C of which is held at rest on a smooth horizontal table. The particles describe the same circle on the table with the same velocity in the same direction, and the angle ACB is right. The point C being released, prove that the radii of curvature of their paths just after the string becomes tight are 5^/57/4 and infinity. Ex. 8. Four small smooth rings of equal mass are attached at equal intervals to a string, and rest on a smooth circular wire whose plane is vertical and whose radius is equal to one-third of the length of the string, so that the string joining the two uppermost is horizontal, and the line joining the other two is the horizontal diameter. If the string is cut between one of the extreme particles and the nearer of the middle ones, prove that the tension in the horizontal part of the string is immediately diminished in the ratio 9 : 5. [Coll. Ex. 1895.] Ex. 9. Six equal rings are attached at equal intervals to points of a uniform weightless string, and the extreme rings are free to slide on a smooth horizontal rod. If the extreme rings are initially held so that the parts of the string ART. 280.] EXAMPLES. 173 attached to them make angles a with the vertical, and then let go, the tension in the horizontal part of the string will be instantaneously diminished in the ratio of cos 2 a to 1 + sin 2 a. [Coll. Ex. 1889.] Ex. 10. Three particles A , B, C are in a straight line attached to points on a string and are moving in a plane with equal velocities at right angles to this line, their masses being m, m', m respectively. If B come in contact with a perfectly elastic fixed obstacle, prove that the initial radius of curvature of the paths which A and C begin to describe is a, where AB BC = a. [Coll. Ex. 1892.] The particle B rebounds with velocity v. By considering the relative motion of A and B we have 4u 2 /a=T/m. By considering the space motion of A, v i \p=T\m. Ex. 11. A tight string without mass passes through two smooth rings A, B, on a horizontal table. Particles of masses p, q respectively are attached to the ends and a particle of mass ?/t to a point O between A and B. If m be projected horizontally perpendicularly to the string, the initial radius of curvature p of its path is given by (m +p + g)//> = pi a -q/b, where OA-a, OB = b. [Coll. Ex. 1893.] Ex. 12. A circular wire of mass M is held at rest in a vertical plane, on a smooth horizontal table, a smooth ring of mass m being supported on it by a string which passes round the wire to its highest point and from there horizontally to a fixed point to which it is attached. If the wire be set free, show that the pressure of the ring on it is immediately diminished by amount -r- :=-,- , where is 3/+4msinH0 the angular distance of the ring from the highest point of the wire. [Coll. Ex. 1897.] Ex. 13. Two particles P, P' of masses m, m' respectively are attached to the ends of a string passing over a pulley A and are held respectively on two inclined planes each of angle a placed back to back with their highest edge vertically under the pulley. If each string makes an angle ft with the plane, prove that the heavier particle will at once pull the other off the plane if m'Im<2 tan a tan /3 - 1. [Coll. Ex. 1896.] Ex. 14. Two particles of masses m, M are attached at the points B, C of a string ABC, the end A being fixed. The two portions AB, BC rest on a smooth horizontal table, the angle at B being a. The particle M has a velocity communi- cated to it in a direction perpendicular to BC. Prove that if the strings remain tight, the initial radius of curvature of the locus of M is a(l + nsin a a), where n=Mlm and BC=a. [Coll. Ex. 1895.] 280. To find the initial radius of curvature when the particle starts from rest. In this problem it may be necessary to use differential coefficients of a higher order than the second. Let x, y be the Cartesian coordinates of a particle, then representing differential coefficients with regard to the time by accents which takes a singular form when the component velocities x', y' 174 INITIAL TENSIONS AND CURVATURE. [CHAP. V. are zero. Putting n = x'y" y'x", we have after differentiation ''" ''" u = xy y x If "*~ OC '// -ii fir ~t~ '// 77 T i/ /> t' */ "^ ^r i/ For the sake of brevity let the initial value of any quantity be denoted by the suffix zero, thus # " represents the initial value of #". Using Taylor's theorem and remembering that #' = 0, y '= 0, we have x ' y " - y' x " = (# V" - < V) * + 4 (*."y? - <* V) + &c. Similarly (x' n - + y'rf = (# //2 + #o" 2 ) f * + &c. If the particle start from rest the initial radius of curvature is therefore zero. But if the circumstances of the problem are such that xj'y"' xd"y<>" = 0> the radius of curvature is given by "11 lv T iv 7/ o yo ^o yo This is the general formula when the axes of x, y have any positions. If the axis of y be taken in the direction of the resultant force # " = 0, and if we then also have x '" = 0, the expression for the radius of curvature takes the simple form -^' If F be the initial resultant force on the particle, X the trans- verse force, the formula when X = 0, X n ' = may be written __ o <> p0 ~i, The corresponding formula for p in polar coordinates may be obtained in the same way. We have when r (r"ff" r'"0") = initially, P where the letters are supposed to have their initial values. If the initial value of r" = 0, this takes the simpler form ART. 283.] SYSTEM STARTS FROM REST. 175 281. Let n particles P 1} P 2 , &c. at rest, be acted on by given forces and be connected by K geometrical relations. To find the initial radius of curvature of the path of any one particle P we proceed in the following manner, though in special cases a simpler process may be used. We differentiate the dynamical equations twice and reduce each to its initial form by writing for all the coordinates (x ly y^), (x 2 , y 2 ), &c. their initial values, and for (#/, yS), &c. zero. We differentiate the geometrical equations four times and reduce each to its initial form. We then have sufficient equations to find the initial values of x", x'", of 1 , &c., R, R', R", &c. where R is any reaction. Lastly solving these for the coordinates of the particular particle under consideration we substitute in the standard formula for p. This process may sometimes be shortened by eliminating the tensions (if these are not required) before differentiation. We thus avoid introducing their differential coefficients into the work. 282. Shorter Methods. We can sometimes simplify the geometrical rela- tions by introducing subsidiary quantities, say 6, , &c. In this way we can express all the coordinates (x lt T/J), &c. in terms of 6, , &c. by equations of the form x=f(6, <(>, &c.), y=F(0, , &c.) ........................ (1), where 6, 0, &c. are independent variables. Substituting in the dynamical equations und eliminating the reactions, we have 2n-K equations of the second order to determine 0, <(>, &c. in terms of t. These eliminations may be avoided and tin- results shortly written down by using Lagrange's equations. Lagrange's method is described in chap. vn. These equations, however obtained, contain 0, 6', 6"; , ', ", &c. and by differentiation we can find as many higher differential equations as are required. Since 0', ', &c. are zero, we find by differentiation where suffixes as usual indicate partial differential coefficients, thus f 6 =dfjde. There are similar expressions for the differential coefficients of y. Substituting in the standard form for p, we obtain the required radius of curvature. 283. We notice that if the partial differential coefficients f e , f^, &c. are zero the initial value of x iv does not depend on any higher differential coefficients of 0, , , which the strings make with the initial position ABC as the independent variables, Art. 283. Ex. 4. Two heavy particles P, Q, are connected by a string which passes through a smooth fixed ring 0, the portions OP, OQ of the string making angles 6, , with the vertical. If the masses m, M of P, Q, satisfy the condition mcos = Mcos , the initial radius of curvature of the path of P is given by M+m sin 2 6 _ sin 2 sin 2 # M ~JT~ r f T^r ' where r= OP and I is the length of the string. Take the polar equations of motion, eliminate the tension and differentiate twice. We thus find the initial values of 0", r", r iv ; since r"=0 the polar formula for p is much simplified. Ex. 5. A uniform rod, moveable about one end O which is fixed, is held in a horizontal position by being passed through a small ring of equal weight; show that if the ring is initially at the middle point of the rod, when it is released the initial radius of curvature of its path is 9 times the length of the rod. [Coll. Ex. 1887.] Taking as origin, the polar equation of motion of the particle shows that the initial values of r", r'" are zero, while that of r*=g0" + 2r0" 2 . Taking moments ART. 285.] SMALL OSCILLATIONS. 177 about O, Art. 261, we have -r -[(Mk^ + mr 2 ) ff} = (Ma + mr) gcos 6. This gives the -r - initial value of 0"=Ggl7a. The length of the radius of curvature follows by the differential calculus, Art. 280. Ex. 6. Three particles whose masses are m l , m 2 , m 3 are placed at rest at the corners of a triangle ABC, and mutually attract each other with forces which vary according to some power of the distance. If m 1 7 2 cF 3 , m 2 m 3 aF l , mgn^bF^ are the forces, prove that the initial radius of curvature p of the path of C is given by o p2 _ = _ m 2 a sin $ { - + mi b sin {- where 6, are the angles CA, CB make with the resultant force on C, P=(m. 2 + m 3 ) aF 1 + m 1 (F s c cos B + F z b cos (7) , Q = (m 1 + m s ) bF 2 + m 2 (F s c cos A + F-^a cos C) , and B is the resultant force on C. Deduce that the initial radii of curvature of the three paths are infinite when the triangle is equilateral. Small oscillations with one degree of freedom. 285. The theory of small oscillations has already been dis- cussed in the chapter on Rectilinear Motion so far as systems with one degree of freedom are concerned. In this section a series of examples will be found showing the method of proceeding in cases somewhat more extended. The particle, or system of particles, is supposed to be either in equilibrium or in some given state of motion. A slight disturbance being given, we express the displacements of the several particles at any subsequent time t from their positions in the state of equilibrium or motion by quantities ac, y, &c. These are supposed to be so small that their squares can be neglected. If required, corrections are afterwards introduced for the errors thus caused. We form the equations of motion either by resolving and taking moments or by Lagrange's method. By neglecting the squares of the displacements these equations are made linear in x, y, z, &c. They are also linear in regard to the reactions be- tween the several particles. Eliminating the latter we obtain linear equations which can in general be completely solved. The solution when obtained will enable us to determine whether the K. D. 12 178 SMALL OSCILLATIONS. [CHAP. V. system oscillates about its undisturbed state or departs widely from it on the slightest disturbance. The principle of vis viva supplies an equation which has the advantage of being free from the unknown reactions, but it has the disadvantage that its terms contain the squares of the velo- cities, that is, the terms may be of the order we neglect. Being an accurate equation, it may sometimes be restored to the first order by differentiating it with regard to t and dividing by some small quantity. Generally the solution is more easily arrived at by using the equations of motion which contain the second differential coefficients with regard to t. 286. Examples. Ex. 1. Two particles whose masses are m, m' are con- nected by a string which passes through a small hole in a smooth horizontal table. The particle m' hangs vertically, while in is projected on the table perpendicularly to the string with such a velocity that m' is stationary. If a small disturbance is given to the system so that m' makes vertical oscillations, prove that the period is 2?r / where c is the mean radius vector of the path of m. v i<7 Let r, be the polar coordinates of m, z the depth of m', I the length of the string and T the tension. The equations of motion after the disturbance are - dt* \dt ~ m' r dt\ dt d?z T W=9--,, r + z = l. The second equation gives r^dOjdt h, where h is a constant whose magnitude depends on the disturbance. Eliminating T, z and dOjdt we find d 2 ?- mh 2 Let r=c + where c is a constant which is as yet arbitrary except that the variable is so small that its square can be neglected. 3fc 2 mfc 2 Let us now choose c to be such that the right-hand side of the equation is zero ; then mh' 2 =m'c 3 g. Substituting for h we find / =A sin(nt + a), n 2 = -. - . m + m' c Since is wholly periodic and has no constant term, its mean value is zero, when taken either for any long time or for the period of oscillation. It follows that r=c is the mean radius vector of the path of m after the disturbance. This is not necessarily the same as the radius of the circle described before disturbance ; whether it is so or not depends on the nature of the disturbance given to the system. Let the particle m before disturbance be describing a circle of radius a with velocity F, then mV 2 la = m'g, each being the tension of the string; and the angular ART. 286.] ONE DEGREE OF FREEDOM. 179 momentum of m is mVa. If the disturbance be given by a vertical blow B ap- plied to the particle m', this reacts on m by an impulsive tension, and, the moment of this about O being zero, the angular momentum of m is unaltered. In this case we have h Va and we find c = a. If the disturbance be given by a transverse blow B applied at ?K, the velocity of m is changed to V where V - V=Blm. In this case h= V'a and c is not equal to a. Ex. 2. A particle of mass m is attached to two points A, B by two elastic strings each having the same modulus E and natural length I. If the particle be displaced parallel to this line, prove that the time of oscillation is 2ir N /(mZ/2.E). [Coll. Ex. 1895.] Ex. 3. A heavy particle hangs in equilibrium suspended by an elastic string whose modulus is three times the weight of the particle. The particle is slightly displaced in a direction making an angle cot" 1 4 with the horizontal and is then released. Prove that the particle will oscillate in an arc of a small parabola terminated by the ends of the latus rectum. [Math. Tripos, 1897.] Ex. 4. A straight rod AB without weight is in a vertical position, with its lower end A hinged to a fixed point, and a weight attached to the upper end B. To B are attached three similar elastic strings equally stretched to a length k times their natural length and equally inclined to one another, their other ends being attached to three fixed points in the horizontal plane through B. Show that, when the strings obey Hooke's law, the condition for stability of equilibrium is that the weight must not exceed that which, when suspended by one of the strings, would cause an increase of length equal to f (2 - Ijk) AB. Show that, when this condition is fulfilled, the system can perform small vibrations parallel to any vertical plane. [Math. Tripos, 1888.] Ex. 5. A smooth ring P can slide freely on a string which is suspended from two fixed points A and B not in the same horizontal line. If P be disturbed, find the time of a small oscillation in the vertical plane passing through A and B. If T be the time, (T/2ir) 2 ' J , g 3 are the roots of C)-BR=0 .................. (8). 388. If B, B' have the same sign, the roots of the quadratic (8) are separated by each of the values p*= -A, p*= - C. To prove this, it is sufficient to notice that the left-hand side of that equation is positive when p 2 = o> and is negative when p 2 has either of the separating values. It is also sometimes useful to notice that the roots cannot be equal unless the two separating values A and C are equal and that the equal roots are then p z =-A=-C. If AC-BB' = Q the biquadratic (5) has two equal zero roots, though the roots of the same equation regarded as a quadratic are unequal. 289. To find the four arbitrary constants L, M, a, /3, we solve the equations (7) with regard to the trigonometrical terms. We thus find - Putting t = 0, we at once have the values of L sin a, M sin p in terms of the initial values of the coordinates. Differentiating with regard to t and again putting < = 0, we find L cos a, M cos p in terms of the initial velocities. 290. Equal roots. The case in which the equation (5) has equal roots has been excepted. This occurs when either (A -C) 2 + iBB' = Q or AC-BB' = 0. When B, B' have the same sign the first alternative requires A = C and either B or B' equal to zero. In the second alternative the equation has two zero roots. Excepting when both B and B' are zero, the solution of the dynamical equations (2) is known to contain terms of the form (Lt + L') e m> . If m is positive or zero (or has its real part positive or zero), this term will increase indefinitely with t. If however the real part of m is negative and not zero, say equal to - r, the maxi- mum value of He'* is Ljre. Since L is so small that its square can be neglected, this term in the solution will always remain small except when r also is small. The existence of equal roots in the determinantal equation (5) does not therefore necessarily imply that the oscillation becomes large. 291. Before disturbance the particle P was in equilibrium at the origin under the influence of the forces X, Y given by (1) Art. 287. When AC-BB', the equations A'=0, T=0 are satisfied by values of x, y other than zero. These lie on the straight line Ax + By = 0. The dynamical significance of the condition AC=BB' is therefore that there are other positions of equilibrium in the immediate neighbourhood of the origin. The roots of equation (8) being 2> 2 = 0, q'= -A-C, the values of .T, y take the form x =L 1 t + 1/2 + M sin (qt + p), By= -A (L 1 t + L 2 )-CMsm(qt + p). The first terms represent a uniform motion along the line of equilibrium, while the trigonometrical terms represent an oscillation in the direction By=- Cx. Whether the particle will travel far or not along the line of equilibrium will depend on the nature of the forces when .r, y become large. ART. 293.] PRINCIPAL OSCILLATIONS. 183 292. Principal oscillations. Let the type of motion be that represented by such equations as (7). By giving the particle the proper initial conditions it may be made to move in either of the ways defined by the following partial solutions a = L sin (pt + a), y = L' sin (pt + a) ......... (10), a; = M sin (qt + /8), y = M ' sin (qt + /3) ......... (11). Each of these is called a principal oscillation and all the modes of oscillation included in (7) are compounded of these two. The dynamical peculiarity of a principal oscillation is the singleness of the period. The solution (10) is sometimes taken as the trial solution instead of the exponential used in obtaining (5). Practically we then begin the solution by finding the principal oscillations and finally combine these into the general solution (7). The paths of the particle when describing the principal oscil- lations are the two straight lines Ly = L'x, My = M'x ..................... (12). In each oscillation the ratio of the coordinates, being equal to L'/L or M'jM, is constant throughout the motion. We have by (7), using the values of p* + q 2 , p*q~, given by the coefficients of the quadratic (8), B' ~&~ "B" It follows that when B, B' have the same sign, the ratios L'jL, M'/M have opposite signs. In one principal oscillation, the co- ordinates x, y increase together ; in the other, when one increases the other decreases. We also notice that when B' = B, the two straight lines (12) are at right angles. The directions of these rectilinear oscillations may be obtained without inves- tigating the motion. The lines must be so placed that if the particle be displaced along either, the perpendicular force must be zero. The lines are therefore given by These lines are real when (A-C) 2 + 4BB' is positive. This condition is satisfied when the roots of the determinantal equation (5) are real or of the form pj-l. 293. When the coordinates are such that only one varies along each principal oscillation, they .are called principal coordinates. 184 SMALL OSCILLATIONS. [CHAP. V. Referring to the equations (9), we see that if we put By + (q* + A)x = %, By + (p* + A)x = >n, , 77 will be the principal coordinates. This transformation of coordinates is always possible, so long as p 2 and q* are real and unequal. We may also discover the principal coordinates without previously finding the values of p z , q*. We deduce from the equations (2) by using an indeterminate multiplier X. If now we write (B + \C)I(A+\B') = \ we see that x + \y will be a trigonometrical function with one period. We have a quadratic to find X; representing the roots by X n X2, the principal coordinates are i)=x + \, i y, or any multiples of these. 294. Conservative forces. When tlie forces which act on the particle are conservative, the solution admits of some simplifica- tions. Let U^be the force function, then, since dUjdx and dU/dy vanish in the position of equilibrium, we have by Taylor's theorem, U=U + 1t(Ax* + 2Bxy + Cy*) + ............... (1). It follows that the equations of motion are *j = X = Ax + By, d ^=Y=Bx + Cy ......... (2). Comparing these with the former values of X, Y, we see that & = 5. If we turn the axes round the origin we know by conies that the equation (1) can be always cleared of the term containing the product xy. Representing the new coordinates by , 17, let the expression for U become Z7-=Z7, + iU' + CV) + .................. (3), where A' + C' = A + C, A 'C' = A G - B 2 . The equations of motion are then The motion is oscillatory for all displacements or for none according as A', C' are both negative or both positive. If A' is negative and C' positive, the motion is oscillatory for a displace- ment along the axis of and not wholly oscillatory for other displacements. ART. 296.] TEST OF STABILITY. 185 The level curves of the field of force are obtained by equating U to a constant ; in the neighbourhood of the position of equili- brium, these become the conies Ax 2 + 2Bxy + Cf = N, or A'? + C'rf = N. The lines of the principal oscillations are the directions of the principal diameters of the limiting level conic, and the periods of the principal oscillations are proportional to the lengths of the diameters along which the particle moves. 295. The representative particle. The investigation of the small oscilla- tions of a particle in a given field of force has a more extended application to dynamical problems than appears at first sight. Suppose, for example, that a system, consisting of several particles connected together by geometrical relations, has two degrees of freedom. Let the position of this system be defined by the two coordinates x, y. The equations giving the small oscillations, after the elimi- nation of the reactions, take the form because the squares of x and y are neglected. If B = B' these are the equations of motion of a single particle moving in the field of force defined by The investigations given in Art. 292 and Art. 294 apply therefore to both problems. To exhibit the motion of an oscillating system to the eye, we take its coordi- nates x, y to be also the Cartesian coordinates of an imaginary particle which moves freely in the field of force U. We represent by a figure the level conies, the path of this representative particle, and sketch the positions of the principal oscillations. The special peculiarities of the motion will then become apparent in the figure. 296. Test of stability*. Let the field of force in which the particle moves be given by the function U. Since dU/dx and dU/dy vanish in the position of equilibrium, U must be at that point a maximum or a minimum. In the neighbourhood we have If AC Br is positive, U is a maximum or a minimum for all displacements according as the common sign of A and C is nega- tive or positive, and if AC B 2 is negative, U is a maximum for * The energy test of the stability of a position of equilibrium is given by Lagrange in the Mecanique Analytique. He gives both this proof and that in Art. 297. The demonstration for the general case of a system of bodies has been much simplified by Lejeune-Dirichlet in Crelle's Journal, 1846, and Liouville's Journal, 1847. See the author's Rigid Dynamics, vol. i. ; the corresponding test for the stability of a state of motion is in vol. n. 186 SMALL OSCILLATIONS. [CHAP. V. some and a minimum for other displacements. It follows from Art. 294 that the motion of the particle, when disturbed from its position of equilibrium, will be wholly oscillatory if U is a real maximum at that point. The particle will oscillate for some dis- placements and not for others if U has a stationary value, and will not oscillate for any displacement if U is a real minimum. We have here assumed that all the coefficients A, B, C are not zero. When this happens the cubic terms in the expression for U govern the series. The equations of motion (2) of Art. 295 will then have terms of the second order of small quantities on their right-hand sides. Besides this if A C & = 0, the quadratic terms of the ex- pression for U take the form of a perfect square, viz. (Ax + ByfjA. In this case the forces X = dlljdx and Y=dU/dy contain the common factor Ax-\- By so that there are other positions of equilibrium in the neighbourhood of the origin, see Art. 291. To determine the motion, even approximately, it is necessary to take account of the powers of x, y of the higher orders. The geometrical theory of maxima and minima has a cor- responding peculiarity, for it is shown in the Differential Calculus that further conditions, involving the higher powers, are necessary for a maximum or minimum. The following investigation shows how far this correspondence extends. 297. Let a particle be in equilibrium at a point P whose coordinates are x , y , and let U=f(x,y) be the work function. Let the particle be projected with a small velocity v^ from a point Pj, whose coordinates are x l ,y l , very near to P . The equation of vis viva gives (Art. 246) f7 1 ) (1), U ) (2), where v 2 = vf + 2(U - UJ (3). Let U be a maximum at the point P for all directions of displacement, then U^ < U and v 2 is a small positive quantity. As the particle recedes from P , U Q U increases, but the equation (2) shows that the particle cannot go so far that U U becomes ART. 299.] BARRIER CURVES. 187 greater than the small quantity |-?v'- The equilibrium is therefore stable for displacements in all directions. Let U be a minimum at P for all directions of displacement, then as the particle moves from P the difference U U increases. So far as the principle of vis viva is concerned, there is nothing to prevent the particle from receding indefinitely from P . Let U be a maximum for some directions of displacement and a minimum for others. The particle cannot recede far from P in the directions for which U is a maximum, but there is nothing to restrict the motion in the other directions. 298. Ex. A particle P is in equilibrium under the action of a system of fixed attracting bodies situated in one plane, the law of attraction being the inverse th power of the distance. Prove that, if /c> 1, the equilibrium of P cannot be stable for all displacements in that plane, though it may be stable for some and unstable for other displacements. If K < 1, the equilibrium cannot be unstable for all displacements in that plane. To prove this let 7j be any particle of the attracting mass, coordinates /, g ; let x, y be the coordinates of P. The potential of ?HJ at P is by definition . , where 1\ is the distance of m 1 from P. We then find by a partial ' differentiation dx* r Summing this for all the particles of the attracting mass and writing U=^U lt we find d*U d*U . , m + - =(K ] ' The right-hand side is positive or negative according as /c>l or Taking the equilibrium position of P for the origin and the principal directions of motions for the axes, Art. 294, we see by Taylor's Theorem where A' = d 2 Uldx 2 , C'=d?Uldy 2 . It is evident that U cannot be a maximum for all displacements in the plane of xy if A' + C' is positive and cannot be a minimum for all displacements in the plane if this sum is negative. The result also follows from Art. 296. 299. Barrier curves. It is clear that this line of argument may be extended to apply to cases in which there is no given position of equilibrium in the neighbourhood of the point of projection. Let the particle be projected from any point P x with any velocity v l in any direction. Throughout the subsequent motion we have - Z7i). 188 SMALL OSCILLATIONS. [CHAP. V. where U is a given function of #, y and U l is its value at the point of projection. If we equate the right-hand side of this equation to zero, we obtain the equation of a curve traced on the field of force at which the velocity of the particle, if it arrive there, is zero. This curve is therefore a barrier to the motion, which the particle cannot pass. If the barrier curve be closed as in Art. 297, the particle is, as it were, imprisoned, and cannot recede from its initial position beyond the limits of the curve. Some applications of this theorem will be given in the chapter on central forces. The right-hand side of the equation will in general have opposite signs on the two sides of the barrier. When this is the case the particle, if it reach the barrier in any finite time, must necessarily return, because the left-hand side of the equation cannot be negative. If the right-hand side of the equation have the same sign on both sides of the barrier, that sign must be positive, and U must be a minimum at all points of the barrier. The particle is therefore approaching a position of equilibrium and arrives there with velocity equal to zero. The particle therefore will remain on the barrier, see Art. 99. The barrier is evidently a level curve of the field of force and, as the particle approaches it, the resultant force must be normal to the barrier. Just before the particle arrives at its position of zero velocity, the tangential component of the velocity must be zero, for this component cannot be destroyed by the force. The path cannot therefore touch the barrier, but must meet it perpendicularly or at a cusp. 3OO. Examples. Ex. 1. Two heavy particles of masses m, m', are attached to the points A, B of a light elastic string. The upper extremity is fixed and the string is in equilibrium in a vertical position. A small vertical disturbance being given, find the oscillations. Let x, y be the depths of m, m' below 0; a, b the nnstretched lengths of OA, AB, E the coefficient of elasticity. The equations of motion reduce to >E E\ E \ E cPy E ART. 300.] SEVERAL DEGREES OF FREEDOM. 189 To solve these we put x-h = Lsin(pt + a), ij - k = M sin (pt + a) ..................... (2), the constants h, k being introduced to cancel the right-hand sides of the equations of motion. Since x = h, y = k make d 2 x/d 2 = 0, d 2 y/dt 2 =0, these constants are the equilibrium values of x, y. We then find / , E E\ ( , E\ E* L m'b , I m 2 --- ) I m 2 - I = 75-, -j> = l 2 ............... (3). \ * a b J \ ' b) b* M E * One principal oscillation is given by (2) and the other by using instead of p 2 , the other root of the quadratic. It follows that in one oscillation the two particles are always moving in the same directions, that is both are moving upwards or both downwards. In the other when one moves upwards the other moves downwards. Ex. 2. Two heavy particles, of masses m, M, are attached to the points A, B of a light inextensible string, the upper extremity being fixed. Prove that the periods of the small lateral oscillations are 2-irjp and 2irlq where p and q are the roots of 1 a + b 1 m a & jo 4 g p 2 M + m g' 2 ~ ' and OA = a, AB = b. Prove also that the magnitudes of the principal oscillations in the inclinations of the upper and lower strings to the vertical are in the ratio (g - bp z )/ap*. Show that in one principal oscillation the two particles are on the same side of the vertical through and in the other on opposite sides. Ex. B. Two particles M, m, are connected by a fine string, a second string connects the particle m to a fixed point, and the strings hang vertically; (1) m is held slightly pulled aside a distance h from the position of equilibrium, and, being let go, the system performs small oscillations; (2) M is held slightly pulled aside a distance k, without disturbance of m, and being let go the system performs small oscillations. Prove that the angular motion of the lower string in the first case will be the same as that of the upper string in the second if M k = (M+m) h. [Math. Tripos, 1888.] Ex. 4. Three beads, the masses of which are m, m', m", can slide along the sides of a smooth triangle ABC and attract each other with forces which vary as the distance. Find the positions of equilibrium and prove that if slightly disturbed the periods 2?r/p of oscillation are given by (P 2 ~ ) (P* ~ P) (P* -y)~ m'm" (p z - a) cos 2 .4 - m"m (p*~P) cos 2 J3 mm' (p 2 y) cos 2 G 2mm' 'm" cos A cos B cos C = 0, where a, /3, y represent m" + m', m + m", m'-t-m respectively. Ex. 5. A particle P of unit mass is placed at the centre of a smooth circular horizontal table of radius a. Three strings, attached to the particle, pass over smooth pulleys A, B, C at the edge of the table and support three particles of masses m 1 ,m 2 ,m B ; the pulleys being so placed that the particle P is in equilibrium. A small disturbance being given, prove that the periods of the oscillations are 2irlp, where ) 2 l X -1 U -=-gaX, ^- u -y + 2 U - = - gb y. To solve these we put x = L sin (pt + a), y=L' COB (pt + a). We then obtain a quadratic for p* and the ratio L'jL. The path of the particle relatively to the moving surface when performing the /a;\ 2 / w \ 2 principal oscillation defined by either value of p- is the ellipse (j) + I jr-, l = 1. The two ellipses are coaxial. 3O2. The insufficiency of the first approximation. In forming the equations of motion in Arts. 287, 294, we have rejected the squares of x and y. ART. 304.] ABOUT STEADY MOTION. 191 But unless the extent of the oscillation is indefinitely small, the rejected terms have some values, and it may be, that they sensibly affect the results of the first approximation. See Art. 141. 3O3. To find a second approximation we include in the equations (2) of Art. 287 the terms of the second order. We write these in the form (5 2 - A)x- By = E l a? + 2ErCy + E 3 y*} - B'x + (S 2 -C)y = F lX * + 2F z xy + F 3 y* f ' ' Taking as our first approximation x = L sin (pt + a) + M sin (qt + /3)| ?/=L'sin (pt + a) + M ' sin (qt + p)\ "" we substitute these in the right-hand sides of (1). The equations take the form ($*-A)x-By = -2PBiu(\t+f,.)\ - B'x + (S 2 - C) y = SQ sin (\t + /x)j " where X may have any one of the values 0, 2p, 2q, pq and P, Q contain the squares of the small quantities L, M, L', M'. To solve these equations, we con- sider only the specimen term of (3) and assume x=L sm(pt + a)+M &iu(qt + p) + R sin(\t + /j.)\ y=L' sin (pt + a) + M r sin(qt + /3) + JJ'sin " We find by an easy substitution '=-P, B'R + R'(\*+C)= PB'-Q(\* R'= It appears that R, R' are very small quantities of the second order, except when X is such that the common denominator is small, and in this case R, R' may become very great. The roots of the denominator are \ 2 =p 2 , X 2 = g 2 , and the denominator is small when X is nearly equal to either p or q. This requires either that one of the two frequencies p, q should be small or that one should be nearly double the other. If for example p is nearly equal to 2q and the numerators of R, R', are not thereby made small, the terms defined by \=p-q and \=2q will considerably in- fluence the motion, the other terms producing no perceptible effect. If p = 2q exactly the denominator is zero and both R, R' take infinite values. The dynamical meaning of the infinite term is that the expressions (2) do not represent the motion with sufficient accuracy (except initially) to be a first approximation. The corrections to these expressions are found to become infinite and if we desire a solution we must seek some other first approximation. 3O4. Oscillation about steady motion. Ex. 1. The constituents of a multiple star describe circles about their centre of gravity with a uniform angular velocity n, the several bodies always keeping at the same distances from ach other. A planet P, of insignificant mass, freely describes a circle of radius a, centre 0, with the same angular velocity, under the attraction of the other bodies. It is required to find the oscillations of P when disturbed from this state of motion. Let r=a(l + x), 6 = nt + y be the polar coordinates of the planet P at any time t. Let the work function in the revolving field of force be -.(I), 192 SMALL OSCILLATIONS. [CHAP. V. at all points in the neighbourhood of the circular motion. Since that motion is possible only in that part of the field in which the force tends to and is equal to n 2 a, it is clear that A = -a?n 2 and B = 0. Substituting the values of r, 6 in the polar equations - - (2) dtj~adx' r dt\ dt)~rdy'" we find the linear equations (a 2 5 2 - aV - A) x - (2a 2 5 + B) y = 0) (2a z n8-B)x + (a?8*-C)y = o} " A principal oscillation is therefore given by x=Lcospt + L' sinpt, y=M cospt + M'sinpt .............. (4), 2a*npL'-BL _- 2a?npL-BL' ' ' .................. (6). The path of the particle when describing a principal oscillation relatively to its undisturbed path is the conic (o V + ^ + a 2 " 2 ) ** + 2Bxy + (ap + C) y* = (L* + L' 2 ) ......... (7), the ratio and directions of the axes being independent of the disturbance. In the limiting case in which n=0 the conic reduces to two straight lines. When the multiple star has tivo constituents A, B, whose masses are M, M', the planet P can describe a circular orbit only when Mp~* sin APO = M'p'~ K BinBPO, where p=AP, p'=BP and the law of force is the inverse *th power of the distance. Since O is the centre of gravity of M, M' this proves that either the angle APO is zero or p=p', except when K= - 1. The planet P must therefore be either in the straight line AB or at the corner C of the equilateral triangle ABC. When the planet P is in the straight line AB at a point C such that the sum of the attractions of A and B on it is equal to n 2 . OC, the planet can describe a circle about with the same periodic time as A and B. This motion is unstable. When the planet P is at the third corner C of the equilateral triangle ABC, the ... , circular motion is stable when ,,,, > 3 MM' \3 - K) These two results may be obtained in several ways. Putting p, p' for the distances of P from the two primaries the work function is 1 / M M' ~-iV-iV- Expressing this in terms of r, 0, and expanding in powers of x, y, including the terms of the second order, the values of A, B, C in equation (1) become known. The periods are then given by (6). Instead of using the work function, we may determine the forces dUjadx and dU/rdy by resolving the attractions of the primaries along and perpendicular to the radius vector of P. This method has the advantage that the task of calculating the terms of the second order becomes unnecessary. Lastly, we may use the Cartesian equations referred to moving axes which rotate round with a uniform angular velocity n, OC being the axis of ; Art. 227. ART. 305.] FINITE DIFFERENCES. 193 In all these methods, the assumption that the mass of the planet P is insignifi- cant compared with that of either of the attracting bodies greatly simplifies the analysis. It does not seem necessary to examine these cases more fully here, as the results and the method of proceeding when this assumption is not made will be considered further on. Ex. 2. If in the last example the attracting primaries either coincide or are so arranged that the field of force is represented by U- E7 =^ ;r + \Ax*; prove that other circular orbits in the immediate neighbourhood of the given one are possible paths for the particle P, Art. 291. Prove also that after disturbance the oscillation of P about the mean circular path is given by x=Lcos (pt + a), py= -2nLem(pt + a), where jj 2 =3w 2 -.4/a 2 , the oscillation having only one period. Ex. 3. Two equal centres of force S, S', whose attraction is pp*, rotate round the middle point of the line of junction with a uniform angular velocity n. A particle in equilibrium at is slightly disturbed, prove that the periods of the small oscillation are given by (p 2 + n 2 -|8) (p* + n 2 - K /3) = 4n2p 2 where p=2nb K ~ 1 and SS'=2b. Thence deduce the conditions that the equilibrium should be stable. Problems requiring Finite Differences. 3O5. Ex. 1. A light elastic string of length nl and coefficient of elasticity E is loaded with n particles each of mass m, ranged at intervals I along it begin- ning at one extremity. If it be hung up by the other extremity, prove that the periods of its vertical oscillations will be given by the formula ' Where i = ' *> 2 n - !* C Math - Tripos, 1871.] Let X K be the distance of the *th particle from the fixed end ; T the tension above, T K ,- i that below, the particle. We then have mx K "=mg + T K+l -T K ................................. (1), and by Hooke's law for elastic strings - i (2)- The equation of motion is therefore x K "-9 = c*(x K+1 -2x K + x K _ l ) .............................. (3), where c 2 =.E/Zni. We assume as the trial solution x K = h K + X K sin(jpt + e) ................................. (4), where h K and X K are two functions of K which are independent of *, and p, e are independent of both K and t. Substituting we find * * The solution is given at greater length than is necessary for this example, in order to illustrate the various cases which may arise. R. D. 13 194 FINITE DIFFERENCES. [CHAP. V. To solve the first of these linear equations of differences we follow the usual rules. Taking X K = Aa K as a trial solution, where A and a are two constants, we get after substitution and reduction Let these values of a be called a and ft. Then X K = Aa* + Bp 1 ' .................................... (8). We notice that when either # = or 2c the equation (6) has equal roots, viz. a=l or - 1. The theory of linear equations shows that the terms depending on these values of p take a different form, viz. X K =(A + BK)(1) K ................................. (9). The complete value of X K may be written in the form X K = h K + A +B K + (A 2c + B 2c K) (- l) K sin (2c + c 2e ) + S (A p a K + B p p?) sin (pt + e p ) ...... (10) , where 2 implies summation for all existing values of p. We have yet to examine the conditions at the extremities of the string. The formula (2) does not express the tension of the highest string unless we suppose that x Q = 0. Again the tension below the lowest particle must be zero and this requires that T n+1 =0. The equation (3) will therefore express the motion of every particle from icl to /c=n only if we make *o=0 *+i- * n = l ................................. (11). Since x Q =Ofor all values of t, it follows from (10) that ho + A^O, A 2C = Q, Ap + Bp = ........................ (12). Since x n+1 -x n =l, we see in the same way that >Wi- / *n + #o = z fc = 0, Apa n + 1 + Bpp n+l = Apa n + Bpp n ......... (13). Eliminating the ratio A P IB P we have a*-|8*M = a-|8 ................................ (14). If p>1c we see by (7) that both a and /3 are real negative quantities. The equation (14) has then one side positive and the other negative, since the integers n, n+1 cannot be both even or both odd. Hence p must be less than 2c, let p = 2c sin 6, hence a = cos 20 + sin 20^-1, 13 = 00820-8^120,^-1 ............... (15). The equation (14) now gives sin(2n.+2) = sin2n0, excluding p = we have ~2n+12' 2c~ where i has any integer value. It is however only necessary to include the values i=0 to i=n-l. The values of 6 indicated by i = i' and 2n-i' are supplementary, while the values of sin indicated by i=i' and i' + 2n + l are equal with opposite signs. The value i=n is excluded because the value p = 2c has been already taken account of. The oscillations of the /cth particle are therefore given by x K =H ic + ^Cpein2 K esm(pt + fp) ........................ (17), where H =h +A + B K, and C P =2A P J-1. ART. 305.] EXAMPLES. 195 The value of h might be determined by solving the second equation of differences (5), using the rules of linear equations adapted to that equation. But it is evident that in the position of equilibrium of the system, when there is no oscillation, every C p = 0, and therefore that position is determined by X K =H K . This enables us to deduce H K from the elementary rules of Statics. We notice that in equilibrium, T n =mg, T n _ 1 =2mg, &c., T K = (n + 1 - K) mg. Hence by Hooke's law Adding these for all values of K from *=1 to KK, and remembering that H = by (12), we find The equation (17) shows that the motion of every particle is compounded of n principal or simple harmonic oscillations. The periods of these are unequal and are represented by 2irlp where p has the values given in (16). Suppose the system to be performing the principal oscillation defined by the value of 0=7r/2y. By considering the signs of sin 2*0 in (17) we see that all the particles determined by K <. y are moving in the same direction' as the highest particle, those determined by K> y but < 2y are moving in the opposite direction, those given by K>2y but <3-y are moving at any time in the same direction, and so on. Ex. 2. A smooth circular cylinder is fixed with its axis horizontal at a height h above the edge of a table. A light string has a series of particles attached to it over a part of its length, the particles being each of mass m and distant a apart. The portion of the string to which the particles are attached is coiled up on the table, and the rest is carried over the cylinder, and a mass M attached to the further end of it. The system is held so that the first particle is just in contact with the table, the free portions of the string being vertical, and is then allowed to move from rest ; prove that if v be the velocity of the system immediately after the nth particle is dragged into motion (na < h), then 2 _ (n - 1) ga 6M 2 - n (2n - l)m 2 3 ' (M+nin)* ' Supposing the string of particles to be replaced by a uniform chain deduce from the above result the velocity of the system after a length x of the chain has been dragged into motion. If I be the length of the chain and /*, the mass, then, if I be less than h, the amount of energy that will have been dissipated by the time the chain leaves the table will be [Coll. Ex. 1887.] If v n represent the velocity required, we deduce from vis viva and linear momentum at the next impact the equation {M+ (n + 1) m} 2 v 2 n+1 - { M+nm} z v\=2ga (H* - nW). Writing the left-hand side # (n+ 1) - < (n), we find (n + 1) - (!) by summing from n 1 to n. Remembering that ^ = 0, this gives v n . The energy dissipated is found by subtracting the semi vis viva, viz. ^(M + n)v z , from the work done by gravity, viz. (M-^n) Ig. 132 196 FINITE DIFFERENCES. [CHAP. V. /'.c. 3. A train of an engine and n carriages running with a velocity //, is brought to rest by applying the brakes to the engine alone, the steam being cut off. There is a succession of impacts between the buffers of each carriage and the next following. Prove that the velocity v of the engine immediately after the rth impact is given by (M + m) 3 (v - w) 2 = J/a/V {2M+ m (r - 1) } , where m is the mass of any carriage, M that of the engine, a the distance between the successive buffers when the coupling chains are tight, / the retardation the brake would produce in the engine alone. [Coll. Ex.] Ex. 4. A heavy particle falls from rest at a given altitude h in a medium whose resistance varies as the square of the velocity. On arriving at the ground it is immediately reflected upwards with a coefficient of elasticity j8. Show that the whole space described from the initial position to the ground at the nth impact If u n be the height described just after the nth rebound, we show Kfi-i)=/3*(n-i)- To solve this equation of differences we put ?< n = 1 + l/w? n . The equation then takes a standard form with constant coefficients. The whole space described is found by taking the logarithm of the product u () u l u, 1 ,...u n _ l . This problem was first solved by Euler in his Mechanica, vol. i. prop. 58, for the case in which /S=l. An extension by Dordoni, Memorie della Societa Italiana, 1816, page 162, is mentioned in Walton's Mechanical Problems, chap. n. page 247. CHAPTER VI. CENTRAL FORCES. Elementary Theorems. 306. To find the polar equations of motion of a particle describing an orbit about a centre of force. Let the plane of the motion be the plane of reference and let the origin be at the centre of force. Let F be the accelerating force at any point measured positively towards the origin. Then by Art. 35, 1 d The latter equation gives by integration where h is an arbitrary constant whose value depends on the initial conditions. This important equation can be put into other forms of which much use is made. Let v be the velocity of the particle, p the perpendicular drawn from the origin on the tangent. Let A be the area described by the polar radius as it moves from some initial position to that which it has at the time t. Then (Art. 7) Remembering that v = ds/dt, we see that the equation (2) may be written in either of the forms h dA , The first of these shows that the velocity at any point of the orbit is inversely proportional to the perpendicular drawn from the centre on the tangent. The second, by integration between the limits t = t to t, shows that the polar area traced out by the radius vector 198 CENTRAL FORCES. [CHAP. VI. is proportional to the time of describing it. We also see that the constant h represents twice the polar area described in a unit of time. Both these are Newtonian theorems. We also infer that in a central orbit, the angular velocity dOJdt always keeps one sign and never vanishes at a finite distance from the origin. The radius vector therefore continually turns round the origin in the same direction. 307. Conversely, we may show that if a particle so move that the radius vector drawn from the origin describes areas propor- tional to the time the resultant force always tends to the origin and is therefore a central force. To prove this let F and G be the components of the accelerating force along and perpendicular to the radius vector. Taking the transversal resolution, we have '- 1 1 ft rdt\' dt) As already explained r^d0 = 2dA, and if the area A bear a constant ratio to the time, say A = at, we have at once r 2 d0/dt = 2a and therefore 6r = 0. 308. If m is the mass of the particle, its linear momentum is mv and this being directed along the tangent to the path, the moment of the momentum about the centre of force is mv.p. The moment of the momentum is called the angular momentum (Art. 79) and we see that in a central orbit the angular momentum about the centre of force is constant and equal to mh. When we are concerned only with a single particle its mass is usually taken to be unity, and h then represents the angular momentum. 309. To find the polar equation of the orbit we must eliminate t from the equations (1). Let r=I/u, then, as in Art. 268, dr _ 1 du d0 , du Substituting this value of d 2 r/dt 2 and the value of d0/dt = hu" given by (2) in j^ ^(-37 ) = ~^> we nave CvC \OtC / F ART. 310.] ELEMENTARY THEOREMS. 199 When the polar equation of the path is given in the form u =f(0) the equation (4) determines F in terms of u and 6. Since the attractive forces of the bodies which form the solar system are in general functions of the distance only we should eliminate by using the known polar equation of the path. We thus find F as a function of u only. Strictly this expression for F only holds for points situated on the given path, but if the initial conditions are arbitrary, the path may be varied and the law of force may be extended to hold for other parts of space. When the force F is given as a function of r or 1/w, the d?u equation (4) is a differential equation of the form ^fi,=f( u )- This differential equation has been already solved in Art. 97. It is evident from dynamical considerations that when the central force is attractive, i.e. when F is positive, the orbit must be concave to the centre of force, and when F is negative the orbit must be convex. By looking at equation (4) we immediately verify the theorem in the differential calculus that a curve is d?u concave or convex to the origin according as ^ + u is positive or negative. 310. To apply the tangential and normal resolutions to a central orbit. Referring to Art. 36 we have the two equations tlii vfi (5), where is the angle behind the radius vector when the particle moves in the direction in which s is measured. Writing dr/ds for cos < and integrating we have v*=C-2/Fdr ........................ (6), where C is a constant whose value depends on the initial con- ditions. This equation is obviously the equation of vis viva, Art. 246. The integral has a minus sign because the central force is, as usual, measured positively towards the origin, while the radius vector is measured positively from the origin. 200 CENTRAL FORCES. [CHAP. VI. If we substitute for v its value h/p given by (3) and differentiate we deduce * **4 dr This expression for the central force F is very useful when the orbit is given in the form p =f(r). 311. Considering the normal resolution (5), we have an ex- pression for v which is useful when both the law of force and the path are known. It has the advantage of giving the velocity without requiring the previous determination of either of the constants C or h. If ^ is one-quarter of the chord of curvature of the path drawn in the direction of the centre of force we may write the equation in either of the forms v 2 = Fp sin < = 2Fx (8). This is usually read ; the velocity at any point is that due to one- quarter of the chord of curvature. When the particle describes a circle about a centre of force in the centre sin < = 1 and p is the radius r. The velocity given by the normal resolution, viz. v 2 /r = F, is often called the velocity in a circle at a distance r from ike centre of force. 312. The velocity acquired by a particle which travels from rest at an infinite distance from the centre of force to any given position P is called the velocity from infinity. Referring to the equation of vis viva (6), let 2ti 1 Now v = when r = oo ; hence, if n is greater than unity, we have (7=0. The velocity from infinity to the distance r = R is 2u, 1 therefore given by v 2 = ^ ^ -. See Art. 181. n 1 R n ~ If n is less than unity the value of G is infinite. Instead of the velocity from infinity we use the velocity acquired by the particle in travelling from rest at the given point P to the origin under the attraction of the central force. In this case v = when 2u, r = R; hence (since w 1 and the velocity from infinity is V 1 , C= V 2 - Fj 2 ; if < 1, C = F 2 + F 2 where 7 is the velocity to the origin. We may obtain another interpretation for the constant C. Selecting any standard distance r = a, the potential energy at a distance r is - --(- M- ' " I * w-lW 1 - 1 r n -*J~ (n-l)a n -^2 2* See Art. 250. It follows that AC plus , -- is equal to the n - 1 a n ~ l whole energy of the motion. Hence by taking the standard position at infinity or the origin according as n is greater or less than unity, we may make ^ C equal to the whole energy. 314. When a point P on the orbit is such that the radius vector OP is perpendicular to the tangent, the point P is called an apse. When OP is a maximum the apse is sometimes called an apocentre, and when a minimum a pericentre. 202 CENTRAL FORCES. [CHAP. VI. 315. Summary. As the formulae we have arrived at are the fundamental ones in the theory of central forces, it is useful to make a short summary before proceeding further. There are three elements to be considered : (1) the law of force, (2) the equations of the path, (3) the velocity and time of describing an arc. Any one of these elements being given, the other two can be deduced by dynamical considerations. There are therefore three sets of equations; firstly, equations (4) and (7) connect the force and path, so that either being known the other can be deduced ; secondly, equation (6) connects the force and velocity ; thirdly, equations (2) and (3) connect the path with the motion in that path. The equations of one of these sets are mere algebraic trans- formations of each other, any one being given the others can be found from it by reasoning which is purely mathematical. But an equation of one set cannot be deduced from an equation of another set in this manner, because each set depends on different dynamical facts. 316. Dimensions. It is important to notice the dimensions of the various symbols used. The accelerating force F, like that of gravity, i.e. g, is one dimension in space and 2 in time. We see this by examining any formula which contains F or g, say s = %gt 2 or Fcos = d?s/dt 2 . The force F will in general vary as some power of the distance from the centre of force, say F=fj,/r n where p is a constant which measures the strength of the central force. The quantity p = Fr n is therefore n + 1 dimen- sions in space and 2 in time. The velocity v = dsfdt is one dimension in space and 1 in time. The constant h = vp is 2 dimensions in space and 1 in time. See Art. 151. 317. Force given, find the orbit. Ex. 1. The force being a particle is projected from an initial distance a, with a velocity which is to the velocity in a circle at the same distance as ^/2 to V 3 > the angle of projection being 45. Find the path described. Putting a = l/c the differential equation of motion is, by Art. 309, ART. 318.] VARIOUS ORBITS. 203 When u = c, the conditions of the question give v 2 =f.F/c and ft=vsin/3/c where sin 2 /3=|, see Arts. 311, 313. We therefore have (7=0, /t 2 =^. The equation now reduces to * du _. e - + (du\_u* \d~0J ~^ Replacing u by 1/r and measuring 6 from the initial radius OA in such a direction that r and increase together, this leads to r = a (1 + 0). From the equation r 2 dOjdt = h, we infer that the time from a distance a to r is (r 3 -a 3 )/3 av /M- Ex. 2. A particle moves under the action of a central force /u(tt 5 -^a 2 w 7 ), the velocity of projection being (25/*/8a 4 )^, and the angle of projection sin" 1 4. Prove that the polar equation of the path is 3a 2 = (4r 2 - a 2 ) (8 + C) 2 . [Coll. Ex. 1892.] Ex. 3. When the central acceleration is ^(w 3 + a 2 tt 5 ) and the velocity at the apsidal distance a is equal to ^n/a, prove that the orbit is r = a en 6 (mod ^/J). [Coll. Ex. 1897.] Ex. 4. The central force being F=2fj.u 3 (l-a 2 u 2 ), the particle is projected from an apse at a distance a with a velocity */M/ a - Prove that it will be at a distance r after a time ^i- -(a 2 log r+ >/( r2 - a ) + rx /( r 2 - a 2 )l . [Math. Tripos.] ^V/* I a Ex. 5. A particle, acted on by two centres of force both situated at the origin respectively F=/j.u 3 and F'=/j.'u 5 , is projected from an initial distance a with a velocity equal to that from infinity, the angle of projection being tan~ 1 v /2. If the forces are equal at the point of projection, the path is ad = (r- a) >J2. Ex. 6. A particle, acted on by the central force F=u 2 f(0), is initially projected in any manner. Prove that the radius vector can be expressed as a function of 6 if the integrals of cos 6f (6) and sin 6f (6) can be found. [Use the method of Art. 122.] 318. Orbit given, find the force. Ex. 1. A particle describes a given circle about a centre of force on the circumference. It is required to find the law of force and the motion. Newton's problem. Let be the centre of force, C the centre of the circle, P the particle at the time t. Let a be the radius of the circle, OP=r. If p = OY be the perpendicular on the tangent, we have (since the angles OPY, OAF are equal) p = r 2 /2a. Hence using (7) of Art. 310, we have ................................. dr p 2 r" If we suppose the magnitude of the force to be given at a unit of distance from the centre of force we write this in the form F=-^, where u is a known constant r sometimes called the magnitude or strength of the force. The constant h is then determined by the equation 8A 2 a 2 =/t .......................................... (2). The velocity at any point P is found by the normal resolution, Art. 310, - . .. . ..... (3). a r 5 2a V 2 r 3 By Art. 312 this velocity is equal to that from infinity. 204 CENTRAL FORCES. [CHAP. VI. To find the time of describing any arc AP, where A is the extremity of the diameter opposite to the centre of force, we use the equation A = %ht, Art. 306. Since the area AOP is made up of the triangle OOP and the sector AGP, we have lit = A = % o 2 (20 + sin 20), where = the angle A OP. Substituting for h /2 t=2a 3 ./ - (20 + sin 20) (4). It appears from this that the particle will arrive at the centre of force after a finite time obtained by writing = ir. The particle arrives with an infinite velocity due to the infinite force at that point. Let the force at all points of space act towards the point and vary as the inverse fifth power of the distance from O. It is required to find the necessary and sufficient condition that a particle projected from a given point P in a given direction PT with a given velocity V may describe a circle passing through 0. It is obvious from (3) that it is necessary that F 2 = ^/i/r 4 where r = OP; we shall now prove that this is also sufficient. Describe the circle which passes through and touches PT at P. The particle which describes this circle freely satisfies the given conditions at P. If then the given particle does not also describe the circle we should have two particles projected from P in the same direction, with equal velocities, acted on by the same forces, describing different paths; which is impossible; Art. 243. We notice that a change in the direction of projection PT affects the size of the circle described, but not the fact that the path is a circle. Ex. 2. A particle moves in a circle about a centre of force in the circum- ference, the force being attractive and equal to p.r n . Prove that the resistance of the medium in which the particle moves is Jju (n + 5) r n sin 6, where cos = ?'/2a. Use the normal and tangential resolutions. [Coll. Ex.] Ex. 3. A particle of unit mass describes a circle about a given centre of force situated on the circumference. If the particle at any point P is acted on by an impulse 2v cos in a direction making an angle IT - with the direction of motion PT, show that the new orbit is also a circle and prove that the ratio of the radii is cos 2 + sin 20 cot 0, where 6 is the angle OPT. Ex. 4. The force being F=fiu*, a particle when projected from a point P with an initial velocity F, equal to that from infinity, describes the circle r=2acos0; investigate the path when the initial velocity is V(l + y), where y is so small that its square can be neglected. Proceeding as in Art. 317, we find The conditions of the question give where c = l/2a and = a initially. Putting u = cseo6 + cij and neglecting the squares of ij and y, we arrive at cos 2 dy cos 3 6 - 2 cos 6 -y y C08 4 sine d0 + sin 2 ' 7 ~sin 2 + cos 4 a sin 2 ' ART. 320.] VARIOUS ORBITS. 205 Each side being a perfect differential, we find cos 2 and K is determined from the condition that 77 = when 6 = a; y .'. K= --/cot cH -- -T (cota + f a + \ sin a cos a). COS Ct Putting u = ljr, we have r=2a cos (1 - 17 cos 6), T y .'. 5- = cos0-KSin 0--7COS0H ~~ (cos + f sin + 4 sin 2 cos 0). It has been assumed that cos a is not small, the point P must therefore not be close to the centre of force. It easily follows that when 6 \tr-K-\-%iry sec 4 o, the distance of the particle from the centre of force is of the order of small quantities neglected above. Ex. 5. Any number of particles are projected in all directions from a given point P each with the velocity from infinity, the central force being F= /im 5 . Prove that their locus at any instant is (6 being measured from OP) sin 3 where OP=c and A is a constant depending on the time elapsed. 319. Ex. 1. A particle describes an equiangular spiral of angle a. under the action of a centre of force in the pole, prove that F= A , h=smafj(j., v = -- -, 2 cos at /Jfj. = rf - ?' 2 , where t is the time of describing the arc bounded by the radii vectores r , 1\. Con- versely, a particle being projected from any point in any direction will describe an equiangular spiral about a centre of force whose law is F=plr 3 , provided the velocity of projection is ^//x/r, i.e. is equal to that from infinity. Assuming p = r sin a we follow the same line of reasoning as in Ex. 1 of Art. 318. Ex. 2. A particle acted on by a central force moves in a medium in which the resistance is /c(vel.) 2 , and describes an equiangular spiral, the pole being the centre of force. Prove that the central force varies as -3e~ 2 " rseca , where a is the r 3 angle of the spiral. [Math. Tripos, I860.] 320. Ex. A particle describes the curve r m = a cos + b sin nQ, under the action of a centre of force in the origin. Prove that __ _ _ ~ r zm+s ~ fS > m + 1 r 87n+2 We notice (1) that the exponents of r are independent of n, (2) that, when m + 1 is positive, the velocity at any point is that due to infinity. Art. 312. Supposing the law of force and the velocity of projection to be given by these formulae, let the particle be projected from any point P in any direction PT. The 206 CENTRAL FORCES. [CHAP. VI. four constants h 2 , n, a, b are determined by joined to the conditions that the curve must pass through P and touch PT. We find that w 2 and ^ - /t'U 2 " 1 cot 2 have the same sign, where E = OP and is the angle of projection. When the sign of n- thus determined becomes negative or zero the curve obviously changes into r m =a'e n6 + Ve- ne , or r m =a + b"0, where 4a'b' = a- - fc 2 and b" is the limit of bn when b is infinite and n zero. It is useful to notice the following geometrical properties of the curve. If p be the perpendicular on the tangent, the angle the radius vector makes with the tangent m In 2 a 2 + 6 J m 2 -7i 2 tan#= -- cot n0, -5 = - H n * * 2 " 12 This example includes many interesting cases. Putting m=2, n = 2, we see that the lemniscate of Bernoulli could be described about a centre of force in the node varying as the inverse seventh power of the distance. Putting m=n, we have the path when the force varies as the inverse (2m + 3)th power and the velocity is that from infinity. Writing m=J, =, we find the path is a cardioid when the central force varies as the inverse fourth power and the velocity is that from infinity. Writing m = l, n=l, the path is a circle described about a centre of force on the circumference. 321. Ex. 1. A particle describes a circle about a centre of force situated in its plane. It is required to find the law of force and the motion. Let be the centre of force, C the centre of the circle, a its radius and C0 = c. Taking the equations of Art. 310, we have =*, V -=F?, .-..*' par a p* Since in a circle 2ap = r 2 + a 2 - c 2 , we can, by substitution, express F and v in terms of r alone. We have where 8a 2 7i 2 =/i and B=a?- c 2 . When JS = 0, the law of force reduces to the inverse fifth power, and the velocity becomes the same as that found in Art. 318. If this law of force be supposed to hold throughout the plane of the circle, the values of /*, and B are given. In order that the orbit may be a circle it is necessary that the velocity of projection should satisfy the above value of v, i.e. should be equal to the velocity from infinity. The direction of projection being also given, the angular momentum h (Art. 313) is also known. The values of a and c follow at once from the equations given above and must be real. Newton, when discussing this problem, supposes that the centre of force lies inside the circle. It follows that B is positive, and at no point of space can either the force or velocity be infinite. When the centre of force is outside the circle, one portion of the orbit is concave and the other convex to the centre of force. We must therefore suppose ART. 322.] VARIOUS ORBITS. 207 that the force is attractive in the first and repulsive in the other part. Writing B= - 6 2 , we have 6'- i =c 2 -a 2 , and therefore b is the length of either of the tangents drawn from the centre of force to the circle, and the force changes sign through infinity when the particle passes the circle whose radius is b. Sylvester, in the Phil. Mag. 1865, points out that the resultant attraction of a circular plate, whose elements attract according to the law of the inverse fifth ttf* power, at an external point P situated in its plane, is where /j. is the mass of the plate, b its radius and r the distance of P from the centre. The circle described by P under the attraction of this plate cuts the rim orthogonally. Let the particle P be constrained to move on a smooth plane under the action of a centre of force situated at a point C distant b' from the plane, the law of force being the inverse fifth power. The component of force in the plane is U,T F= r~s TTTTT. , where r is the distance of P from the projection of the centre of ^ '^ force on the plane. Putting B = b, it appears from what precedes that, if the velocity of projection is equal to that from infinity, the path of the particle on the plane is a circle. The length of the chord bisected by the point is constant for all the circles and equal to 26'. Ex. 2. A particle moves under the action of a centre of force F=/j,u 5 . Prove that all the circles which can be described either pass through a fixed point or have a fixed point for centre. 322. Ex. 1. A particle moves under the action of a centre of force whose attraction is F= ^ and the velocity at any point is equal to that from infinity. It is required to find the path. The equation of vis viva (Art. 310) gives (1). Since this formula is independent of the path and it is given that v is zero when r is infinite we see that (7=0. Substituting for v its value hjp, the equation of the path becomes r 2 + B = ip 2 , i/i 2 =M ................................. (2). The curve required is therefore such that a linear relation exists between p* and r 2 . There are several species of curves which possess this property distinguished from each other by the values of B and i. One such curve is known to be an epicycloid. Supposing the radii of the fixed and rolling circles to be a and b, we have at the cusp r=a, p = and at the vertex p and r are each equal to a + 26. We thus find M (a + 26) 2 -a 2 a ' P- 1 - (a + 26) 2 .......................... - (3) ' The law of force and the conditions of projection being given both B and 7i 2 are known. If the force is attractive, B negative, and y//i 2 less than unity, the path is an epicycloid, the values of a and 6 being given by (3). Changing the sign of b the epicycloid becomes a hypocycloid and in this case we learn from (3) that i and jt are negative. When therefore the force is repulsive, and B negative, the path is a hypocycloid. 208 CENTRAL FORCES. [CHAP. vi. The remaining species are more easily separated by putting the equation (2) into the form p=ip, a result which follows at once from the identity p=rdr/rfp. Remembering that p=p + d z pjd^ the differential equation becomes J.72~ V 1 ~ i)p = (*). When i is less than unity or is negative we easily deduce the cycloidal species given above. If (P= 1 - i, we find p L sin fi\l/ + M cos j8^. If the axis of x pass through the cusp, we have p = when ^=0 and p = a + 2b when ^ = ^ir. Hence L = a + 26 and M= 0. When i is greater than unity we have the forms where a?=i-l and the second form occurs when ?' = !. Since in any curve the projection of the radius vector on the tangent is dpjd\f/, we find by elementary geometry \d\f/ J ' dp where

, y = 6 sin <, where < is the eccentric angle of the particle. It immediately follows that the particle so moves that Jfit. When has increased by lir the particle has made a complete circuit and returned to its former position. The periodic time is therefore 2?r/\/A t - It appears from this that the periodic time is independent of all the conditions of projection and is the same for all ellipses. It depends solely on the strength fi of the central force. In general the time of describing any arc PP' is the difference of the eccentric angles at P and P' divided by V/*- 142 212 LAW OF THE DIRECT DISTANCE. [CHAP. VI. When the orbit is a hyperbola we have x = \ a (e? + e-*'), y = %b (e*' - e~*'), where (f> is an auxiliary angle. It immediately follows that where JJL is positive and equal to p. 327. When the velocity V and angle /3 of projection as well as the initial distance R are given, the semi-axes a, b of the conic described may be deduced from the equations 2/0 /i 2 Flfrsin'/S F 2 a z b- = = , a- + b- = R- H -- - . p p fi These give real values to a 2 and 6 2 . The angle 6 which the major axis makes with the initial distance is given by cos 2 sin 2 ^_J L s/ . _6 2 a 2 -.R 2 "V ~&T"~I? ; ' -tf'W^v- Since F = \fnR' t it is evident that the problem of finding the particular conic described when R and F are given is the same as the geometrical problem of constructing a conic when two semi- conjugate diameters R, R are given in position and magnitude. This useful construction is given in most books on geometrical conies. 328. Eeferring to the equations (1) of Art. 325 we see that the motion in an ellipse about a centre of force F=far is the resultant of two rectilinear harmonic oscillations along two arbitrary directions Ox, Oy represented by X= - /JLX, Y= - ny. The resultant of any number of rectilinear harmonic oscillations (performed in equal times) along arbitrary straight lines OA, OB, &c. may be found by resolving the displacements of each along two arbitrary axes and compounding the sums of the components. The resulting motion is therefore an elliptic motion with for centre. Ex. Investigate the conditions that the resultant of two rectilinear harmonic oscillations, of equal periods, whose directions make an angle 6, should be (1) a rectilinear, (2) a circular motion. Prove that in the first case their angles or phases must be equal; in the second their amplitudes must be equal and their phases differ by TT - 6. The radius is a sin 6. 329. Ex. 1. If OP, OQ are conjugate diameters of an ellipse, prove that the time from P to Q is one-quarter of the whole periodic time. This follows at once from the fact that the area POQ is one-quarter of the area of the ellipse. Ex. 2. Prove that in a hyperbolic orbit the time from the extremity of the major axis to a point whose distance from that axis is equal to the minor axis is the same for all hyperbolas. ART. 330.] POINT TO POINT. 213 Ex. 3. If the circle of curvature at any point P of an ellipse cut the curve again in Q, and A is the extremity of the major axis nearest to P, prove that the time from Q, to A is three times the time from A to P. Since (f> = jtj.t, Art. 326, the theorems in conies which, like this one, are con- cerned with eccentric angles may at once be translated into dynamics. Ex. 4. Two tangents TP, TQ are drawn to an ellipse, prove that the velocities at P and Q are proportional to the lengths of the tangents. [For these tangents are known to be proportional to the parallel diameters. ] 33O. Point to Point. To find the directions in which a particle must be projected from a given point P with a given velocity V, so as to pass through another given point Q. Let rj, r 2 be the distances of P, Q from the centre of force 0. Let OP be produced to D where D is such that the velocity V of projection at P is equal to P' D 7 that acquired by a particle starting from rest at D and moving to P under the action of the centre of force. Let OD = k. Then since V 2 =p(a 2 + b 2 - r x 2 ), the sum of the squares of any two semi-conjugates of the trajectory is ft 2 . Bisect PQ in N and let ON=x, NP=NQ = y. From the equation of the ellipse, x 2 // 2 3 + prrf =1; Since x, y, k are given, this quadratic gives two values of a 2 , showing that there are two directions of projection which satisfy the given conditions. Let these directions of projection from P intersect ON produced in T and T', then since a 2 =ON . OT, the quadratic gives the positions of T and T'. We also have OT . OT" = ft 2 , and NT . NT' = y*. The roots of the quadratic (1) are imaginary if x + y>k. Produce PO to P' where OP'=OP, the roots of the quadratic are imaginary unless Q lie within the ellipse whose foci are P, P' and semi-major axis a'=k. This ellipse is the boundary of all the positions of Q which can be reached by a particle projected from P with the given velocity. It is also the envelope of all the trajectories. Ex. 1. If two circles be described having their centres at and N and their radii equal to k and y respectively, prove (1) that their radical axis will intersect ON produced in the middle point R of TT' ; (2) that RT 2 is equal to the product of the segments of any chord drawn from R to either circle. 214 LAW OF THE DIRECT DISTANCE. [CHAP. VI. Ex. 2. Show that the greatest range r=PQ on any straight line PQ making a given angle 9 with OP=r l is determined by (fc'--r 1 8 )/r=4-r, cos0. Show also that in this case OT = k, and NT=NP = NQ. Thence deduce that the common tangent at Q to the trajectory and the envelope intersects the direction of projection from P at right angles in a point T which lies on the circle whose centre is and radius k. The first part follows from the focal polar equation of the ellipse and the second from known geometrical properties of the ellipse. 331. Examples. Ex. 1. If the sun were broken up into an indefinite number of fragments, uniformly filling the sphere of which the earth's orbit is a great circle, prove that each would revolve in a year. [Coll. Ex.] The attractions of a homogeneous solid sphere on the particles composing it are proportional to their distances from the centre. Ex. 2. A particle moves in a conic so that the resolved part of the velocity perpendicular to the focal distance is constant, prove that the force tends to the centre of the conic. [Math. Tripos.] Ex. 3. A particle describes an ellipse, the force tending to the centre ; prove that if the circle of curvature at any point P cut the ellipse in Q, the tunes of transit from Q to P through A and P to Q through B are in the same ratio as the times of transit from A to P and P to B, where A and B are the extremities of the major and minor axes and P lies between A and />'. Ex. 4. A particle is attracted to a fixed point with a force /u times its distance from the point and moves in a medium in which the resistance is k times the velocity ; prove that, if the particle is projected with velocity v at a distance a from the fixed point, the equation of the path when referred to axes along the initial radius and parallel to the direction of projection is k tan" 1 2anyl(2v- aky) + n log (x 2 /a 2 + ny-lv- - kxyjav) = 0, where ri*=n - fc 2 /4. [Coll. Ex. 1887.] Ex. 5. Three centres of force of equal intensity are situated one at each corner of a triangle ABC and attract according to the direct distance. A particle moving under their combined influence describes an ellipse which touches the sides of the triangle ABC. Prove that the points of contact are the middle points of the sides, and that the velocities at these points are proportional to the sides. [Math. Tripos, 1893.] Ex. 6. If any number of particles be moving in an ellipse about a force in the centre, and the force suddenly cease to act, show that after the lapse of (l/2s-)th part of the period of a complete revolution all the particles will be in a similar concentric and similarly situated ellipse. [Math. Tripos, 1850.] Ex. 7. A particle moves in an ellipse under a centre of force in the centre. When the particle arrives at the extremity of the major axis the force ceases to act until the particle has moved through a distance equal to the semi-minor axis ; it then acts for a quarter of the periodic time in the ellipse. Prove that if it again ceases to act for the same time as before, the particle will have arrived at the other end of the major axis. [Art. 325.] [Math. Tripos, I860.] ART. 331.] EXAMPLES. 215 Ex. 8. An elastic string passes through a smooth straight tube whose length is the natural length of the string. It is then pulled out equally at both ends until its length is increased by ^2 times its original length. Two equal perfectly elastic balls are attached to the extremities and projected with equal velocities at right angles to the string, and so as to impinge on each other. Prove that the time of impact is independent of the velocity of projection, and that after impact each ball will move in a straight line, assuming that the tension of the string is proportional to the extension throughout the motion. [Math. Tripos, I860.] Ex. 9. A point is moving in an equiangular spiral, its acceleration always tending to the pole S ; when it arrives at a point P the law of acceleration is changed to that of the direct distance, the actual acceleration being unaltered. Prove that the point P will now move in an ellipse whose axes make equal angles with SP and the tangent to the spiral at P, and that the ratio of these axes is tan ^ a : 1 where a is the angle of the spiral. Ex. 10. A series of particles which attract one another with forces varying directly as the masses and distance are under the attraction of a fixed centre of force also varying directly as the distance; prove that if they are projected in parallel directions from points lying on a radius vector passing through the centre of force with velocities inversely proportional to their distances from the centre of force, they will at any subsequent time lie on a hyperbola. [Math. Tripos, 1888.] Ex. 11. A particle starting from rest at a point A moves under the action of a centre of force situated at S whose magnitude is equal to p . (distance from S). It arrives at A after an interval T and the centre of force is then suddenly transferred to some other point S' without altering its magnitude. If the particle be at a point B at the termination of a second interval T equal to the former, prove that the straight lines SS' and AB bisect each other. If at this instant the centre of force be suddenly transferred back to its original position S, prove that at the end of a third interval T the particle will be at S'. If at that instant the centre of force ceased to act, the particle will describe a path which passes through its original position A. Ex. 12. If the central force is attractive and proportional to w 2 /(cw + cos 6) 3 , prove that the orbit is one of the conies given by the equation (c + cos0) 2 =a + &cos2(0 + a). [Coll. Ex. 1896.] Putting CM + cos0=J7, the differential equation of the path becomes the same as that for a central force varying as the distance I/ U. The solution is therefore known to be the form given above. Ex. 13. A particle moves under a central force F=/m 2 (l + fe a sin 2 0)~^. Find the orbit and interpret the result geometrically. [Math. Tripos.] Ex. 14. A smooth horizontal plane revolves with angular velocity u about a vertical axis to a point of which is attached the end of a weightless string, extensible according to Hooke's law and of natural length d just sufficient to reach the plane. The string is stretched and after passing through a small ring at the point where the axis meets the plane is attached to a particle of mass m which moves on the plane. Show that, if the mass be initially at rest relative to the plane, it will describe on the plane a hypocycloid generated by the rolling of a circle of radius a { 1 - u (md\~ 1 )^ } on a circle of radius a, where a is the initial extension and \ the coefficient of elasticity of the string. [Math. Tripos, 1887.] 216 THE INVERSE SQUARE. [CHAP. VI. The accelerating tension is Xr/md = /*r (say). The path in space is therefore an ellipse having a and b = wa/^/fi for semi-axes. To find the path relative to the rotating plane we apply to the particle a velocity ur transverse to ; backwards. If p' be the perpendicular from the centre on the resultant of v and ur, we have by taking moments about the centre (w 2 - 2vup + wV) p' 2 = (op - cor 2 ) 2 . Substituting for t> 2 and vp their values in elliptic motion we find This is a linear relation between r 2 and p'' 2 and the curve will be an epicycloid if the radii of the corresponding circles are real (Art. 322). To find the radius of the fixed circle, we put p'=0; this gives the radius r=a. To find the radius c of the rolling circle, we put p' = r, and r=a + 2c; this gives the required value of c. If c is negative the curve is a hypocycloid. Law of the inverse square of the distance. 332. A particle is acted on by a centre of force situated in the origin whose acceleration is F = /*w 2 where u is the reciprocal of the radius vector. It is required to find the possible orbits. We have the differential equation (Art. 309) d?u F^ _^ dffi + ~hW~K" .'. u = jj- 2 + A cos(0- a), where A and a are the constants of integration. Comparing this with the equation of a conic lu= 1 + ecos(0-a) (2), where I is the semi-latus rectum, we see that the orbit is a conic having one focus at the centre of force. We also have h 2 = fd. Conversely, if the orbit is a conic with the centre of force in one focus, the law of force must be the inverse square. To prove this, we let (2) be the given equation of the orbit; substituting in the left-hand side of equation (1) we find F=f*u*, where p, has been written for the constant h 2 /l. 333. The velocity. The relations between the conic and the force are more easily deduced from the equation ' ART. 333.] THE PATH AND MOTION, the force being attractive in the standard case, 217 where C is the constant of integration. The p and r equation of an ellipse having a focus 8 at the origin is r a where I = b 2 /a is the semi-latus rectum. Comparing these equations, we have the standard formula? ' 2 1N -(A). h* = r a We change from the ellipse to the hyperbola by making the centre C pass through infinity to the other side of the origin 8, we therefore put a' for a ; also 6 2 becomes b 1 ' 2 , the semi-latus rectum remaining positive and equal to b' 2 /a'. We now have r a (B). In passing from that branch of the hyperbola which is concave to the centre of force to the convex branch, the radius vector r changes sign through infinity from positive to negative. Before comparing the equation of the orbit with that of the hyperbola we should write r' for r in the latter. Also since this branch is convex to the origin the force is repulsive and /j, is negative, let us put fjL = fj>'. Comparing the formulae we have 2 !_ ~ r' + a" IL / 2 1 N a' ' \ r' a') In the parabola, a is infinite, and = fil, (7=0, v 2 = /i- .(C). .(D). 218 THE INVERSE SQUARE. [CHAP. VI. All these formulae may be included in the standard form (A) of the ellipse if we understand that on the concave branch of the hyperbola the major axis is by interpretation negative; on the convex branch, the radius vector being made positive, the major axis is positive while the semi-latus rectum I and the strength /* are negative. 334. Construction of the orbit. When the velocity V and the distance R are known at any point P of the orbit (say, the initial position), we may determine the curve in the following manner. Let the force be attractive. The orbit is now concave to the centre of force and n is positive. Comparing the formulae (A), (B) and (D) and remembering that the velocity V l from infinity to the initial position is given by V? = 2/i/jRj (Art. 312), we see that the orbit is an ellipse, parabola or the concave branch of a hyperbola according as the velocity is less than, equal to, or greater than that from infinity. We notice that this criterion is independent of the angle of projection at P. Let the force be repulsive. Since the path is convex to the centre of force the orbit is the convex branch of a hyperbola. 335. Having ascertained the nature of the orbit we have next to determine the lengths of the major axis and latus rectum. Supposing the ellipse to be the standard case, we have by (A), 1 2 V z - = -~ . We notice that the length a is independent of the angle of projection. If then particles are projected from the same point ivith equal velocities the major axes of the orbits described are equal. If /3 be the angle of projection (Art. 313) we have p = R sin /? and h= Vp. The constant h and the semi-latus rectum I are therefore found from h VR sin ft, h 2 = fil. 336. The position in space of the major axis may be found in various ways. Let S be the focus occupied by the centre of force and A the extremity of the major axis nearest to S. We may find 6 from the analytical equation of the curve l/r = l+e cos 6, where 6 is the angle the initial radius vector SP makes with SA. ART. 337.] CONSTRUCTION OF THE ORBIT. 219 We may also use a geometrical construction. The focus S and the tangent PT at P being known, we can draw a straight line PH so that SP, PH make equal angles with PT, the direction of PH depending on whether the curve is an ellipse or hyperbola. If the point H is then determined so that SP + PH 2a, where a has been already found, it is clear that H is the empty focus. If the curve is a hyperbola, these lengths (as already explained) must have their proper signs. The position of the major axis is then found by joining 8 and H, and a being known the eccentricity e is equal to SH/2a. 337. Ex. 1. The initial distance of a particle from the centre of force being r, and the initial radial and transverse velocities being V 1 and F 2 , prove that the latus rectum 21 and the angle which the radius vector r makes IV 2 V F, with the major axis are given by -5 = 2 - , tan 0= T _ 2 , . r 2 n F 2 2 -/ii/r Ex. 2. Prove that there are two directions in which a particle can be projected from a given point P with a given velocity V, so that the line of apses may have a given direction Sx in space, and find a geometrical construction for these directions. Since V is given, a is known. With centre P and radius 2a-r describe a circle cutting Sx in H, H'. The required directions bisect externally the angles SPH, SPH'. Let /3 be either of the angles the direction of projection at P makes with SP, Art. 313. The quadratic giving the two values of tan /3 is 5t 2 jS + ( 2 - r -\ cot 6 cot /3+^ - 1 = 0, where is the angle PSx. This follows from Ex. 1 by writing F 1 = Fcos/3, F 2 = Fsin /3. The quadratic may also be written in the form -0-0 tan(0 + /3) = (--l } tan/3. Ex. 3. Three focal radii SP, SQ, SR of an elliptic orbit and the angles between them are given. Show that the ellipticity may be found from the equation &A = aA', where A is the area PQR, A' the area of a triangle whose sides are 2SQ* . SR^ sin QSR and two similar expressions. [Math. Tripos, 1893.] Let P', Q', R' be the points on the auxiliary circle which correspond to P, Q, R. We first find by elementary conies the length of the side Q'R' in terms of SQ, SR and the contained angle. The result shows that the side Q'R' is equal to the corresponding side of the triangle A' after multiplication by ajb. Since the areas of the triangles PQR, P'Q'R' are known to be in the ratio bja, the result follows at once. Ex. 4. Two particles P, Q describe the same orbit about a centre of force 0. Prove that throughout the motion the area contained by the radii vectores OP, OQ is constant. 220 THE INVERSE SQUARE. [CHAP. VI. Thence deduce that if a ring of meteors (not attracting each other) describe a closed orbit, the angular distance between consecutive meteors varies inversely as the square of their distance from 0. Ex. 5. Two particles P, Q describe adjacent elliptic orbits of small eccentricity in equal times, the centre of force being in the focus and the major axes coincident in direction. Supposing the particles to be simultaneously at corresponding apses, prove that the angle ^ which PQ makes with the line of apses is given by cot ^= - 3 cosec 2nt + cot 2nt, and find when ^ is a maximum. 338. Elements of an orbit. To fix the position in space of an elliptic orbit described about a focus we must know the values of six constants, called the elements of the orbit. These are (1) the angle which the radius vector from the given focus to the nearer extremity of the major axis makes with some determinate line in the plane of the orbit, the angle being measured in the positive direction; (2) the length of the major axis ; (3) the eccentricity ; (4) a constant usually called the epoch to fix the longitude of the particle at the time t = 0. This con- stant will be considered later on. To determine the plane of the orbit we require two more constants. Taking the focus as origin, let some rectangular axes be given in position. Let the plane of the orbit intersect the plane of xy in the straight line N'SN. This line is called the line of nodes, and that node at which the particle passes to the positive side of the plane of xy is called the ascending node. We require (5) the angle the radius vector to the ascending node makes with the axis of x, and (6) the inclination of the plane of the orbit to the plane of xy. 339. Point to Point. To project a particle with a given velocity V from a given point P so that it shall pass through another given point Q. H' <2 N Let r lt r 2 be the distances SP, SQ. The velocity at P being given, /2 1\ the major axis 2a is also known from the formula F 2 = u, ( ) . \r, aj ART. 341.] POINT TO POINT. 221 With centres P and Q, describe two circles of radii 2a r l} 2a n; these intersect in two points H, H'. Either of these may be the empty focus. The three sides of the equal triangles PQH, PQH f are therefore known. There are two directions of projection which satisfy the given conditions. These directions are the bisectors of the supplements of the angles SPH, SPH'. Let /9, /3' be the angles of projection at P (measured behind the radius vector SP, see Art. 313), then /3 + #' is equal to the supplement of SPQ, and /3 /3' is equal to the known angle HPQ. The range PQ on a given straight line is the greatest possible when H, H' coincide and lie on the straight line PQ. We then have PQ = PH+QH=4a-r l - n. This equation requires that the semi-major axis should be one- quarter of the perimeter of the triangle SPQ. Since two consecutive trajectories whose foci are in the neigh- bourhood of PQ intersect in Q, the locus of Q as the range PQ turns round P is the envelope of all trajectories from a given point P with a given velocity. Since PQ + QS = 4a n this locus is another ellipse having its foci at P and S. Each trajectory touches the enveloping ellipse in the point where the straight line joining P to the empty focus of the trajectory cuts either curve. 34O. Ex. 1. Prove that the semi-major axis a', the eccentricity e' and the semi-latus rectum I' of the enveloping ellipse are given by 2a' = 4a-r,, e'- A TI , r 2 =2a (2a-r,). 4a-r 1 Ex. 2. If the variation of gravity is taken account 'of and the resistance of the air neglected, prove that the least velocity with which a shot could be projected from the pole so as to meet the earth's surface at the equator is about 4 miles per second, and that the angle of elevation is 22. [Coll. Ex. 1892.] Ex. 3. If a particle when projected from Pj passes through two other points P 2 , P 3 , prove that the semi-latus rectum I is given by either of the equalities ina sin asin a 3 , where r lt r 2 , r s , are the distances SP lt SP 2 , SP 3 ; A lt A z , A 3 are the areas of the triangles P 2 /$} where a is the earth's radius. [Coll. Ex. 1895.] 345. Orbits of small eccentricity. The equations (A) and (B) of Arts. 342, 343 determine the time of describing any given angle v in an elliptic orbit of any eccentricity, the equation (B) giving u when v is known while the equation (A) then determines t. The converse problem of finding the polar coordinates r and v when t is given is usually called Kepler's problem. One solution by which u and v are expressed in terms of t by series arranged in ascending powers of e will be presently considered. It is enough here to notice that in a planetary orbit, where e is small, the value of u when t is given can be found by successive approxima- tion. The value of v then follows from (B) by using the trigonometrical tables. 346. To solve (u) = u-e sinu-m = by Newton's rule, when m, i.e. nt, is given. Supposing MJ , 2 to be two successive approximations to the value of u, that rule gives u _- ' (MJ) 1 - e cos u^ ' where m^ =w 1 -esinu 1 . To find a first approximation we notice that u lies between m and me, the upper or lower sign being taken according as m is . We choose some value of u, lying between these limits, which is an integer number of minutes so that its trigonometrical functions can be found from the tables without interpolation. By Fourier's addition to Newton's rule this first approximation should be such that (u) and " (u) have the same sign. Substituting this first approximation for w x , the formula gives a second approxi- mation. Substituting again this second approximation for MJ , we obtain a third, and so on. When e is very small the first computed value of the denominator is sometimes sufficiently accurate for all the approximations required. See Encke, Berliner Astronomisches Jahrbuch, 1838. Gauss, Theoria Motus &c., translated by C. H. Davis. Adams's Collected Works, vol. i. p. 289. ART. 347.] TIME OF DESCRIBING AN ARC. 225 Ex. Prove that if we choose 1^ = 111 + e as the first approximation, the error of the value of u. 2 is of the order e 3 . 347. Ex. 1. Leverrier's rule. If terms of the order e* can be neglected, prove 1 / e sit I 2 \l-ec V 1-ecosm 2\l-ecos7W/ Glaisher remarks that if we replace the third term by - \ (e sin m) 3 (1 - e cos m)~ the formula is correct when terms of the order e s are neglected. He also gives a series for u correct up to e 9 . Monthly Notices of the Astronomical Society, 1877. Ex. 2. Prove that cot u = cot m - -T-. r where f(u-m) x 1 . 3 . . 15 ., f(x)= . =1 + 7; sin-'x + TT: sm 4 a; + sin 6 ar+&c. sin a; 6 40 336 Putting u=m + e on the right-hand side of the first equation we obtain an approximation for cot u whose error is of the order e 3 . This is Zenger's solution of Kepler's problem. He has tabulated the values of f(e) for the eight principal planets. Some improvements of the method have been suggested by J. C. Adams. Both papers are to be found in the Monthly Notices of the Astronomical Society, 1882, vol. XLII. p. 446, vol. XLIII. p. 47. Ex. 3. Prove the following graphical solution of Kepler's problem. Construct the curve of sines y = ainx, measure a distance OM=m along the axis of x and draw MP making the angle PMx equal to cot" 1 e. If MP cut the curve in P, the abscissa of P is the value of M. This method was described by J. C. Adams at the meeting of the B. Association in 1849. It is also given by See in the Astronomical Notices, 1895, who also refers to Klinkerfues and Dubois. Another graphical solution, using a trochoid, is given by Plummer, Astronomical Notices, 1895, 1896. Ex. 4. The equation u-esmu=m has only one real value of u when m is given. This follows from the graphical construction. If the ordinate MP could cut the curve in a second point Q, move the straight line PQ parallel to itself until P and Q coincide. We should then have a tangent to the curve making an angle tan" 1 1/e with the axis of x. But if e < 1 this is impossible, for in the curve of sines the greatest value of the angle is 45. Ex. 5. By using Lagrange's theorem we may expand /(M) in a series of ascending powers of the eccentricity, the coefficients being functions of m. Prove that if the form of the function /(u) be so chosen that the coefficient of e 2 is zero, we obtain the series cot w=cot m - e cosec m + 3 sin m + &c., which takes a very simple form, when the cubes of e can be neglected. This equation is due to Rob. Bryant, Astronomical Notices, 1886. Ex. 6. Prove that when e 4 can be neglected sin i (M - m) = e sin m + e 2 sin 2m + & e 3 sin 3m + &c. [R. Bryant.] Ex. 1. If 0' be the longitude of a planet seen from the empty focus and measured from an apse, prove that tf=.nt + J e z sin 2nt + &o., the error being of the order e*. It follows that the angular velocity round the empty focus is very nearly constant. R. D. 15 226 THE INVERSE SQUARE. [CHAP. VI. 348. We may apply the method of Art. 342 to find the time of describing an arc of the concave branch of the hyperbola. Taking the focus as origin the equation of a hyperbola may be written where u is an auxiliary quantity and / a constant which will be immediately chosen to be the base of the Napierian logarithms ; /. hdt = 2dA=x'dy'-y'dx' = ab (/+/-)_ l du. Since A 3 = /i& 2 /o we have, putting /x/a s =H 2 , nt= -w + esinhu .................................... (A). Again, as in Art. 343, we have x =CN= ~ (/"+/-); m ae-x v /e + 1. ,u .: cost> = -- ; /. tan s = / -tanh- ............... (B), ex -a 2 -V e-1 2 where v= L ASP. If we eliminate u, we have To find a geometrical interpretation for the auxiliary quantity u, let us describe a rectangular hyperbola having the same major axis and produce the ordinate NP to cut the rectangular hyperbola in Q. Then tan QCW=tanh u. Ex. A particle describes the convex branch of the hyperbola, and /*= -/t'is negative. Prove v /e 1 u nt=u + esrnhu, tan- = . / - tanh-, where v=ASP, ft, t la 3 =n*. 349. The time in a parabolic orbit may be more easily found by using the equation r z d0 = hdt. Putting l/r = 1 + cos v where I is the semi-latus rectum, and A 2 = fd, we have This formula gives the time t of describing the true anomaly v = ASP. If c be the radius of the earth's orbit, and p the perihelion distance of the particle expressed as a fraction of c, we have / = 2pc. To eliminate fi, let T = 2?r v/c 3 /^ be the length of a year. Then 7r/2 & r v i , , _^- . $ = p* jtan g + ^ tan 3 ^ If we write T= 365*256 this gives t in days. ART. 350.] EULER'S THEOREM. 227 When a formula like this has to be frequently used we construct a table to save the continual repetition of the same arithmetical work. Let the values of {tan ^v + % taa s v} be calculated for values of v from to 180, with differences for interpolation. When p is known for any comet moving in a parabolic orbit, the table can be used with equal ease to find the time when the true anomaly is given or the true anomaly when the time is known. 350. Euler's theorem. A particle describes a parabola under the action of a centre of force in the focus S. It is required to prove that the time of describing an arc PP' is given by 6 v>< = (r + r' + k)* -(r + r'~ kf, where r,r are the focal distances of P,P' and k is the chord joining P,P'. Let so, y; x', y be the coordinates of P, P', then since y 2 = k* = (x- xj + (y- yj = (y- yj As we wish to make the right-hand side a perfect square, we put y + y' = 4>a tan 6, y y = 4ta tan ............ (1). We shall suppose that in the standard case y is positive and y' numerically less than y ; then and are positive, .'. k = 4atan< sec 6 ........................ (2). Also r + r' = 2a + x + x' = 2a (sec 2 6 + tan'<) ; .'. r + r' + k = 2a (sec + tan <) 2 j _ r + r - k = 2a (sec - tan ) 2 } ' = (2a) f {(sec + tan <) 3 - (sec - tan ) 3 } = 2 (2a) f {3 + 3 tan 2 + tan"<} tan 0. Drawing the ordinates PN, P'N', we see that area PSP' = APN- AP'N' + SP'N' - SPN = l(xy- x' = |a a tan {3 tan 2 + tan z < + 3}. Since the area PSP'= ^ht = ^*J(Zap)t the result to be proved follows at once. 152 228 THE INVERSE SQUARE. [CHAP. VI. The arc PP' gradually increases as P f moves towards and past the apse. The quantity r + r' k decreases and vanishes when the chord passes through the focus. To determine whether the radical changes sign we notice that this can happen only when it vanishes. We can therefore without loss of generality so move the points P, P', that, when the chord crosses the focus, PP' is a double ordinate. We then have 6 V/< = (2r + 2?/) 1 - (2r - 2 are the acute angles made by the chords PP', PQ' with the axis of y. Ex. 2. Show that there are two parabolas which can pass through the given points P, P', and have the same focus. Show also that in using Euler's theorem to find the time P to P', the second radical has opposite signs in the two paths. To find the parabolas we describe two circles, centres P, P' and radii SP, SP. These circles intersect in S and the two real common tangents are the directrices. These tangents intersect on PP' and make equal angles with it on opposite sides. The concavities of the parabolas are in opposite directions, and the angles described are PSP' and 360 - PSP'. If then one angle is greater than 180, the other must be less. Ex. 3. A parabolic path is described about the focus. Show that the squares of the times of describing arcs cut off by focal chords are proportional to the cubes of the chords. 352. Lambert's Theorem*. If t is the time of describing any arc PP of an ellipse, and k is the chord of the arc, then nt = (- sin - sin #'), /r+r' + k /r+r'-k where sm^=i ^/ ^ , sm\' = ^ ^ , (A). Let u, u' be the eccentric anomalies of P, P', /. A 2 = a 2 (cos u - cos u') 2 + a 2 (1 - e 2 ) (sin u - sin u') 2 =4a 2 sin 2 (w-w'){l-e 2 cos 2 i(u+u')} (1), * This proof of Lambert's theorem is due to J. C. Adams, British Association Eeport, 1877, or Collected Works, p. 410. He also gives the corresponding theorem for the hyperbola, using hyperbolic sines. In the Astronomical Notices, vol. xxix., 1869, Cayley gives a discussion of the signs of the angles , '. The theorem for the parabola was discovered by Euler (Miscell. Berolin. t. vii.), but the extension to the other conic sections is due to Lambert. ART. 354.] LAMBERT'S THEOREM. 229 r + r' = 2a- ae cos u - ae cos u' = 2a {1-ecos (u + u')cos^ (u- u')} (2), nt = u - u' - e (sin u - sin u') = u-u'-2e cos (M + M') sin \(u-u') (3). Hence we see that if a, and therefore also n, are given, then r + r 1 , k, and t are functions of the two quantities u - u', and e cos \ (u + u 1 ). Let u -u'=2a, ecos (u + u') = cos j8 (4). /. fc=2asina sin )3 (5), r + r' + fc=2a{l-cos(/3 + a)} (6), r + r'-fc = 2a{l-cos( / 8-a)} (7), wt=2a-2sinacos0 (8). If we put /3 + a=, /3-a=0', the equations (6) and (7) lead to the expressions for sin , sin % ' given above, while (8) when put into the form nt = { + a - sin (|3 + a)} - { - a - sin (fi - a)} gives at once the required value of nt. 353. Let us trace the values of as the point P travels round the ellipse in the positive direction beginning at a fixed point P 1 . We suppose that u increases from u' to 2?r + u'. The positive sign has been given to the square root k. Since k can vanish only when P coincides with P', and a begins positively, we see that both a and /S lie between and IT for all positions of P. The latter is also restricted to lie between cos -1 e and IT - cos" 1 e. We have by differentiating (4) d = d{i + da.= \du {l + ecosec/3 si d is always positive and d continually increases from /S to 2n--j3 and ' decreases from A, to -j8 . When = 7r, r + r' + k = 4a, and the chord P'P passes through the empty focus H. Let it cut the ellipse in Q. It follows that

' is positive or negative according as P lies in the arc P'R or EP'. The values of tf>, ' are determined by the radicals (A). Each of these gives more than one value of the angle, thus may be greater or less than IT and ' may be positive or negative. This ambiguity disappears (as explained above) when the position of P on the ellipse is known. Thus sin and sin 0' have the same sign when the two foci are on the same side of the chord PP' and opposite signs when the chord passes between the foci. 354. Ex. 1. Prove that the time t of describing an arc P'P of a hyperbola is given by t . / ~ 3 = - 230 THE INVERSE W th POWER. [CHAP. VI. i- # /r + r' + k ,' /r + r'-k where sinn - = * / , sinn = . / , and I; is the chord of the arc. [Adams.] Ex. 2. The length of the major axis being given, two ellipses can be drawn through the given points P, P' and having one focus at the centre of force. Prove that the times of describing these arcs, as given by Lambert's theorem, are in general unequal. To find the ellipses we describe two circles with the centres at P, P 1 and the radii equal to 2a - SP, and 2a - SP 1 . These intersect in two points H, H', either of which may be the empty focus, and these lie on opposite sides of the chord PP'. 355. Two centres of force. E.r. 1. An ellipse is described under the action of two centres of force, one in each focus. If these forces are F 1 (rj and F 2 (r 2 ), prove that 2 (r^Fj) = ^ (r 2 2 .F 2 ). If one force follow the Newtonian r i *"i r z dr^ law, prove that the other must do so also. These results follow from the normal and tangential resolutions. E.r. 2. A particle describes an elliptic orbit under the influence of two equal forces, one directed to each focus. Show that the force varies inversely as the product of the distances of the particle from the foci. [Coll. Ex.] Ex. 3. A particle describes an ellipse under two forces tending to the foci, which are one to another at any point inversely as the focal distances ; prove that the velocity varies as the perpendicular from the centre on the tangent, and that the periodic time is IT (a 2 + 6 2 )//ta6, ka, kb being the velocities at the extremities of the axes. [Coll. Ex.] Ex. 4. A particle describes an ellipse under the simultaneous action of two centres of force situated in the two foci and each varying as (distance)" 2 . Prove that the relation between the time and the eccentric anomaly is _ _ __ _ dt a 3 (l-ecosu)" a? (1 + ecosu) 2 ' [Cayley, Hath. Messenger, 1871.] The inverse cube and the inverse w th powers of the distance. 356. The law of the inverse cube. A particle projected in any given manner describes an orbit about a centre of force whose attraction varies as the inverse cube of the distance. It is required to find the motion*. * The orbits when the force F=fj.u 3 were first completely discussed by Cotes in the Harmonia Mensurarum (1722) and the curves have consequently been called Cotes' spirals. The motion for F=fj.u n when the velocity is equal to that from infinity is generally given in treatises on this subject. The paths for several other laws of force are considered by Legendre (Thiorie des Fonctions Elliptiques, 1825), and by Stader (Crelle, 1852) ; see also Cayley's Report to the British Association, 1863. Some special paths when F=/j.u n , for integer values of n from n = 4 to n=9, are discussed by Greenhill (Proceedings of the Mathematical Society, 1888), one case when n = 5, being given in Tait and Steele's Dynamics. ART. 357.] THE INVERSE CUBE. 231 Let attraction be taken as the standard case and let the accelerating force be F={JLU S . We have dhi F M jte + u = d 2 u " dffi + ( 1 ' The solution depends on the sign of the coefficient of u. Let V be the velocity of the particle at any point of its path (say the point of projection), /3 the angle and R the distance of projection, then h = VRsmj3; (Art. 313). Let F a be the velocity from in- finity, then V^ = fijR 2 . It follows that h? is > or or At, we put lfi/h^ = n 2 , then n< 1 or >1 according as the force is attractive or repulsive. The equation of the path is (Art. 119) u = a cos n(d a). The curve consists of a series of branches tending to asymptotes, each of which makes an angle ir/n with the next. When the curve is given the motion may be deduced from the following relations (Art. 306), Also by integrating dOjdt = fiu?, and putting a = l/b, we find that the time of describing the angle 6 = o. to 6, i.e. r = b to r, is given by . hnt , 357. Case 2. Let //. be positive and > A 2 , we put 1 p/h? = n 2 . The equation of the path is then u = Ae n0 + Be" 9 . The values of the constants A, B are to be deduced from the initial values of u and dujdO. Two cases therefore arise, according as A and B have the same or opposite signs. In the former case, u cannot vanish and therefore the orbit has no branches which go to in- finity; in the latter case there is an asymptote. If we write = l + a and choose a. so that Ae na = + Be""*, we may reduce 232 THE INVERSE W th POWER. [CHAP. VI. the equation to one of the three standard forms u = (e*** e-^), u = where 2na = log(+ B/A), a= 2\/(+ .45), the upper or lower signs being taken according as A, B have the same or opposite signs. The third case occurs when .5 = 0; the orbit is then the equi- angular spiral already considered in Art. 319. When the curve is given the motion may be deduced from the following relations where C is determined by making t vanish when r has its initial value and b = I/a. When A and B have the same sign the two branches beginning at the point #i = 0, i.e. 6 = a, wind symmetrically round the origin in opposite directions. When A and B have opposite signs the two branches begin at opposite ends of an asymptote, whose distance from the origin is y = 1 /an, and then wind round the origin. As the particle approaches the centre of force, the convo- lutions of either branch become more and more nearly those of an equiangular spiral whose angle is given by cot $ = + n, the upper or lower sign being taken according as = + oo . The particle arrives at the pole with an infinite velocity at the end of a finite time. 358. Case 3. Let /* be positive and = A 2 . The orbit is u = a (0 a). When the path is known the motion is given by h? = n, v* = IJL (a? + a 2 ), t^/p = br, where t is the time from a distance r to the centre of force and b = I/a. We notice that the radial velocity is constant. ART. 360.] A VELOCITY FROM INFINITY. 233 Beginning at the opposite extremities of an asymptote the two branches wind round the origin and ultimately when = oo cut the radius vector at right angles. If OZ is drawn perpen- dicular to the radius vector OP to meet the tangent at P in Z, we may show that OZ is constant and equal to I/a. 359. Ex. The motion for a force F=f(u) being known, show how to deduce that for a force F=f(u) + /j.u 3 and give a geometrical interpretation. [Newton.] The differential equations are These may be reduced to the forms used when F=f(u) by writing c0 = 0', ch = h', where c 2 = 1 - pj h 2 . To construct the path u =

(9) is known, we make the axis of x together with the latter curve revolve round the centre of force with an angular velocity dujdt, where cO = 6 - w. The axis of x therefore advances or regredes according as c is less or greater than unity. 36O. Law of the inverse nth power. It is required to find the path of a particle when the central force F=fjut n . See Art. 320. We have *!? + -_?!._.-. S***~SE? I?" ' except when n=l, for then the right-hand side takes a logarithmic form. The integration of this equation can be reduced to elementary forms when C=0; this requires that n>l for otherwise v 2 would be negative. The equation then shows that at every point of the orbit the velocity is equal to that from infinity, Art. 312. If V be the velocity, R and p the distance and angle of projection, we have Representing _ = by c-', we have .(3), where the upper or lower sign is to be taken according as duldd is initially negative or positive, i.e. according as the angle /3 is acute or obtuse. To integrate this put CU=X K where K is to be chosen to suit our convenience. Taking the logarithmic differential we find duju = K dxjx, and the integral equation (3) becomes Kdx 234 THE INVERSE 71 th POWER. [CHAP. VI. We now see that if we put K (n - 3) = - 2 the integration can be effected at once, but this supposition is impossible if n = 3. We find n-S (r \ * 71 3 - ) =cos- r (0-a). C J i Conversely, when the path is given, we have = -- _ _. n - 1 c n ~ 3 ' n - 1 r""" 1 It appears that the orbit takes different forms according as n> or <3. In the former case the curve has a series of loops with the origin for the common node and r = c for the maximum radius vector. In the latter case the curve has infinite branches, and rc for the minimum radius vector. 361. If the force is repulsive, we write F= -/*'"" We then have If (7=0, we must have n This can be reduced to elliptic integrals as explained in Cayley's Elliptic Functions, Art. 400, or Greenhill, The Elliptic Functions, Art. 70. The integration can be effected in two cases : (1) when velocity of projection is equal to that from infinity, and (2) when the initial conditions are such that h* = 2/aC. In the latter case the right-hand side of (1) is a perfect square. Ex. 1. Prove that the integration when h*=2nC leads to the curves tanh (0/^2)= r/c or c/r, which have a common asymptotic circle r = c where c=*Jnlh. Prove also that the velocity V of projection is given (Art. 313) by F 2 sin 4 j8 = 2F' 2 { 1 ^(1 - sin 4 ft) } , where V is the velocity from rest at infinity, and the upper or lower sign is to be taken according as the path is outside or inside the asymptotic circle. Ex. 2. Prove that, if the central force F=/j.u 5 , the inverse of any path with regard to the origin is another possible path provided the total energy of the motion exceed the potential energy at infinity by a positive constant E reckoned per unit mass and also that for the two paths Eh'*=E'h*. Prove that when 7t 4 >4/i>0 the path is of the form r = asn f K - modulus k or the inverse form. [Math. Tripos, 1894.] According to the notation of Art. 313, 2E = C. 365. The inverse fourth power. The equation (1) of Art. 360 is /&.y w _ \de) 3ft 2 \ This cubic can always be written in the form &.y w _w sex de) 3ft 2 \ %i*%J" and the integration can be reduced to forms similar to those in Art. 364 by writing a= 2 . The integration can be effected when the initial conditions are such that = 3/j?C. In this case the right-hand side has the factor (u- ft 2 //*) 2 . Ex. Show that the integration leads to the curves u= ---- , ,- , the upper signs being taken together and the lower together. These curves have a common asymptotic circle r=.u//t 2 , one curve being within and the other outside. 236 THE INVERSE W th POWER. [CHAP. VI. 366. Other powers. K.r. If the force F=/iU 7 , and the initial conditions are such that 2ft 3 = 8C > /M) prove that the equation (1) of Art. 360 takes the form where 6 2 =^/ v //t. Thence deduce the integrals " 2 = T , having a common asymptotic circle. The Lemniscate can also be described under this law of force, if the velocity is equal to that from infinity ; Arts. 320, 360. 367. Nearly circular orbits. To find the motion approsri- mately, when the central force F = /*u n and the orbit is nearly circular. Beginning as in Art. 360 with the equation d z u F _p d& + ~u*~h* U we put u = c(l+ai) where c is some constant to be presently chosen but subject to the condition that a; is to be a small fraction. We thus find We see now that the right-hand side of the equation will be simplified if we choose c so that the constant term is zero, i.e. we put h* = fic n ~ 3 . The equation then becomes fj2 T ^ + * = (>i-2)#+Hrc-2)(7 i -3)* 2 +&c ....... (3). As a first approximation, we assume x = M cos ( pO + a) ........................ (4), where M is a small quantity. Substituting and rejecting the squares of M we find (1 -p*) Mcos (pO + a) = (n - 2) M cos (pB + a) ...... (5). The differential equation is therefore satisfied to the first order, if we put p 2 = 3 n. In this case we have as the equation of the path u = c{l + Mcos(p0 + a)} ................. (6). If n< 3, the equation (6) represents a real first approximate solution of the differential equation (1). We notice that the partiole oscillates between the two circles u = c(l+M) and u = c (1 M). The meaning of the constant c is now apparent ; geometrically, 1/c is the harmonic mean of the radii of the bound- ing circles; dynamically, 1/c is the radius of that circle which ART. 369.] NEARLY CIRCULAR ORBITS. 237 would be described about the centre of force with the given angular momentum h. The positions of the apses are found by equating du/dB to zero. This gives pO + a = i-n; the angle at the centre of force between two successive apses is therefore tr/p. If n > 3, the value of p is imaginary, and the trigonometrical expression takes a real exponential form, Art. 120. The quantity x therefore becomes large when 6 increases, and the particle, instead of remaining in the immediate neighbourhood of the circumference of the circle, deviates widely from it on one side or the other. As the square of x has been neglected the expo- nential form of (6) only gives the initial stage of the motion and ceases to be correct when x has become so large that its square cannot be neglected. It follows from this that the motion of a particle in a circle about a centre of force in the centre is unstable ifn>3. 368. Ex. If the law of force is F=u 2 f(u), and the orbit is nearly circular, prove that a first approximation to the path is Thence it follows that the apsidal angle is independent of the mean reciprocal radius, viz. c, only when F=/j.u n , i.e., when the lauf of force is some power of the distance. 369. A second approximation. The solution (6) is in any case only a first approximation to the motion, and it may happen that, when we proceed to a second or third approximation, the value of p is altered by terms which contain M as a factor. Besides this, we shall have x expressed in a series of several trigono- metrical terms whose general form is Ncoa(q0 + fi), where N contains the square or cube of M as a factor together with some divisor K introduced by the integration, Arts. 139, 303. Representing the corrected value of p by jp + A, the error in p9 + a, i.e. 0A, increases by 2?rA after each successive revolution of the particle round the centre of force. The expression (6) will therefore cease to be even a first approximation as soon as 0A has become too large to be neglected. On the other hand the additional term to the value of u may be comparatively unimportant. The magnitude of the specimen term is never greater than N and, unless K is also small, we can generally neglect such terms. In proceeding to a higher approximation we should first seek for those terms in the differential equation which contain cos (p0 + o) ; these being added to the terms of the same form in equation (5) will modify the first approximate value of p. We should also enquire if any term in the differential equation acquires by integration a small divisor K and thus becomes comparatively large in the solution. 238 DISTURBED ELLIPTIC MOTION. [CHAP. VI. 37O. To obtain a second approximation we substitute the first approximation (6) in the small terms of the differential equation (3). Writing (3), for brevity, in the form (7), where /3 = (n-2), 7=$ (n-2) (n-4), &c., we find after rejecting the cubes of M d?x - 2 = (n-3) {* + i/31/*(l + cos22>0)} ........................ (8), where p6 has been written for p6 + a for the sake of brevity. This equation shows (Art. 303) that the second approximate value of x has the form x=M cosp0 + M*(G + Acos2pO) ........................... (9), where G and A are two constants whose values may be found by substitution, and p has the same value as before. To obtain a third approximation, we retain the term yx 3 in (7) and assume x=M cosp6 + M' 2 (G+Acos2pO) + M 3 JBcos3p6 ............... (10). To find the values of p, G, A and B we substitute in (7), express all the powers of the trigonometrical functions in multiple angles and neglect all terms of the order M*. Equating the coefficients of eospO, cos2p6, cos3p0 and the constants on each side, we find = (n - 3) {M*B + M 3 Ap + J M^}, Solving these equations, and remembering that p 2 differs from 3- by terms of the order M 2 , we find G=-(n-2), ^=^(n-2), B = ^(n-2) (n-3), ^3=(3-n){l- T \j(n-2)(n + l)Jtf 2 } (11). The three first are correct when M 2 is neglected and the last when M* is neglected. We notice that up to and including the third order of approximation the terms G, A, B in equation (10) do not contain any small denominators, so that if M be small enough all these terms may be neglected. The motion is then represented very nearly by u=c {1 + M cos (p0 + a)} (12), and this approximation holds until 6 gets so large that M 4 cannot be neglected. We notice also that the additional term in the value of p vanishes only when the law offeree it either the inverse square or the direct distance. Disturbed Elliptic Motion. 371. Impulsive disturbance. When a particle is describing an orbit about a centre of force it may happen that at some particular point of that orbit the particle receives an impulse and begins to describe another orbit. We have to determine ART. 372.] IMPULSIVE DISTURBANCE. 239 how the new orbit differs from the old, for example how the major axis has been changed in position and magnitude, and in general to express the elements of the new orbit in terms of those of the undisturbed orbit. Let the unaccented letters a, e, I, &c. represent the elements of the undisturbed orbit, while the accented letters a', e', I', &c. represent corresponding quantities for the new. We first express the velocity v and the angle ft at the given point of the orbit in terms of the undisturbed elements. Thus v and ft are given by /* - -"-\ A J" Tf r^ w /n \ v 2 =u , sm/3 = -= -- (1), \r aj r vr when the undisturbed orbit is an ellipse described about the focus. We next consider the circumstances of the blow. Let m be the mass of the particle, mB the blow. The particle, after the impulse is concluded, is animated with the velocity B in the given direction of the blow, together with the velocity v along the tangent to the original path. Compounding these the particle has a resultant velocity v' and is moving in a known direction. Since the position of the radius vector is not changed by the blow we may conveniently refer the changes of motion to that line. If P, Q are the components of B along and perpendicular to the radius vector and ft' is the angle the direction of motion makes with the radius vector, we have v' cos ft' = vcosft + P, v' sin ft' = v sin ft + Q (2). Having now obtained v', ft', the formulas (1), writing accented letters for the old elements, determine the new semi-major axis a' and the new semi-latus rectum I'. The position in space of the major axis follows from Art. 336. 372. We may sometimes advantageously replace the second of the equations (1) by another formula. We notice that mh is the moment of the momentum of the particle about the centre of force. Since just after the impulse the velocity v' is the resultant of v and B, the moment of v' is equal to that of v together with the moment of B. Hence h' = h + Bq (3), where q is the perpendicular on the line of action of the blow. Since h? = pi, when the law of force follows the Newtonian law, 240 DISTURBED ELLIPTIC MOTION. [CHAP. VI. this equation leads to (4). Thus the change in the latus rectum is very easily found. As a corollary, we may notice that when the blow acts along the radius vector, the angular momentum mh and therefore the latua rectum of the orbit are unchanged. We also observe that if the magnitude of the attracting force or its law of action were abruptly changed, the value of h is unaltered. 373. Ex. 1. Two particles, describing orbits about the same centre of force, impinge on each other. Prove where m^, wt 2 ft 2 ; hV m 2 V are their angular momenta before and after impact. Ex. 2. A particle P of unit mass is describing an ellipse about the focus S. A circle is described to touch the normal to the conic at P whose radius PC represents the velocity at P in direction and magnitude. Prove that if the particle is acted on by an impulse represented in direction and magnitude by any chord MP of the circle, the length of the major axis is unaltered by the blow. Since .B=2vcos 0, the velocity in the direction of the blow is simply reversed. Hence v' = v and a'=a by Art. 335. 374. If the direction of the blow does not lie in the plane of motion, the plane of the new orbit is also changed. For the sake of the perspective, let the radius vector SP be the axis of x and let the plane of xy be the plane of the old orbit ; then v cos /3, v sin /3 are the components of velocity parallel to the axes of x and y. Let the components of the blow be mX, mY, mZ ; then just after the blow is concluded the components of velocity parallel to the axes are vcoafi + X, v sin /3 + F, and Z. The inclination i of the planes of the two orbits is therefore ? given by tan.i= ; = r ,. The particle begins to move in its new orbit with a v sin /3 + Y velocity v' in a direction making an angle /3' with the radius vector SP given by The problem is now reduced to the case already considered. If mh' is the angular momentum in the new orbit, its components about the axes of x, y, z are 0, - mh' sin i, mh' cos i. Hence h' cos i = h + Yr, h' sin i = Zr, where r = SP. 375. Examples. Ex. 1. A particle is describing a given ellipse about a centre of force in the focus, and when at the farther apse A', its velocity is suddenly increased in the ratio 1 : n. Find the changes in the elements. The direction of motion is unaltered by the blow and since this direction is at right angles to the radius vector from the centre of force, the point A' is one of the apses of the new orbit. Let a, e ; a', e' be the semi-major axes and eccentricities of the orbits. Then since SA' is unaltered in length r=a' (l + e') = a(l + e) .................................... (1). ART. 375.] EXAMPLES. 241 We have here chosen as the standard figure for the new orbit an ellipse having A' for the further apse. A negative value of the eccentricity e' therefore means that A' is the nearer apse. Also since v'=:nv, we have where a' must be regarded as negative if the new orbit is a hyperbola, Art. 333. From these equations we find a' 1 + e , . . - = s 27^ x' e' = l-n 2 1-e). a 2-n 2 (l-e) The point A' is therefore the farther or nearer apse according as n 2 (l-e) is < or >1 ; if equal to unity the new orbit is a circle, if equal to - 1, a parabola. The new orbit is an ellipse or hyperbola according as n 2 (l-e)< or >2. Ex. 2. A particle describes an ellipse under a force tending to a focus. On arriving at the extremity of the minor axis, the force has its law changed, so that it varies as the distance, the magnitude at that point remaining the same. Prove that the periodic time is unaltered and that the sum of the new axes is to their difference as the sum of the old axes to the distance between the foci. [Math. Tripos, I860.] By Art. 325 the new orbit is an ellipse having the centre of force S in the centre. Let the new law of force be fj.'r. Then when r=a, the forces are equal, hence /*'o=/t/a 2 (1). Measure a length SD parallel to the direction of motion at B, such that the velocity v at B is Jp'.SD. Then SD is the semi-conjugate of SB in the new orbit. Equating the velocities at B in the old and new orbits, we have when r=a (---\=fj,>, \r aj SD = a (2). The conjugates SB, SD are equal diameters, the major and minor axes are therefore the internal and external bisectors of the angle BSD. Kepresenting the semi-axes by a', b', we have a' 2 + &' 2 =SB 2 + SD 2 =2a 2 , a'b'=SB . SDsinBSD = ab (3). The internal bisector of the angle BSD is clearly the major axis. If the change in the velocity had been made at any point of the ellipse, we proceed in the same way. By drawing SD parallel to the direction of motion we arrive at the known problem in conies, given two conjugate diameters in position and magnitude, construct the ellipse. The periodic times in the two orbits are respectively 2ir/ N //*' and 27r v /a 3 //i. The equality of these follows from the equation (1). The rest of the question follows from (3). Ex. 3. A particle is describing an ellipse under a force jtt/r 2 to a focus : when the particle is at the extremity of the latus rectum through the focus this centre of force is removed and is replaced by a force /*V at the centre of the ellipse. Prove that if the particle continue to describe the same ellipse /j.'b*=fM. [Coll. Exam. 1895.] K. D. 16 242 DISTURBED ELLIPTIC MOTION. [CHAP. VI. Ex. 4. A planet moving round the sun in an ellipse receives at a point of its orbit a sudden velocity in the direction of the normal outwards which transforms the orbit into a parabola, prove that this added velocity is the same for all points of the orbit, and if it be added at the end of the minor axis, the axis of the parabola will make with the major axis of the ellipse an angle whose sine is equal to the eccentricity. [Coll. Exam. 1892.] Ex. 5. A particle describes a given ellipse about a centre of force of given intensity in the focus S. Supposing the particle to start from the further extremity of the major axis, find the time T of arriving at the extremity of the minor axis. At the end of this time the centre of force is transferred without altering its intensity from S to the other focus H, and the particle moves for a second interval T equal to the former under the influence of the central force in H . Find the position of the particle, and show that, if the centre of force were then transferred back to its original position, the particle would begin to describe an ellipse whose eccentricity is (3e - c 2 )/(l + e). [Math. Tripos, 1893.] Ex. 6. A body is describing an ellipse round a force in its focus S, and HZ is the perpendicular on the tangent to the path from the other focus H. When the body is at its mean distance the intensity of the force is doubled, show that SZ is the new line of apses. [Coll. Ex.] Ex. 7. A particle describes a circle of radius c about a centre of force situated at a point on the circumference. When P is at the distance of a quadrant from 0, the force without altering its instantaneous magnitude begins to vary as the inverse square. Prove that the semi-axes of the new orbit are %c^2 and Jc^/3. Ex. 8. Two inelastic particles of masses m 1 , m 2 , describing ellipses in the same plane impinge on each other at a distance r from the centre of force. If a,, 1 1 ; Oj, I?; are the semi-major axes and semi-latera recta before impact, prove that in the ellipse described after impact Ex. 9. A planet, mass M, revolving in a circular orbit of radius a, is struck by a comet, mass m, approaching its perihelion ; the directions of motion of the comet and planet being inclined at an angle of 60. The bodies coalesce and . . (M+mfa proceed to describe an ellipse whose semi-major axis is ,, , ., - j r . Prove M {M + (4-^/2) m\ that the original orbit of the comet was a parabola ; and if the ratio of m to M is small, show that the eccentricity of the new orbit is (7^ - 4^2)* (m/M). [Coll. Ex. 1895.] 376. Continuous forces. We may apply the method of Art. 371 to find the effects of continuous forces on the particle. Lety, g be the tangential and normal accelerating components of any disturbing force, the first being taken positively when in- creasing the velocity and the second when acting inwards. We divide the time into intervals each equal to 8t and consider ART. 378.] VARIATION OF THE ELEMENTS. 243 the effect of the forces on the elements of the ellipse at the end of each interval. We treat the forces, in Newton's manner, as small impulses generating velocities fSt and gBt along the tangent and normal respectively. The effect of the tangential force is to increase the velocity at any point P from v to v + Bv, where v =fBt, the direction of motion not being altered. To find the effect of the normal force we observe that after the interval Bt the particle has a velocity gSt along the normal, while the velocity v along the tangent is not altered. The direction of motion has therefore been turned round through an angle S/3 = gSt/v. If the disturbing force were now to cease to act, the particle would move in a conic whose elements could be deduced from these two facts, (1) the velocity at P is changed to v + Bv, (2) the angle of projection is (3+ Bft. The conic which the particle would describe if at any instant the disturbing forces were to cease to act is called the instantaneous conic at that instant. 377. To find the effect on the major axis, we use the formula (1). r a Since v is increased to v + Bv, we see by simple differentiation ,..(2). a p In differentiating the formula for v 2 we are not to suppose that dv represents the whole change of the velocity in the time dt. The particle moves along the ellipse and experiences a change of velocity dv in the time dt given by vdv= -- z dr ....................................... (3). Taking dt = 8t, the change of velocity in the time dt is Sv + dv, the part Sv being due to the disturbing forces and the part dv to the action of the central force. 378. To find the changes in the eccentricity and line of apses. We may effect this by differentiating the formulae l=a(I-e>), h^pl, -=l+ecos<9 ............... (4). Since mh is the angular momentum, the increase of mh, viz. mBh, is equal to the moment of the disturbing forces about the origin (Art. 372). Let /3 be the angle the direction of motion at P makes with the radius vector, 162 -n cos/3. V* 244 DISTURBED ELLIPTIC MOTION. [CHAP. vi. We deduce from equations (4) 81 = (1 e 2 ) 8a 2ae8e and the values of 8e and 80 follow at once. = (1 e 2 ) 8a 2ae8e, = cos 08e e sin 080, 379. Herschel has suggested a geometrical method of finding the changes of the eccentricity and the line of apses in his Outlines of Astronomy*. He considers the effect of the disturbing forces /, g on the position of the empty focus. The effect of the tangential force / is to alter the velocity v and therefore to alter a. Since SP + PH=2a, the empty focus H is moved, during each interval St, along the straight line PH a distance HH' = 2Sa, where 5a is given by (2). The effect of the normal force g is to turn the tangent at P through an angle 5p=g$tlv. Since SP, HP make equal angles with the tangent, the empty focus H is moved perpendicularly to PH, a distance HH"=2PH . 5/3. X Y B' Consider first the tangential force/, we have SH=2ae, SH' = 2 (ae + Sae). Hence projecting on the major axis 28 (ae) =HH' . cos PHS = 26a , , where r'=HP=a + ex, and x is measured from the centre ; x - e 2 x Sa 2a (1 - e 2 ) xv ,, -j = - rjot r a n r J oe= Let or be the longitude of the apse line HS measured from some fixed line through S, .-. 2aeSnr = HH'sinPHS = 2Sa^, r y Sa 2a yv ,, A e8iff=+ = y f5t. r a n r Consider secondly the normal force g. We have SH=2ae. SH" .-. 28 (ae) = - HH" sin PHS = - 2r'5/3 HH" cos PHS 2aeSia= -, = 2r'5/3 1 y 1 x + ae a v a v , gdt. * See also some remarks by the author in the Quarterly Journal, 1861, vol. iv. It should be noticed that Herschel measures the eccentricity by half the distance between the foci, a change from the ordinary definition which has not been followed here. ART. 382.] VARIATION OF THE ELEMENTS. 245 38O. The expressions for Se, 5nr should be put into different forms according to the use we intend to make of them. Let \j/ be the angle the tangent at P makes IPx with the major axis, then tan $= We easily find by elementary conies a y b x ay sm^=- ,,--7,, cosi/>= r - - . a J(rr') b J(rr') (2 1 \ UT' --- ) = . It immediately follows that ? a J ar fcos\f/St, br ar x + ae . Se = -- 77 -gcos\p8t, e5'&=- r r . r- gain $ St. J These formulae give the changes of e and or produced by any tangential or normal force. 381. Draw two straight lines OX, OY parallel to the principal diameters situated as shown in the figure. Since /cos \[/, /sin \f/ are the components of the tangential disturbing force parallel to the principal diameters, we see that when the force acts towards OX the eccentricity is increased, and ichen towards OY the apse line is advanced ; the contrary effects taking place when the force tends from these lines. The same rule applies to the normal disturbing force so far as the eccentricity is concerned. It applies also to the motion of the apse except when the particle lies between the minor axis and the latus rectum through the empty focus, and the rule is then reversed. When the eccentricity is small, - = very nearly when the particle is near the minor axis; so that the effects of the tangential force in this part of the orbit may be neglected and the rule applied generally. 382. Examples. Ex. 1. The path of a comet is within the orbit of Jupiter, approaching it at the aphelion. Show that each time the comet comes near Jupiter the apse line is advanced. This theorem is due to Callandreau, 1892. The comet being near the aphelion and Jupiter just beyond, both the normal and tangential disturbing forces act towards OY; the apse therefore advances. Ex. 2. A particle is describing an elliptic orbit about the focus and at a certain point the velocity is increased by I/nth, n being large. Prove that, if the direction of the major axis be unaltered, the point must be at an apse, and the change in the eccentricity is 2 (1 e)jn. [Coll. Ex. 1897.] Ex. 3. An ellipse of eccentricity e and latus rectum I is described freely about the focus by a particle of mass in, the angular momentum being mh. A small impulse mu is given to the particle, when at P, in the direction of its motion ; prove that the apsidal line is turned through an angle which is proportional to the intercept made by the auxiliary circle of the ellipse on the tangent at P, and which cannot exceed lujeh. [Math. Tripos, 1893.] Ex. 4. A body describes an ellipse about a centre of force S in the focus. If A be the nearer apse, P the body, and a small impulse which generates a velocity T act on the body at right angles to SP, prove that the change of direction of the 246 DISTURBED ELLIPTIC MOTION. [CHAP. VI. T /2 \ apse line is given approximately by T- I + cos ASP ) SPsmASP, where e is the ft \6 / eccentricity of the orbit and h twice the rate of description of area about S. [Math. Tripos.] Ex. 5. A particle describes an ellipse about a centre of force in the focus S. When the particle has reached any position P the centre of force is suddenly moved parallel to the tangent at P through a short distance .r, prove that the major axis of the orbit is turned through the angle ^-^ sin sin (0 - 0) where G is the point at oCr which the normal at P meets the original major axis, 6 the angle SGP and the angle the tangent makes with SP. [Coll. Ex. 1895.] Ex. 6. A particle describes an ellipse about a centre of force M/ 7 " 2 an /r 2 , where r is the instantaneous distance of that planet from the sun, and //, is a constant ; Art. 332. It is proved in Art. 341 that when the central force is /j,u 2 , the periodic time in an ellipse is T = %7ra* / <\/ p, where a is the semi-major axis. Now Kepler's third law asserts that for all the planets T 2 is proportional to a 3 ; it follows that jjt, is the same for all the planets. Laws corresponding to those of Kepler have been found to hold for the systems of planets and their satellites. Each satellite is therefore acted on by a force tending to the primary and that force follows the law of the inverse square. It has been possible to trace out the paths of some of the comets and all these have been found to be conies having the sun in one focus. These bodies therefore move under the same law of force as the planets. 250 LAW OF GRAVITATION. [CHAP. VI. 389. The laws of Kepler, being founded on observations, are not to be regarded as strictly true. They are approximations, whose errors, though small, are still perceptible. We learn from them that the sun, planets and satellites are so constituted that the sun may be regarded as attracting the planets, and the planets the satellites, according to the law of the inverse square. We now extend this law and make the hypothesis that the planets and satellites also attract the sun and attract each other according to the same law. Let us consider how this hypothesis may be tested. Let m lt w 2 , &c. be certain constants, called the masses of the bodies, such that the accelerating attraction of the first on any other body distant r x is m^rf, the attraction of the second is ra2/r 2 2 , and so on. Let //, be the corresponding constant for the sun. Assuming these accelerations, we can write down the differen- tial equations of motion of the several bodies, regarded as particles. For example, the equations of motion of the particle m, 1 may be obtained by equating d 2 x/dt 2 , &c., to the resolved accelerating attractions of the other bodies. The equations thus formed can only be solved by the method of continued approximation. Kepler's laws give us the first approximation ; as a second approximation we take account of the attractions of the planets, but suppose that m 1} m 2 , &c. are so small that the squares of their ratio to /* may be neglected. This problem is usually discussed in treatises on the Planetary theory. The solution of the problem enables us to calculate the positions of the planets and satellites at any given time and the results may be compared with their actual positions at that time. The comparison confirms the hypothesis in so extraordinary a way that we may consider its truth to be established as far as the solar system is concerned. 390. Extension to other systems. The law of gravitation being established for the solar system, its extension to other systems of stars may be only a fair inference. But we should notice that this extension is not founded on observation in the same sense that the truth of the law for the solar system is established*. The constituents of some double stars move round * Villarceau, Connaissance des temps for the year 1852 published in 1849; A. Hall, Gould's Astronomical Journal, Boston, 1888. ART. 393.] EXTENSION TO OTHER SYSTEMS. 251 each other in a periodic time sufficiently short to enable us to trace the changes in their distance and angular position. We may thus, partially at least, hope to verify the law of gravitation. What we see, however, is not the real path of either constituent, but its projection on the sphere of the heavens. We can deter- mine if the relative path is a conic and can verify approximately the equable description of areas ; but since the focus of the true path does not in general project into the focus of the visible path, an element of uncertainty as to the actual position of the centre of force is introduced. We cannot therefore use Kepler's first law to deduce from these observations alone that the law of force is the inverse square. 391. Besides this, there are two practical difficulties. First, there is the delicacy of the observations, because the errors of observations bear a larger ratio to the quantities observed than in the solar system. Secondly, a considerable number of observations on each double star is necessary. Five conditions are required to fix the position of a conic, and the mean motion and epoch of the particle are also unknown. Unless therefore more than seven distinct observations have been made, we cannot verify that the path is a conic. These difficulties are gradually disappearing as observations accumulate and instruments are improved. 392. Besides the motions of the double stars we can only look to the proper motions of the stars in space for information on the law of gravitation. Some of these velocities are comparable to that of a comet in close proximity to the sun and yet there is no visible object in their neighbourhood to which we could ascribe the necessary attracting forces. At present no deductions can be made, we must wait till future observations have made clear the causes of the motions. 393. Other reasons. The law of gravitation is generally deduced from Kepler's laws, partly for historical reasons and partly because the proof is at once simple and complete. It is however useful and interesting to enquire what we may learn about the law of gravitation by considering other observed facts. Ex. 1. It is given that for all initial conditions the path of a particle is a plane curve : deduce that the force is central. Consider an orbit in a plane P, then at every point of that orbit the resultant force must lie in the plane. Taking any point A on the orbit project particles in all directions in that plane with arbitrary velocities, then since the plane of motion of each must contain the initial tangent at A and the direction of the force at A, each particle moves in the plane P. It follows that at every point of the plane P traversed by these orbits the resultant force lies in the plane. If these orbits do not cover the whole plane we take a new point B on the boundary of the area covered, and again project particles in all directions in that plane with arbitrary velocities. By continually repeating this process we can traverse every point of the plane, provided no points are separated from A by a line along which the 252 THE HODOGRAPH. [CHAP. VI. force is infinite. It follows that at every point of the plane P the force lies in that plane. Next let us pass planes through any point A of one of these orbits and the direction AC of the force at A. Then by the same reasoning as before the direction of the force at points in each plane must lie in that plane and must therefore intersect AC. Thus the force at every point intersects the force at every other point. It follows that the force is central. An observer placed at the sun, who noticed that all the planets described great circles in the heavens, would know from that one fact that the force acting on each was directed to the sun. Halphen, Comptes Rendus, vol. 84, Darboux's Notes to Despeyrous' Mecanique. Ex. 2. If all the orbits in a given plane are conies, prove that the force is central. If a particle P be projected from any point A in the direction of the force at A, the radius of curvature of the path is infinite at A. Since the only conic in which the radius of curvature is infinite is a straight line, the path of the particle P is a straight line and therefore the force at every point of this straight line acts along the straight line. The lines of force are therefore straight lines. These straight lines could not have an envelope, for (unless the force at every point of that curve is infinite) we could project the particles along the tangents to the envelope past the point of contact so as to intersect other lines of force. The directions of the force would not then be the same at the same point for all paths. Bertrand, Comptes Rendus, vol. 84. Ex. 3. If the orbits of all the double stars which have been observed are found to be closed curves, show that the Newtonian law of attraction may be extended to such bodies. Bertrand has proved that all the orbits described about a centre of force (for all initial conditions within certain limits) cannot be closed unless the law of force is either the inverse square or the direct distance. By examining many cases of double stars we may include all varieties of initial conditions, and if all these orbits are closed the law of the inverse square may be rendered very probable. See Arts. 370, 426. Bertrand when giving this theorem in Comptes Rendus, vol. 77, 1873, quotes Tchebychef. The Hodograph. 394. A straight line OQ is drawn from the origin parallel to the instantaneous direction of motion and its length is propor- tional to the velocity of a particle P, say OQ = kv. The locus of Q has been called by Sir W. R. Hamilton the hodograph of the path of P. Its use is to exhibit to the eye the varying velocity and direction of motion of the particle. See Art. 29. By giving k different values we have an infinite number of similar curves, any one of which may be used as a hodograph. ART. 396.] EXAMPLES. 253 It follows from Art. 29 that, if s' be the arc of the hodograph, ds'/dt represents in direction and magnitude the acceleration of P. 395. If the force on the particle P is central and tends to the origin 0, it is sometimes more convenient to draw OQ per- pendicularly instead of parallel to the tangent. If OF be a perpendicular to the tangent, the velocity v of P is A/OF; hence if OQ = kv, we see that the hodograph is then the polar reciprocal of the path with regard to the centre of force, the radius of the auxiliary circle being *J(hk). If F be the central force at P, the point Q travels along the hodograph with a velocity kF. 396. Examples. Ex. 1. The path being an ellipse described about the centre C, and OQ being drawn parallel to the tangent, prove that the hodographs are similar ellipses. Let CQ be the semi-conjugate of CP, then v = ^n.CQ, Art. 326. Hence if k = lM/j,, the hodograph is the ellipse itself. The point Q then travels with a velocity *JfA . CP. Ex. 2. The path being an ellipse described about the focus S, prove that a hodograph is the auxiliary circle, the other focus H being the origin and HQ drawn perpendicular!}' to the tangent at P. Let SY, HZ be the two perpendiculars on the tangent, then v = h\SY=HQIk y also SY.HZ=IP, :. HQ=HZ if k = b*lh. Since the locus of Z is the auxiliary circle the result follows at once. Ex. 3. The path being a parabola described about the near focus S, prove that a hodograph is the circle described on AS as diameter, where A is the vertex and SQ is drawn perpendicularly to the tangent. Ex. 4. The hodograph of the path of a projectile is a vertical straight line, the radius vector OQ being drawn parallel to the tangent. If the tangent at P make an angle ^ with the horizon, the abscissa of Q is kv cos ^. This is constant because the horizontal velocity of P is constant. The point Q travels along this straight line with a uniform velocity kg. Ex. 5. An equiangular spiral is described about the pole, show that a hodo- graph is an equiangular spiral having the same pole and a supplementary angle. See Art. 30. Ex. 6. A bead moves under the action of gravity along a smooth vertical circle starting from rest indefinitely near to the highest point. Show that a polar equation of a hodograph is r' = b sin \6', the origin being at the centre. Ex. 1. The hodograph of the path of a particle P is given, show that if the path of P is a central orbit, the auxiliary point Q must travel along the hodograph with a velocity v'=\p' 2 p', where p' is the perpendicular from the centre of force on the tangent to the hodograph and p' is the radius of curvature. Show also that the central force F=v'lk and the angular momentum h=ll\k 3 . The condition that the path is a central orbit is v*lp = Fpjr. Writing p = c 2 // and r=c 2 /jp', we find F and thence v'. 254 THE HODOGRAPH. [CHAP. VI. Ex. 8. The hodograph of the path of P is a parabola with its focus at 0, and the radius vector OQ = r' rotates with an angular velocity proportional to r'. Prove that the path of P is a circle passing through 0, described about a centre of force situated at 0. Since the angular velocity of OQ is nr', we find by resolving v' perpendicularly to OQ that v' = nr' s lp'. In a parabola Jr' = 2p' 2 , and since />' r'dr'ldp' we see that v' = \p'*p' where \ = njl. The path is therefore a central orbit. But the polar reciprocal of lr' = 2p' 2 (obtained by writing p' = c*lr, and /=c 2 /p) is r z =p (2c 2 /J), and this is a circle passing through 0. Ex. 9. A particle describes a curve under a constant acceleration which makes a constant angle with the tangent to the path ; the motion takes place in a medium resisting as the nth power of the velocity. Show that the hodograph of the curve described is of the form b~ n e - ne cot* = ,.-n _ a -n_ j- CoU- E x -j Ex. 10. A particle, moving freely under the action of a force whose direction is always parallel to a fixed plane, describes a curve which lies on a right circular cone and crosses the generating lines at a constant angle. Prove that the hodo- graph is a conic section. [Coll. Ex.] 397. Elliptic velocity. Since the velocity is represented in direction and magnitude by the radius vector of the hodograph we may use the triangle of velocities to resolve the velocity into convenient directions. Thus when the path is an ellipse described about the focus S, the velocity is represented perpendicularly by HZjk, where k = b*/h and H is the other focus. If C be the centre this may be resolved into the constant lengths HC, CZ, the former being a part of the major axis and the latter being parallel to the radius vector SP. Hence the velocity in an ellipse described about the focus S can be resolved into two constant velocities one equal to ae/k in a fixed direction, viz. perpendicular to the major axis, and the other equal to afk in a direction perpendicular to the radius vector SP of the particle, where k = b z /h. [Frost's Newton, 1854.] 398. The hodograph an orbit. We have seen that when the force is central a hodograph of the path of P is a polar reciprocal. It follows that if the hodo- graph is the path of a second particle P', each curve is one hodograph of the other. Ex. 1. Let r, r' be the radii vectores of any two corresponding points P, Q of a curve and its polar reciprocal, the radius of the auxiliary circle being c. If these curves be described by two particles P, P with angular momenta h, h', prove that fe 2 &' 2 the central forces at the two points P, Q are connected by FF' = g rr'. Ex. 2. Prove that the two particles will not continue to be at points which correspond geometrically in taking the polar reciprocal, unless the orbit of each is an ellipse described about the centre. [The necessary condition is that the velocity v' = kF in the hodograph should be equal to the velocity v' = h'jp' in the orbit. Since p' = c' 1 lr, this proves that F varies as r.] ART. 401.] TWO ATTRACTING PARTICLES. 255 Motion of two or more attracting Particles. 399. Motion of two attracting particles. This is the problem of finding the motion of the sun and a single planet which mutually attract each other. To include the case of two suns revolving round each other, as some double stars are seen to do, we shall make no restriction as to the relative masses of the two particles. The problem can be discussed in two ways ac- cording as we require the relative motion of the two particles or the motion of each in space. Let M, m be the masses of the sun and the planet, r their instantaneous distance. The accelerating attraction of the sun on the planet is M/r 2 , that of the planet on the sun m/r 2 . Initially the sun and the planet have definite velocities. Let us apply to each an initial velocity (in addition to its own) equal and opposite to that of the sun ; let us also continually apply to each an acceleration equal and opposite to that produced in the sun by the planet's attraction. The sun will then be placed initially at rest, and will remain at rest, while the relative motion of the planet will be unaltered. See Art. 39. The planet being now acted on by the two forces M/r* and m/r 2 , both tending towards the sun, the whole force is (M + m)/r 2 . The planet therefore, as seen from the sun, moves in an ellipse having the sun in one focus. The period is 2?r where a is the semi-major axis of the relative orbit. In the same way the sun, as seen from the planet, appears to describe an ellipse of the same size in the same time. 400. We notice that the periodic time of a double star does not depend on the mass of either constituent, but on the sum of the masses. The time in the same orbit is the same for the same total mass however that mass is distributed over the two bodies. 401. Consider next the actual motion in space of the two particles. We know by Art. 92 that the centre of gravity of the two bodies is either at rest or moves in a straight line with 256 TWO OR MORE ATTRACTING PARTICLES. [CHAP. VI. uniform velocity. It is sufficient to investigate the motion relatively to the centre of gravity, for, when this is known, the actual motion may be constructed by imposing on each member of the system an additional velocity equal and parallel to that of the centre of gravity. Let S and P be the sun and planet, G the centre of gravity, then M .SP = (M + m) GP. The attraction of the sun on the planet is M M 3 1 W_ . &P 2 ~ (M + mf GP* ~ GP* ' The attraction of the sun on the planet therefore tends to a point G fixed in space and follows the law of the inverse square. The planet therefore describes an ellipse in space with the centre of gravity in one focus, and the period is ,.-, a , where a is the semi-major axis of its actual orbit in space. The actual orbits described by the sun and planet in space are obviously similar to each other and to the relative orbit of each about the other. If a, a' be the semi-major axes of the actual orbits of the planet and sun, a that of the relative orbit, we have by obvious properties of the centre of gravity, a/M = a'/ra = af(M+ m). 402. To find the mass of a planet which has a satellite. Since the mean accelerating attractions of the sun on the two bodies are nearly equal, their relative motion is also nearly the same as if the sun were away. Taking the relative orbit to be an ellipse, let a' be its semi-major axis. If m, mf are the masses of the 4>7T~ planet and satellite, T' the period, we have T' 2 = , a /s . When m + m T' and a' have been found by observation, this formula gives the sum of the masses. The masses in this equation are measured in astronomical units, i.e. they are measured by the attractions of the bodies on a given supposititious particle placed at a given distance. It is therefore necessary to discover this unit by finding the attraction of some known body. Consider the orbit described by the planet round the sun. Since we can neglect the disturbing attraction of the satellite, ART. 404.] MASS OF A PLANET. 257 we have, if a is the semi-major axis of the relative orbit and T 42 the period, T 2 =-^ a 3 . M+m Dividing one of these equations by the other, we find a'\* M + m~\T'J \aJ This formula contains only a ratio of masses, a ratio of times and a ratio of lengths. Whatever units these quantities are respec- tively measured in, the equation remains unaltered. Since m is small compared with the mass M of the sun, and m small com- pared with the mass m of the primary, we may take as a near m iT \ z /a'\ 3 approximation >"* (&)( ) I n this way the ratio of the mass JXL \2 / \Q// of any planet with a satellite to that of the sun can be found. 403. The determination of the mass of a planet without a satellite is very difficult, as it must be deduced from the pertur- bations of the neighbouring planets. Before the discovery of the satellites of Mars, Leverrier had been making the perturbations due to that planet his study for many years. It was only after a laborious and intricate calculation that he arrived at a determina- tion of the mass. After Asaph Hall had discovered Deimos and Phobos the calculation could be shortly and effectively made. According to Asaph Hall the mass of Mars is 1/3,093,500 of the sun, while Leverrier made it about one three-millionth. This close agreement between two such different lines of investigation is very remarkable; see Art. 57. The minuteness of either satellite enables us to neglect the unknown ratio m'/m in Art. 402 and thus to determine the mass of Mars with great accuracy. 4O4. Examples. Ex. 1. Supposing the period of the earth round the sun and that of the moon round the earth to be roughly 365J and 27J days and the ratio of the mean distances to be 385, find the ratio of the sum of the masses of the earth and moon to that of the sun. The actual ratio given in the Nautical Almanac for 1899 is 1/328129. Ex. 2. The constituents of a double star describe circles about each other in a time T. If they were deprived of velocity and allowed to drop into each other, prove that they will meet after a time T/4^2. Ex. 3. The relative path of two mutually attracting particles is a circle of radius b. Prove that if the velocity of each is halved, the eccentricity of the sub- sequent relative path is 3/4 and the semi-major axis is 4fc/7. R.D. 17 258 TWO OR MORE ATTRACTING PARTICLES. [CHAP. VI. Ex. 4. Two particles of masses m, m', which attract each other according to the Newtonian law, are describing relatively to each other elliptic orbits of major axis 2a and eccentricity e, and are at a distance r when one of them, viz. m, is suddenly fixed. Prove that the other will describe a conic of eccentricity <' such that . [2 (m + m')(l-e^)) /2 1\ (m + m') |- - - .. * . I =m ( --- ) . ' [r a?n(l-e 2 ) j \r aj It is supposed that the centre of gravity had no velocity at the instant before the particle m became fixed. [Coll. Ex. 1895.] . Ex. 5. Two particles move under the influence of gravity and of their mutual attractions: prove that their centre of gravity will describe a parabola and that each particle will describe relatively to that point areas proportional to the time. [Math. Tripos, I860.] Ex. 6. The coordinates of the simultaneous positions of two equal particles are given by the equations x=aff- 2a sin0, y = a-acos0; x l = a0, y l = -a + acoa9. Prove that if they move under their mutual attractions, the law of force will be that of the inverse fifth power of the distance. [Math. Tripos.] Ex. 7. Two homogeneous imperfectly elastic smooth spheres, which attract one another with a force in the line of their centres inversely proportional to the square of the distance between their centres, move under their mutual attraction, and a succession of oblique impacts takes place between them; prove that the tangents of the halves of the angles through which the line of centres turns between successive impacts diminish in geometrical progression. [Math. T. 1895.] Consider the relative motion. The blow at each impact acts along the line joining the centres, hence the latera recta of all the ellipses described between successive impacts are equal. The normal relative velocity is multiplied by the coefficient of elasticity at each impact. The radius vector of the relative ellipse is the same at each impact, being the sum of the radii of the spheres. The result follows immediately from Ex. 1, Art. 337. 4O5. /,'./ . 1. Herschel says that the star Algol is usually visible as a star of the second magnitude and continues such for the space of 2 days IB. 1 , hours. It then suddenly begins to diminish in splendour and in 3 hours is reduced to the fourth magnitude, at which it continues for about 15 minutes. It then begins to increase again and in 3 hours more is restored to its usual brightness, going through all its changes in 2 d. 20 hr. 48 min. 54-7 sec. This is supposed to be due to the revolution round it of some opaque body which, when interposed between us and Algol, cuts off a portion of the light. Supposing the brilliancy of a star of the second magnitude to be to that of the fourth as 40 to 0-3 and that the relative orbit of the bodies is nearly circular and has the earth in its plane, prove that the radii of the two constituents of Algol are as 100 : 92 and that the ratios of their radii to that of their relative orbit are equal to -171 and -160. If the radius of the sun be 430000 miles and its density be 1-444, taking water as the unit, prove that the density of either constituent of Algol (taking them to be of equal densities) is one-fourth that of water. The numbers are only approximate. [Maxwell Hall, Observatory, 1886.] ART. 407.] THREE ATTRACTING PARTICLES. 259 Ex. 2. The brightness of a variable star undergoes a periodic series of changes in a period of T years. The brightness remains constant for mT years, then gradually diminishes to a minimum value, equal to 1 - fc 2 of the maximum, at which minimum it remains constant for nT years and then gradually rises to the original maximum. Show that these changes can be explained on the hypothesis that a dark satellite revolves round the star. Prove also that, if the relative orbit is circular, and the two stars are spherical, the ratio of the mean density of the double star to that of the sun is sinHJP F(l + fc) 2 cos 2 nir - (1 - fc) 8 cos 2 T 2 (1 + k 3 ) {_ cos 2 mr- cos 2 mir where D is the apparent diameter of the sun at its mean distance. [Math. T. 1893.] 406. Three attracting Particles. The problem of deter- mining the relative motions of three or more attracting particles has not been generally solved. The various solutions in series which have as yet been obtained usually form the subjects of separate treatises, and are called the Lunar and Planetary theories. Laplace has however shown that there are some cases in which the problem can be accurately solved in finite terms*. 407. Let the several particles be so arranged in a plane that the resultant accelerating force on each passes through the com- mon centre of gravity of the system and that each resultant is proportional to the distance of the particle from that centre. It is then evident that if the proper common angular velocity be given to the system about 0, the centrifugal force on each particle may be made to balance the attraction on that particle. The particles of the system will then move in circles round with equal angular velocities, the lines joining them forming a figure always equal and similar to itself. Each particle also will describe a circle relatively to any other particle. Let us next enquire what conditions are necessary that the particles may so move that the figure formed by them is always similar to its original shape, but of varying size. Let the distances * Laplace's discussion may be found in the sixth chapter of the tenth book of the Mdcanique Celeste. The proposition that the motion when the particles are in a straight line is unstable was first established by Liouville, Academie des Sciences, 1842, and Connaissance des Temps for 1845 published in 1842. His proof is different from that given in the text. The motion when the particles are at the corners of an equilateral triangle is discussed in the Proceedings of the London Mathematical Society, Feb. 1875. See also the author's Rigid Dynamics, vol. i. Art. 286, and vol. n. Art. 108. There is also a paper by A. G. Wythoff, On the Dynamical stability of a system of particles, Amsterdam Math. Soc. 1896. 172 260 TWO OR MORE ATTRACTING PARTICLES. [CHAP. VI. of the particles from the centre of gravity be r n r 2 , &c. We then have for each particle the equations _ __ dt* \dt ~ ' rdt\ dt)~ Since the figure is always similar, these equations are to be satisfied when dBjdt is the same for every particle, and r lt r 2 , &c. have the ratios a,, a 2 , &c., where 3 f = - 2,111111 \3 K 409. The line arrangement. Three mutually attracting particles whose masses are M, m', m are placed in a straight line. It is required to determine the conditions that throughout their sub- sequent motion they may remain in a straight line. Let the law of attraction be the inverse /cth power of the distance. Let M, m, be the two extreme particles, m' being between the other two. Let a, b, c be the distances Mm, Mm', m'm ; then a = b + c. A necessary condition is that the resultant accelerating forces on the particles must be proportional to their distances from the centre of gravity (Art. 407). We therefore have M fa" + m'/c" _ M/b" - m/c" _ m/a* + m'/b" Ma + m'c Mb me ma + m'b where the numerators express the accelerating forces on the particles and the denominators are proportional to the distances from 0. The equalities (1) are equivalent to only one equation, for if we multiply the numerators and denominators of the three frac- tions by m, m', M respectively, the sum of the numerators and also that of the denominators are zero. Putting a = b(l+p), c = bp, we arrive at The left-hand side is negative when p = and positive when p is 262 TWO OR MORE ATTRACTING PARTICLES. [CHAP. VI. infinitely large, the equation therefore has one real positive root, whatever positive values M, m', in may have. Putting p = l, the left side becomes (M w)(2* +1 1); since we may take M as the greater of the two extreme particles we see that the real positive value of p is less than unity, provided tc + l is positive. If K + 1 were negative the root would be greater than unity. Whatever the masses of the particles may be it follows that if they are so placed that their distances have the ratios given by this value of p, and their parallel velocities are proportional to their distances from 0, they will throughout their subsequent motion remain in a straight line. When the attraction follows the Newtonian law, the equation (2) becomes the quintic (M + m')p 5 + (3M + 2m')p* + (3M+m')p* - (m' + 3m)p* - (2m' + 3m) p - (m + m') = 0. . .(3). The terms of this equation exhibit but one variation of sign, and there is therefore but one positive root. It may be shown in exactly the same way that in the general case, when K has any positive integral value, the equation (2) has only one positive root; all the terms from p 2 ** 1 to p K+l being positive, while those from p" to p are negative. 410. When the positions of two of the masses are given, there are three possible cases ; according as the third is between the other two or on either side. Since the analytical expression for the law of the inverse square does not represent the attraction when the attracted particle passes through the centre of force, Art. 135 ; these three cases cannot be included in the same equation. We thus have three equations of the form (3), one for each arrangement. 411. In the case of the sun, earth, and moon, M is very much greater than either m or m'. Since p vanishes when m and m' are zero, we infer that p is very small when m/M and m'/M are small. The equation (3) therefore gives Sp* = (m + m')/M, or, using the numerical values of m, mf and M, p = 1/100 nearly. If the moon were therefore placed at a distance from the earth one hundredth part of that of the sun, the three bodies might be projected so that they would always remain in a straight ART. 412.] THREE PARTICLES IN A LINE. 263 line. The moon would then be always full, but at that distance its light would be much diminished. This configuration of the sun, earth and moon however could not occur in nature because this state of steady motion is unstable. On the slightest dis- turbance the whole system would change and the particles would widely deviate from their former paths. 412. Three mutually attracting particles whose masses are M, m', m describe circles round their common centre of gravity and are always in a straight line. Prove that if the force vary as any inverse power of the distance this state of motion is unstable. Reducing the particle M to rest we take that point as the origin of coordinates. Let (r, 6) be the coordinates of m, (r r , 6') those of m'. The particle m is acted on by (M+ )//" along the straight line mM, and m'/r'" in a direction parallel to m'M. The polar equations of the motion of m are d6\2 M+m m' m' -5-- 1 = cos cos dt/ r r '" jj .(1), 1 d I O d0\ m' . - j- I r 2 = sn r dt\ dtj r 'x m r sin w sin w-t R* R where u, are the angles at M, m of the triangle formed by joining the particles and R is the side mm'. In the same way the polar equations of the motion of m' are '"" vo " ' m m ,\ cos w -{ cos ,.K DK dV ./(W'Y M+m' m m - - r I -r- I = d? \dt (2), 1 d f ,,d6'\ m . m . -. ( r 2 -r- I = sin w sm r' dt\ dt J ,* # where ' is the external angle of the triangle at m'. In forming these equations the standard case is that in which 6' >& and ''/r = sin u/R we find after some reduction (52 _ n a_ K E) x - 2n5y + m'icB + Q. 77 = 0, 2n8x + (S 2 + m'B) y + . - m'Btt = 0, mxAx + . ?/ + (3 2 - n 2 - K.F) - 2n5r> = 0, 0.x- mAy + 2n5| + (S 2 + 7?^) t\ = 0, where for brevity we have written S for rf/dt, and c = a-b, a \c K+1 m F= | O ic+l c ic+l' The steady motion has been already found in Art. 409, but it may also be deduced from the first and third of the equations (1) and (2) by equating the constants. We thus find n 2 = E - m'B, ri > =F-mA. 264 TWO OR MORE ATTRACTING PARTICLES. [CHAP. VI. We notice that the constants E, F are positive. When K + 1 is positive, it has been shown in Art. 409 that a > b > c, and therefore A , I) and E + F - 2n 2 are positive. Lastly whatever K may be E + F-n? is positive To solve the four equations, we put x = Ge M , y = He M , = Ke M , i)~Le M . Sub- stituting and eliminating the ratios G, H, K, L we obtain a determinantal equation whose constituents are the coefficients of x, y, , 17, with X written for 5. This determinant is of the eighth degree in X. To find its factors we must before expansion make some necessary simplifications which we can only indicate here. We first add the column to the x column and the i] column to the // column. The second column may now be divided by X. Multiplying the second column by 2n and subtracting from the first, we see that X 2 - (K - 3) n 2 is another factor which we divide out. Subtracting the first row from the third and the second from the fourth, the first column acquires three zeros and the second column two. The determinant is now easily expanded and we have \2 |x 2 - ( K - 3) H 2 } { (X 2 + C) (X 2 - CK - (K + 1) ?i 2 ) + 4?t 2 X 2 } = 0, where C=E + F-2n 2 . If x>3, this equation gives a real positive value of X and the motion is therefore unstable. If K have any positive value C is positive, and the third factor has the product of its roots negative ; one value of X 8 is real and positive and the other real and negative. The motion is therefore unstable for all positive values of K. 413. Ex. 1. Three mutually attracting particles are placed at rest in a straight line. Show that they will simultaneously impinge on each other if the initial distances apart are given by the value of p in the equation of the (2*c + l)th degree of Art. 409. [This equation expresses the condition that the distances between the particles are always in a constant ratio.] Ex. 2. Three unequal mutually attracting particles are placed at rest at the corners of an equilateral triangle and attract each other according to the inverse *th power of the distances. Prove that they will arrive simultaneously at the common centre of gravity. If the law of attraction is the inverse square, the time of transit is \ IT (a s /2/x)^ where ^ is the sum of the masses and a the side of the initial triangle, Art. 131. 414. A swarm of particles. Let us suppose that a comet is an aggregation of particles whose centre of gravity describes an elliptic orbit round the sun. The question arises, what are the conditions that such a swarm could keep together*? Similar conditions must be satisfied in the case of a swarm consolidating * The disintegration of comets was first suggested by Schiaparelli who proved that the disturbing force of the sun on a particle might be greater than the attraction of the comet. He thus obtained as a necessary condition of stability mjb 3 > 2Mja 3 . The subject was dynamically treated by Charlier and Luc Picart on the supposition of a circular trajectory. They arrived at the condition mjb 3 > 3 Mja 3 ; Bulletin de I'Academie de S. Petersbourg, Annales de V Observatoire de Bordeaux, Tisserand, M6c. Celeste, iv. The condition of stability was extended to the case of an elliptic trajectory by M. 0. Callandreau in the Bulletin Astronomique, 1896. The brief solutions here given of these problems are simplifications of their methods. ART. 414.] A SWARM OF PARTICLES. 265 into a planet in obedience to the Nebular theory. The following example will illustrate the method of proceeding. We shall suppose the sun A to be fixed in space, Art. 399. Let B be the centre of the .swarm, G any particle. Let r, 6 be the polar coordinates of B referred to A, and f, 77 the coordinates of G referred to B as origin, the axis of f being the prolongation of AB. Let M be the mass of the sun. Supposing, as a first approximation, that the swarm is homogeneous and spherical, its attraction at an internal point G is p,p, where p = BG. If m be the mass and b the radius of the swarm, pb = m/b 2 . The equations of motion are, by Art. 227, dt d 2 r^_ fd0\" 1 ^ dt 2 \dt J r+ % dt These equations also apply to the motion of the particle at B, where f = 0, 17 = 0. Hence when we expand in powers of , 77, all the terms independent of , 77 must cancel out. We thus have ^ ^_ O _ ^_ n\ C I .. "t. -I . *l 1 .0. t I dt 2 dt dt ' dt 2 * \dt , , /0 , r \^)- d 2 r) nd][d0 ^_ /^V_ZJ^7_ If the centre of gravity of the swarm describe a circle about the sun, we write r = a, ddjdt = n. The equations then become (3). Putting f = A cos (pt + a), 77 = B sin (pt + a), we immediately ob- tain the determinantal equation (p 2 -/j, + 3n 2 )(p 2 -fjL)-4!p 2 n 2 = ............... (4). The condition that the particles of the swarm should keep together is the same as the condition that the roots of this quadratic should be real and positive. The left-hand side is positive whenp 2 = + oo , and negative when^) 2 = p. and p 2 = p. '3n 2 . The required condition 266 TWO OR MORE ATTRACTING PARTICLES. [CHAP. VI. is therefore fi > 3w a , Art. 288. The condition that the swarm is , ji ..i ^ m n M stable is therefore 77 > 3 . o 3 a 3 Unless therefore the density of the swarm exceed a certain quantity the swarm cannot be stable. If the mass of the sun were distributed throughout the sphere whose radius is such that the swarm is on the surface, the density of the swarm must be at least three times that of the sphere. The path of the particle C when describing either principal oscillation is (relatively to the axes B%, By) an ellipse with its centre at J3. Substituting the values of , tj in the equations of motion and using the quadratic, we find B -2np' B* ' 4 p z ' B-J3. 2 Since p lies between the values of p 2 , the first equation shows that A l \B l and A^B.} have opposite signs, and accordingly the radical is negative. It follows that the oscillation which corresponds to the smaller value of p has the major axis directed along B, while in the other that axis is along By. The particle also describes the ellipses in opposite directions, in the former case the direction is the same as that of the swarm round the sun, in the latter, the opposite. If the centre of gravity of the swarm describe an ellipse of small eccentricity, we may obtain an approximate solution of the equations of motion. Assuming the expansions 6 = nt + 2e sin nt + e 2 sin 2nt, it is evident that all the coefficients of the differential equations (2) can be at once expressed in terms of t, including all terms which contain e 2 . It is however unnecessary for our present purpose to write these at length. It is easy to see that the equations become eX = 4ew cos nt -^ - 2eu 2 sin nt rj + Wen 2 cos nt + &c., (tt eY= - 4en cos nt ^ + 2en 2 sin nt % + en 2 cos ntrj + &c. CLt As a first approximation we neglect eX, eY. Comparing the equations (5) and (3) we see at once that we shall have the quadratic }-4pn t =0 .................. (6). The condition that the swarm is stable is then /t > 7t 2 (3 + 5e 2 ) ; .'. 3 > 3 (3 + 5e 2 ). It appears therefore that the gradual dissipation of a comet is more probable when the trajectory is elliptical thun when it is circular. ART. 415.] TISSERAND'S CRITERION. 267 As a second approximation, we substitute % = A cos (pt + a), r/ = B sin (pt + a) in the expressions X and F. By Art. 303 the only important terms are those which become magnified by the process of solution. These terms are of the form Pcos(\t + L) where \=pn or p2. Unless therefore the roots p, p' of the quadratic (6) or (4) are such that pp' is nearly equal to n or 2n, the terms derived from X, Y remain respectively of the order e or e 2 . This relation between the roots cannot occur when e is small. 415. Tisserand's criterion*. When a comet describing a conic round the sun passes very near to a planet, such as Jupiter, its course is much disturbed. When it emerges from the sphere of perceptible influence of the planet, it may again be supposed to describe a conic round the sun, but the elements of the new path may be very different from those of the old. Since Jacobi's integral (Art. 255) holds throughout the motion, the elements of both the conies must satisfy that equation. Let (, 1 ), (ai,li) be the semi-major axis and semi-latus rectum before and after passing through the sphere of influence of the planet. Let i , t\ be the inclinations of the planes of the comet's orbit to the plane of the planet's motion. Let the sun be taken as the origin of coordinates, and let the axis of pass through the planet P. Let r, p be the distances of the comet Q from and P respectively and c = OP. Let M, m be the masses of the sun and planet, then, reducing the sun to rest (Art. 399), we regard the comet as acted on by the resultant attraction of the sun and planet together with a force m/c 2 acting parallel to PO. The field of force is therefore defined by JT _ M , m m % L 1 . r p c 2 We suppose that the planet P describes a circular orbit relatively to with a constant angular velocity n, where n 2 = (M + m)/c 3 . The Jacobian integral takes the form i ir M m m ^V 2 -nA +-j = C, r p c 2 * Tisserand's criterion may be found in his Note sur Pintegrale de Jacobi, et sur son application a la theorie des cometes, Bulletin Astronomique, Tome vi. 1889, also in his Mecanique Celeste, Tome iv. 1896. M. 0. Callandreau's addition is given in the second chapter of his Etude sur la theorie des cometes p^riodiques, Annales de VObservatoire de Paris, Memoires, 1892, Tome xx. There are also some investigations by H. A. Newton on the capture of comets by planets, especially Jupiter, American Journal of Science, vol. XLII. pages 183 and 482, 1891. 268 TWO OR MORE ATTRACTING PARTICLES. [CHAP. VI. where V is the space velocity of the comet and A its angular momentum referred to a unit of mass. Since (Art. 333) F 2 = M (---}, A= cos \7* CL/ the integral becomes 1 /1 m /I \ 1 /I, mf\ f,\ \-ncosi OA / if>+ -- ^ =s -- hncost lA /irr+nrr -- ) , a V M M\p (?) 2ai V M M\pi c-J 2a where O>PO', %i>pi> ar e the values of , p when the comet is respec- tively entering and leaving the sphere of influence of the planet. We obviously have p = pi, and since the comet does not stay long within the sphere, we may neglect when multiplied by the very small quantity m/M. Writing then n 2 = Mfc? as a close approximation, Art. 341, we obtain the criterion 1 cos i *J1 _ 1 cos i t K^I! 2a cVc 2i c\Jc 416. Tisserand uses this criterion to determine whether two comets both of which are known to have passed near Jupiter could be the same body. If the criterion is not satisfied by the known elements of the two comets, they cannot be the same body. If it is satisfied it is then worth while to examine more thoroughly how much the elements of either body have been altered by the attraction of Jupiter. This must be done by using the method of the planetary theory and is generally a laborious process. In Tisserand's criterion the orbit of Jupiter is considered to be circular, which is not strictly correct. This defect has been corrected by M. 0. Callandreau. Taking account only of the first power of the eccentricity he adds a small term containing that eccentricity as a factor. This term, unlike those in Tisserand's criterion, depends on the manner in which the comet approaches Jupiter. 417. Stability deduced from Vis Viva. The Jacobian integral has been used by G. W. Hill * to determine whether the moon could be indefinitely pulled away from the earth by the disturbing attraction of the sun. In such a problem as this, it is convenient to take the origin at the earth P and the moving axis of if directed towards the sun 0. Reducing the earth to rest, the moon Q is acted on by (?ft + m')/p 2 along QP and M[c z parallel to OP. The Jacobian equation for relative motion, Art. 255 (3), takes the form * G. W. Hill's researches in the Lunar theory may be found in the American Journal of Mathematics, vol. i. 1878. ART. 418.] STABILITY OF THE MOON. 269 where p=PQ, r=OQ, c=OP and /j. is the sum of the masses m, m' of the earth and moon. We treat the sun's orbit as circular and put as a near approximation j/c 3 =M a . Since p 2 = 2 + ij 2 , this equation becomes W = + * + ?{(e-tf + f\-C'. Since the left-hand side is essentially positive it is clear that the moving particle Q can never cross the surface defined by equating the right-hand side to zero, and can only move in those parts of space in which the right-hand side is positive. Art. 299. If the initial circumstances of the motion make C' negative, the right-hand side is always positive and the equation supplies no limits to the position of Q. The form of the surface when C' is positive has been discussed by Hill. When C' exceeds a certain quantity the surface has in general three separate sheets. The inner of these is smaller than the other two and surrounds the earth. The second is also closed but surrounds the sun, the third is not closed. When the constants are adapted to the case of the moon, that satellite is found to be within the first sheet. It must therefore always remain there, and its distance from the earth can never exceed 110 equatorial radii. Thus the eccentricity of the earth's orbit being neglected, ice have a rigorous demonstration of a superior limit to the radius vector of the moon. 418. Ex. 1. If the moon Q move in the plane of motion of the earth P and c s / 2 \ if also the sun is so remote that we may put + j- 2 =f c 2 ( 1 + -^) when the left- hand side is expanded in powers of /c and t\\c, the bounding surface degenerates into the curve - + i 7i a *=C". It is required to trace the forms of this curve for P different positive values of C". The curve has two infinite branches tending to the asymptotes f w 2 | 2 =C". If C" is greater than the minimum value of M/l + l" 2 ? 2 there is also an oval round the body S. If the particle Q is within the oval, it cannot escape thence and its radius vector will have a superior limit. If the particle is beyond either of the infinite branches, it cannot cross them and the radius vector will have an inferior limit. The velocity at any point of the space between the oval and the infinite branches is imaginary. [Hill.] Ex. 2. A double star is formed by two equal constituents S, P whose orbits are circles. A third particle Q whose mass is infinitely small moves in the same plane and initially is at a distance from P on SP produced equal to half SP, starting with such velocity that it would have described a circular orbit about P if ,S* had been absent. Show that the curve of no relative velocity is closed, and that the particle being initially within that curve cannot recede indefinitely from the attracting bodies S and P. This example is discussed by Coculesco in the Comptes Rendus, 1892. He also refers to a memoir of M. de Haerdtl, 1890, where the revolution of Q round P is traced during two revolutions and it is shown that at the end of the third the particle is receding from A *. * Since writing the above the author has received Darwin's memoir on Periodic Orbits, Acta Mathematica, xxi. in which the motion of a planet about a binary star 270 THEORY OF APSES. [CHAP. VI. Theory of Apses. 419. When the law of force is a one-valued function of the distance, every apsidal radius vector must divide the orbit sym- metrically. Let be the centre of force, A an apse (Art. 314). The argument rests on two propositions. (1) If two particles are projected from A with equal velocities, both perpendicularly to OA but in opposite directions, it is clear that (the force being always the same at the same distance from 0) the paths described must be symmetrical about OA. (2) If at any point of its path, the velocity of the particle were reversed in direction (without changing its magnitude), the particle would describe the same path but in a reverse direction. If then a particle describing an orbit arrive at an apse A, its subsequent path when reversed must be the same as its previous path. Hence OA divides the whole orbit symmetrically. We may notice that if the law of force were not one-valued, say F = /J, [u *J(u 2 a-}}, where the apsidal distance OA = a, the first proposition is not true, unless it is also given that the radical keeps one sign. 420. There can be only two apsidal distances though there may be any number of apses. Let the particle after passing an apse A arrive at another apse B. Then since OB divides the orbit symmetrically, there must be a third apse C beyond B such that the angles AOB, BOG are equal and OC= OA. Since 00 divides the orbit sym- metrically, there is a fourth apse at D, where OD = OB and the angles BOG, GOD are equal. The apsidal distances are therefore alternately equal, and the angle contained at by any two con- secutive apsidal distances is always the same. has been more thoroughly studied. Taking a variety of initial conditions he has traced the subsequent paths of a particle of insignificant mass. Some of the paths thus presented to the eye have such unexpected and remarkable forms that the paper is full of interest. ART. 422.] TWO APSIDAL DISTANCES. 271 421. Examples. Ex. 1. Show that an ellipse cannot be described about a centre of force whose attraction is a one-valued function of the distance unless that centre is situated on a principal diameter and is outside the evolute. By drawing all the tangents to one arc EF of the evolute we see that they cover the whole area of the quadrant ACB of the ellipse. It follows that a normal to the ellipse can be drawn through any point P situated in this quadrant, and this normal does not divide the ellipse symmetrically, unless P lies between E and A or between F' and JB. Ex. 2. If the path is an equiangular spiral and the central force a one-valued function of the distance, prove that the centre of force must be situated in the pole. Ex. 3. If a particle of mass m be attached to a fine elastic string of natural length a and modulus X, and lie with the string unstretched and one extremity fixed on a smooth horizontal plane ; prove that, if projected at right angles to the string with velocity v, the string will just be doubled in length at its greatest extension if 3mv 2 =4a\. [Coll. Ex.] Ex. 4. A particle is projected from an apse with a velocity v, prove that the apse will be an apocentre or a pericentre according as the velocity v is less or greater than that in a circle at the same distance. 422. The apsidal distances. To find the apsidal distances when F=/jbU n , and n is an integer. The equation of vis viva, viz. v 2 = C 2 fFdr, gives > to f /**?.!. ^ v ' =/i |UJ +Mh= -! .(l). Let V be the velocity at the initial distance R, ft the angle of projection, then O.. / 1 \ n i Sf * '** h=VR8m/3 (2). n - 1 \RJ - Thus both h and C are known quantities, at an apse u is a max- min, and therefore du/dd = 0. The apsidal distances are therefore given by If an equation is arranged in descending powers of the unknown quantity, we know by Descartes' theorem that there cannot be more positive roots than variations of sign. The arrangement of the terms of equation (A) will depend on whether n 1 is greater 272 THEORY OF APSES. [CHAP. VI. or less than 2 ; but, since there are only three terms, it is clear that in whatever order they are placed there cannot be more than two variations of sign. The equation cannot therefore have more than two positive roots. This is an analytical proof that there cannot be more than two real apsidal distances. 423. If n is a fraction, say n = pjq in its lowest terms, we write M=?C'; the indices of w are then integers and ID and therefore u can have only two positive values. It is assumed that if q is an even integer the sign of F is given by some other considerations, for otherwise F would not be a one- valued function of u. 424. The propositions proved in Arts. 420 and 422 are not altogether the same. The complete curve found by integrating (A) may have several branches separated from each other so that the particle cannot pass from one to the other. In 420 it is proved that the actual branch described cannot have more than two unequal apsidal distances. In 422 it is proved that when F=fjiu n all the branches together cannot have more than two unequal apsidal distances. If the force be some other one-valued function of the distance the complete curve may have more than two unequal apsidal distances. (du\^ j^\ = A (u - a) (u - b) (u - c) be the differential equation of an orbit, prove that the central force is a one-valued function of the distance. Prove also that the curve has two branches and three unequal apsidal distances, and that either branch may be described if the initial conditions are suitable. See Arts. 309, 441. Ex. 2. If the central force is F=(M n , where n>3 and the velocity is greater than that from infinity, prove that the apsidal distances lie between p and q, where 2fj.=h z (n- l)p n ~ 3 and h?=Cq 2 . [This follows from a theorem in the theory of equations applied to equation (A) of Art. 422.] 426. The apsidal angle. To find the apsidal angle when F= ftu n , where n<'3, and the orbit is nearly circular. The equation of the path with these conditions has been found by continued approximation in Arts. 367 to 370. Taking the first approximation, we see by referring to the equation (6) of those articles that dufdd is zero only when pO + a. = ITT, where i is any integer. These values of B therefore determine the apses and the reciprocals of the two corresponding apsidal distances are c (1 M). The apsidal angle described between two consecutive apses is therefore fr/p, where p z = 3 n. ART. 427.] THE APS1DAL ANGLE. 273 Taking the higher approximations, we use the equations (12) and (13) in the same way. The apsidal angle is therefore tr/p, where The reciprocals of the apsidal distances are very nearly c (1 + M ). 427. There is another method of finding the apsidal angle which is founded on a direct integration of the equations of motion*. Beginning with we have, as in Art. 422, n-l let u=a, u = b be the reciprocals of the inner and outer apsidal distances. Since the right-hand side of the equation must vanish for each of these values of u, we have 2 M ~ fttl ] 7=0. n- J. Eliminating ft 2 and C we find f d$ ,du W n - J , M 2 , 1 a"- 1 , a 2 , 1 To find the apsidal angle we have to integrate the value of d6 from u = b to a. To simplify the limits we put a = c(l + M), b=c(l-M) and u = c (l + Hx) ; the limits of integration are then x - 1 to + 1. Also since the orbit is nearly circular, we suppose M to be a small quantity. It now becomes necessary to expand A in powers of M. This may be effected by using some simple properties of determinants. If we subtract the upper row from each of the other two, the determinant is practically reduced to a determinant of two rows. Noticing that where C=i(-2), D = J(n-2) (n-3), E=-fa (n-2) (-3)(n-4), we see that the new determinant is | 1 + CM (x - 1) + &c., 2 + M (x - 1) Subtracting one row from the other and performing some evident simplifications, we find A= 2 (a; 2 - 1) {1 + J (n - 2) Mx + T V (n - 2) M 2 ((n - 4) x*+n - 6)}, where 2 = 2c n+1 M 3 (n - 1) (n - 3). We thence deduce * The method of finding the apsidal angle by a direct integration of the apsidal equation was first used by Bertrand, Comptes Rendus, vol. 77, 1873. An improved version was afterwards given by Darboux in his notes to the Cours de MScanique by Despeyrous, 1886. K. D. 18 274 THEORY OF APSES. [CHAP. VI. In the same way we find after some reductions (a"- 1 - &-!)*= {2C"- 1 M (n - 1)}* {1 + & ( - 2) (n - 3) M 2 }. Remembering that du=cMdx t these give dO 1 1 The integrations can be effected at sight by putting #=sin0. Taking the limits to be 0= TT to make the apses adjacent, we find that the apsidal angle is (n-2) (n + 1) 24 428. Closed orbits. An orbit is described about a centre of force whose attraction is a one-valued function of the distance. Prove that if the orbit is closed, for all initial conditions within certain defined limits, the law of force must be the inverse square or the direct distance. [Bertrand, Comptes Rendus, vol. 77, 1873.] If the path is closed and re-entering it must admit of both a maximum and a minimum radius vector. The orbit therefore has two apsidal distances and must lie between the two circles which have these for radii and their centres at the centre of force. By varying the initial conditions we may widen or diminish the space between the circles, yet by the question the orbit is always to be closed so long as the radii of the circles remain finite. Representing the first approximation to the reciprocals of the radii by c (1 M) the apsidal angle will be irfp, where p can be expressed in some series of ascending powers of M. The orbit cannot be closed unless the apsidal angle is such that, after some multiple of it has been described, the particle is again at the same point of space and moving in the same way. Hence p must be a rational fraction for all values of M whether rational or not. The coefficients of all the powers of M must therefore be zero, while the term independent of M must be a rational fraction. When F=fjiu n the series forp is (Art. 426) p = ^(3 - n ) (1 -fo( n - 2) (n + 1) M 2 + &c.}. Since the coefficient of M z must be zero we see that n = 2 or 1, i.e. the law of force must be the inverse square or the direct distance. In either case the condition that V(3 n) should be a rational fraction is satisfied. If we take the most general form for the force, we have F=u*f(u). We know by Art. 368 that the first term of the ART. 430.] CLOSED ORBITS. 275 series for p is, in general, a function of c, i.e. of the reciprocal of the mean radius. Since this can be varied arbitrarily the apsidal angle cannot be commensurable with TT unless this first term, viz. cf (c)//(c), is independent of c. Putting this equal to a constant m we find by an easy integration that f (c) = /*c m . Hence F = fjM m+2 . The general case is therefore reduced to the special case already considered. 429. Classification of orbits. The force being F=/j.u n it is required to classify the various forms of the orbit according to the number of the apsidal distances *. We suppose p. to be positive and h not to be zero. Arranging the apsidal equation (A) (Art. 422) in descending powers of u, it takes one or other of the three following forms -- n-1 l-7i according as n>3, n lies between 3 and 1, and ra 2 +F 2 , where F x is the velocity from infinity, and F the velocity to the origin. The first equation is to be used when Fj is finite, i.e. when >1; the second when F is finite, i.e. when n3, C is negative or zero, i.e. the velocity v is less than or equal to that from infinity ; when n lies between 3 and 1, C must be positive or zero, i.e. the velocity v is greater than or equal to that from infinity. Lastly we see from the third form of the equation (A) that when n (u) cannot change sign as u varies from to oo . Since $ (u) must have the same sign as the highest power of it, its sign is positive or negative according as ?! > or < 3. We notice that if n is a fraction, say n = p/q, we replace the factor u - a by w - b where u=w q , a=b q ; Art. 423. As in most cases the force F varies as some integral power of the distance, it will be more convenient to retain the form given above. Since the left-hand side of (2) is necessarily positive, the whole of the curve must lie inside the circle u a if n>3, and must lie outside that circle if n<3. Suppose the particle, as it moves round the centre of force, to have arrived at the apse. It will then begin to recede from the circle and must always continue to recede because dujd6 is not again zero. The orbit has therefore two branches extending from the apse to the centre of force or to infinity according as n> or< 3. The apse is an apocentre in the first case and a pericentre (as in a hyperbola described about the inner focus) in the second case. The motion in the neighbourhood of the apse may be found by writing u=a + x and retaining only the lowest powers of x. We then have where &Ah?=(a). The path is therefore such that the particle describes a finite angle while it moves from uu to u = a. Since d6jdt=hu- is finite, the time of describing this finite angle is also finite. 431. Cases II. and III. To find the conditions that there may be either two apsidal distances or none. The apsidal equation must have two positive roots or none. The condition for this is that the right-hand side of (A) must have the same sign when w=0 and w=oo . First. Let n>3, this condition requires that C should be positive and not zero. The velocity at every point must therefore be greater than that from infinity. To distinguish the cases we find the max-min value M of the right-hand side by equating to zero its differential coefficient. We thus find M //i 2 V Ti-1 !/=-- I ) +C7, K= -- -. K \M/ -3 Taking the second differential coefficient we find that H is a minimum when n > 3 and a maximum when n < 3. We notice that when n>3, the two terms of M have opposite signs and that we can make either predominate by giving h or C small values. Thus M may have any sign if the initial conditions are suitably ctiosen. The path may there- fore have either two apsidal distances or none; there will be two if M is negative and none if M is positive. If M=0 the apsidal distances are equal. Secondly, let 3>7i>l. The right-hand side of (A) cannot have the same sign when u = and w = oo unless C is negative. The velocity at every point must there- fore be less than that from infinity. ART. 433.] CLASSIFICATION OF ORBITS. 277 Writing as before , -, 3-71 we shall prove that M is necessarily positive and has zero for its least value. Then since the right-hand side of (A) is negative when u=0 and w=oo and is equal to the positive quantity M for some intermediate value, there must be two apsidal distances which can be equal only when M = 0. To prove that M is positive, we notice that M is least when h is greatest. Since h=vrsinfi (Art. 313) this occurs when h = vr, i.e. when the particle is projected perpendicularly to the radius vector. Substituting this value of h and remembering that C=v 2 -V 1 2 , we can see by a simple differentiation that M is again least when v 2 = M/r" 1 , that is, when the velocity is equal to that in a circle. This value of v is less than the velocity from infinity (n being < 3), and is there- fore admissible here. Substituting this value of v we find that the minimum value of M is zero. The value of M is therefore positive and is zero only when the path is a circle. We may also prove that the orbit has two apsidal distances by observing that since the velocity is insufficient to carry the particle to infinity, the orbit must have either an apocentre or must approach an asymptotic circle. In either case the apsidal equation has one positive root and therefore has another. Thirdly, let l>n. Since C=v 2 + F 2 we notice that C must be positive. We now have we may prove in the same way as before that M is least when h=vr and v s =nlr n ~ 1 and that then M=-~-- r l ~ n +V 2 ^0 by Art. 312. Thus M is always positive J. 71 and the curve has two apsidal distances which can be equal only in a circle. We verify this result by noticing that since an infinite velocity is required to carry the particle to infinity (n being <1, Art. 312), the orbit must have an apocentre or approach an asymptotic circle. The apsidal equation must therefore have two positive roots. 432. It follows from what precedes that the curve defined by the apsidal equation (A) can be without an apse only when n>3. In that case the orbit extends from the centre of force to infinity. We arrive at the same result by noticing that if there is no apse, the velocity must be sufficient to carry the particle to infinity. If 1 > n this condition cannot be satisfied (Art. 312). If n>l this condition requires C to be positive and it is evident that the second form of the apsidal equation has then a positive root. It also follows that there can be an asymptotic circle only when n > 3. For if the orbit be ultimately circular the constant M must be zero, and this cannot happen when n < 3 unless the orbit is circular throughout. See also Art. 447. 433. To find the motion when the orbit has tivo apsidal distances. If a, b be the reciprocals of these distances, the apsidal equation (A) takes the form 278 THEORY OF APSES. [CHAP. VI. where (u) is positive or negative according as n > or < 3. Since the left-hand side is necessarily positive we see that u cannot lie between the limits a and b if $ (u) is positive but must lie between them if (u) is negative. The whole curve must therefore lie outside the annulus defined by the circles u=a, u = b if n>3, and must lie within that annulus if n < 3. It appears that when n>3 the full curve defined by the differential equation (A) contains two distinct branches, either of which can be described by the particle with the given energy \C and the given angular momentum h. These, being separated by the empty annulus, do not intersect, so that when the point of pro- jection is given the particular branch described by the particle is determined. We notice also that this branch has only one apsidal distance though the complete curve has two. When n<3 the path of the particle undulates between the two circles u=a, u=b, touching each alternately and being always concave to the centre of force. 434. Case IV. To find the motion ichen the apsidal distances are equal. The apsidal equation now takes the form The motion as the particle approaches the circle u=a may be found by putting u=a + x and retaining only the lowest powers of x. We then have where m 2 =(/>(a)//i 2 . The particle therefore approaches the limiting circle in an asymptotic path and arrives at the circle only when = oo. Since dOjdt (being ultimately equal to ha 2 ) is finite, the time of describing an infinite number of revo- lutions round the centre of force is infinite. The conditions that the right-hand side of the apsidal equation (A) may have a square factor and be positive are (1) the coefficients of the highest and lowest powers must be positive, and (2) we must have M=0, Art. 431. If n>3, G must be positive, i.e. the velocity at every point must be greater than that from infinity. If n<3 the coefficient of the highest power of u is negative, and there can be no asymptotic circle. (See also Art. 432.) 435. When w>3 and it is known that the path has an apse, we may prove that that apse is a pericentre or apocentre according as the velocity of projection is greater or less than the velocity in a circle at the same distance. Let v be the velocity of the particle, F 2 the velocity in a circle at the same distance r, F a the velocity from infinity ; then (Art. 313) V^ v% ** 2 v*4-Cl l\\ K i- n _l r n-i y z- r n-i> -K!+O .................. (i), .: v 2 -F 2 2 =-(n-3)F 1 2 +C' .............................. (2). If r=rj represent any apsidal distance, we have at that apse v 2 fp=F, F 2 2 /r 1 =.F. At a pericentre the orbit lies outside the circle of radius r lt hence p>r and /. 7> 2 > F 2 2 . At an apocentre the orbit lies inside the circle and v 2 < F 2 2 . It follows by inspection of (2) that at a pericentre both sides of that equation are positive, and, since F a decreases when r increases, both sides must continue to be positive as the particle recedes from the origin. The particle also cannot arrive at a second apse, for this requires the left side to become negative. In the same way at an apocentre the two sides of (2) are negative and must continue to be negative as the particle approaches the origin. The conclusion is that the velocity ART. 438.] CLASSIFICATION OF ORBITS. 279 at any point is greater or less than that in a circle at the same distance according as the path has a pericentre or apocentre. It follows also that the path described cannot have both a pericentre and an apocentre. 436. The following table sums up the possible orbits when F=^u n . n>3, v^V-i {one apsidal distance, path inside the circle. v > F! (two apsidal distances, path inside or outside both circles M negative \ according as v is < or > F 2 . v>V 1 (no apsidal distance, the path extends from the centre of force M positive \ to infinity. v> Fj (an asymptotic circle, approached from within or from without M=0 ( according as v is < or > F 2 . 3>ra>l, VS^F! {one apsidal distance, path outside the circle. v< F! {two apsidal distances, path between the circles. 1 > n, v < Fj {two apsidal distances, path between the circles. Here F 2 is the velocity in a circle at the distance of the point of projection. Ex. When the force F=/j.u n is repulsive show that the path, if not rectilinear, has a pericentre with branches stretching to infinity. 437. The motion in the neighbourhood of the origin is found by retaining the highest powers only of u. We thus have by (A), Art. 429, according as w>3 or <3, where (n-l)B 2 =2/j.. The first alternative gives after integration, supposing the particle to be approaching the origin, where p = |(n-3), q = ^(n + l); showing that the particle (except when n=3) describes a finite angle in a finite time when the radial distance decreases from r=r to zero. The negative sign in the second alternative shows that, when n<3, the particle cannot reach the origin unless h=0, i.e. unless the path is a radius vector. 438. The motion at an infinite distance from the origin is found by retaining the lowest powers only of u. We then have according as n> or <1. The negative sign in the second alternative shows that when n < 1 the curve can have no branches which extend to infinity. When C is positive, i.e. when the velocity v of projection is greater than that from infinity, the first alternative leads to h(u-u )=-ejC, r-r =tJC, showing that when the particle travels from r=r to infinity it describes a finite angle 6 round the origin, and that the time is infinite. The path therefore tends to a rectilinear asymptote whose distance from the origin is - d0/du=hl*/C. If however (7=0, i.e. the velocity v of projection is equal to that from infinity, the lowest existing power of u in the apsidal equation (A) is u 2 or u"" 1 . We 280 THEORY OF APSES. [CHAP. VI. then have /dr\ 2 ,, /dw\ 2 2/u I 15- ) = ft 2 I I = - 7i 2 w 2 , or H j tt"" 1 , \dty \dej n-1 according as n>3 or n<3 but >1. The first alternative shows that (except when h = Q) there are no branches leading to infinity. The second alternative, i.e. n<3, gives, supposing the particle to recede from the origin, rP-r p =-p 0, i J i-r ' 1 =Bqt, where (n-l)B 2 =2/j,, p -\ (3-n), q = %(n+l). These equations show that as the particle proceeds from r=r to infinity it describes a finite angle in an infinite time. The path tends to a rectilinear asymptote at an infinite distance from the origin. 439. Stability of the orbits. Referring to Art. 436 we see that when n>3 the orbit extends to the origin or to infinity except when the particle is approaching an asymptotic circle. The existence of such a circle depends on the equality of the factors of the right-hand side of the apsidal equation, and a slight change in the constants C, h may render the factors unequal or imaginary. In either case the new path will lead the particle either to the centre of force or to infinity. Such orbits may be called unstable. When n<3 and the velocity of projection less than that from infinity, the path is restricted to lie between the two circles u = a, u=b, and the values of a and b depend on the constants C and h. Any slight disturbance will alter the values of these constants, but the orbit will still be restricted to lie between two circles though the radii will not be exactly the same as before. Such orbits may be called stable. 440. Ex. Prove that any small decrease of the angular momentum h or increase of the energy %C will widen the annulus within which the particle moves; that is, will increase the oscillation of the particle on each side of the central line. 441. Apsidal boundaries when F=f(u). When the law of force contains several terms the argument becomes more complicated. Let F=~2,A n u n , then Transposing the terms, the apsidal equation is = (u-a 1 )(u-a 2 ) ...(u-a K )(u), where a a , a 2 , ... are positive quantities arranged in descending order, and (u) contains all the factors which do not vanish between u=0 and u=oo . The factor (u) keeps one sign, viz. that of the highest power of u. Let us divide the plane of motion into annular portions by circles whose common centre is at the centre of force and whose radii are the reciprocals of a l , a 2 , &c. Then since (dujd0) 2 changes sign when u passes any one of these boundaries, it is clear that the curve defined by the differential equation (B) can have branches only in the alternate annuli, the intervening ones being vacant. The space between u = a- L and M=QO being occupied or vacant according as (u) is positive or negative. ART. 444.] APSIDAL BOUNDARIES. 281 If the initial position of the particle lie between any two contiguous circles, the subsequent path is restricted to lie between these circles and touches each alternately. If the initial position lie outside the greatest circle or inside the least, the subsequent path must also lie outside or inside these circles and must therefore extend to infinity or to the centre of force. 442. Next, let some of the factors of the apsidal equation be equal, say where / (u) has been written for the remaining factors. To determine the motion in the neighbourhood of the circle u = a, we write u = a + x and retain only the lowest powers of x. We then have, supposing m>2, /pttl __ -J- h 2 X K < h where B 2 =f(a) (a), and K=^(m-2). The case in which m = 2 is discussed in Art. 434. We see that the circle u=a is asymptotic. The particle arrives at the circle after describing an infinite number of revolutions round the centre of force and at the end of an infinite time. 443. Let us trace the surface of revolution whose abscissa is r and ordinate z = Fr 3 , and let the ordinate z be perpendicular to the plane of motion of the particle. We notice that this surface is independent of the initial conditions and that its form depends solely on the law of force. It is easy to see that the ordinate z corresponding to any value of r represents the square of the angular momentum in a circular orbit described with radius r. It will therefore be useful also to trace the plane whose ordinate is z 7i 2 , where h is the angular momentum of the path described. By describing circles whose radii are the abscissas of the maximum and minimum ordinates of the surface, we may divide the plane of motion into annular portions in which the function z = Fr 3 is alternately increasing or decreas- ing outwards from the centre of force. These we may call the ascending or de- scending portions of the surface. 444. If r represent any apsidal distance, we have at the corresponding apse v 2 lp=F and v = hjr; hence h z =Fpr 2 . At a pericentre the orbit lies outside the circle of radius r, hence p > r, and the angular momentum h of the path must be greater than that in a circle of radius r. In the same way, at an apocentre the orbit lies inside the circle, and the angular momentum h is less than that in a circle of radius r. Referring to the surface z=Fi 3 , we see that a pericentral distance r=OA must have an ordinate A A' less than that of the plane z = h z , and an apocentral distance OB must have an ordinate BB' greater than that of the plane. It immediately follows that if A, B are the pericentre and apocentre of the same path, both the points A', B', cannot lie on the same descending portion of the surface. This con- clusion does not apply if A, B are the pericentre and apocentre of different branches of the complete curve ; (Art. 441). We infer from this result that an annular space on the plane of motion (Art. 443) in which Fr 3 decreases outwards has this element of instability, viz. that a path having both a pericentre and an apocentre cannot be described within the space. If the path have a pericentre the particle will leave the space on its outer margin ; 282 THEORY OF APSES. [CHAP. VI. if an apocentre it will move out of the space on its inner boundary. We see also that when the particle has left the annular space it must proceed to infinity or to the centre of force, unless it come into some other external annular space in which Fr 3 has increased sufficiently to exceed the 7t 2 of its own path or into some internal space in which Fr 3 has become less than W. 445. We may also deduce this result very simply from the radial resolution. Wehave As the particle approaches and passes an apocentre r increases to a maximum and decreases, hence dr\dt changes sign from positive to negative and d^r/dP is negative. In the same way, when the particle passes a pericentre, (Pr/dt* is positive. It immediately follows that at an apocentre Fr*>h? and at a pericentre Fr 3 3, the surface z = Fr 3 has only a descending portion, there can therefore be one and only one asymptotic circle. Also the path described cannot have both an apocentre and a pericentre, though different branches of the same curve may have one an apocentre and another a pericentre. See Arts. 444, 446, 436. If n<3, the surface z=Fr s has only an ascending portion. Hence there cannot be an asymptotic circle, but the path can have both an apocentre and a pericentre. 448. Ex. Discuss the properties of the surface Z = Fr-v 2 , where the velocity v is a known function of r given in Art. 441. Prove that (1) the abscissae of its max-min ordinates are the same as those of the surface z=Fr 3 , so that the ascending and descending portions of each correspond (Art. 443) ; (2) each asymptotic circle must be one of the intersections of the surface with the plane of motion; (3) conversely, if at any intersection we also have z=h?, that intersection is an asymptotic circle. The first result follows from , = -3 -=- . To prove the second and third we dr r 2 dr ART. 450.] LAW OF FORCE IN A CONIC. 283 notice that when Z=0, the velocity is equal to that in a circle; and when z = h 2 , the angular momentum is equal to that in a circle. 440. Examples. Ex. 1. Find the law of force with the lowest index of u such that an orbit can be described having two given asymptotic circles whose radii are the reciprocals of a and b, and find the path. Find also the conditions of projection that the path may be described. Referring to Art. 441 we see that the right-hand side of the apsidal equation (B) must be n (u - a) 2 (u - 6) 2 . We then find F= im* (u -a)(u- b) (2u-a-b) + (j.'u?, and the angular momentum at projection must be Jfj.'. Ex. 2. Let F=fj.u* {(u-a) (Bu-a -6) + CM}, where F is the central force. If the conditions of projection are such that h 2 =fjLC and the velocity v when u=a is v 2 =/j.ca?, show that the path is - = (tanh#) 2 , where c/c 2 =2(a-6). Show also that the curve has two infinite branches tending to the same asymptotic circle 74 = a, with an apse at a distance 1/6. Ex. 3. A particle arrives at an apse distant r from the centre of force with a velocity v equal to that in a circle at the distance r. If the velocity be reversed in direction, will the particle describe the same path in a reverse order or will it travel along the circle? See Art. 419. At such an apse the radius of curvature p of the path must be equal to r. But since-=w + 5 at any apse this requires that d?uld6 2 =Q. The apsidal equation p ao j (B) of Art. 441 must therefore have equal roots, and the apse is at the extremity of a path with an asymptotic circle. The particle therefore can never arrive at such an apse in any finite time (Art. 442). If the particle be projected from a point on the asymptotic circle with the given values of v and h it may be said to describe either orbit, for the deviation of one from the other is indefinitely small at the end of any finite time. Boussinesq, Comptes Eendus, vol. 84, 1877, considers the circular motion to be a singular integral of the differential equation. Korteweg and Greenhill have also discussed this problem. On the law of force by which a conic is described. 450. Newton's theorem*. An orbit is described by a particle about a centre of force C whose law is known: it is required to find the law of force by which the same orbit can be described about another centre of force 0. * Newton's theorem is given in Prop. vn. Cor. 3 of the second section of the first book of the Principia. The application to the motion of a particle in a circle acted on by a force parallel to a fixed direction follows in the next proposition. Sir W. R. Hamilton's paper, giving the law F=/j.rjp 3 , is in the third volume of the Proceedings of the Irish Academy, 1846. Villarceau in the Connaissance des Temps 284 FORCE IN A CONIC. [CHAP. VI. Let F, F' be the forces of attraction iending respectively to C and 0. Let CY, OZ be the perpendiculars on the tangents at any point P;CP = r,OP = r'. Then since 8mCPY=CY/r, we have v* ^ _ = Tf __ p r ' ET = Similarly = If we draw C are similar, and 71 = h - CY' h'V F' CY* parallel to OP, the triangles OPZ and OZ OP CY CG' r'V If then F is given as a function of r, the law of force F' tending to any assumed point is also known, when we have deduced CG as a function of r and r' from the geometrical properties of the curve. Remembering that the area A \ht t we see that the periodic times in which the whole curve is described about G and respectively are inversely as the arbitrary constants h and h'. By choosing these properly we can make the ratio of the periodic times have any ratio we please. We also notice that if the time of describing any arc PQ is known when the central force tends to C, the area PCQ is known. Now the area POQ differs from this by a rectilinear for 1852, using Cartesian coordinates, arrived at two possible laws of force. Afterwards Darboux and Halphen investigated two laws equivalent to these, and proved that there is no other law in which the central force is a function only of the coordinates of its point of application. Their results may be found in vol. 84 of the Comptes Eendus, 1877. The investigations of Darboux were reproduced by him at somewhat greater length in his notes to the Cours de Mecanique by Despeyrous, 1884. There is a third paper by Glaisher in vol. 39 of the Monthly Notices of the Astronomical Society, 1878, who also gives the expression (2w/ x //i) -aft for the periodic time. Darboux uses chiefly polar coordinates, while Halphen employs Cartesian, beginning with the general differential equation of all conies: Glaisher simplifies the arguments by frequently using geometrical methods. There is also a paper by S. Hirayama of Tokyo in Gould's Astronomical Journal, 1889. ART. 453.] HAMILTON'S LAW OF FORCE. 285 figure whose area can therefore be found. Hence the area POQ and therefore the time of describing the same arc PQ when the central force tends to can be found. 451. Suppose the orbit is a conic, then the force tending to the centre G is F=fjur, and h = *J/j,,ab. It immediately follows h' 2 CQ 3 that the force tending to any point is F' = ^ . ^- . If, for Ct 0" IT example, is a focus, it is a known geometrical property of a conic that G lies on the auxiliary circle and that therefore CG = a. We then have F' = ///r' 2 , where h' 2 = pi 2 / a. 452. Parallel forces. To find the force parallel to a given straight line by which a conic can be described. See Art. 323. Let the point be at an infinite distance, then in Newton's formula PO and CG remain parallel to the given straight line throughout the motion. Also the length r' = OP is constant. The required law of force is therefore F' = yu, . CG 3 , where p is some constant. If the direction PO of the force at P cut the diameter con- jugate to CG in N, we have CG.PN=b' 2 , where b' is the semi- diameter parallel to CG. The law of force may therefore also be written F' = A/PN 3 , where A = pb' 6 . To find the constant fi, we notice that in any central orbit, the velocity being v = h/p, the component of the velocity per- pendicular to the radius vector r' is hjr'. In our case when the force acts parallel to a given straight line this component is con- stant. Representing this transverse velocity by V, the Newtonian V 2 formula of Art. 451 becomes F' = y- CG 3 . a 2 b 2 453. Hamilton's formula. A particle describes a conic about a centre of force situated at any point 0. It is required to find the law of force. Taking the same notation as in Newton's theorem, we let F, F' be the forces tending respectively to the centre C and the point 0. Then (Art. 450) F' h' 2 OP CY\ 3 It is a geometrical property of a conic that, if p and or are the perpendiculars drawn from P and the centre C on the polar line 286 FORCE IN A CONIC. [CHAP. VI. of 0, =7Tir It follows that the law of force tending to is BT U JL f"=-q: 2 (} r', where p and r' vary from point to point of the curve and h', a, b and vr are constant. If we write the Hamiltonian expression for the force in the form F'=/jfr'/p 3 ,vfe see that the angular momentum k'^t/fjk'.aKf*', where as before cr is the perpendicular from the centre on the polar line. From this we easily deduce the periodic time in an elliptic orbit. Remembering that the whole area is irab, the formula A = \h't gives as the time of describing a complete ellipse 454. To find the time of describing any portion of the ellipse luith Hamilton's law of force. The coordinates of any point P referred to an origin at the centre of force with axes parallel to the principal diameters are x = a cos $ f, y = b sin (f> g, where is the eccentric angle of P and f, g the coordinates of the centre of force referred to the centre of the curve. Then, if h be the angular momentum, hdt = xdy ydx = (ab fb cos ga sin <) d$, :. ht = ab(j> fb sin +ga cos < ga, where the time is measured from the passage through the apse from which is measured. This, if required, can be expressed in terms of x and y, ht = ab fy +gx ga. This result can be deduced at once from the formula A = %ht, by equating A to the excess of the area of the sector ACP (viz. ^ab(f>) over the sum of the triangles A CO, OOP. * The following is a short analytical proof: Let the conic be Ax z + By 2 =l and let /, g be the coordinates of 0. The polar line of O and the tangent at P are respectively Af!- + Bgri = l, Ax+Byi) = l. The perpendiculars from P and 0, viz. p and OZ, are therefore _l-Afx-Bgy l-Afx-Bgy ~ ** ' ~J(Ax* + By*) The perpendiculars from the centre, viz. w and CY, are found by replacing the numerators by unity. It follows that ART. 457.] THE TWO LAWS OF FORCE. 287 The time of describing an arc of a hyperbola or parabola may be found by proceeding as in Arts. 348, 349. 455. Examples. Ex. 1. Deduce from Hamilton's expression (1) the central force to the focus of a conic, and (2) that to the centre. [In the latter case ta and p are both infinite but their ratio is unity.] Ex. 2. A particle describes an ellipse whose centre is C under the action of a centre of force F situated at a point R in the major axis. If the tangent at P cut the major axis in T, prove that the force F varies as RP . (CTfRT) 3 . 456. The Hamiltonian expression for the force may be put into two different forms. First, we have the form F=/j,r/p s (Art. 453). Secondly. Let OA, OB be two tangents drawn to the conic from the centre of force 0, and let PL = ct, PM=j3, PN=y; these being the three perpendiculars drawn from any point P on the sides of the triangle OAB. By a property of conies we have a/3 = n}lt=re, (2) ,3, ^-r/ From these equations, when the" path is known, we can find the angular momentum h and the two components of velocity ; conversely we can deduce the path when the circumstances of projection are given. These equations follow from the preceding propositions. An independent proof may be obtained by differentiating the equation of the conic twice and writing for d 2 a;/dt 2 , dPy/di 2 their values -fj&jp 3 , -py/p*. We thus obtain three equations which may be transformed into those given above by simple processes. Ex. 2. Prove that the conic described is an ellipse, parabola or hyperbola according as /t (2p - c)/p z -p^ is positive, zero or negative, where p is the distance of the point of projection from the polar line and p' the resolved initial velocity. Ex. 3. If Ax* + 2Cxy + By* + 2Dx + 2Ey + G = Q is the conic described, show that the periodic time in an ellipse is T=-^ < a v j j.n 2 Ex. 4. A particle is acted on by a central force F=H - tending to the origin where r is the radius vector and p the distance from a fixed straight line. Prove that the equation of the path is c/r = sin0+/(0), where c/r=/(0) is the polar equation of the path when the force tending to the origin is F=/=/J.K%, is the angle at the corner of the triangle occupied by the central force, and c is the perpendicular from the centre of force on the polar line AB. The negative sign shows (what is indeed obvious from the figure) that the force is repulsive on the side of the polar line nearest to the centre of force, i.e. /u, is negative. Ex. 2. A particle is projected from the point P with a velocity V and the tangent GPH intersects the given straight lines OA, OB in G and H. Prove that the areal equation of the path, referred to the triangle OGH, is where l=GP, m=HP, A is the area of the triangle, and the radius of curvature /> of the path at P is given by V 2 lp=FsinGPO. It follows that the conic is inscribed or escribed according as F is positive or negative, i.e. according as the force is attractive or repulsive. 464. There are no other laws of force besides OP OP (PL.PM)*' which, being a one- valued function of the coordinates (except as regards sign), are such that a conic will be described with any initial conditions. To prove this consider two conies intersecting in the four points A, B, C, D, which it is convenient to take as real. It follows from Hamilton's theorem that for points on any one conic the force to a given point must be F = pr/p 3 . Hence if the force is to be one-valued, i.e. the same at the same point of space for all paths through that point, we must have at each of the four points A, B, C, D, p s j/j. = p' 3 lf*', where p, p' are the perpendiculars on the two polar lines of 0. We now require the following geometrical theorem*. If two conies intersect in four points A, B, C, D and the ratios of the perpendiculars from each of these points on the polar lines of a point are equal, then either the polar lines are coincident or two common tangents (real or imaginary) can be drawn from 0. In the former case the common law of force for the two conies is given by the first form of F, in the latter case by the second form. * Let the conies be, see Art. 457, ax* + 2yxy + $/ 2 = (Dx + Ey + G) 2 , a'x 2 + Zy'xy + p'y 2 = ( D ' x + E 'y + G ') 2 - Since Dx + Ey + G=Q, D'x + E'y + G'=0 are the polar lines of the origin, we must have at the points of intersection az 2 + 2yxy + Py 2 = m (a'x* + 2y'xy + p'y 2 ). This quadratic equation gives only two values of yfx for the same value of m. The equation cannot therefore be satisfied at four points unless either a, /J, y are respectively proportional to a', /3', y', or the four points lie on two straight lines (say OAB, OCD) passing through 0. In the former case the two conies have a pair of common tangents, in the latter the polar line of is common to the two conies. This common polar line can be constructed by dividing OAB, OCD har- monically in E, F and then joining EF. 192 292 SINGULAR POINTS. [CHAP. VI. Singular Points in Central Orbits. 465. Singular Points. It has already been pointed out in Art. 100 that cases present themselves in our mathematical pro- cesses in which either the force, the velocity or both become infinite. Such infinite quantities do not occur in nature and if we limit ourselves to problems which have a direct application to natural phenomena these are only matters of curiosity. Never- theless it is useful to consider them because they call our attention to peculiarities in the analysis which we might otherwise pass over. The utility of such a discussion is perhaps shown by the differences of opinion which exist regarding the subsequent path of a particle on arriving at a singular point*. 466. Points of infinite Force. Let us suppose that a particle P, describing an orbit about a centre of force 0, arrives at a point B where the tangent passes through the centre of force and therefore coincides with the radius vector. At first sight we might suppose that the particle would move along the straight line BO and proceed in a direct line to the centre of force. But this is not necessarily the case. Supposing B to be at a finite distance from and the curvature to be finite, we see from the equations (Art. 306) tf T,p h d& h - = F^, v = -, r-j- = -, p r p dt r that both v and F are infinite at the point B. We shall also suppose that when the particle passes on the force changes its direction and reduces the velocity again to a finite quantity. At the same time the component of the velocity perpendicular to the radius vector OP, viz. rd6/dt, remains finite however near the particle approaches B. Since there is no force to destroy this transverse velocity, the particle must cross the straight line OB and proceed to describe an arc on the opposite side. * The singularity of the motion when the particle describes a circle about an external centre of force is discussed in Frost's Newton, 1854 and 1863. The same result is independently arrived at by Sylvester in the Phil. Mag. 1866. Other cases are considered by Asaph Hall in the Messenger of Mathematics, 1874. There are several papers also in the Bulletin de la Societe' Math^matique de France, such as Gascheau in voL x. 1881, and Lecornu in vol. xxn. ART. 468.] POINTS OF INFINITE FORCE. 293 467. To simplify the argument, let us suppose that the particle describes a circle about a centre of force external to the circumference. By Art. 321, the circumstances of the motion are given by p- fir tf^V L_ *--- (r 2 -^) 3 ' 2(r 2 -6 2 ) 2 ' 8a 2 ' where b is the length of each of the tangents OB, OB' drawn from to the circle. Describe a second circle having a radius equal to that of the given circle and touching OB at B on the opposite side. If a second particle, properly projected along the second circle, arrive at B simultaneously with the given particle P, but moving in the opposite direction, both the velocity v and the transverse velocity h/r of the two particles will be equal and opposite each to each. If the velocity of the second particle be reversed, Art. 419, it will retrace its former path in a reverse order and this must be also the subsequent path of the particle P. The particle will therefore describe in succession a series of arcs of equal circles. The points of discontinuity at which the particle changes from one circle to the next lie on a circle whose centre is and radius OB = b, and the successive arcs are alter- nately concave and convex to the centre of force. The particle will thus continually move round the centre of force in the same direction in an undulating orbit, but the curve will not be re- entering after one circuit unless the angle BOB' is a submultiple of four right angles. The same arguments will apply to other orbits. When a conic is described about an external centre of force as ex- plained in Art. 462, the particle by a proper projection can be made to describe either of the arcs contained between the tangents drawn from 0. On arriving at the point of contact B, it will cross the tangent and describe an arc of a conic equal to the undescribed arc of the original conic. 468. The particle arrives at the centre of force. When the particle P arrives at the centre of force in a finite time, the determination of the subsequent path presents some other peculiarities. Taking first the Newtonian case in which the particle describes a circle about a centre of force on its circumference, we notice that the transverse velocity hjr (as well as the velocity v) becomes infinite at 0. To understand how the particle can 294 SINGULAR POINTS. [CHAP. VI. have an infinite velocity in a direction perpendicular to what is ultimately a tangent to the path, we observe that, since 2op = r 2 , the transverse velocity ft/r is infinitely less than the tangential velocity hip. When the particle has passed through the origin, the central force, changing its direction, reduces the velocity again to a finite quantity. Meantime the transverse velocity carries the particle across the tangent to the circle. By the same reasoning as before, the subsequent path is an equal circle which touches the original circle at the centre of force. On arriving a second time at the centre of force, the particle returns to the original circle, and so on continually. 469. One peculiarity of this case is that the radius vector of the particle while describing the second circle moves round the centre of force in the opposite direction to that in the first circle. Let P, P' be two positions of the particle, equidistant from the centre of force, just before and just after passing through that point. The transverse velocity being unaltered the moments of the velocity at P and P' taken in the same direction round are equal and opposite. Since this moment is r^ddjdt, it follows that at the point of discontinuity h changes its sign. 470. When the particle moves in an equiangular spiral about a centre of force whose law is the inverse cube, it describes an infinite number of continually decreasing circuits and arrives at the centre of force at the end of a finite time, Art. 319. The subsequent path is another equiangular spiral, Art. 357, having the same angle. To determine its position we consider the conditions of motion at the point of junction. Let us construct a second equiangular spiral obtained from the first by producing each radius vector PO backwards through the origin to an equal distance OP'. If two particles P, P 1 describe these spirals so as to arrive simul- taneously at the centre of force 0, the particles are always in the same straight line with 0, and at equal distances from it. Their radial and transverse velocities are also always equal and opposite each to each. If the velocity of P' be reversed, it will retrace its former path in a reverse order, and this must therefore be the subsequent path of P. On passing the centre of force the particle will recede from the origin and describe the spiral above constructed. We notice also that the radius vector of the particle moves round the centre of force in the opposite direction to that in the first spiral. 471. Limiting Problems. We may sometimes simplify the discussion of some singularities by replacing the dynamical problem by another more general one of which the given problem is a limiting case. But the use of the method requires some discrimination. For example the motion of a particle attracted by a centre of force at a point whose law of force is the inverse cube, may in some cases be regarded as a limit of the motion when the particle is constrained to move in a smooth fixed plane and is attracted by an equal centre of force situated at a point C outside the plane, where CO is perpendicular to the plane and is equal to some small quantity c. The method requires that the limiting motion should be the same whether we put the radius vector r = first and then c = 0, or c = first and then r=0. We know by the principles of the differential calculus that the order in which the variables r and c assume their limiting values is not always a matter of indifference. ART. 473.] KEPLER'S PROBLEM. 295 The component of force in the direction of the radius vector PO is /ur/(r 2 + c 2 ) 2 when the centre of force is at C, and is /u/?- 3 when the centre is at 0. As long as the particle is at a finite distance from the origin, these components are sub- stantially the same, but when the particle is in the immediate neighbourhood of 0, the former is yitr/c 4 and therefore zero when the particle passes through 0, while the latter is infinite. In the former case, though the orbit at a distance from is very nearly an equiangular spiral, it becomes elliptical in the neighbourhood of 0. The force is not sufficient to draw the particle into the centre ; the path has a pericentre and the particle retires again to an infinite distance. See also Art. 322. 472. Examples. Ex. 1. A particle describes one branch of the spiral r6=a under the action of a centre of force in the origin (Art. 358). Show that after passing through the centre of force it will describe another spiral of the same kind, obtained from the first by producing each radius vector backwards through the origin to an equal distance. Since the tangent to the curve is ultimately perpendicular to the radius vector, the two branches of the spiral may have a common tangent, and it might therefore be supposed that the particle would describe the second branch. But this argu- ment requires that the particle should not pass through the origin, so that the radial velocity dr/dt (which is known to be constant) has its direction altered with- out any change in the direction of the force. Ex. 2. A particle describes an epicycloid with the centre of force in the centre of the fixed circle (Art. 322). Supposing the force to become repulsive when the particle enters that circle, show that the path on passing the cusp is a hypocycloid. Kepler's Problem. 473. A particle describes an ellipse about a centre of force in one focus, it is required to express in series the two anomalies and the radius vector in terms of the time. If we require only the first few terms of the series it is convenient to start from the equations ...... (1), where v is the true anomaly. Eliminating r, we have = (1 - f e 2 + &c.) (1 - 2e cos v + 3e 2 cos 2 v - &c.) = 1 2e cos v + f e 2 cos 2v + &c. Remembering that v=0 a, where a is the longitude of the apse nearest to the centre of force, we have nt + = - 2e sin (6 - a) + \& sin 2 ((9 - a) + &c (2), where ?i 2 = u/a 3 . 296 KEPLER'S PROBLEM. [CHAP. vi. We notice that when the planet makes a complete revolution, 6 increases by 2-Tr and that the corresponding increment of t is 27r/n. It follows immediately that n represents the mean angular velocity, the mean being taken with regard to the time ; see Art. 341. The equation (2) may be extended to higher powers of e, and therefore when e is small it may be used to determine the time of describing any angle 0. 474. To find 6 in terms of t, we reverse the series. Writing it in the form 6 = nt + e + 2e sin (0 - a.) - \& sin 2 (0 - a), we have as a first approximation 6 = nt + ; a second approximation gives 6 = nt + + 2e sin (nt + e a). Writing v = nt + e a, a third approximation gives 6 a = v + 2e sin (v + 2e sin v ) f e 2 sin 2v ; /. 6 = nt + e + 2e sin (nt + e - a) + f e 2 sin 2 (nt + e - a). . .(3), and so on, the labour of effecting the successive approximations increasing at each step. As the eccentricity of the earth's orbit is about l/60th it is obvious however that the terms become rapidly evanescent. 475. For the sake of clearness we recapitulate the meaning of the letters in the important equation we have just investigated; 6 is the true longitude of the planet measured from any axis of x in the plane of the orbit ; a is the longitude of the apse nearest the centre of force or origin ; n is the mean angular velocity, the mean being taken with regard to time for one complete revolution; e is a constant whose magnitude depends on the instant from which the time t is measured. To define the epoch e. Let a particle P move round the centre of force in such a manner that its longitude is given by the equation = nt + e. It follows that this planet moves with a uniform angular velocity n and has therefore the same periodic time as the true planet P. When the radius vector of the particle P passes through an apse a and therefore nt + e a is an ART. 477.] THE SERIES FOR THE LONGITUDE. 297 integral multiple of TT. It immediately follows from (2) that 6 = nt + e. Hence the radii vectores of the two planets coincide when the true planet passes through either apse. The definition of P may be shortly summed up thus. Let an imaginary planet move round the centre of force with a uniform angular velocity in the same period as the true planet and let their radii vectores coincide at one apse and therefore at the other. This planet is called the Dynamical Mean Planet. Its longitude at the time t = is the constant e and is called the epoch. 476. To express the mean anomaly and radius vector in terms of the time. Since both the mean and true planets cross the nearer apse at the time given by nt + e = a, the mean anomaly may be repre- sented by m = nt + e. If u be the eccentric anomaly we have by Art. 342, u = m + e sin u. Proceeding as before we have for the three first approximations, u = m, u = m + e sin m, u = m + e sin (m + e sin m) = m + e sinm + ^e 2 sin 2m (4). Again, as in Art. 343, r /y __ ^/y> n ^_ no r*c\o. i/ ~~* \,v d'y \M LCC' \s\Jij Cv = a ae cos (m + e sin m) = a {1 e cos m + ^6 2 (1 cos 2m)} (5). The series for the longitude and radius vector are given here only to the second power of the eccentricity. Laplace in the Mecanique Celeste (page 207) and Delaunay in his Theorie de la Lune (vol. i. pages 19 and 55) give the series up to the sixth power. Stone has continued the expansion up to the seventh power in the Astronomical Notices, 1896 (vol. LVI. page 110). Glaisher has given the expansion of the eccentric anomaly up to the eighth power in the Astronomical Notices, 1877 (vol. xxxvu. page 445). 477. When the eccentricity e is very nearly equal to unity, as in the case of some comets, the formulse giving the relations between t and v must be modified. Starting as before (Art. 473) from the equations dO f n - " 2x dt we put the perihelion distance a (1 - e) = p. 1 * ' ~ = ' (1 + ecosr) 2 ' 298 KEPLER'S PROBLEM. [CHAP. vi. Let (1 - e)l(l + e) =f and put tan i v =x for the sake of brevity ; All+1)- Al + * 2 )<** r V P ~J (i+/* 2 )' When v is given this formula determines the time t measured from perihelion. If / is small the term independent of / is the one requiring the most arithmetical calculation and this can be abbreviated by using the tables constructed for that purpose ; see Art. 349. Conversely when t is given and v is required the same tables give a first approximate value of x. Representing this by tan w, it is usual to expand the correction v - u in terms of in a series ascending in powers of /. For these formulae we refer the reader to Watson's Astronomy and Gauss, Theoria, &c. 478. When the eccentric anomaly is given, the true and mean anomalies and the radius vector are expressed by the equations m=u-esinu, v /! + . u ... ^ = . / - tan- (1), i ^y X e i r=a-ex=a(l -e cosw) .............................. (2). When any one of the other quantities is taken as the independent variable, the corresponding equations can be deduced from these in the form of series. Two methods are used to find the general term of these series. First we may have recourse to Lagrange's theorem, viz., when y = z+x(y), where Li = l . 2 . 3...i, and the S implies summation from i=l to oo . By the second method the general term is expressed by a definite integral which is usually a Bessel's function. 479. Ziagrange's theorem. To express the eccentric anomaly u and the radius vector r in terms of the time. Since u = m + e sin u, we have by Lagrange's theorem e* d { ~ 1 M = m-f S - , (sin mY. Li dm'" 1 The expansion of (sin m)' in cosines of multiple angles when i is even and in sines when i is odd is given in books on trigonometry ; (see Hobson's Trigonometry, Art. 52). The (i - l)th differential is always a series of sines and is easily seen to be . In the same way, expanding cos u by Lagrange's theorem, i.e. writing /(j/) = cos y, we find r e i+1 d'- 1 1 = - e cos u= - e cos m + S -=-r -5-7, (sin m) i+1 , a Li dm'- iv where as before Z implies summation from i=l to CD . 48O. Bessel's functions. We shall now briefly examine the second method by which we express the general term in a definite integral. We know by Fourier's theorem that we can expand any function

(m) is a periodic function having the period 2ir, the expansion will hold for all values of m. If tf> (m) does not change sign with m we may omit the second line of the expansion, while if it does change sign with m, we omit the first line. To find At we use Fourier's rule; multiply both sides by cos m and integrate from m= - TT to +TT. Eemembering that Jcos f MI cos i'm dm = 0, Jcos im sin i'm dm = 0, Jcos 2 im dm = Jsin 2 im dm = TT, we find J$ (m) cos im dm = irA t , J< (m) dm = 2irA . Similarly multiplying by sin im and integrating between the same limits, we find J0 (m) sin im dm = irBt . 481. To expand u-m = esinu in a series of sines of multiples of m. We put .'. irBi = J(u - m) sin im dm, the limits being m= - IT to IT. Integrating by parts, iriBi = - (u- m) cos im + Jcos im (du - dm). The integrated part is zero, for u and m are equal when u = TT. We thus have iriBi = Jcos im du Jcos im dm. The second integral is zero ; substituting for m its value in terms of u, 71-1.6;= Jcos i (u - e sin u) du. This definite integral when taken between the limits and TT is written irJ^ie). We have u = m + SB; sin im, iB i =2J i (ie). 482. The series thus obtained is convergent, for ., [du du . . [d?u . . jn 2 Bi = I d sin im = - sin im- I r n 8iu.imdm. J dm dm J am 2 The integrated part vanishes at both the limits m TT. Also d?u e sin u j ;/i -- r< > dm 2 (1 - e cos M)- J and since e' 2 , &c. also form a convergent series, provided the eccentricity e is less than unity. But if the series is arranged according to the powers of e, the positive and negative terms are added together in a different way. It may then be that the series of coefficients of e, e 2 , &c. are only made convergent by more limited values of e. The condition of convergency is given in Art. 488. 300 KEPLER'S PROBLEM. [CHAP. vi. 484. The expression for B< may be written iriBi = Jcos zu . { 1 - i t a e 2 sin 2 u + &c. } du + Jsin iu . {ie sin u - &c. } du. If we expand sin 2 M, sin 4 , &c. in cosines of multiple angles and remember that Jcos iu cos i'udu = 0, we see that every term in the first integral will be zero in which the power of e is less than i. A similar remark applies to the second integral. Hence the lowest power of e which accompanies the term sin im is e i . 486. To express rja = \-e cos u in a series of cosines of multiples of m, we put .: vA t = - ejcos u cos imdm, where the limits of integration are m= -v to ir. Integrating by parts to change dm into du, we have iriA^ - e cos u sin im - ejsin im sin udu. The integrated part vanishes between the limits. Writing m=M-csinu, the integral becomes viA t = - ejsin i (u - e sin u) sin udu = \e Jcos {(i + 1) u - ie sin u} du - Je Jcos {(i - 1) u - ie sin u} du] .: iA i = e{J i+l (ie) - J^ (ie) } . Similarly 2?r^ = - ejcos u dm= - ejcos M . (1 - e cos u) du. Integrating between limits u = - IT to IT, we find A =$e": .: r/a = 486. That this series is convergent may be proved in the same way as before. We have fd cos M , fd 2 cos u = -e I ; - a cos im = e \ 35 cos tm dm, J dm J dm 4 by integrating by parts. Since u = m + e sin u, we find by differentiation <2 2 cos u e- cos ,,.,,., . m , - , = .^ -- rs . This has obviously a maximum value, say k. Then since dm 2 (1-ecosw) 2 cosim(z) we have (1) to distinguish which root we expand in powers of x, (2) to determine the test of convergency. It is shown that if a contour exist enclosing the complex point z=m, such that at every point of the boundary the 302 KEPLER'S PROBLEM. [CHAP. vi. modulus of x< !^ z ' is less than unity, the given equation has but one root within z m the area and the Lagrangian expansion for that root is convergent *. To apply this theorem to Kepler's problem we put (z) = sin z and let x repre- sent the eccentricity of the ellipse, Art. 478. We measure a real length OA = m from an assumed origin 0, and with A for centre describe a circle with an arbitrary radius r. Representing the complex line OP by z, the Lagrangian series will be convergent if r can be so chosen that the modulus of - is less than unity for all positions of P on the circle. Since (mod) 2 of ( + r,i) = $ (f + it) . z = m + re , where e is the base of Napier's logarithms, we have 2 x sin z _ /.r\ 2 sin (m + re 6i ) s z-m ~\r) e 6i e 1 /x\ 2 = - ( - I {cos (2rtsin 9) -cos(2m + 2rcos 0)} -rsin0 )2 _ Cos2 {m + r cos 0) } . be a continuous one- valued function over the area of a circular contour whose centre is x = a, then Cauchy's theorem asserts that f(x) can be expanded by Taylor's theorem in a convergent series of powers of x - a for all points within the contour ; (see Forsyth's Theory of Functions, Art. 26). When z = m + x

(z) = (regarded as an equation to find z when x is given) has two roots. To determine whether this is so, we use another theorem of Cauchy's (see Burnside and Panton, Theory of Equations). We measure OA = m from the assumed origin O and with A for centre describe a circle of radius r. Let a point P describe this circle once, then by Cauchy's theorem if log.F(z) is increased by 2niri, the equation F (z) has n roots within the contour. Hermite writes log F (z) = log (z - m) + log ( 1 - ?&} . \ z mj (1) The equation z-m=0 has but one root and that root lies within the contour, hence as P moves round, log (z - m) is increased by 2iri. (2) If the modulus of u= ' is less than unity at all points of the circle, * Z -~ ffl the value of log(l-u), (being the same on departing from and arriving again at any point of the contour) increases by zero when P moves round the contour. It follows that log F(z) increases by 2iri when P makes one circuit, that is the equation z = m + xd>(z) has but one root within the contour if the modulus of --- z m is less than unity at all points on the circumference. ART. 489.] CONVERGENCY OF THE SERIES. 303 Now, putting e rsine +e~ rsinfl =w + - , we see that the first term of this ex- v pression continually increases from v = l, or 0=0 to v = cc, and is therefore greatest when 6=.\ir. The least value of the second term is zero. The modulus ix is therefore less than - - (e r + e~ r ). The Lagrangian series is therefore convergent 2 T for all values of the eccentricity x less than 2rl(e r + e~ r ). To find the maximum value of this function of r, we equate its differential coefficient to zero. This gives V= e r (r - 1) - e~ r (r + 1) = 0. Since dVjdr is positive for all values of r this equation has hut one positive root, and this root lies between 1 and 2. Using the value of e r given by the equation F=0, we find that the maximum value of the eccentricity is v /(? %2 -l), which reduces to '66. CHAPTER VII. MOTION IN THREE DIMENSIONS. The four elementary resolutions and moving axes. 490. The Cartesian equations. The equations of motion of a particle in three dimensions may be written in a variety of forms all of which are much used. The Cartesian forms of these equations are ftx d* d?z where x, y, z are the coordinates of the particle and X, F, Z the components of the accelerating forces on the particle. These equations are commonly used with rectangular axes, but it is obvious that they hold for oblique axes also, provided X, Y, Z are obtained by oblique resolution. 491. The Cylindrical equations. From these we may deduce the cylindrical or semi-polar forms of the equations. Let the coordinates of the particle P be p, <, z, where p, (f> are the polar coordinates in the plane of xy of the projection N of the particle P on that plane, and z = PN. By referring to Art. 35, we see that the first two of the equations (A) change by resolu- tion into the first two of the following equations (B), while the third remains unaltered. We have where P, Q are the components of the accelerating forces respec- tively along and perpendicular to the radius vector p. ART. 493.] ELEMENTARY RESOLUTIONS. 305 492. Principle of angular momentum. Since the moments of the components P and Z about the axis of z are zero, the moment of the whole acceleration about the axis of z is equal to Qp. In the same way the moment of the velocity about Oz is equal to the moment of its component perpendicular to the plane POz, and this is p 2 d/dt. Introducing the mass m of the particle as a factor, the second of the equations (B) may be written in the form d /moment of \ _ /moment of dt \momentumy \ forces The moments may be taken about any straight line which is fixed in space, such a line being here represented by the axis of z. The moment of the momentum is also called the angular momentum of the particle (Arts. 79, 260). When the forces have no moment about a fixed straight line the angular momentum about that straight line is constant throughout the motion. 493. The polar equations. We may immediately deduce from the semi-polar form (B), ike polar equations (C). Let r, 6, be the polar co- ordinates of P, where r = OP, 6 is the angle OP makes with the axis Oz, and

increases. If JR, 8, T are the components of the acceleration of the particle respectively in the directions of (1) the radius vector OP, (2) the perpendicular to OP in the plane of zOP, and (3) the per- pendicular to the plane zOP, taken positively when they act in the directions in which r, 0, $ are respectively increasing, we have d*r (d6\* /dd>\* . de- r (dt)-p(dt) 8ine = l rIt( r ^)- p (^J C Se = S ^ ' P o i We notice that p = r sin 6. 404. Ex. If v be the velocity, show that the radial acceleration is 495. Reducing a plane to rest. Referring to the semi- polar equations (B), we notice that if we transfer the term p (d^jdff to the right-hand side of the first equation and include it among the impressed accelerating forces, the first and third equations become the same as the Cartesian equations of motion of a particle moving in a fixed plane zOP (Art. 31), while the second equation determines the motion perpendicular to that plane. We may therefore replace the first and third resolutions by any of the other forms which have been proved to be equivalent to them. Art. 38. For example, if we replace these two resolutions by their polar forms (Art. 35) we obtain at once the equations (C). The process of regarding p (d/dt) 2 as an impressed accelerating force acting at P and tending from the axis of z is sometimes called reducing the plane zOP to rest. See Arts. 197, 257. ART. 497.] SOLUTION OF THE EQUATIONS. 307 496. The intrinsic equations. To find the intrinsic equa- tions of motion, due to the tangential and normal resolutions. Let P, P' be the positions of the particle at the times t, t + dt ; v, v + dv the velocities in those positions, dty the angle between the tangents. In the time dt, the component of velocity along the tangent at P has increased from v to (v + dv) cos dty. Writing unity for cos dty, the acceleration along the tangent, i.e. the rate of increase of the velocity, is dvjdt. The component of velocity along the radius of curvature at P has increased from zero to (v + dv) sin d-fy, which in the limit is vdty. The acceleration along the radius of curvature is therefore vd^r/dt, or which is the same thing v 2 /p. The osculating plane by definition contains two consecutive tangents. The component of velocity perpendicular to that plane is zero and remains zero. The acceleration along the perpendicular to the osculating plane, i.e. the binormal, is therefore zero. If F and G are the component accelerations measured posi- tively in the directions of the arc s, the radius of curvature p and H the component perpendicular to the osculating plane, the equations of motion are 497. Show that the solution of the equations of motion of a particle in polar coordinates can be reduced to integrations when the work function has the form ivhere / a (r), / 2 (6) and f 3 ($) are arbitrary functions. The third of the equations (C) gives, with this form of U, the mass being unity, 1 <* / ) 1 rsind dt\ ' dt J rsin 6d r 2 sin 2 ^' (1). The second of the equations (C) gives 1 d I de\ . /dcA 2 df 2 (6) 1 Substituting for dfjdt, we obtain 202 308 MOTION IN THREE DIMENSIONS. [CHAP. VII. The equation of vis viva is After substituting from (1) and (2) this becomes > These are the first integrals of the equations of motion. Since the variables are separable in all the equations, they can be reduced to integrations. Substitut- ing for dt from (4) in (2), that equation gives in terms of r. Substituting again in (1), we find in terms of r. Lastly (4) determines t in terms of r. 498. Moving axes. To find the equations of motion of a particle referred to rectangular axes which move about the origin in an arbitrary manner. Let us suppose that the moving axes Ox, Oy, Oz are turning round some instantaneous axis 01 with an angular velocity which we may call 0. Let lt 2 , 3 be the components of about the instant- aneous positions of Ox, Oy, Oz. Then in the figure l represents the rate at which any point in the circular arc yOz is moving along that arc, 2 is the rate at which any point of the circular arc zOx is moving along the arc, and so on. Let us represent by the symbol V any directed quantity or vector such as a force, a velocity, or an acceleration. Let V x , V y , V z be its components with regard to the moving axes. Let Of, Ov, 0% be three rectangular axes fixed in space and let Vi, V 2 , V s be the components of the same vector along these axes. Let a, /3, 7 be the angles the axis makes with Ox, Oy, Oz. Then V s = V x cos a + V y cos (3 + V z cos 7, dV X dV da. Let the arbitrary axis of coincide with Oz at the time t, i.e. let the moving axis be passing through the fixed axis. Then a = TT, AET. 500.] MOVING AXES. 309 yg = ^TT, ry = 0. HcnCC dV 3 _dV z v da. v d@ ~dT = ~~ dt ' x dt~ y ~dt' Now da/dt is the angular rate at which the axis Ox is separating from a fixed line 0% momentarily coincident with Oz, hence da/dt = 2 . Similarly d/3/dt = 6^. Substituting dV 3 _ dV z _y a . y Q dt " dt Vx ^ + Vy 1 ' Similarly ^ = * - V y 6 3 + V Z 2 , fJV JV 113 _ av y Y o , Y Q dt " dt Vz ^ + Vx ^' When the moving axes momentarily coincide with the fixed axes, the components of the vector V are equal, each to each, i.e. V X =V 1 , V y =V z , V Z =V 3 . As the moving axes pass on, this equality ceases to exist. The rates of increase of the components relatively to the moving axes are dV x /dt, dV y jdt, dV z /dt; while the rates of increase relative to the fixed axes are dV^dt, dV 2 /dt, dV 3 /dt. The relations which exist between these rates of increase are given by the equations just investigated. 499. If the vector V is the radius vector of a moving point P, the components V x , V y , V z are the Cartesian coordinates of P, and the rates of increase are the component velocities. If the vector V is the velocity of P, the rates of increase are the com- ponent accelerations. Let then x, y, z be the coordinates of a point P ; u, v, w the components of its velocity in space ; X,Y,Z the components of its accelerations. Then == dt~ zBl + x0 *' Y= di dz n n dw 5OO. If the origin of coordinates is also in motion these equations require some slight modification. Let p, q, r be the resolved parts of the velocity of the origin in the directions of the axes. In order that u, v, w may represent the 310 MOTION IN THREE DIMENSIONS. [CHAP. VII. resolved velocities of the particle P in space (i.e. referred to an origin fixed in space), we must add p, q, r respectively to the expressions given for u, v, w in Art. 499. These additions having been made, u, v, w represent the component space velocities of P, and the expressions for the space accelerations X, Y, Z are the same as those given above. See Art. 227. The theory of moving axes is more fully given in the author's treatise on Rigid Dynamics. The demonstration here given of the fundamental theorem is founded on a method used by Prof. Slesser in the Quarterly Journal, 1858. Another simple proof is given in the chapter on moving axes at the beginning of vol. u. of the treatise just referred to. 5O1. Moving field of force. When the field of force is fixed relatively to axes moving about a fixed origin we may obtain the equation corresponding to that of vis viva. If T be the semi vis viva, we know that dTjdt is equal to the sum of the virtual moments of the forces divided by dt. Hence, the mass being unity, =Xu+Yv + Zw dt If A l , A 2 , A 3 are the angular momenta about the axes (Art. 492), A^yw-zv, A a =zu-xw, A 3 =xv-yu, and, taking moments about the axes, dA 1 ldt=yZ-zY, dA 2 ldt=zX-xZ, dA 3 ldt=xY-yX. The equation of vis viva therefore becomes ^_/ ^i_ fl ^3_/i ^s_^ dt x dt 2 dt 3 dt ~ dt ' where U is a function of the coordinates x, y, z only. If 6 lt 6 2 , 6 3 are constant, this, when integrated, reduces to the equation of Art. 256. 5O2. Ex. 1. Show how to deduce the polar forms (C), Art. 493, from the equations for moving axes. Let the moving axes be represented by 0, Oij, Of. Let the axis of move so as always to coincide with the radius vector OP ; let Or) be always perpendicular to the plane zOP. The angular velocity d&jdt of the radius vector may therefore be represented by 6 2 =d&ldt about Orj. The plane zOP has an angular velocity d&dt about Oz, and this may be resolved into & 1 = cosddjdt. Also the coordinates of P are r, 17 = 0, f=0. It immediately follows from the equations of moving axes that u = drldt, v = 6 3 r, w= - 0%r. Substituting these in the expressions for X, Y, Z we obtain the com- ponents of acceleration already written at length in Art. 493. Ex. 2. If (a^iVi), (a-^p 2 y^), (os/VXs) are ^ e direction cosines of a system of orthogonal axes moving about the origin, prove that where 3 is positive when the rotation is from the first axis to the second. To prove this we notice that 6 3 measures the rate at which the axis of y is separating from the position of the axis of x at the time t. Hence - 6 3 dt is the cosine of the angle the new axis of y makes with the old axis of x. ART. 504.] LAGRANGE'S EQUATIONS. 311 Ex. 3. A particle is describing an orbit about a centre of force which varies as any function of the distance, and is acted on by a disturbing force which is always perpendicular to the plane of the instantaneous orbit and is inversely pro- portional to the distance of the particle from the centre of force. Prove that the plane of the instantaneous orbit revolves uniformly round its instantaneous axis. [Math. Tripos, I860.] Lagrange's Equations. 503. Lagrange has given a general theorem by which we can form the equations of motion of a particle, or of a system of particles, in any kind of coordinates*. The expression " coordinate " is here used in a generalized sense. Any quantities are called the coordinates of a particle, or of a system of particles, which determine the position of that particle or system in space. In using Lagrange's equations, it will be found convenient to represent by some special symbols, such as accents, all total differential coefficients with regard to the time; thus of, x" represent respectively dx/dt and d 2 x/dt 2 . 504. Lemma. Let L be a function of any variables x, y, &c., their velocities x', y', <&c., and the time t. If we express x, y, &c. as functions of some independent variables 6, (f>, &c. and the time t, say x=f(t,d,$,&c.\ y = F(t,e,$,&c.\ z = &c (1), then will d dL dL /d dL dL\ dx fd dL dL\ dy / | I / . I / _1_ fan dtdd' dd~ \dtdx dx)d0^ \dtdy' dy) dd^ Representing partial differential coefficients by suffixes, we have by differentiating (1), x'=f t +f e e'+f 4> ' + &c (2). Since 6 enters into the expression L through both x, y, &c. and their velocities x', y', &c. while 6' enters only through x', y', &c., * The Lagrangian equations are of the greatest importance in the higher dynamics and are usually studied as a part of Rigid Dynamics. We give here only such theorems as may be of use in the rest of this treatise. The application of the method to impulses, to the cases in which the geometrical equations contain the differential coefficients of the coordinates, the use of indeterminate multipliers, the Hamiltonian function, &c., are regarded as a part of the higher dynamics. 312 LAGRANGE'S EQUATIONS. [CHAP. vn. we have the partial differential coefficients dL _ dL dx dL dx' .,. dd dx dd dx' dO dL _ dL dx' - , . . 'jfu j 7 j/j7 I OCC .................. . ......... \-T ), dd dx dd where in each case the &c. represents the corresponding terms for y, z, &c. dec dec By differentiating (2) we see that ,/#/ = / = j# Hence d dL _ dtdff dL fd dL dL\ dx __ __ I __ __ _ I _ _1_ /vp d6~ \dtdx' dxjde^ By differentiating f e totally with regard to t, we have -rto ...................... (6). The right-hand side of this equation is seen by differentiating (2) dx' to be equal to -^ . It therefore follows that all the terms in the Cvv second line of (5) vanish. The lemma has therefore been proved. 505. By using this lemma we may deduce Lagrange's equations from the Cartesian equations of motion. For the sake of generality, let there be any number of particles, of any masses m 1 , m 2 , &c., and let their coordinates be (x^y^z-^, (x 2 , y 2 , z 2 ), &c. Let T be the semi vis viva of the system, then 2Z 7 = 2m(tf /2 + 2 /' 2 + / 2 ) .................. (7). Let U be the work function of the impressed forces, then U is a function of the coordinates only. Let R x , R y , R z be the com- ponents of any forces of constraint which act on the typical particle ra. We have as many Cartesian equations of motion of the form dU p dU dU ^ ~^o =Rx ' my ~W = v> ^ ~~dz =R as there are particles. The particles may be free or connected together, or constrained by curves and surfaces, but after using all the given geometrical relations, the position of the system may be made to depend on some independent auxiliary quantities or coordinates. Let these ART. 506.] THE LAGEANGIAN FUNCTION. 313 be 0, , &c. ; then writing L = T + U, we have for the particle m, d dL dL_d '__D dt dx' dx dt dx with similar forms for y and z. Hence using the lemma, d dL dL v / p dx dy dz where 2 implies summation for all the particles. The right-hand side of this equation (after multiplication by BO) is the virtual moment of the forces of constraint for a geometrical displacement &0. This by the principle of virtual work is known to be zero. Since the variations of the coordinates x, y, &c. due to the displacement 66 are deduced from the partial differential coefficients dx/dO, dyjdf), &c., t not varying, the displacement given to the system is one consistent with the geometrical relations as they exist at the instant of time t. Taking the various kinds of forces of constraint it has been proved in Art. 248 that the virtual moment of each for such a displacement is zero. Consider the case of a particle constrained to rest on a curve or surface, the virtual moment is zero for any displacement tangential to the instantaneous position of the curve or surface. The restriction that the geometrical equations must not contain the time explicitly is not necessary in Lagrange's equations. If some of the particles are connected together so as to form a rigid body, the mutual actions and reactions of the molecules are equal. Their virtual moments destroy each other because each pair of particles remain at a constant distance from each other. The Lagrangian equations may therefore be applied to rigid bodies. 506. The Lagrangian equations of motion are therefore d_dL_dL_ d L dI^_dL dtde' de~ dtd$ d The function L=T+U and is therefore the sum of the kinetic energy and the work function. If we use the function V to repre- sent the potential energy, we have, by definition, U+V equal to a constant. We then put L= T V, so that L is the difference between the kinetic and potential energies. Substituting these values for L, and remembering that U and V are functions of the coordinates and not of their velocities, we may also write the Lagrangian equations in the two typical forms d L dT^_dT^_dU ddT^_dT dV _ dt dff dd ~ dd ' dt dff d0 + ~d0 ~ where 6 stands for any one of the coordinates. It should be 314 LAGRANGE'S EQUATIONS. [CHAP. vu. noticed that in these equations, all the differential coefficients are partial, except those with regard to t. The function L is sometimes called the Lagrangian function. We see that when once it has been found, all the dynamical equations, free from all unknown reactions, can be deduced by simple differentiation. 5O7. Virtual moment of the effective forces. If we substitute for L in the lemma of Art. 504 the value of T given by (7) we have d_ did0 i ~~dl)~ ( IX d0 my d8 ] The right-hand side (after multiplication by 50) is the sum of the virtual moments of the effective forces mx", my", &o. It follows therefore that the Lagrangian expression on the left-hand side (after multiplication by 50) represents tlie sum of the virtual moments of the effective forces, when expressed in terms of the generalized coordinates 0, 0, &c. In the same way writing T for the arbitrary function L in (4), we have by (7) The left-hand side (after multiplication by 50) therefore represents the sum of the virtual moments of the momenta of the several particles of the system for the displace- ment 50. It is often called the generalized 6 component of the momentum. 5O8. Meaning of the lemma. The fundamental equation represented by the lemma has been deduced from the principles of the differential calculus without reference to any mechanical theorem. Analytically, it expresses the fact that the Lagrangian operator symbolized by _d__d_d_ ~d0~dide' follows the same law as the differential coefficient d/dO, i.e. which may also be written where 50, Sx, dy, &c. are any small arbitrary variations consistent with the geometrical relations which hold at the time t. If we interpret the lemma dynamically (Art. 506), the equation asserts that the sum of the virtual moments of the effective and impressed forces for a displacement 50 has the same value whatever changes are made in the coordinates. 5O9. Working rule. When we solve a dynamical problem we begin by writing down the equation of vis viva, viz. T=U+C. It appears that when we have done this, Lagrange's method enables us to write down all the equations of motion of the second order by performing certain differentiations on the quantities on each side of the equation (Art. 506). ART. 511.] THE FUNCTION T. 315 We shall presently show that before performing these differentiations, we may remove certain factors from one side to the other by making a change in the independent variable t ; Art. 524. 510. The ftinction T. We have assumed that the Cartesian coordinates x, y, z of every particle of the system can be expressed in terms of the generalized coordinates 6, <, &c. by means of equations of the form *=/(*, 0,4>,&c.) (1); these equations may contain t, but not 6', ', &c. (Art. 504). In choosing therefore the Lagrangian coordinates, we see that they must be such that the Cartesian coordinates of every particle could be expressed if required in terms of them by means of equations which may contain the time, but do not contain differential co- efficients with regard to the time. Differentiating the geometrical equations (1) as in Art. 504 *'=ft + fiP +/*' + &c., 2/' = &c (2), and substituting in the expression for the vis viva 22 T = 2m(V 2 + 2/' 2 + / 2 ) (7), given in Art. 505, we observe that 2T takes the form 2T = A n 0'* + 2A l2 6'' +...+ BJ' + B 2 ' + ... + C, where the coefficients A n , &c., B 1} B 2 , &c., and C are functions of t, 9, , &c. In most dynamical problems, the geometrical equations do not contain the time explicitly, i.e. t does not enter into the equations (1) except implicitly through 9, <, &c. The term/ t will therefore be absent from the equation (2), Art. 504. Hence x', y', z are homogeneous functions of 9', ', &c. of the first order. When substituted in (7), we find that 2T is a homogeneous function of 6', (j>', &c. of the second order, viz. 2T =A U 0'* + 2 A 12 6' $'+..., where A u , A 12 , &c. are functions of the coordinates 9, <, &c. but not of t. 511. Examples of Lagrange's equations. Ex. Two particles, of masses M, m, are connected by a light rod, of length I. The first A is constrained to move along a smooth fixed horizontal wire, while the other B is free to oscillate in the vertical plane under the action of gravity. It is required to find the motion. To fix the positions of both the particles in space, we require two coordinates, say, the distance of the point A from some origin and the inclination of AB 316 LAGRANGE'S EQUATIONS. [CHAP. vn. to the vertical. The Cartesian coordinates of B are then x = + 1 sin 6 and y = I cos 0. The serai vis viva and work functions are then ........................ (1), U=mglcoa0 ............................................................ (2). Substituting in the Lagrangian equations, d_dT_'. Since these three directions are orthogonal, we have Substituting in the Lagrangian equation d L dT_, we obtain j (mr') - m (rO' z + r sin 2 0$'*) = ^ , (mr 3 *?') - mr 2 sin cos W 2 = which evidently reduce to the forms given in Art. 493. ART. 512.] VARIOUS EXAMPLES. 317 Ex. 2. To deduce the accelerations for moving axes from Lagrange's equations when the component velocities are known. We have given by Art. 499, u=x' - y9 3 + z0 2 , v=y' - Also the mass of the particle being unity. Since x' enters into the expression for T only through u, while x enters through both v and iv, we have dT_dTdu_ dT__dT_dv_ dTdw_ dx' ~ du dx' dx ~ dv dx dw dx~ 3 2 ' d dT dT dU The Lagrangmn equation __-_ = _ becomes - v0 s + w9 2 = X. dt dx' dx dx du lit Ex. 3. To deduce the equation of vis viva from Lagrange's equations. Multiplying the Lagrangian equations d dT dT dU d dT dT dU _ - . fyQ dt d6' de dd ' dt d<(> d d' by 6', $', &c. respectively and adding the results, we have Id dT\ dT\ dT_ dU where 2 implies summation for all the coordinates. If the geometrical equations do not contain the time explicitly, T is a homo- geneous function of 6', d>', &c., Art. 510, and byEuler's theorem 20' , = 2T. Also da since T and U are not functions of t, Substituting in the expression given above, we have 3 dT_dT = dU m . dt dt dt ' where C is an arbitrary constant, usually called the constant of vis viva. Ex. 4. The position of a moving point is determined by the radii l/, I/T;, 1/f of the three spheres which pass through it and touch three fixed rectangular coordinate planes at the origin. Find the component velocities u, v, w of the point in the directions of the outward normals of the spheres, and prove that the com- ponent accelerations in the same directions are dujdt + v^u-^v) - w (%w - fu), and two similar expressions. [Coll. Ex. 1896.] Writing D = ^ 2 + 7j 2 + f 2 we deduce from the equations of the spheres that x=2ID, &c. Noticing that the spheres are orthogonal, we find, by resolving the velocities x', y', z' along them, u= -#'/ v= ~yn'hi w ~ z t'lt' Hence T= 4 (u 2 + tf + w z ) = 2 (f ' 2 + T?' 2 + ' 2 )/Z> 2 . Also the acceleration along the axis is dU/udt or -^DdUjd^. Substituting in the Lagrangian formula -^ = T -^ - -TJ- , we obtain the required result. It may QJ(^ at a% t* also be deduced from the formulae of Arts. 499, 502, Ex. 2. 318 LAQRANGE'S EQUATIONS. [CHAP. vu. 513. To apply the Lagrangian equations to determine the small oscillations of a system of particles about a position of equilibrium, when the geometrical equations do not contain the time explicitly. Let the system have n coordinates and let these be 6, , &c. Let their values in the position of equilibrium be a, yS, &c., and at any time t, let = a. + x, = P + y, &c. The vis viva being a homogeneous function of 6', <', &c. (Art. 510), we have 2T = P0' 2 + 2Q0'' + R' 2 + &c., where P, Q, &c. are functions of 0, , &c. and when expanded takes the form 2 U = 2 U + 2B& + 2 J5# + &c. + B u x* + ZB^y + &c. . . .(2). We assume that these expansions are possible. Since the system is in equilibrium in the position defined by x = 0, y 0, &c., we have by the principle of virtual work, ^=0, ^ = 0,&c.; .-. , = 0, B 2 = 0, &c (3). If the position of equilibrium is not known beforehand, the values of o, /3, &c. may be obtained by solving the n equations (3). To find the equations of motion we substitute in the n Lagrangian equations typified by d dT dT dU ,_ ^ _ _ r ,_,.. ( 4 ) dt dx' dx dx ' Since the expansion for T does not contain the coordinates x, y, &c., we have dTjdx = 0, dT/dy = 0, &c. The equation (4) there- fore becomes A u x" + A l2 y" + A 13 z" -f &c. = B n x + B^ + B 13 z + &c.] " + &c. = B^ + B^y + B.^z + &c. I . . .(5). &c. = &c. ART. 516.] SMALL OSCILLATIONS. 319 To solve the equations (5) we follow the rules given in Art. 292. Let any principal oscillation be represented by x = G sin (pt + a), y = H sin (pt + a), &c .......... (6), where G, H, &c. are constants. We find by an easy substitution ... = ......... (7). &c. = OJ Eliminating the ratios G : H : &c., the n values of p* are given by the Lagrangian equation &c. &c. &c. &c. &c. = (8). 514. It is shown in the higher dynamics that, because the vis viva 2T is necessarily positive for all real values of x' , y r , &c., the values of p 2 given by this determinantal equation are real. If all the roots are positive the values of p are real, and the system of particles then oscillates about the position of equi- librium. If any or all the values of p 2 are negative, some or all the values of p take the form + q V 1. The corresponding trigonometrical terms in (6) become exponential and the system does not oscillate. See Art. 120. 515. If a value of p 2 is zero p has two equal zero values, and the corresponding term in (6) takes the form A + Bt. In such a case the coordinate may become large and the system will then depart so far from the position of equilibrium that it will be necessary to take account of the small terms in (1) and (2) of higher orders than the second. 516. Rule. When applying Lagrange's equations to any special case of oscillation about a position of equilibrium we begin by writing down the expressions for the vis viva and work function for the system in its displaced position, and express these in the quadratic forms (1) and (2) (Art. 513). If the whole motion is required we follow in each special case the process described in the general investigation. But if, as usually happens, only the periods are required, we omit the intervening steps and deduce the determinant (8) immediately from the expansions (1), (2). 320 LAGRANGE'S EQUATIONS. [CHAP. vn. To help the memory, we notice that, if we drop the accents in the expression for T, the determinant (8) is the discriminant of the quadric Tp* + U. 517. To apply Lagrange's equations to determine the initial motion of a system. The method has been already explained in Art. 282. The Lagrangian equations give the values of 6" ', ", &c. in the initial position without introducing the unknown reactions. Differen- tiating the Lagrangian equations of Art. 506 we obtain 6"', <}>"', &c., and any higher differential coefficients. If x, y, z are the Cartesian coordinates of any point P of the system, we have by Art. 510, and therefore by differentiation the initial values of x', x", &c., i/, y", &c., /, &c. may be found. The initial radius of curvature follows from the formulae of the differential calculus, Art. 280. 518. Let, for example, the initial accelerations be required when the system starts from rest The initial position being 6 = a, = /3, &c. we put, as in Art. 513, = a + x, = fi + y, &c. Since the system starts from rest, the velocities x', y', &c. are small and we can make the expansions (1) and (2) as before. Since the initial position is not one of equilibrium, we no longer have .fij = 0, 5 2 = 0, &c. Retaining only the lowest powers of x, y, &c. which occur in the equations of motion, we have &c. = &c. These determine the initial accelerations of the coordinates and therefore the component accelerations of every point of the system. 519. Ex. 1. Let us apply the Lagrangian equations to find the small oscilla- tions of the two particles described in Art. 511. The quantities , 6 represent the deviations of the rod from its position of equilibrium. The vis viva and work function expressed in quadratic forms are The determinant is the discriminant of Tp 2 + U % (M + m) p 2 ^ 2 + mlp^e + %ml (lp* - g) 6- ; , =- mlp*, ml(lp*-g) ' ART. 521.] EXAMPLES. 321 One principal motion is given by , B=H sin (pt + a). The other is determined by p- = 0; this implies that one coordinate takes the form A + Bt. It is evident that the rod could be so projected along the horizontal wire that has this form while = 0. The student should apply Lagrange's equations to the problems on small oscil- lations and initial motions already considered in the chapter on motion in two dimensions. He will thus be able to form a comparison of the advantages of the different methods. Ex. 2. Three uniform rods AB, BC, CD have lengths 2a, 2b, 2a and masses HI, m', 7. They are hinged together at B and C, and at A, D are small smooth rings which are free to move along a fixed fine horizontal bar. The rods hang in equilibrium, forming with the bar a vertical rectangle. When a slight symmetrical displacement is given, the period of a small oscillation is given by 4mop 2 = 3# (m + m') . Find also the periods when the displacement is unsymmetrical. [Coll. Ex. 1897.] Ex. 3. Two equal strings AC, BC have their ends at the fixed points A, B, on the same horizontal line, and at C a heavy particle is attached. From C a string CD hangs down with a second heavy particle at D. Find the periods of the three small oscillations. [The two periods of the oscillations perpendicular to the verti- cal plane through A and B are given in Art. 300, Ex. 1.] 520. Solution of Lagrange's Equations. Our success in obtaining the first integrals of the Lagrangian equations will greatly depend on the choice of coordinates. When the position of the system is determined by only one coordinate, the equation of vis viva is the first integral, and this is sufficient to determine the motion. When there are two or more coordinates, integrals can be found only in special cases. The general problem of the solution of the Lagrangian equations is too great a subject to be attempted here. It is sufficient to state a few elementary rules which may assist the student. 521. We should, if possible, so choose the coordinates that some one of them is absent from the expression for the work function U. For example, if there be any direction such that the component of the impressed forces is zero throughout the motion, we should take the axis of z in that direction and let z be one of the co- ordinates. Again if the moment of the forces about some straight line fixed in space, say Oz, is always zero, the angle which the plane POz makes with xOz will be a suitable coordinate. In that case d U/d = and U is independent of . These, or similar, R. D. 21 322 LAGRANGE'S EQUATIONS. [CHAP. vn. mechanical considerations generally enable us to make a proper choice. Let be the coordinate absent from the work function, then if 6 is also absent from the expression for T, though the differential coefficient & is present, the Lagrangian equation d dT dT dU . dT where A is the constant of integration. Thus a first integral, different from that of vis viva, has been found. 522. liiouville's integral. Liouville has given an integral of Lagrange's equations which has the advantage of great simplicity when it can be applied. This may be found in vol. xi. of his Journal, 1846 ; the following is a slight modi- fication of his method. Let us suppose that the vis viva has the form '2T=M(P6'*+Q' 2 + Rt'* + &c.) ........................... (1), where the products d', '$', &c. are absent. The method requires that the co- efficient P should be a function of 6 only, while Q, R, d-c., are not functions of 6. We notice that M may be a function of all or any of the coordinates, and Q, R, &c. functions of any except 6. It is also necessary that the impressed forces should be such that the work function U has the form M(U+C) = F l (e) + F(f,-if< Using the equation of vis viva, this takes the form iwj-ortif*f-$ Substituting on the right-hand side from (2) and multiplying by ', we have Since F l (6) is a function of and not of any of the other coordinates, this gives by an easy integration Returning to the coordinate 6, we have the integral (4). When the initial conditions are given, the value of C can be found by introduc- ing these conditions into the equation of vis viva. If a solution is required for all ART. 524.] LIOUVILLE'S INTEGRAL. 323 initial conditions C is arbitrary and in that case the condition (2) requires that both MU and M should have the general form indicated on the right-hand of that equation. If and Q, R, &c. are respectively functions of , \f/, &c. only, it is evident that the method supplies all the first integrals. Ex. If T=Jf(P0' 2 + Q0' 2 ), M=f l (6) +/ 2 (p), MU=F 1 (0) + F 2 (), integrate the Lagrangian equations by Liouville's method. The integrals are ^M t PO'=F 1 (9) + Cf, (6) + A, , ^M*Q^=F a (0) + Cf a (4>) + A a , adding these and using the equation of vis viva we see that A 1 + A 2 =0. The paths are then given by M Multiplying these by/ x , f. 2 and adding, the time is found by where all the variables have been separated. 523. Jacobi's integral. If T be a homogeneous function of the coordinates 6, , &c. of n dimensions and U a homogeneous function of the same coordinates of - (n + 2) dimensions, then one integral is where C is the constant of vis viva and A an arbitrary constant. To prove this, we multiply the Lagrangian equations by 6, , &c. and add the products. Remembering Euler's theorem on homogeneous functions, we have e%- ctt The left-hand side is the same as since T is a homogeneous function of 0', ', &c. of two dimensions. Remembering that T-U=C, we have %. i04a+Ao.l =(n + 2) C. dt [ d6 \ Ex. A free system of particles moves under the influence of their mutual attractions, the law of force being the inverse cube: show that 2mr 2 =A+Bt + Ct 2 where r is the distance of the particle in from the origin. [Vorlesungen iiber Dynamik.] Some developments of these results are given in the first volume of the author's treatise on Rigid Dynamics. 524. Change of the independent variable. It is sometimes useful to be able to change the independent variable in Lagrange's equations from t to some other quantity T so that dr=Pdt, where P is any function of the coordinates. We suppose that the geometrical equations do not contain the time explicitly, so that T is a homogeneous function of the form *' a + .............................. (!) 21 _ 2 324 LAGRANGE'S EQUATIONS. [CHAP. vn. Let suffixes applied to the coordinates mean differentiations with regard to r just as accents denote differentiations with regard to /. Then e r =pe lt #'=p' d~ d 1 + ...... )P ........................... (3). r Supposing that P is a function of the coordinates only, not of 6', ', &c., we have dT_ldA u _n -~ ~ The Lagrangian equation therefore becomes after a slight reduction d_dT^_dT^_ _T*dP_ 1 dU dr 'dfa ~ ~d$~ ~T d$ + P ~d$ ........................... * ''. If we use the equation of vis viva, viz. T=U+C, and notice that T = PT 2 , the right-hand side of this equation becomes -r- = - . The typical Lagrangian form therefore takes the form d_dT 2 _dT J _d_ U+C dr d^ d~d P .............................. * '' We notice that though T=PT 2 , they are differently expressed. To obtain the partial differential coefficients of T 2 , the quantities 6', ', &c. must be replaced by P0 lt P0 1; &c. before differentiation. Suppose for example that the equation of vis viva (Art. 509) is T=M {^0' 2 + &c.} = U+ C, and that we wish to remove the factor M before deducing the Lagrangian equations. Changing the independent variable so that dr=Pdt, we deduce the Lagrangian equations by operating on Choosing 3/P=l, we have r 2 =iyf0 1 2 + &c., U a =M(U+C). The factor M has thus been transferred from the expression for the vis viva to the work function. Here 31 is a function of the coordinates only. We may now change the suffixes into accents if we remember that the differen- tiations are to be taken with regard to r instead of t. This difference is of no importance if we require only the paths of the particles and not their positions at any time. If the time also be required, we add the equation dt = Hdr. 525. Orthogonal Coordinates. The Lagrangian equations are much sim- plified when the expression for T can be put into the form where the products 0'', <&c. are absent. We shall now prove that this will be the case when the coordinates of the particle are the parameters of systems of curves or surfaces at right angles. ART. 526.] ORTHOGONAL COORDINATES. 325 Let the equations of three systems of surfaces which intersect at right angles be f^x, y, z)= Pl , f a (x, y, z) = p it f a (x, y, z)= Ps ............ (1), where />j , p 2 , /> 3 are three constants or parameters whose values determine which surface of each system is taken. These parameters may be regarded as the co- ordinates of the point of intersection of the three surfaces. Such coordinates are called sometimes orthogonal coordinates and sometimes curvilinear coordinates. Their theory was given by Lam6 in his Lemons sur les coordonnees curvilignes, 1859. In what follows we adopt his notation as far as possible. As an example of orthogonal coordinates we call to mind a system of confocal ellipsoids and hyperboloids of one and tico sheets, the lengths of the major axes being usually taken as the parameters. These are called elliptic coordinates. We may also notice that all the coordinates in common use, whether Cartesian, cylindrical or polar, are orthogonal. In the first the point is defined as the intersection of three orthogonal planes, in the second we use a cylinder cut by two planes, and in the third a sphere cut by a right cone and a plane. Let (oj, fcj, Cj) be the direction cosines of a normal to the surface whose parameter is p lt then a _i^i b- 1 dpl Ol -!<**' Let ds l be an elementary arc of the intersection of the two surfaces p. 2 , p 3 ; then ds 1 is also an elementary length measured along the normal to the surface p 1 . As we travel along this arc x, y, z and /> : vary, while p. 2 , p 3 are constant. Hence dx dy ' dz :. a 1 dx + b 1 dy + c 1 dz = dp 1 /h 1 ........................... (4). But the left side is the sum of the projections of dx, dy, dz on the normal and is therefore ds 1 ; hence d8 1 = dp 1 jh 1 . It follows that the component v : of velocity along the normal to the surface p l is v 1 = r- -^ . In the same way the components h>-\ clt of velocity normal to the other two surfaces may be found, and since these are at right angles, where accents denote differential coefficients. In order to use this expression, it will be necessary to express h^, h 2 , h 3 , in terms of the new coordinates p lt p z , p 3 . To effect this we solve the equations (1) and determine x, y, z as functions of p lt p 2 , p 3 ; finally substituting these values in the expressions (3) for ftj , h 2 , h 3 . This is sometimes a lengthy process. Motion on a Curve. 526. Fixed Curves. To find the motion of a particle on a smooth curve fixed in space. To find the velocity, we resolve the forces along the tangent to the curve. If F be the component of the impressed forces 326 MOTION ON A CURVE. [CHAP. vn. X, F, Z, this gives as in Art. 181, dv r, ,,dx ,,dy ~dz mv -v = F = X -T- + F/ + Z j - . a* as a* as If Z7 be the work function, F=dU/ds, and we have which is the equation of vis viva. To find the pressure, we resolve in any two directions which may suit the problem under consideration. Supposing that we choose the radius of curvature and binormal, we have where G, H are the components of the impressed forces; R lf R z the corresponding components of the pressure on the particle. These equations show that the pressure of the particle on the curve is the resultant of two forces, (1) the statical pressure due to the forces urging the particle against the curve, (2) the centri- fugal force mv*/p acting in the direction opposite to that in which p is measured, Art. 183. 527. Ex. 1. A plane is drawn through the tangent at P making an angle i with the osculating plane. If p' be the radius of the circle of closest contact to the curve in this plane, then ,- =G' + E' where G' and E' are the components of the impressed accelerating force and of the pressure respectively. This follows from the theorem on curves p'cosi = p, corresponding to Meunier's theorem on surfaces. Ex. 2. A helix is placed with its axis vertical, and a bead slides on it under the action of gravity. Find the motion and pressure. Let a be the radius of the cylinder, a the inclination of the tangent to the horizon. Drawing PL perpendicular to the axis of z, the radius of curvature is a length measured along PL equal to a sec 2 a. If PT is the tangent, the osculating plane is LPT. If the helix is smooth we have a m = o cos eH - . m If the particle start from rest at a height h, we see that C=2gh. Since v= -dsjdt and dssiua=dz, we find that the time of descending that height is cosec a , ART. 528.] MOVING CURVES. 327 If the helix is rough, the friction is /* V /(.R 1 2 + -R 2 2 )- Supposing that the coefficient of friction is /u = tan a, the resolution along the tangent becomes dv sin a ... v -=-= - a sin a H J(v* cos 2 a + a j <7 2 ) ; ds a writing v 2 cos a = for brevity, we find s sin 2a : +C. To integrate this we multiply the numerator and denominator of the fraction on the left-hand side by the denominator with the minus sign changed. We then find _ ag + J(v* cos 2 a + a V) _ s sin 2a ' cos a a To find C we require the initial value of v. If this were zero the particle would remain at rest because /x=tan a. Ex. 3. A rough helical tube of pitch a and radius a is placed so as to have its axis vertical and the coefficient of friction is tan a cos e. An extended flexible string which just fits the tube is placed in it : show that when the string has fallen through a vertical distance ma its velocity is (ag sec a sinh 2^, where JJL is determined by the equation cotetanhjn=tanh(^sin e + ^mcosasin 2e). [Math. Tripos, 1886.] Ex. 4. Two small rings of masses m, m' can slide freely on two wires each of which is a helix of pitch p, the axes being coincident and the principal normals common ; the rings repel one another with a force equal to p.mm'r when they are at a distance r from one another. Prove that if be the angle the plane through one ring and the axis makes with the plane through the other ring and the axis, the time [p in which increases from a to /S is / {Atp? - 2B cos + C} ~ * d, where J a . ab . and a, b are the radii of the cylinders on which the helices are drawn. [Coll. Ex. 1896.] 528. Moving curves. Ex. 1. A particle P is constrained to move on the plane curve z=f(x), which rotates about a straight line Oz in its plane with an angular velocity u. It is required to form the equations of motion. Applying to P an acceleration w 2 x tending from the axis of rotation, we treat the curve as if it were fixed, Art. 495. Taking the tangential and normal resolutions, we have dv F v 2 G R v -r- = + t>rx COS \f/, = orx sin \f/ + . ds m p m m where v is the velocity of the particle relatively to the curve, \f/ the angle the tangent at P makes with the axis of x, and p is the radius of curvature. Also F and G are the components of the impressed forces along the tangent and radius of curvature at P. We may replace the first of these equations by the integral of vis viva, viz. 328 MOTION ON A CURVE. [CHAP. VII. The second equation then gives the component E of pressure in the plane of the curve. The component R' of pressure perpendicular to the plane of the curve is given by m- (* 2 x at where H is the corresponding component of the impressed force, and x is the distance of the particle from the axis of rotation. Ex. 2. A circular wire is constrained to turn round a vertical tangent Oz with a uniform angular velocity u. A heavy smooth bead, starting from the highest point A without x any velocity relative to the curve, descends under the action of gravity. Find the velocity and pressure. Let C be the centre, OC=a; let P be the particle, the angle ACP=0, v = ad0/dt. We re- duce the plane to rest by applying to P an accele- rating force measured by w 2 x, where x = a + a sin 0, and acting parallel to OC. The equation of vis viva then gives /x xdx; a .: v 2 = 20 (a - a cos 6) + a> 2 a 2 (2 sin 6 + sin 2 6). The components E, E' of the pressure on the particle respectively along PC and perpendicular to the plane are given by v 2 R 1 d , E' = m r ) are capable of sliding on a smooth circular wire of radius a, whose vertical diameter is fixed, the rings being below the centre and connected by a light string of length a ^/2 : prove that if the wire is made to rotate round the vertical diameter with an angular velocity -! ^ - / ! > the rings can be in relative equilibrium on opposite sides of the [a^/3 m-m \ vertical diameter, the radius through the ring m being inclined at an angle 60 to the vertical. Show also that the tension of the string is - , g. 7ft *~ Wt s ~ [Coll. Ex. 1897.] Ex. 4. A smooth circular cone of angle 2ct has its axis vertical and its vertex, pierced with a small hole, downwards. A mass M hangs at rest by a string which passes through the vertex and a mass m attached to the upper extremity describes a horizontal circle on the inner surface of the cone. Find the time T of a complete revolution, and prove that small oscillations about the steady motion take place in the time T cosec a {(M+ mJ/Sm}*. [Coll. Ex. 1896.] Ex. 5. A smooth plane revolves with uniform angular velocity u about a fixed vertical axis which intersects it in the point 0, at which a heavy particle is placed at rest. Show that during the subsequent motion v*=p'W + 2gz; where z is the depth of the particle below O, p its distance from the axis and v the speed with which the path is traced on the plane. [Coll. Ex. 1893.] ART. 529.] FREE MOTION, TWO CENTRES. 329 529. A case of free motion with two centres of force. Ex. 1. A particle P, of unit mass, is constrained to move along an elliptic wire without inertia which can turn freely about its major axis. The particle is acted on by two centres of force, situated in the foci S, H, ivhich attract according to the law of the inverse square. Prove that the pressure on the curve is zero in certain cases. We take the major axis as the axis of z and the origin at the centre. Let w be the angular velocity of the wire. Representing the distance of the particle P from the major axis by y, the component R' of pressure on the particle perpendicular to the plane of the curve is given by But since the wire is without inertia, i.e. without mass, the wire moves so that the pressure R' on it is zero, Art. 267. We therefore have throughout the motion where B is the constant of angular momentum about the axis of rotation. Let the distances of the particle from the foci S, H be r lt r%; and let the central forces be t/r-f, nr 2 . To find the motion in the plane zOP, we apply to P an acceleration w 2 j/ = tending from the major axis, and then treat the curve as if it were fixed. We notice that the particle could freely describe the ellipse under any one of the forces Mi/^i 2 , /W r 2 2 > B 2 ly 3 if properly projected; see Arts. 333, 323. It immediately follows that if all the three forces act simultaneously, the pressure on the particle will be a constant multiple of the curvature, Art. 272. The pressure will be zero, if the square of the velocity of projection is equal to the sum of the squares of the velocities when the particle describes the curve freely under each force separately ; Art. 273. We find therefore that if v^ be the velocity relatively to the curve, the pressure is zero, if If v be the resultant velocity of the particle in space, we have v 2 =v 1 2 + w 2 ?/ 2 . Hence t> 2 = (---}+ ( 2 1 V B* a *~ b * SMl \ri S/ M2 \'2 / b * When the pressure is zero, the wire may be removed and the particle describes its path freely in space under the action of the two given centres of force. The general path under all circumstances of projection has not been found. If the particle is projected along the tangent to any ellipse having S, H for foci with a velocity whose component in the plane of the ellipse is v l , and whose component perpendicular to the plane is v' = uy = Bjy, it will continue to describe the ellipse freely, while the ellipse itself moves round the straight line SH with a variable angular velocity ta = B/y 2 . Ex. 2. If the particle is also acted on by a third centre of force situated at the centre and attracting according to the direct distance, prove that the pressure on the revolving wire is zero in certain cases. 330 MOTION ON A CURVE. [CHAP. VII. 53O. Ex. A particle P of unit mass moves on a smooth curve which is con- strained to tarn about a fixed axis with an angular velocity w. It is required to find the relative motion. Let the axis of rotation be the axis of z and let the axes of x, y be fixed to the curve and rotate round Oz with the angular velocity w. Let us refer the motion to these moving axes. Since 0i = 0, 2 = 0, 3 = w, the equations of Art. 499 become du dx V(t3j U ?/W dv - dz (1), where E 1} R 2 , E 3 are the components of the pressure on the particle. Eliminating u, v, w, (2). du , dx The resultant of the two accelerating forces X 1 = u*x, Y^u^y is a force tending directly from the axis of rotation and whose magnitude is F 1 = wV, where r is the distance of the particle P from the axis. The resultant F z of the two forces X 2 =ydwjdt, F,= - xdujdt is F 2 = -rdujdt, and it acts perpendicularly to the plane containing the axis of rotation and the particle in the direction in which the angular velocity w is measured. To find the resultant .F 3 of the forces X 3 = 2udyjdt, Y 3 = -Zudxjdt, we notice that the component along the tangent to the curve, viz. X 3 dxlds + Y 3 dy/ds, is zero. The resultant acts perpendicularly to the given curve, and may be compounded with and included in the reaction. When only the motion of the particle is required, the force F 3 may be omitted. Beasoning as in Art. 197, we see that the equations of motion (2) become the same as if the particle were moving on a fixed curve, provided we impress on the particle (in addition to given forces X, Y, Z) two accelerating forces, viz. (1) a force F^i^r and (2) a force F 2 = -rdujdt. The process of including the two forces F lt F 2 among the impressed forces is sometimes called reducing the curve to rest. The curve having been reduced to rest, the velocity of the particle relatively to the curve is found either by the equation of vis viva or by resolving along the tangent. We find a z rdr- r. l< where U represents the work function. If the angular velocity is uniform, this reduces to The velocity thus found is the velocity relative to the curve. The actual velocity in space is the resultant of velocity v and the velocity wr of the point of the curve instantaneously occupied by the particle. ART. 534.] A CHANGING CURVE. 331 531. The pressure of the fixed curve on the particle is not the same as the actual pressure of the moving curve. Representing the first by R' and the second by R, we see that R' is the resultant of R and the two forces X B = 2udyjdt, Y 3 = -2udx/dt. We may compound these two forces into a single force F 3 . We project the moving curve on a plane perpendicular to the axis of rotation. If P' be the projection of P, dxjdt and dyjdt are the component velocities of P. The resultant is then evidently F 3 =2uv' where v' is the velocity of P' relatively both to the curve and its projection. The direction of F 3 is perpendicular not only to the given curve but also to its projection. The components along and perpendicular to the radius vector are + 2urd0[dt and - 2wdrjdt. 532. Ex. A small bead slides on a smooth circular ring of radius a which is made to revolve about a vertical axis passing through its centre with uniform angular velocity u, the plane of the ring being inclined at a constant angle o to the horizontal plane. Show that the law of angular motion of the bead on the ring is the same as that of a bead on the ring of radius a/sin a revolving round a vertical diameter with angular velocity u sin a. [Coll. Ex.] 533. A changing curve. A bead of unit mass moves on a smooth curve whose form is changing in any given manner. It is required to find the motion. Let the equations of the curve be written in the form where 6 is an auxiliary variable. We may regard the position of the particle at any given time t as defined by some value of 6. Our object is to find 6 in terms of the time. Let us use Lagrange's equations. We have T=lZ(/,0'+/ t ) .................................... (2), where S implies summation for all the coordinates, and partial differential coeffi- cients are indicated by suffixes. The Lagrangian equation is d_ dT _ dT _ dU It d6' ~d6~~de .................................... ( '' 2|w+/ < )/ fl -S(//+/P(/,/+/ M )-^ ............... (4). This is a differential equation of the second order from which 6 may be found. The three components of the pressure on the particle in the directions of the axes may be found by differentiating the equations (1). If X, Y, Z, be the com- ponents of the impressed forces; B lt R 2 , R 3 those of the pressure, the Cartesian equations of motion are Since the pressure must be perpendicular to the tangent to the instantaneous position of the curve, we do not necessarily require all these equations, though it may be convenient to use them. 534. Ex. A helix is constrained to turn about its axis Oz, which is vertical, with a uniform angular velocity u. Find the motion of a particle of unit mass descending on it under the action of gravity. Let the axes OA , OB move with the curve and let OA make an angle wt with some axis of x fixed in space. Let the angle AON =6. See the figure of Art. 527. 332 MOTION ON A SURFACE. [CHAP. vii. The equations of the helix referred to axes fixed in space are ), z = a0tana; Substituting in Lagrange's equation, we find after a little reduction aO"= -gem a cos a, which admits of easy integration. It should be noticed that this result is inde- pendent of the angular velocity of the guiding curve, provided only it is constant. A similar result holds for any curve on a right circular cylinder turning uniformly about its axis. To find the pressure of the helix on the particle we use cylindrical coordinates, Art. 491. Let P, Q, R be the components of the pressure, then since in the helix p=a, (f>=9 + ut, we find by substitution P=-a(0' + w) 2 , Q = a6", Z - g = a tan aO". These show that the pressure on the particle is equivalent to a sustaining force g cos a acting perpendicularly to the osculating plane together with the radial pressure P. Motion on a Surface. 535. Any Surface. To find the motion of a particle on a fixed surface. Let P be the position of the particle at the time t, Pij a tangent to the path, P a normal to the surface, and P% that tangent to the surface which is perpendicular to the path. Let PC be the radius of curvature of the path, PA the bi- normal, then PA, PC lie in the plane . Let x be the angle Let X, Y, Z be the resolved impressed forces parallel to any axes x, y, z fixed in space. Let the equation of the surface The resolved accelerations of the particle in the directions PA, Prj, PC are known to be zero, vdv/ds and v*/p respectively, Art. 496. Hence resolving in the direction Py, dv v dx , T which if U be the work function at once reduces to $mtf This is the equation of vis viva. (1). ART. 538.] ELEMENTARY RESOLUTIONS. 333 Let R be the pressure of the constraining surface on the particle measured positively inwards. Then resolving along the normal, mv- rr D cos x = n + R, where H is the component of the impressed forces. If p' be the radius of curvature of the normal section i)P of the surface made by a plane through the tangent to the path, it is proved in solid geometry that p = p cos %. We therefore have (2). 536. If a, b are the radii of curvature of the principal sections of the surface at P, <(> the angle the tangent to the path makes with the section a, we have by Euler's theorem 1 cos 2 sin 2 p'~ a b Let v 1 , v 2 be the resolved velocities of the particle along the tangents to the principal sections, then v^=v cos and v 2 = v sin <. The equation (2) then takes the form fv 2 v 2 \ + -r )=H+jFJ. 537. If the forces are conservative, the velocity of the particle is given by the equation (1) in terms of its coordinates at any instant and of the initial conditions. To determine the velocity at any point we do not require to know the path by which the particle arrived at that point (Art. 181). The pressure R is given by (2) in terms of the velocity at that point, the normal component of force and the radius of curvature of the normal section of the surface through the tangent. The pressure is therefore also independent of the path. The whole energy C being given, the pressure depends on the position of the particle and the direction of motion. The equation (2) shows that the acceleration of the particle normal to the surface is v*/p'. It is therefore independent of the position of the osculating plane but depends on the direction of motion. 538. To find the path of the particle we resolve in some third direction. Choosing the direction P%, we have (3), 334 MOTION ON A SURFACE. [CHAP. VII. where F is the component of the impressed force along that tangent to the surface which is perpendicular to the path. This may also be written in the forms rav 2 , mv* r, 7- tan x=^> Y = ' where p" is the radius of curvature of the projection of the path on the tangent plane. It is also called the geodesic radius of curvature. 539. Geodesic path. If the only impressed forces acting on the particle are normal to the surface, F = 0, and the third equation shows that either sin ^ = or that the path is a straight line. The path is therefore necessarily a geodesic line. If the surface is rough, the friction acts opposite to the direction of motion, and F would still be zero. So also if the particle moves in a resisting medium the resistance is opposite to the direction of motion. Generally we conclude that the path of a particle on a rough surface in a resisting medium when acted on by forces normal to the surface is a geodesic. Conversely, iftJiepath is a geodesic line we must have sin % = and therefore F=0. The component of the impressed force tan- gential to the surface must then also be tangential to the path. 540. To find the radius of curvature of the path and the position of the osculating plane when the position and direction of motion of the particle are given. To effect this we use the two equations . sin y r, 1 cos 2 sin 2 < cosy = F, - = ----- s + r-Z = --- * . ' The particle being in a given position, v 2 , a and b are known. Since is the angle the direction of motion makes with the principal section whose radius of curvature is a, we have F= A cos , where A and B are the given components of impressed force along the tangents to the principal sections. Thus the values of both sin %/p and cos %/p follow at once. ART. 541.] A SURFACE OF REVOLUTION. 335 541. Motion on a surface of revolution. When the surface on which the particle moves is one of revolution, it is generally more convenient to use cylindrical coordinates. Let the axis of figure be the axis of z and let f be the distance of the particle P from that axis. Let the equation of the surface be z =/ (). Let U be the work function, and let the mass be unity. The equation of motion obtained by resolving perpendicularly to the plane zOP is dU We have also the equation of vis viva T=i{^ + 2 '- + *'*} = U+C ............... (2), which, by using the equation of the surface, may be written in the form Here accents denote differentiations with regard to the time. By solving (1) and (3) we determine the two coordinates , (j> in terms of the time. In certain cases the solution can be effected. The equation (1) gives Let the impressed forces be such that PU-RW+FM.Z) ..................... (4), where F 1} F z are arbitrary functions. We then have W % (f y ) = d ^ *' 5 -V i <" = F, (<) + A . . . (5). Substituting this value of ' in (3) we find ,(f,*) + Cf'-.i ............ (6). Since z is a known function of , the variables in this equation are separable. The determination of as a function of t has therefore been reduced to integration. The differential equation of the path is found by Dividing (5) by (6). It is evident that here also the variables are separable. 336 MOTION ON A SURFACE. [CHAP. VII. Since the expression for the vis viva, given in (3), can be written in the form where P is a function of only, this solution is an example of Liouville's method of solving Lagrange's equations ; Art. 522. 542. Motion on a sphere. When the surface on which the particle moves is a sphere, we may use polar coordinates, the centre being the origin. The equations corresponding to (1) and (3) of Art. 541 are found by putting =/sin#, where I is the radius ; we then have ^( sin2 0') = xp ^ 2 {0 /2 + sin 2 0 /2 }= U+C. These admit of integration when U. expressed in polar coordinates, has the form sin 2 U = F 1 (r, <) + F 2 (r, 6}. The resulting integrals are l* sinW 2 = F a (I, 0) + C sin 2 - A 543. Examples. Ex. 1. A particle of mass m moves on the inner surface of a cone of revolution, whose semi-vertical angle is a, under the action of a repulsive force mnjr 3 from the axis ; the angular momentum of the particle about the axis being m^/yutana; prove that its path is an arc of a hyperbola whose eccentricity is sec a. [Math. Tripos, 1897.] Resolve along the generator and take moments about the axis, thus avoiding the reaction, Art. 541. These prove by integration that the path lies on a plane parallel to the axis. The angle between the asymptotes is therefore equal to the angle of the cone. Ex. 2. A particle P moves on a sphere of radius I under the action both of gravity and a force X=fj.lx 3 tending directly from a vertical diametral plane taken as the plane of yz. Show that the determination of the motion can be reduced to integration. If the particle is projected horizontally from the extremity of the axis of x, show that when next moving horizontally, it is in a lower position. Ex. 3. A particle is acted on by a force the direction of which meets an infinite straight line AB at right angles and the intensity of which is inversely proportional to the cube of the distance from AB. The particle is projected with the velocity from infinity from a point P at a distance a from the nearest point of the line in a direction perpendicular to OP and inclined at an angle a to the plane AOP. Prove that the particle is always on the sphere the centre of which is O, that it meets every meridian line through AB at the angle o, and that it reaches the line AB in the time a 2 sec a/^/tt, where p is the absolute force. [Math. Tripos, I860.] ART. 545.] VARIOUS SURFACES. 337 Ex. 4. A particle moves on a spherical surface of unit radius, its position being determined by its polar distance and its longitude . If the tangential acceleration is always in the meridian, and &in 2 0dldt = h, cot0 = w, prove that its value is /t 2 (1 + w 2 ) ( u + -=-s ) . Prove also that the law of force perpendicular to the equatorial plane under which the sphero-conic . = . , + . ,; can be described is that of the inverse sin 2 sm 2 a sin 2 6 cube of the distance. [Math. Tripos, 1893.] Ex. 5. A particle moves on a smooth helicoid, z = atj>, under the action of a force fj.r per unit mass directed at each point along the generator inwards, r being the distance from the axis of z. The particle is projected along the surface perpendicularly to the generator at a point where the tangent plane makes an angle a with the plane of xy, its velocity of projection being a*Jp.. Prove that the equation of the projection of its path on the plane of xy is l + 2 /r 2 = sec 2 a {cosh (0/cosa)} 2 . [Math. Tripos, 1896.] 544. Cylinders. Ex. 1. A particle moves on a rough circular cylinder under the action of no external forces. Prove that the space described in time t is a sec 2 a, / /J.V cos 2 at\ , ,, ,. , , log I 1 H 1 where the particle has initially a velocity V m a direc- tion making an angle a with the transverse plane of the cylinder. [Math. Tripos, 1888.] Ex. 2. A heavy particle moves on a rough vertical circular cylinder and is projected horizontally with a velocity V. Prove that at the point where the path cuts the generator at an angle 0, the velocity v is given by agjv' 2 sin 2 = ag\ V 2 + 2/u log (cot + cosec ) , and that the azimuthal angle 6 and vertical descent z are ag6=fo 2 d and gz=fo* cot0d<, the limits being =^ir to . [Math. Tripos, 1888.] The cylindrical equations of motion give (v sin <(>)= - - u a sin 3 , (v cos )=g--v 2 sin 2 cos . at a (it (t First eliminating dt and putting v = l[w we obtain the first result. Secondly eliminating p we obtain the others. Ex. 3. A smooth cylinder whose cross section is a cardioid is placed with its generators inclined at an angle a to the vertical and having the generator through the cusp in its highest position, and a particle is projected from the cusp line with velocity V along the inner surface of the cylinder inclined at an angle )3 to the generator ; show that it will leave the surface if F 2 < - : , where 2a is the 5 sin-* p breadth of any section through the cusp. [Math. Tripos, 1887.] 545. String on a surface. Ex. I. A string, one end of which is fastened at a point of the surface of a smooth circular cylinder whose axis is vertical, winds round the cylinder for part of its length, and terminates in a straight portion of length c at the end of which a particle is tied. Show that when the particle is projected in the direction horizontal and perpendicular to the string it begins to rise or fall according as the velocity is greater or less than sin a (gc sec a)*; a being the angle at which the string cuts the generators. B. D. 22 338 MOTION ON A SURFACE. [CHAP. VII. Prove also that during the ensuing motion - j- (r 2 u) + aur^O; r being at any T (tt time the length of the projection of the straight portion of the string on a horizontal plane, w the angular velocity of the vertical plane drawn through the string and a the radius of the cylinder. [Coll. Ex. 1895.] Ex. 2. A string is wound round a vertical cylinder of radius a in the form of a given helix, the inclination to the horizon being i. The upper end is attached to a fixed point on the cylinder, and the lower, a portion of the string of length I sec / having been unwound, has a material particle attached to it which is also in contact with a rough horizontal plane, the coefficient of friction being /x. Supposing a horizontal velocity V perpendicular to the free portion of the string to be applied to the particle so as to tend to wind the string on the cylinder, determine the motion and prove that the particle will leave the plane after the projection of the unwound portion of the string upon the plane has described an angle 5 T = Io 8 o MI 2 - o i* [Math. T. I860.] 2/4 tan i 2/*K 2 tan 2 t-2ju0Hant + 0a Ex. 3. A fine string of length I is fastened to a point A of a smooth cylinder of radius o, and, being wound round the cylinder, has a particle of given mass attached to the free end. Show that, if the particle is projected in any direction, it will, so long as the string is tight and some portion of it remains wound on the cylinder, describe a geodesic line on the surface 1 a; cos -(Jl*-z z - x* + y*-a*) + y sin - (J^l?- J x * + y* - a 2 ) = a, a a where the axis of the cylinder is the axis of z, and the axis of x is the radius through A. Show also that the particle cannot be so projected that the string shall not slip on the cylinder, except when the path lies in the plane of the circular section of the cylinder drawn through A. [Math. Tripos, 1893.] 546. Gauss' coordinates. The motion of a particle on a surface may also be investigated by using the geodesic polar coordinates of Gauss. In this method every surface has a geometry of its own, in which all the lines under consideration are drawn on the surface. The geodesies on the surface correspond to straight lines on a plane, and the properties of the figures are discussed by reasoning analogous to that of two dimensions. Let be any origin, p the length of the geodesic drawn from O to any moving point P. Let w be the angle OP makes with some fixed geodesic Ox. Let OP' be a neighbouring geodesic, PL the perpendicular to OP'. Then in the limit LP' dp, PL = Pdu. The theorem that OP=OL is proved in Salmon's Solid Geometry, Art. 394, edition of 1882. The quantity P is a function of p and u, whose form depends on the particular surface under consideration. On a plane P p, and on a sphere of radius a, P=asinp/a. On an ellipsoid when the origin O is at an umbilicus, P=y cosecw, where u is the angle the geodesic OP makes with the arc containing the four umbilici. The difficulty of finding the value of P for any surface prevents this method from coming into general use. The vis viva 2T of a particle of unit mass is given by where accents as usual denote differential coefficients with regard to the time. ART. 549.] GAUSS' COORDINATES. 339 Let U be the work function ; F, G the accelerations at P along and perpendicular to the geodesic radius vector OP. We have by Lagrange's theorems, d_dT_dT = dU . F n_ p dP u>z dtdp' dp dp dp u d_dT__dT dU . Id (pv) _^ w , 2 . dt du' du du P dt v ' du dP dP , dP . bince -7- = -j- p + - if, this reduces to at dp du G = Pu" + '^-u'* + 2~u'p'... ,...(2). du dp 547. We may also arrive at these results without using Lagrange's equations. Let u, v be the component velocities of P along and perpendicular to the tangent PT at P to the geodesic OP. Let P'T' be the projection of the tangent to OP' on the tangent plane at P. Since the tangent planes at P, P' make an indefinitely small angle with each other the component velocities at P along and perpendicular to P'T' are u + du and v + dv. If d6 be the angle PT makes with P'T', the accele- rations along and perpendicular to PT are (as in Art. 225), v _du dO r _d dO ~di~ v di' r ~di +u di' Now u=p', v = Pu', and by a theorem proved in Salmon's Solid Geometry, Art. dP 392, dO = -r- du. We therefore have dp These reduce to the same forms as before. 548. Ex. A particle P, constrained to move on an ellipsoid, is attached to an umbilicus by a string of given length, which also lies on the surface. Prove that the particle describes a geodesic circle with a uniform velocity V, and that the angular velocity of the string about the umbilicus is Vsinujy. Prove also that the accelerating tension is F 2 cos j3/y, where /3 is the angle the tangent at P to the string makes with the axis of y. 549. Developable surfaces. When the surface on which the particle moves is developable, we may sometimes fix the position of the particle by using the edge as a curve of reference. Let s be the arc of the edge measured from some fixed point A to a point Q such that the tangent at Q passes through P. Let QP=u measured positively in the same direction as s. We then have - The form of the surface being given, the radius of curvature p of the edge at Q is known as a function of s. When U is given as a function of u and s the Lagrangian method supplies two equations to find the coordinates u and s. Ex. A heavy particle moves on a developable surface whose edge is a helix with its axis vertical. Obtain two integrals by which s' and u' may always be found in terms of u and *. Show also that if the particle is projected along a tangent to the helix, it will continue to describe that tangent. 222 340 MOTION OF A HEAVY PARTICLE. [CHAP. VII. Motion of a heavy particle on a surface of revolution. 550. To find the motion of a heavy particle on a surface of revolution the axis of which is vertical. Let the axis of z be the axis of the surface and let z be measured upwards. The velocity v is then given by v* = 2g(h-z} ........................... (1), where h is a constant depending on the initial conditions. Let the plane z = h be called the level of no velocity. Let be the distance of the particle P from the axis of figure, and < the angle the plane zOP makes with the plane zOx. Then where mA is the constant angular momentum and its value is known when the initial values of f and d/dt are given ; Art. 492. The velocity v at any point being given by (1), the angular mo- mentum A must lie between zero and v%. It is the former when the particle is moving in the plane zOP and the latter when moving horizontally. The particle therefore can occupy only those points of the surface at which v%> A, i.e. those points at which 2g (h z)%*> A 2 . If then we describe the cubic surface (h-*)?=A'/2g ........................ (3), the of the particle for any value of z must be greater than the corresponding of the cubic surface. This cubic divides the given surface of revolution into zones, separated by horizontal circles, and the particle can move only in those zones which are more remote from the axis of figure than the corresponding portions of the cubic. The zone actually moved in is determined by the point of projection. The particle moves round the axis of figure and must continue to ascend or to descend until it arrives at a point at which the vertical velocity can be zero, that is, until it reaches one of the boundaries of the zone. If the particle is projected horizontally it is on the boundary of two zones. It will move on that neighbouring zone which is the more remote from the axis than the corresponding portion of the cubic. If the cubic touch the surface of revolution, the particle is situated on an evanescent zone and will then describe ART. 553.] SURFACE OF REVOLUTION. 341 a horizontal circle. The path is stable or unstable according as the neighbouring zones a.re less or more remote from the axis of figure than the cubic surface. 551. Ex. A particle is projected horizontally with a velocity V at a point whose coordinates are , z. Will it rise or fall? If mR be the pressure on the particle, \f/ the angle the radius of curvature makes with the vertical, we see by resolving vertically, that the particle if inside and ^ < T will rise or fall according as R cos i// is greater or less than g. To find R we resolve along the normal to the surface. Since the particle is moving along that principal section whose radius of curvature is the normal n, we have V*/n = R- g cos ^, Art. 536. Since n sin ^=, we see that the particle will rise, fall, or describe a horizontal circle according as V 2 is greater, less, or equal to g tan \//. If z=/() be the equation of the surface of revolution, tan ^ = dzjd^. To find the level to which the particle will rise or fall we use the cubic surface described in Art. 550, the constants A and h being known from the equations F = A , V 2 = 2g (h - z). The intermediate motion may be deduced from the equations (1), (2) of the same article. 552. Ex. To find the pressure on the particle when in any position. We use the formula given in Art. 536. The principal radii of curvature of the surface are the radius of curvature p of the meridian and the normal n. The velocity perpendicular to the meridian being v 2 =%d(f>/dt, the velocity v l along the meridian is given by ^ 2 =v 1 2 + v 2 2 . The formula v 2 v 2 + = R-g cos \b, p n shows that 2g(h-z) A 2 -^ - '- 2 /I 1\ --- ). \ P/ p This problem has a special interest because we can use it to represent experi- mentally the path of a particle under the action of a centre of force. If Q be the projection of the particle on a horizontal plane, the motion of Q is the same as that of a particle moving under the action of a central force whose magnitude is R sin ^. If then a surface is so constructed that the generating curve satisfies the differential equation R sin ^ = M/ 2 , where R has the value given above, the path of Q should be a conic with a focus at the origin. The experiment cannot be property tried with a particle, for the surface must then be very smooth. It is better to replace the particle by a small sphere which is made to roll on a rough surface, but in that case, the theory must be modified to allow for the size of the particle. Nature, 1897. 553. Small oscillation. Ex. A heavy particle P, describing a horizontal circle on a surface of revolution, is slightly disturbed. It is required to find the oscillations to a first approximation. The plane z OP may be reduced to rest if we apply to the particle a horizontal acceleration (djdt=:A, this acceleration is equal to Resolving along the meridian, we have where \J/ is the angle PGO which the normal to the surface makes with the axis. 342 MOTION OF A HEAVY PARTICLE. [CHAP. VII. Let the radius of the mean circle be N 1 P l = c and let the normal to the surface at any point of its circumference make an angle P 1 G l O=y with the vertical. Since s may be taken to be the arc of the meridian between the particle and the mean circle, we have where p is the radius of curvature of the meridian at its intersection with the mean circle. d 2 s Substituting, we find by Taylor's theorem -j-^=F- p*s, F=- cos 7-0 sin 7, p*= 3.d 2 cos 2 7 jdt A; d=2a N /2, (1 -n) tan0 = l + n. [Math. T. 1871.] Ex. 3. A heavy particle moves on a paraboloid with its axis vertical, the equation of the surface being x 2 la + y-j^ = 4z. Show that the particle when moving o/3/l A (B 1\ z(h-z) horizontally must lie on the quartic surface -3- I s 4 1 1 s --- 5)= ; 5 - 4 Vy ) \2g v*) & 1 a; 2 y 2 1 /re' 2 i/' 2 \ where a = -5 + ^, + 4, and B is the initial value of , I -- h ^ + 2o I . Show also p z a 2 /S 2 r\*' ft / that when the paraboloid is a surface of revolution, the intersection reduces to two horizontal planes and two coincident planes at the vertex. 555. The Conical Pendulum. To find the motion of a heavy particle P on a smooth sphere*. It will be convenient in this problem to take the origin of coordinates at the centre of the sphere and to measure Oz vertically downwards. Let I be the length of the string OP and 6 the angle it makes with Oz. Let $ be the angle the vertical plane zOP makes with some fixed plane zOx. Let r be the * The problem of the conical pendulum has been considered by Lagrange in the second volume of his Mecanique Analytique. He deduces equations equivalent to (1) and (3) of Art. 555 from his generalized equations, and notices that the cubic has three real roots. He reduces the determination of t and to integrals, and makes approximations when the bounding planes are close together. He refers also to a memoir of Clairaut in 1735. There is an elaborate memoir by Tissot in Liouville's Journal, vol. xvn. 1852. He expresses t, z, and the arc s in elliptic integrals in terms of u. A long communication by Chailan may be found in the Bulletin de Soc. Math, de France, 1889, vol. xvn. There is a brief discussion of this problem in Greenhill's Applications of Elliptic Functions, 1892, Art. 208. 344 MOTION OF A HEAVY PARTICLE. [CHAP. VII. distance of P from Oz. Let h be the altitude above of the level of zero velocity. We now proceed as in Art. 550. J By the principles of angular momentum and vis viva, Eliminating d/dt and writing r = I sin 6, //7#\ (2). Putting z I cos 6, this may also be written in the form '(Z 2 -.* 2 )-^ 2 (3). To find the positions of the horizontal sections between which the particle oscillates (Art. 550), we put dz/dt = 0. We thus have the cubic (jfc+*)(p-*)-.A"/20-0 (4). Since the initial value of z must make (dz/di) 2 positive, the left-hand side of the cubic (4) is positive for some value of z lying between z 1. When z = + I the left-hand side is negative, hence the cubic has two real roots lying between + I and separated by the initial value of z. Let these roots be z = a and z = b. Lastly when z is very large and negative the left-hand side is positive, the third root of the cubic is therefore negative and numerically greater than 1. Let this root be z = c. The particle oscillates between the two horizontal planes defined by z = a, z = b. Since the cubic can be written in the form z 3 + hz 2 - I 2 z + (A 2 /2g -^Fh) = 0, we have the obvious relations a + b-c = -h, (a + b)c-ab = l 2 , abc = A?/2g - I 2 h. Conversely, when the depths a and b of the two boundaries of the motion are given, the values of the other constants of the motion, viz. c, h, and A, follow at once. We have P + ab A 2 (l 2 -a?)(l*-b*) c=~j-, -5- = -V- , h = c a b. a + b 2g a + b 656. Ex. Prove (1) that one of the two horizontal planes bounding the motion lies below the centre ; (2) that the plane equidistant from the two bounding planes also lies below the centre ; (3) that both the bounding planes lie below the centre if 2ghP or or '=A, the values of cos xlp and sin xlp are evident. Ex. 2. A particle is projected with velocity V horizontally from a point on the surface of a smooth sphere. Prove that the radius of curvature of its path is where I is the radius of the sphere and a the inclination to the vertical of the radius at the point. [Coll. Ex. 1881.] Ex. 3. A particle is projected inside a smooth sphere of radius I with a velocity J2gl along a tangent to the horizontal equator, prove that at first the radius of curvature is 21/^5. [Coll. Ex. 1897.] 56O. Ex. Prove that the projection of the path of the particle on a horizontal or plane is a central orbit described under a force Esin 9 = --^ {2h + 3^(l- - r 4 )}, where the radical changes sign when r=l. Show also that if the two roots a and 6 of the cubic in Art. 555 have the same signs, the central path is a spiral curve touching alternately two circles whose radii are ^/(P-a 2 ) and *J(P-b*), the curve being always concave to the centre of force. If a and b have opposite signs the central path after touching each bounding circle, touches the circle r=l and then touches the other bounding circle. There will be a point of inflexion only if R vanishes and changes sign. ART. 562.] CONICAL PENDULUM. 347 561. Ex. If we write ^h + lcoa = KCOS, the general equation of motion of a conical pendulum may be reduced to the form - sin* * = (cos 30 - cos 80), by properly choosing the constants K and a. Show that these values are 9 oj2 a a K = 5V(ft + 8P), -K3 CO s3a = + 0^ q ftP ' o g &i o Find also the positions of the bounding planes when the constants K and a of the motion are given. 562. Time of passage. The motion of the particle as it travels from one boundary to the other may be found by an elliptic integral. We write the equation (3) of Art. 555 in the form where the limits are z = a and z = b, and a > 6. Putting z = a j~ s , the integral takes a standard form which is reduced to an elliptic integral by writing = sin ^r *J(a b}, i.e. we write z = a cos 2 - + b sin 2 - a-b where 2 = - - , c = - r . a+ c a+b If the time of passage from one boundary to the other is required, the limits are and \ir. If the two bounding planes are close together, K is small. By expanding in powers of K and effecting the integrations we find that the time from one boundary to the other is given by If the two bounding planes are also close to the lowest point, we put a=Zcosa = (l-a 2 ), b = lcos/3 = l(I - /3 2 ). We then find that the time of passage from one boundary to the other is f _ ~" A/ ' 16 348 MOTION OF A HEAVY PARTICLE. [CHAP. VII. the fourth powers of a and # being neglected. This result is given by Lagrange. Let u= I m - T. r v and K be the value of u when = I Jo i/r = ^TT. Let t be the time of passage from the lower boundary to the depth z defined by any value of i/r, and T the time from one boundary to the other, then t/T = ujK. 563. Ex. 1. Prove that when half the time of passing from the lower to the upper boundary has elapsed, the particle is above the mean level between the two boundaries. Prove also that the depth of the particle is then ('a + &)/(*' + 1), where K^l-K 2 . [Tissot.] Ex. 2. Prove that when a quarter of the time has elapsed, the depth z of the particle is 564. The apsidal angle. To find tJie change in the value of as the particle moves from one bounding plane to the other. Eliminating dt between (1) and (3) of Art. 555 we find dz Al J^(a-z) *J(z - b) VO + c) (P - z*) ' where the limits of integration are z = b and z = a, and a > b. Putting a = ra + //,, b = m /*, z = m + % so that m is the middle value of z and p the extreme deviation on each side of the middle, we have f ] J(u? - where the limits are = p and p. 565. When the bounding planes are close to each other, the range p, of the values of is small. If also the planes are not near the lowest point, the two last factors in the denominator are not small for any value of f. We may therefore expand these in powers of % and thus put the integral into the form d % (P+Q + E 2 ) = 7r(P After calculating P and R, this gives irl f, 3(3Z 2 + 13w 2 )m> 2 u> = 3m 2 ) I 4 (Z 2 - m 2 ) (Z 2 + 3m 2 ) ART. 567.] CONICAL PENDULUM. 349 566. If both the bounding planes are near the lowest point of the sphere, I and z are nearly equal, and the last factor in the denominator of (Art. 564), may be so small that its changes in value are considerable fractions of itself. We write the integral in the form Al JJ(a-z)J(z-b)(l-z)' The two factors in the denominator of the second fraction are not small and these may be expanded in powers of some small quantity properly chosen. We shall make the expansion in powers of l-z = ij. Eemembering the values of A and c found in Art. 555, we have all these integrals are common forms. To find the first we put I - z = I/M. We have dz 1 f du where a and /3 are two constants which we need not calculate. For since the limits of the first integral, viz. z = a, z = b, make the denominator vanish, the limits of the other must be u=a, w=/3. Putting w = (a + /3) + we see at once that the value of that integral is IT. Since -i\=l-z the values of the remaining integrals have just been found. Hence where we have written for c + 1 its value given in Art. 555. If p, q be the radii of the circles which bound the oscillation, we have and in the small terms which contain the product pq as a factor, we can write a=l, b = l; hence (see Art. 562) I ( l j> 2 +<7 2 ) ~ The first of these results differs from that given by Lagrange. The correction was first made by M. Bravais in a note to the Mtcanique Analytique. 567. Ex. A simple spherical pendulum of length I is drawn out to the horizontal position and is then projected horizontally with a velocity 2pl. Show that, if is the angle that the string makes with the vertical, and the azimuthal angle of the vertical plane through the string, sin sin ( -pt) = - */2 cos 6, where is equal to Jjjl. [Math. Tripos, 1893.] 350 MOTION ON AN ELLIPSOID. [CHAP. VII. Motion on an Ellipsoid. 568. Cartesian coordinates. To find the motion of a particle of unit mass on an ellipsoid*. Let X, Y, Z be the components of the impressed forces in the directions of the principal axes. Let R be the pressure on the particle measured positively inwards. Since the direction cosines of the normal are px/a z , &c., the equations of motion are x" = X-R P ^, y"=Y-Rp^, z" = Z-R P z -...(l), where accents denote differential coefficients with regard to the time. We also have from the equation of the surface x 1 i/* z 2 xx yy zz 1 xx" yy" zz" a/ 2 y'* z" 2 _ ~oy"~w + ~? + ~tf + + 7~ ............ w Multiplying the dynamical equations (1) by x', y', z', adding and integrating, we have ...... (4); where U is the work function and C is a constant. This is of course the equation of vis viva. Substituting from (1) in (3), we find y' 2 z' 2 \ (Xx Yy - * The motion of a particle constrained to remain on an ellipsoid is discussed by Liouville in his Journal, vol. xi. 1846. He uses elliptic coordinates and shows that the variables can be separated when U (/j? - v-) = F 1 (/j.) - F 2 (v). There is also a paper on the same subject by W. B. Westropp Eoberts in the Proceedings of the Mathematical Society, 1883. He also uses elliptic coordinates and especially treats of the case in which the path is a line of curvature. The case in which the particle is attracted to the centre by a force proportional to the distance is solved in Cartesian coordinates by Painleve, Lemons sur I 'integration des Equations diffe- rentielles de la M^canique, 1895. He also treats separately the limiting case of a heavy particle moving on a paraboloid whose axis is vertical. There is a short paper by T. Craig in the American Journal of Mathematics, vol. i. 1878. He discusses the same problem as Painleve, beginning with Cartesian coordinates, but passing quickly to Elliptic coordinates. He shows that the path is a geodesic when the central force is zero and the particle is acted on by what is equivalent to a force tangential to the path and varying as / ( () + F (*) v where is the arc described. This result follows also from Art. 539. ART. 569.] CARTESIAN COORDINATES. 351 In an ellipsoid we have 1 _x* f z? l__l! , 2 ,^! (&\ ^~^ + 6 4+ ?' D 2 ~a 2 + 6 2 c 2 " where D is the semi-diameter of the ellipsoid whose direction cosines are (I, m, n). Also the radius of curvature of the normal section whose tangent is parallel to D is p = D z /p. Taking D to be parallel to the tangent to the path l = ac'/v, m = y'fv, n = z'/v. The equation (5) is therefore the Cartesian equivalent of where N is the inward normal component of the impressed force. 569. In certain cases we may find another integral. Differ- entiating (5) and remembering (6), we have d ( R \ - 9 ( x ' x " 4. y'y" 4. z '^'\ 4. d ( Xx 4. *y 4. Zz v dXx\ -^ : +T- - =- (2afX + jr ) \ ft/ at \pj a? \ at J If then the forces acting on the particle are such that we have R = Ap 3 (10). Substituting in (5) or (7), we have the third integral which may be written in either of the forms rt/2 a/a 9/2 V ', V*. V~ \ J ** / ' J ~' I n 1 tt " ^ *-jy-,i 3^ ^ (11) ' P If only the direction of motion is required, we eliminate v between the equations (4) and (7). Remembering that p = D 2 /p, we see that the direction of motion at any point of the path is parallel to that semi-diameter D whose length is given by 352 MOTION ON AN ELLIPSOID. [CHAP. VII. Supposing the condition (9) to be satisfied we notice that when the initial velocity and direction of motion are such that the equation (11) gives A 0, it follows by (10) that the pressure R is zero throughout the motion. The particle is therefore free and moves unconstrained by the ellipsoid. Conversely, if the particle, when properly projected, can freely describe a curve on the ellipsoid, the condition (9) is satisfied. If it can describe the same curve when otherwise projected, the pressure varies as p 3 . If the components X, Y, Z do not satisfy the condition (9), we may sometimes make them do so by adding to them the components of an arbitrary normal force F and subtracting F from the reaction R. The condition (9) then becomes where F is an arbitrary function of x, y, z and p is a function of sc, y, z given by (6). The equation (10) then becomes R = F + Ap 3 . It is only necessary that the condition (9) should hold for the path of the particle, but as this is generally unknown, the con- dition should be true for every arc on the ellipsoid. 57O. Ex. A particle is acted on by a centre of attractive force situated at the centre of the ellipsoid, the force being KI: If D is the semi-diameter parallel to the tangent to the path, prove that These reduce to the ordinary formulae of central forces when .4 = 0. Since X= - KX, &c. the condition (9) is satisfied. The first of the results to be proved then follows from (11), for N=xp. 571. Ex. A particle P moves on the ellipsoid under the action of a force Y= - K/y 3 , whose direction is always parallel to the axis of y, and is projected from any point P with a velocity i; 2 = r/z/ 2 in a direction perpendicular to the geodtsic joining P to an umbilicus. Prove that the path is a geodesic circle having the umbilicus for centre, i.e. the geodesic distance of P from the umbilicus is constant *. We see by substitution that the condition (9) is satisfied by this law of force. The path is therefore given by i> 2 N L-=A?+-, t> 2 = D 2 p where, as before, D is the semi-diameter parallel to the tangent to the path. Since the cosine of the angle the normal makes with the axis of y is pyjb'*, we have * This result is due to W. E. W. Eoberts, who gives a proof by elliptic co- ordinates in the Proceedings of the London Math. Soc. 1883. ART. 572.] CARTESIAN COORDINATES. 353 1 A 1 N= Kpjb 2 y 2 . The conditions of projection show that (7=0. Hence -^ = If p, pff = abclp. If also D, D' are two semi-diameters at right angles of the same plane Substituting for p and r their Cartesian values x z z*\ ~ ~ Using the equation to the surface, this becomes 7j2)'2 a 2 C 2 I d 2 & 2 C 2 K f b* ' Since the particle is projected perpendicularly to the geodesic defined by pD' = ac, the coefficient of i/ 2 must be zero. It then follows that throughout the subsequent motion pD'=ac, and the path cuts all the geodesies from the umbilicus at right angles. These geodesies are therefore all of constant length. Let w be the angle which the geodesic joining the particle P to an umbilicus U makes with the arc joining the umbilici. If ds be an arc of the orthogonal trajectory of the geodesies, ds = Pdw, where P=7//sin w (Art. 546). Since v 2 = /c/i/ 2 , it follows that the angular velocity w' of the geodesic radius vector is given by u'=^ sin w. When the ellipsoid reduces to a disc lying in the plane xij, the geodesies become straight lines and the geodesic circle reduces to a Euclidian circle having its centre at H (Art. 576). The theorem is then identical with one given by Newton, viz. that a circle can be described under the action of a force F= K/y 3 . The motion of a particle in a geodesic circle under the action of a force, or tension, along the geodesic radius is given in Art. 548, where the result is deduced from Gauss' coordinates. 572. Ex. 1. A particle, moving on the ellipsoid, is acted on by a centre of force situated at any given point E. If the force F is such that the condition (9) is satisfied, prove that F^/w/P 3 , where r and P are the distances of the particle from E and from the polar plane of E respectively. Thence show that, if the initial conditions are such that the constant .1=0. the path is a conic and the velocity at any point is given by v*=pN. To prove this we put X=G(x-a), Y=G(y-p), Z=G(z-y), where G=F/r and (a, 0, 7) are the coordinates of E. Substituting in the equation (9) and remembering (2) Art. 568, we have an easy differential equation to find G. When A = 0, the particle moves freely on the ellipsoid under the action of a central force. The path is a plane curve and is therefore a conic. The equation of vis viva fails to give the velocity, but this is determined by (11) Art. 569, when the direction of motion is known. R. D. 23 354 MOTION ON AN ELLIPSOID. [CHAP. VII. Ex. 2. A particle moving on a prolate spheroid is acted on by a central force tending to one focus and attracting according to the Newtonian law. Prove that the integrals of the equations of motion are + Ct v2= 2 2 r where j> is the perpendicular from the centre on the tangent plane, r the distance from the focus, and A, B the constants of integration. 673. Ex. 1. A particle under the action of no external forces is projected from an umbilicus of an ellipsoid, prove that the path is one of the geodesies defined by pD = ac. Ex. 2. A particle is projected with a velocity v along the surface of an indefinitely thin ellipsoidal shell bounded by similar ellipsoids. Prove that wheii it leaves the ellipsoid the perpendicular p from the centre on the tangent plane is given by MP 2 R' i =v*p*abc, where R is the radius vector parallel to the initial direction of motion, P the perpendicular on the initial tangent plane, M the attracting mass and a, b, c the semi-axes of the ellipsoid. [Math. Trip. I860.] 674. Ex. Let the forces be such that -y(Xd\+ Ydp + Zdv) is a perfect differential, say dS, for all displacements on the ellipsoid, where X, /*, v are the direction cosines of the normal, i.e. \=pxja' i , &o. Prove that where B is the constant of integration. Divide (8), Art. 569, by p 2 and integrate by parts. The integrals of the equations of motion are then obtained by using (6) and (7), remembering that p 676. In order to include in one form all the different cases of paraboloids, cones, and cylinders, it may be useful to state the results when the quadric on which the particle moves is written in its most general form (x, y, z) = 0. 4 Writing p=^'+#/+& f i where suffixes denote partial differential coefficients, let the forces satisfy the condition for all displacements on the quadric. We then find that the pressure R=Ap s . The three components x', y', z' of the velocity may be deduced from the equations (4), ............... (5), where the numbers appended to the equations correspond to those in Arts. 568, &c. 576. Elliptic coordinates. Preliminary statement. The position of the particle P in space is defined by the intersection of three quadrics confocal to a given quadric. In the figure ABC, A'MM', A"NN' are respectively the ellipsoid, hyperboloid of one sheet and that of two sheets; only that part of each being drawn which lies in the positive octant. Let their major axes OA \, OA'=n, ART. 577. ELLIPTIC COORDINATES. 355 OA"=v. Let a, b, c be the three axes of any confocal. If a 2 -5 2 = ft 2 , a 2 -c 2 = fc 2 , then OH=h, OK=k are the major axes of the focal conies. The quantities X, /j., v are the elliptic coordinates of P; the first X is always positive and greater than k; the second n is less than k and greater than h; the third v is less than ft, and changes sign when the particle crosses the plane of yz. The y axes of the quadrics are .^/(X 2 - h 2 ), *J([j? - ft 2 ), N /(f 2 - ft 2 ) ; two of these are real and the third is imaginary. These radicals are positive when the particle lies in the positive octant, but the second or third vanishes and changes sign when the particle crosses the plane of xz, according as it travels along PN or PM. Similar remarks apply to the z axes. The major axes of the three confocals which intersect in any point (x, y, z) are given by the cubic x 2 y 2 z* i ** I _, __ 1 a 2 a 2 - ft 2 a 2 - k 2 where ft and k are the constants of the system. Clearing of fractions and arrang- ing the cubic in descending powers of a 2 , we see that the three roots X 2 , (i?, v 2 are such that (1). \/j.v = hkx From the third equation we infer by symmetry - * 2 ) = - A 2 ) y\ ~ ^) *l .(2). 577. To prove that the velocity v of a particle in elliptic coordinates is given by . (X 2 -ft 2 ) (X 2 -fc 2 ) (M 2 - ft 2 ) (/* 2 - We notice that the three quadrics confocal to a given quadric cut each other at right angles at P, so that the square of the velocity 232 356 MOTION ON AN ELLIPSOID. [CHAP. VII. is the sum of the squares of the normal components of velocity. It is therefore sufficient to prove that the first term is the square of the component normal to the ellipsoid, the other terms follow- ing by symmetry. If p is the perpendicular on the tangent plane to the ellipsoid, the normal component is p'. Let (I, m, n) be the direction cosines of p, then p* = X 2 f + (x 2 - h 2 ) m* + (X 2 - & 2 ) w 2 = X 2 - h*m* - W ; .-. pp = XX'. If D lt D 2 are t ne semi-diameters of the ellipsoid respectively normal to the tangent planes at P to the two hyperboloids, we know that - (X a See also Salmon's Solid Geometry, Art. 410. 578. To find the motion of a particle on an ellipsoid in elliptic coordinates. Let the ellipsoid on which the particle moves be defined by a given value of X. The mass being taken as unity the vis viva is determined by _ 2 , = ^ } te} ~ (*- /> 2 ) ( - A; 2 ) ' This we write for brevity in the form 2T=M{P^+Q^} ........................ (5). If we express the work function U in terms of (X, /it, v), we have (since X is constant) the Lagrangian function T + U expressed in terms of two independent coordinates /*, v. Comparing (5) with Liouville's form, Art. 522, we may obviously solve the Lagrangian equations by proceeding as in that article. The results are that when the forces are such that the work function takes the form (f-v*)U = F,(ti + F z ( v ) .................. (A), the integrals are ART. 581.] ELLIPTIC COORDINATES. 357 There is also the equation of vis viva (C). Dividing one of the equations (B) by the other, and remembering that \ is constant, the equation of the path takes the forms *- '"* '' in which the variables are separated. 57O. Ex. 1. Let v 1 and v 2 be the components of the velocity of the particle in the directions of the lines of curvature defined by /* = constant and v= constant respectively. Prove that 1 ~ Prove also that the pressure R on the particle is given by p i AT _ MM + W + A F z ( v )-Cv*-A\ 2p - where j> is the perpendicular on the tangent plane and N the normal impressed force. The value of p in elliptic coordinates is given in Art. 577. See Art. 568. Ex. 2. Supposing that the equation (D) of Art. 578 is written in the form Pdn=Qdv in which the variables are separated, show that the time t=\Pfj?dn-\Qv z dv. [Liouville, XL] The equations (B) become (t*?-v z )PdfJi. = dt, G" 2 - v 2 ) Qdv=dt. Multiplying these by /j?, v 2 respectively and subtracting we obtain the result. 58O. To translate the elliptic expressions into Cartesian geometry we use the equations (1) and (2) of Art. 576. Let the normals at the four umbilici U l , U 2 , &c. intersect the major axis in the two points E lt E 2) which of course are equally distant from the centre 0. We easily find that OF - hk FTT _^_V(* 0^--, E&--- The equations (1) Art. 576 give Let r lt r 2 be the distances of the particle from the points E lt E 2 , and let m be the distance of E 1 from the umbilicus U^ ; then (M-J^V-m 2 , (M + ') 2 =r 2 3 -m 2 ........................ (2). From these /u, v may be found in terms of x, y, z and the constant X. 581. Ex. Show that the equation U (fjf - v^) = F 1 (fj.) + F 2 (v) is equivalent to d 2 d 2 (Up l p 2 ) = --(U Pl p 2 ), where p^J^-m?), p^v/^-m 2 ). We have U ( 2 - v 2 ) = 0, and by (2) Art. 580 d _ d d d _ d d d/j. dp 2 dp 1 ' dv dp. 2 dp 1 ' The result follows at once. 358 MOTION ON AN ELLIPSOID. [CHAP. VII. 583. The condition (A) of Art. 578, viz. .............................. (A), can be satisfied by several laws of force. 1. Let the force tend to the centre of the ellipsoid and vary as the distance. Eepresenting the force by Hr, we have, by (1) Art. 576, 17= - J-ffr^ - i-ff {fj? + 2 + (X 2 - fc 2 - fc 2 )} ; Substituting these in the equations (B), the motion is known. 2. Let the direction of the force be parallel to the axis of x, and X= - 2Hjx 3 . Then TT H Hh*k* . , _ HIM* ( 1 ^ 2 = Xy^> MM 2 -.' 2 )^ X2 --|-- f TT 3. Let the work function tf=-77-s -- ,\, where 7s is the distance of the V(i - *r) particle from the point E a , Art. 580. We then have To find the force we notice that since dUld\=0, the direction of the force is tan- gential to the ellipsoid. Also - 2xhkj\ - \ 2 + A 2 + fc 2 ; , dU_ H ( hk fhkx \ -te~~ (T^o 5 r x + V"^" ~ / with similar expressions for Y and Z. Now the equation to the ellipsoid being X = constant, the last term of each of the three expressions represents the compo- nent of a normal force. This normal force has no effect on the motion. Taking only the remaining terms ice see that X, Y, Z are the components of a central force tending to the point E whose magnitude is - ; . When the ellipsoid is reduced (rf - 7?t 2 )* to a disc, \=k (Art. 576), and m=0 (Art. 580). The point E 1 becomes a focus and the law of force is the inverse square. 683. Ex. 1. Show that a particle can describe the line of curvature defined TT by M=MO under the action of the central force - l j tending to the point E r (r-f - 7Jt 2 ) 8 f 2 1) Show also that the velocity at any point is then given by v 2 =H < - r -- V . (fa 2 - m 2 )* **o) We notice that when the ellipsoid reduces to a plane, 7=0, and this becomes the common expression for the velocity under the action of a central force varying as the inverse square. Eeferring to the general expressions marked (A) and (B) in Art. 578, we see that the particle will describe the line of curvature if both /x' = and fj." = when /u=/u . This will be the case if we choose the constants C and A so that Fj (p) + C/x 2 + A = (M - MO) "* (/*) where 0(/x) is some function of /t. Supposing this done, we have, when /x=Mo (Art. 579) W ART. 584.] SPHEROIDS. 359 In the special case proposed U=H/(n-v). We have therefore to make Cfj? + Hn + A = (n-/j. ) z C. This gives -2CfjL =H, A = Cn 2 . Also F z (v) = Hv. ...^gU-.-lUgJ 2 -11. U*o-" Mo) ((r^-m 2 )* Mo) 2?;r. 2. A particle is constrained to move on the surface y = xt&nnz. By putting x = fji cos nz, y=n sin nz, we have Hence show that when the forces are such that the Lagrangian equations can be integrated. The path is given by u.' 2 z^ a B + l){f ( ) + C( % 2 + l) 41 = f (z \_ A ' [Liouville, 1846.] If the particle is acted on by a force tending directly from the axis of z and varying as the distance from that axis, find the components of velocity along the lines of curvature. 684. Spheroids. When the ellipsoid on which the particle moves becomes a spheroid either prolate or oblate, the formulae (A) and (B) of Art. 578 require some slight modifications. Let (X, b, c), (M, b', c'), (v, b", c") be the semi-axes of the three quadrics which intersect in P; then also a = X, a' = p, a"=v, In a prolate spheroid b = c, h = k, and the focal conies become coincident with CU Prolate. OH and HA. The axes of the hyperboloid of one sheet are n=h, b'=0, c'=0; it therefore reduces to the two planes t/ 2 /b' 2 + z 2 /c' 2 =0, the ratio b'/c' being indeter- minate. Art. 576. In an oblate spheroid \=b, ft=0; one focal conic becomes coincident with OC, while the other is a circle of radius fc. The axes of the hyperboloid of two sheets are v=0, b"=0, c" 2 = - fc 2 ; it therefore reduces to the two planes a; 2 /j/ 2 + ^ 2 /6" 2 =0, the limiting ratio vjb" being indeterminate. In the figure the positions of the focal conies just before they assume their limiting positions are represented by the dotted lines, while PM or PN represents one of the planes assumed by the hyperboloid. Before taking the limits of the equations (A) arid (B) we shall make a change of variables. In the prolate spheroid we replace M by a new variable , such that ~u 2 -^' ' BW **~Frp' " -*' 2 =/ 360 MOTION ON AN ELLIPSOID. [CHAP. VII. Thus tan varies between the limits and x as ^ varies between k and h. Since b*=n*-h*, c' 2 =ju 2 -A: 2 , andi/ 2 /6' 2 +* 2 /c' 2 =0, it is clear that is ultimately the angle the plane PM makes with the plane AB. Putting fj. = h, the formulae (A) and (B) become - i (" 2 - X 2 ) v'*=F a (") - CV 2 - A. In the oblate spheroid, we replace v by the variable

2 - ft 2 v' z tan 2 ^.= -- -^- , /. v = ft cos 0, /. - ^' 2 =- 2 ^p , thus tan varies between and oo as varies between ft and 0. Also since a; 2 // 2 + j/ 2 /6" 2 =0, is ultimately the angle the plane PM makes with the plane AC. Putting j> = 0, ft=0, the limiting forms of the equations (A), (B) are CHAPTER VIII. SOME SPECIAL PROBLEMS. Motion under two centres of force. 585. To find the motion of a particle of unit mass in one plane under the action of two centres of force*. Let the position of a point P be defined as the intersection of two confocal conies, the foci being H lt H 2 , and let OH 1 = h. Let the semi-major axes be OA = //,, OA' = v: the semi-minor axes are therefore VO 2 - h 2 ), VO 2 - h*). x* y 2 Since + , t , = l, we have fj? /j? h? /i 4 - (# 2 + y 2 + A 2 )/* 2 + /iV = ............... (1). The relations between the elliptic coordinates p, v of any point P and the Cartesian coordinates x, y are therefore _ - _ y - where r is the distance from the centre. We also have r z = p + v, r l = fji v, where ?\ , r 2 are the distances of P from the foci. * Euler was the first who attacked the problem of the motion of a particle in one plane about two fixed centres of force, Mtmoires de VAcademie de Berlin, 1760. Lagrange, in the Mecanique Analytique, page 93, begins by excusing himself for attempting a problem which has nothing corresponding to it in the system of the world, where all the centres of force are in motion. He supposes the motion to be in three dimensions and obtains a solution where the forces are a/r 2 + 2yr and Plr"* + 2yr. Legendre in his Fonctions elliptiques pointed out that the variables used by Euler were really elliptic coordinates, and Serret remarks that this is the first time these coordinates were used. Jacobi took this problem as an example of his principle of the least multiplier, Crelle, xxvu. and xxix. Liouville in 1846 and 1847 gives two methods of solution, the first by Lagrange's equations and the second by the Hamiltonian equations. Serret extends Liouville's first method to three dimensions, Liouville's Journal, xiu. 1848, and gives a history of the problem. Liouville in the same volume gives a further communication on the subject. 362 TWO CENTRES OF FORCE. [CHAP. VIII. Proceeding as in Art. 577, the velocity v of the particle ex- pressed in elliptic coordinates is where the accent represents dfdt. Comparing this with Liouville's form 2T = M (P// 2 + Qz/ 2 ) in Art. 522, we may obviously solve the Lagrangian equations by proceeding as in that Article. The results are that when the work function has the form (r*-i?)U=fM + F t (v) .................. (3), we have the two integrals There is also the equation of vis viva which may be deduced from these by simple addition, viz. ........................... (5). 586. Let the central forces tending to the foci be respectively i/r^ and H^rf. We then have The integrals (4) then become ..'2 I/O 9\9 r t> W w _,, 2)2 ^ where K 1 = H 1 + H 2 , K Z H^H^ To find the path we eliminate t, (drf -(dv)* 2 (dtp ( f jf-K i )(CiJ?+K l/ ji + A)~(v 2 -h 2 )(-C^ + K^-A) (M 1 -" 2 ) 2 " The initial values of /z, p, v, v being given, the equations (7) determine the constants A, G. Another constant is introduced by the integration of (8) which is also determined by the initial values of //,, v. A fourth constant makes its appearance when the time is found in terms of either or v. ART. 588.] ELLIPTIC COORDINATES. 363 587. Ex. 1. Show that the particle will describe the ellipse defined by /X=MO> if the particle is projected along the tangent at any point with a velocity v given by Mo To prove this we notice that if the particle describe the ellipse, p. is constant throughout the motion, and the values of p', /j." given by (7) must be zero. The right-hand side of that equation must take the form C (/j. - /* ) 2 , and therefore - 2(7yu =.K 1 . Substituting for C in the equation of vis viva (5) the result follows at once. See also Art. 274. Ex. 2. A particle is projected so that both the constants A and C are zero. Show that the velocity is that due to an infinite distance and that the path is given by +B K 2 where /u= h sec 2 , v = h cos 2 and B is a constant. Ex. 3. A particle moves under the action of two equal centres of force, one attracting and the other repelling like the poles of a magnet. The particle is projected with a velocity due to an infinite distance. Show that if the direction of projection be properly chosen the particle will oscillate in a semi-ellipse, the two poles being the foci. If otherwise projected the path is given by d where v = h cos 2 + /3 sin 2 0, 2k = 1 - pjh and A = 2Hp. Ex. 4. Prove that the lemniscate, rr' = c 8 , can be described under the action of two centres of force each Hjr 3 tending to the foci, provided the velocity at the n /2F7 node is ^ A/ -3-- See Art. 190, Ex. 11. 588. To find the motion of a particle of unit mass in three dimensions under the action of two centres of force attracting according to the Newtonian law. Let the two centres of force H 1} H?, be situated in the axis of z and let the origin bisect the distance HJH.^. Let be the angle the plane zOP makes with zOx and let p be the distance of P from Oz. Since the impressed forces have no moment about Oz, we have by the principle of angular momentum (Art. 492), /*/>' = * (1). We now adopt the method explained in Art. 495. We treat the particle as if it were moving in a fixed plane zOP under the influence of the two centres of force together with an additional force p' 2 = IP/p s tending from the axis of z. This problem has been partly solved in Art. 585 ; it only remains to consider the 364 TWO CENTRES OF FORCE. [CHAP. VIII. effect of the additional force. This force adds the term B 2 /2p 2 to the work function U. Taking H lf H 2 as the foci of a system of confocal conies, let /A, v be the elliptic coordinates of P. As before, we suppose that the work function U of the impressed forces satisfies the condition d*-*)U=FM + F t (v) .................. (2). Since p is the ordinate of the conies [Art. 585], . , _ -h* p* ~/j?-h* v*-h?" The term to be added to U has therefore the same form as those already existing in U and shown in (2). To obtain the integrals we have merely to add the terms given in (3), (after multiplication by - $&) to the functions F 1} F z . In this way, we find the integrals When the central forces follow the Newtonian law, ' where K l = H 1 + H 2 , K 2 = H^ H 2 , as in Art. 586. We therefore write in the solution (4), F^ (//,) = K^, F z (v) = K 2 v. If the particle is acted on by a third centre of force situated at the origin and attracting as the distance, we add to the expression for U the term - %H 3 r* = -^H 3 (^ + v 2 - h?). The effect of this is to increase the functions F 1} F 2 by ^H 3 (/i 4 /?V 3 )> and ^H 3 (v* h?v z ) respectively. In the same way if the particle is also acted on by a force tending directly from the axis of z and equal to K/p 3 , or a force parallel to z and equal to KJz 3 , the effect is merely to give additional terms to the functions F^ and F 2 . See Art. 582. 689. Ex. A particle P moves under the attraction of two centres of force at A and B, If the angles PAB, PBA be respectively O lt 2 , the distances AP, BP be r lt r 2 , and the accelerations be Mi/ r i 2 > /*2/ r 2 2 > prove that where AB = a, C is a constant and the motion is in one plane. ART. 591.] BRACHISTOCHRONES. 365 If the motion is in three dimensions, prove that t de, 1 + ft 2 cot Q l cot 6. 2 = a (yLtj cos O l + /u 2 cos 2 ) + C, where h is the areal description round the line of centres. [Coll. Ex. 1895.] On Bracliistochrones. 59O. Preliminary Statement. Let a particle P, projected from a point A at a time t with a velocity v , move along a smooth fixed wire under the influence of forces whose potential U is a given function of the coordinates of P, and let the particle arrive at a point B at a time x with a velocity v^ . Let us suppose that the circumstances of the motion are slightly varied. Let a particle start from a neighbouring point A' at a time t + St with a velocity v + 5v . Let it be constrained by a smooth wire to describe an arbitrary path nearly coincident with the former under forces whose potential is the same function of the coordinates as before, and let it arrive at a point B' near the point I? at a time tj + 5t x with velocity v 1 + 8v 1 . According to the same notation, if x, y, z; x', y', z', are the coordinates and resolved velocities at any point P of the first path at the time t, then x + 8x, &c.; x' + Sx', &c., are the coordinates and resolved velocities at any point P' of the varied path occupied by the particle at the time t + St. Let P, Q be any two points on the two paths simultaneously occupied at the tune t. Let the coordinates of Q be x + Ax, y + Ay, &c. Then dx exceeds Ax by the space described in the time 8t, .'. Ax = Sx (x' + dx') St = dx x'St when quantities of the second order are neglected. We may regard dx, Sy, Sz, as any indefinitely small arbitrary functions of x, y, z, limited only by the geometrical conditions of the problem. We here consider two independent changes of the coordinates. There are (1) the differentials dx, dy, dz when the particle travels along the undisturbed path, and (2) the variations 3x, Sy, Sz when the particle is displaced to some neighbouring path. It follows from the independence of these two displacements that d8x = Sdx. 591. The Brachistochrone. A particle of unit mass moves under the action of forces so that its velocity v at any point is given by ^v 2 = U + C, where U is a known function of the coordinates, the constant C being also known. Supposing the initial and final positions A, B to lie on two given surfaces, it is required to find the path the particle must be constrained to take that the time of transit may be a minimum*. * An account of the early history of this problem is given in Ball's Short History of Mathematics. Passing to later times, the theorem v = Ap for a central force is given by Euler, Mechanica, vol. n. There is a memoir by Roger in Liouville's Journal, vol. xui. 1848 ; he discusses the brachistochrone on a surface 366 ON BRACHISTOCHRONES. [CHAP. VIII. The time t of transit being t = fds/v, we have to make this integral a minimum. Since a variation is only a kind of dif- ferential, we follow the rules of the differential calculus and make the first variation of t equal to zero. Let the curve AB be varied into a neighbouring curve A'B', each element being varied into a corresponding element. Since the number of elements is not altered, the variation of the integral is the integral of the variation. Writing for 1/v to avoid fractions, we have Bt =/8 (ds) =f(d$s + dsB(j>). Since (ds)* = ( d /, dx\\ . ) , = A (-r- Sac + &c. + I] \-~ - -r ( T- ) &c + &C.V ds, r \ds ) ] \\_dx as V rts/J j where the part outside the sign of integration is to be taken between the limits A to B. We notice that in this variation, G has not been varied. If C were different for the different trajectories, we should have dd> dd> d _, 8 = -^ Sx + -f- dy + -f- 5z + -, SC. dx dy * dz dC There would then be an additional term inside the integral. It follows that v* is regarded as the same function of x, y, z for all the trajectories. Since the time t is to be a minimum for all variations con- sistent with the given conditions, it must be a minimum when the ends A, B are fixed (Art. 144). We then have at these points &e = 0, By = 0, Sz = 0, and the part outside the integral vanishes. The required curve must therefore be such that the integral is zero whatever small values the arbitrary functions 8x, Sy, Sz may have. It is proved in the calculus of variations (and is and generalises Euler's theorem that the normal force is equal to the centrifugal force. Jellett in his Calculus of Variations, 1850, proves these theorems and deduces from the principle of least action that the brachistochrone becomes a free path when v = fc 2 /u'. Tait has applied Hamilton's characteristic function to the problem in the Edinburgh Transactions, vol. xxiv. 1865, and deduces from a more general theorem the above relation to free motion. Townsend in the Quarterly Journal, vol. xrv. 1877, obtains the relation v = v' in free motion, and gives numerous examples. There are also some theorems by Larmor in the Proceedings of the London Mathematical Society, 1884. ART. 593.] THE GENERAL EQUATIONS. 367 perhaps evident) that the coefficients of Bx, 8y, Bz must separately vanish. We therefore have, writing l/v for <, d_ (1\ _ d (I doc\ d/l\_d_/ldy\ _ dx\v) ds \vdsJ' dy\v) ds\vds)' dz\v ds\v ds These are the differential equations of the brachistochrone. These three equations really amount to only two, for if we multiply them by dx/ds, dylds, &c. and add the products, we find df 1 d W*Y*rfVfK\ i 4.rf/*M +Ts = 2d* rw + * (TS) + +(TS)\' which is an evident identity. 592. Supposing these differential equations to have been solved, it remains to determine the constants of integration. To effect this we resume the expression for Bt, now reduced to the part outside the integral sign. We have (dx* dy* ni fa which is to be taken between the limits A to B. Since we may vary the ends A , B of the curve, one at a time, along the bounding surface (Art. 144), this expression for Bt must be zero at each end. The variations Bx, By, Bz are proportional to the direction cosines of the displacement of the end, and dx/ds, &c. are the direction cosines of the tangent to the brachistochrone. This equation therefore implies that the brachistochrone meets the bounding surface at right angles. The expression for Bt may be put into a geometrical form which is sometimes useful. Let B(7 1} 8 due to the variation of the ends. If x , y , z ; x l ,y l , z lt are the coordinates of the ends A, B, we then have ds + &c. + S^ [^ ds + &c., J dx 1 where the &c. indicate terms with y and z respectively written for x. The conditions at the ends are then found by equating this expression to zero. 694. The equations of the braclristochrone are found by equating the first variation of the time to zero. To determine whether this curve makes the time a maximum, a minimum, or neither, it is necessary to examine the terms of the second order. For this we refer the reader to treatises on the calculus of variations. lu most cases there is obviously some one path for which the time is a minimum, and if our equations lead to but one path, that path must be a true brachistochrone. In other cases we can use Jacobi's rule. Let AB be the curve from A to B given by the calculus of variations. Let a second curve of the same kind but with varied constants be drawn through the initial point A and make an indefinitely small angle at A, with the curve AB. If they again intersect in some point C, the curve satisfies the conditions for a true minimum only if C be beyond B. 595. Theorem I. When the only force on the particle acts (like gravity) in a vertical direction, < = I/v is a function of z only, and the first two differential equations of the curve (Art. 591) admit of an immediate integration. Remembering that dx/ds = cosa, dy/ds = cos/3, it follows that the brachistochrone for a vertical force is such a curve that at every point v = a cos a, v = b cos /3, where a, ft are the angles the tangent makes with any two horizontal straight lines, and a, b are the two constants of integration. By equating the two values of v and integrating, we see that the brachistochrone is a plane curve. 596. Theorem II. Let X, Y, Z be the components of the impressed forces, the mass being unity; then since ^v*=U + C, we have X = ^dv^/dx, &c. The differential equations of the brachistochrone therefore become ds Let X, ft, v be the direction cosines of the binormal, then since the binormal is perpendicular both to the tangent and the radius of curvature dx dy dz d*x d' 2 y d*z . \j- + ft -j- + V j-=0, \-j-+LL J~ + V f- = ...... (2). ds r ds ds ds' 2 r ds* ds 2 ART. 598.] THREE THEOREMS. 369 Using the values of X, Y, Z given in (1) we find \X + fiT+vZ=0 ........................ (3), the resultant force is therefore perpendicular to the binomial, and its direction lies in the osculating plane. d z x d 2 y Let I = p -= . m = p ~- , &c. be the direction cosines of the r ds 2 r as 2 positive direction of the radius of curvature, then IX + mY+nZ Il 2 + m 2 + ri> d l (,dx -- -y- _ v 3 v p ds \vj ( ds Since the radius of curvature is at right angles to the tangent, the last term is zero, and we have - - ..................... (4). This equation proves that in any brachistochrone the component of the impressed forces along the radius of curvature is equal to minus the component of the effective forces in the same direction. 597. To find the pressure on the constraining curve. Let F lt F 2 be the components of the impressed forces in the directions of the radius of curvature and binomial. Let R 1} R 2 be the pressures on the particle in the same directions. Then by Art. 526 In a brachistochrone F^ = and F t = v*/p, hence R 2 = and ^ = -2^. 598. To find a dynamical interpretation of Theorem II. We see by referring to the equations of motion in Art. 597, that if we changed the sign of F lt the component of pressure R! would be zero, and the path would then be free. We also suppose the tangential component of force to remain unchanged so that the velocity is not altered. It follows immediately, that a brachistochrone and a free path may be changed, either into the other, by making the resultant force at each point act at the same angle to the same direction of the tangent as before, but on the other side, and still in the osculating plane. In this comparison the velocities of the particle, when free and when constrained, are equal at the same point of the path, i.e. v' = v. R. D. 24 370 ON BRACHISTOCHRONES. [CHAP. VIII. 599. Theorem III. The equations of motion of a particle P constrained to describe the brachistochrone are ^(\dx\ = ^ /1\ d_ (I dy\ _ d_ /1\ & ds \v dsj ~ dx \v/ ' ds\vds/ dy\vj' _ dy If we now write vv' = Jc 2 or, which is the same thing vds = k*dt', where v' = ds/dt', the first of these equations becomes d f ,dx\ dv 1 ___ | at' _ 1 . _ ds \ ds) dx' Now v'dx/ds being the x component of the velocity, is equal to dx/dt'. Multiplying by v' or ds/dt', the equations take the form _ _ dt'*~2 dx' W z ~Z~dy' These are the equations of motion of a free particle f moving along the same path with a velocity v' and occupying the position x, y, z at the time t'. It follows that the brachistochrone from point to point in a field U + C is the same as the path of a free particle k* 1 jfc 2 in a field U' + C', provided U' + C' = -^ -^ : i.e. v = . 4 U + C v To understand better the relation between the two fields of force we notice that if X, X' be the components of force in any the same direction at the same point, X = X' = X' = - dx ' dx ' We also notice that dt'/dt = v/v'. 600. This theorem is useful, as it enables us to apply to a brachistochrone the dynamical rules we have already studied for free motion. It also enables us to express at once the fundamental differential equations in polar or other co- ordinates. The first theorem (Art. 595) follows at once from the third, for when the force is vertical we see by resolving horizontally that v' cos a is constant. Since v' = & 2 /t>, this gives the result. To deduce the second theorem, we notice that in the free motion v' 2 /p=F 1 ', where F^ is the component of force along the radius of curvature. Using the theorems t/ = fc 2 /v, X'= -X(klv) 4 , (where X is here Fj) this becomes v 2 /p= - JFj. 601. Ex. 1. To find the brachistochrone from one given curve to another, the acting force being gravity and the level of no velocity given. The motion is supposed to be in a vertical plane. Let the axis of x be at the level of no velocity and let y be measured down- wards ; then v 2 = 2gy. By Art. 595 the curve is such that v = a cos o. This gives j/ = 2bcos 2 a, where b is an undetermined constant. This is the well-known ART. 603.] VERTICAL FORCE. 371 equation of a cycloid, having its cusps at the level of no velocity. The radius of the generating circle and the position of the cusps on the axis are determined by the conditions that the cycloid cuts each of the bounding curves at right angles ; Art. 592. Ex. 2. If in the last example the bounding curves are two straight lines which intersect the axis of no velocity in the points L, L' ; and make angles /3, j3' with the horizon, prove that the diameter 26 of the generating circle is LL'K/i - ft') and the distance of the cusp from L is 2fy3. Explain the results when the lines are parallel. 602. Ex. Show by using Jacobi's rule that the cycloid from one given point A to another B is a real minimum, the level of zero velocity being given (Art. 594). The cycloid found by the calculus of variations passes through A and B and there is no cusp between these points. Describe a neighbouring cycloid passing through A and having its cusps on the same horizontal line, the radii of the generating circles being b and b + db. Since the base of a cycloid from cusp to cusp is 2irb, it is easy to prove that the next intersection of the two curves lies in a vertical which passes between the two next cusps. The cycloids therefore cannot again intersect between A and B and the time from A to B must be a minimum. See also Art. 654. 603. Ex. Find the brachistochrone from one given curve to another when the acting force is gravity and the particle starts from rest at the upper curve. Fixing the ends, it follows, from Art. 601, that the brachistochrone is a cycloid having a cusp on the higher curve. To determine the constants of the curve, we examine the part of St due to the variation of the two ends. Let x , j/ ; x lf y t be the coordinates of the upper and lower ends, then v*=2g (y-y )- By Art. 593 we have (dx du -where Ijv and the expression is taken between limits. Now in our problem d d d I dy\ dy dy ds\ ds J ' by using the differential equation of the brachistochrone in Art. 591. We there- fore have Eemembering that = 1/u and v = a cos o, this takes the form [dx + tan a 5y]J - 5y [tan et]J = 0. When we fix the lower end, we have, since y is measured downwards, 5x 1 = 0, (/j = 0. Hence -(5x + tana 5?/ )-5y (tana 1 -tana ) = (1). When we fix the upper end, dx = 0, 5(/ = 0; /. Szj + tanctjSy^O (2). The last of these two equations proves that the brachistochrone cuts the lower curve at right angles, while the first, giving 6y () l5x = dy 1 l5x 1 , proves that the tangents to the bounding curves at the points where the brachistochrone meets them are parallel. 242 372 ON BRACHISTOCHRONES. [CHAP. VIII. 6O4. Ex. 1. A particle falls from rest at a fixed point A to a fixed point C, passing through another point B ; find the entire path when the time of motion is a minimum, (1) supposing B to be a fixed point, (2) supposing B constrained to lie on a given curve. [Math. Tripos, 1866.] The paths from A to B, B to C are cycloids having their cusps on a level with the point .1. It is supposed that there is no impact at B in passing from one cycloid to the next. The particle describes a small arc of a curve of great curva- ture and moves off along the next cycloid without loss of velocity. We have yet to find the position of B when it is only known to lie on a given curve. Taking the origin at A, and the axis of z vertically downwards, we have t> 2 =2#2. The time is given by where accents refer to the lower cycloid. \fdx f dy , dz , \-|* r 1 fdx' dy' dz'\~\ c r (8x + -f-8y + - r dz)\ + -^ I -5-7 dx + -'- Sy + , Sz }\ = 0, Jz \ds ds * ds JJ A \_^fz'\ds ds ' ds JJ B by Art. 592. Let (a, 0, 7), (a', /3', y'), (0, , $) be the direction angles of the tangents at B to the two cycloids and to the constraining curve. Then remember- ing that A and C are fixed points and that B is varied on the curve, we have (cos a cos 6 + cos |8 cos + cos y cos \//) - (cos a' cos 6 + cos /3' cos + cos y' cos ^) = 0. It follows that the tangent to the locus of B makes equal angles ivith the tangents to the two cycloids AB, EC. This determines the point B. Ex. 2. Find the curve of quickest descent from a fixed point A to another C, supposing that a screen is interposed between A and C having a given finite aperture through which the path must pass. [So long as the curve AC can be arbitrarily varied the minimum curve is found by Arts. 591, 601. Hence if the single cycloid A C does not pass through the aperture, the minimum curve must pass through a point B on the boundary of the aperture. The curve then consists of two cycloids AB, BC, and the position of B is found by Ex. 1.] [Todhunter.] 6O5. Ex. 1. If the brachistochrone is a parabola when the force is parallel to the axis, prove that the magnitude of the force is inversely proportional to the square of the distance from the directrix. [This follows from the equation v = acosa.] Prove also that the time of describing any arc PQ varies as the area contained by the focal radii, SP, SQ. [For cos a varies as 1/p, therefore dt varies as pds.] See also Art. 649. Ex. 2. A point moves in a plane with a velocity always proportional to the curvature of the path, prove that the brachistochrone of continuous curvature between any two given points is a complete cycloid. [Math. Tripos, 1875.] We here have $pds=fodx a minimum, where = (l + y' 2 ) 2 ly". The curve can be immediately found by using two rules in the calculus of variations. First, we have 5j< dx = 8x + ( Y, - Y,/) u + Y,, u' + J(- Y/ + Y H ") u dx, where Y,, Y tl , are the partial differential coefficients of with regard to y', y" ; u = Sy-y'Sx, and the part outside the integral sign is to be taken between limits. Also accents denote differentiation with regard to x. The extreme points being given, 5a:=0, Sy = at each end. Hence exactly as in Art. 591, 592, the differential equation of the curve is Y/-Y tl " = and Y tt =0 at each end. This gives F,-r,/ = A. ART. 606.] CENTRAL FORCE. 373 Secondly, the calculus of variations gives also the integral 4> = (Y l -Y H ')y' + Y ll y'' + B. Eliminating Y t ' between our two first integrals we find =Ay' + Y tt y" + B, which contains two arbitrary constants A, B. Substituting for and Y t/ , this leadsto (1 + 2/T/2/" = Ml/' + 2 # ; .-. pds = %Ady + \Bdx. Taking the straight line Ay + Sx = as an axis of , this is equivalent to /> = C sin ^ where sin \l/=di)[ds and C is a constant. This is the known equation of a cycloid. The condition Y lt =0 at each end gives y" infinite and therefore />=0. The cycloid is therefore complete. Ex. 3. Prove that the differential equation of the brachistochrone from rest at one given point A to another point B, when the length of the curve is also given, is [Airy's Tracts.] To make \ds\v a minimum subject to the condition that Jds is a given quantity we use a rule supplied by the calculus of Variations. We make J(X/t? + 1) ds a minimum without regard to the given condition and finally determine the constant X so that the arc has the given length. 6O6. Central force. Ex. 1. Prove that the brachistochrone for a central force F is given by v = Ap, where %v 2 =$Fdr and p is the perpendicular from the centre of force on the tangent. The mass is unity, as is usual in these problems. The brachistochrone is a free path for a particle moving about the same centre but with such a law of force that the velocity v' Wjv. Since v'p = ft by Art. 306, we have v = Ap. When F=/M n , and the velocity is equal to that from infinity, the differential equation v=Ap can be integrated exactly as in Arts. 360, 363. Ex. 2. Prove that the same path will be a brachistochrone for F=/j.u n and a free path for F' = fj.'u n ' if n+n' = 2, provided the velocity in each case varies as some power of the distance. For the brachistochrone and the free paths respectively, we have v* = 2fM n - l l(n - 1), v' 2 =2(*.'u n '- 1 l(ri- 1). These satisfy the condition vv'=k z if n + n' = 2, (Art. 599). Ex. 3. Prove that the ellipse is a brachistochrone for a central force tending from the focus and equal to /j.j(2a -r) 2 . [Townsend.] The conic is a free path for a force p/SP 2 tending to the focus S. Hence making the force act on the other side of the tangent as described in Art. 598, the conic is a brachistochrone for an equal force tending from the other focus H. Ex. 4. Prove that the central repulsive force for the brachistochronism of a plane curve varies as d (p*)jdr, the circle of zero velocity being given by the vanishing of p. Prove that the cissoid x (x z + y 2 ) = 2ay 2 is brachistochronous for a central repulsive force from the point (-a, 0) which at the distance r from that point is proportional to r/(r 2 + 15a 2 ) 2 , the particle starting from rest at the cusp. [Math. Tripos, 1896.] 374 ON BRACHISTOCHRONES. [CHAP. VIII. Ex. 5. Prove that the lemniscate of Bernoulli can be described as a brachis- tochrone in a field of potential jur 8 , r being measured from the node of the lemniscate, and find the necessary velocity. [See Arts. 320, 606, Ex. 2.] [Math. Tripos, 1893.] Ex. 6. A particle, acted on by a central attractive force whose accelerating jLtT* effect at a distance r is . , =-5 , a being a constant, is projected from a given point (a + T*)" with the velocity from infinity. Prove that the form of the groove in which it must move in order to arrive at another given point in the shortest possible time is a hyperbola whose centre coincides with the centre of force. [Math. Tripos.] Ex. 1. Show that the force of attraction towards the directrix of a catenary, along perpendiculars to it, for which the catenary is a brachistochrone, will vary as the inverse cube of the perpendicular. [Coll. Ex. 1897.] 607. Brachistochrone on a surface. To find the brachis- tochrone on a given surface we require only a slight modification in the argument of Art. 591. Proceeding as before, we find = i (& + &c.) + f(PBx + QSy + R8z)ds, where P=^ ---- r I T-II with similar expressions for and ax v as \v as] R. Since 8t is zero for all variations of the curve on the surface, we must have If f(x, y,z) = is the equation of the surface, the variations are connected by the one equation where suffixes imply partial differential coefficients. We must therefore have P/f x = Qff y = R/f z . The equations of a brachisto- chrone on the surface f(x, y,z)=0 are therefore given by d^l _ d dx\ I f _fd 1 _ d dy\ I ' _/d \__d_ dz\ If dx v ds vds) f \dy v ds vds/ ^ v \dz v ds vds) / If the brachistochrone is to begin and end at given bounding curves drawn on the surface, we equate to zero the integrated part of Bt, taken between the limits. Fixing the ends in turn, we see that at each end the cosine of the angle between the tangents to the curve and to the boundary is zero (Art. 592). The brachis- tochrone therefore cuts the boundaries at right angles. 6O8. By writing v = fc 2 /t>' as in Art. 599 these equations may be put into the form (ffix__dU\ I _(#y__*V \dt' 2 dx)/ J *~ \dt'* dy _ dt' 2 dz ART. 609.] ON A SURFACE. 375 These are the equations of motion of a particle moving freely on the constraining surface. It follows that the brachistochrone from point to point on a constraining surface in a field U+ C is a free path on the same surface in a field U' + C', where lv*=U+C, i*' 2 =*7' + C", vv'=t?. /fe\ 4 The relation between the component forces in any direction is F'= - F ( - I . Ex. If the particle is constrained by a smooth wire to describe the brachisto- chrone on the surface without a change in the field of force, prove that -v^sinx/p-G, v z cosx/p = H+R, -2G=R 2 , where H, G are the components of the impressed forces along the normal to the surface, and that tangent to the surface which is perpendicular to the path, and R , R 2 are the components of the pressure in the same directions. Also p is the radius of curvature of the path, and x the angle the osculating plane makes with the normal to the surface. The first is obtained by transforming the equation of motion of a free particle P', viz. v' 2 sinxlp=G' by the rule given above, the others then follow from the ordinary equations of motion of the particle P. 6O9. We may also sometimes find the brachistochrone on a given surface by making a comparison with the brachistochrone on some other more suitable surface. Let us derive a second surface from the given one by writing for the coordinates x, y, 2 of any point P some functions of , 77, f, the coordinates of a corresponding point Q. Let these functions be such that where /j. is a function of , 77, f. Geometrically this equation implies that every elementary arc ds drawn from a point P on the surface bears the same ratio to the corresponding arc dCB'B, the sides of the triangle BCB' will then be elementary and the triangle may be regarded as recti- linear. It follows that the arc CB' > CB. The time of describing CB' is > than that of describing CB because the velocity at every point in the neighbourhood of C is ultimately the same. The time of describing the line ACB is therefore less than that of describing AB' or AB. The path AB could not then be a brachisto- chrone. This proof is the same as that used by Salmon in his Solid Geometry, Art. 394, to prove the corresponding theorem for geodesies. Bertrand's theorem is now generally enunciated in a generalized form and to this we proceed in the next article. 611. A surface 5^ being given, let us draw from every point A on it that brachistochrone which starts off at right angles to the surface. Let lengths AB be taken along these lines so that the time t of transit from the surface along each is equal to a given quantity. The locus of the extremities B traces out a second surface which we may call S 2 . By Art. 592, we have St = 30- 2 COS 2 /l> 2 - Sff 1 COS #!/! . By construction cos ff 1 = for each line and, since the times of describing neigh- bouring lines are equal, dt=0. It follows that the surface S 2 a '* cuts the lines at right angles. If the surface S l is an infinitely small sphere all the brachistochrones diverge from a given point A. The locus of the other extremities of the arcs drawn from A and described in equal times is therefore an orthogonal surface. This proof may be applied to brachistochrones drawn on a given surface by expressing the conditions at the limits in Art. 607 in a form similar to that in Art. 592. This theorem though enunciated for a brachistochrone applies generally to problems in the calculus of variations. The time t may stand for any integral of the form J0 . ds where is a given function of x, y, z, and the curve is such that the integral is a minimum between any two points taken on it. 612. Ex. 1. Prove that the equations of a brachistochrone on a surface of revolution for a heavy particle with a given level of zero velocity are r^-^Av, ART. 613.] MOTION RELATIVE TO THE EARTH. 377 v 2 =2gz, where r, , z are cylindrical coordinates, z being measured downwards from the zero level. Prove also that the brachistochrone touches the meridian at the zero level. Ex. 2. A heavy particle is projected from a given point along a smooth groove cut on the surface of a right circular cone whose axis is vertical and vertex upwards, with a velocity due to the depth from the vertex. Prove that, if it reach another given point not more than half-way round the cone in the least possible time, the curve of the groove must be such as would, if the cone were developed, become a parabola with the point corresponding to the vertex as focus. [Math. Tripos, 1873.] Ex. 3. Prove that the brachistochrone on a vertical cylinder for a heavy particle with a given level of zero velocity becomes the brachistochrone on a vertical plane when the cylinder is developed on the plane. [Roger.] Ex. 4. Find the brachistochrone when the velocity at any point of space is proportional to the distance from a given straight line. Prove that the curve lies on a sphere and cuts all the circles whose planes are perpendicular to the given straight line at a constant angle, i.e., the curve is a loxodrome. [Tait.] Motion of a particle relative to the earth. 613. Let be any point on the surface of the earth and let X be its latitude. Then \ is the angle which the normal to the surface of still water at makes with the plane of the equator. Let OL = b be a perpendicular from on the axis of rotation. Let b opposite to the direction of motion of 0, i.e. in a direction due westwards from 0. When the particle has been projected from the earth it is acted on by the attraction of the earth and the applied force co 2 b. The force usually called gravity is not the attraction of the earth, but is the resultant of that attraction and the centrifugal force. The form of the earth is such that at every point of its surface this resultant acts perpendicularly to the surface of still water. Let g be this force at the point 0, then when the particle is at 0, and has been reduced to rest, the resultant force is represented by g. 378 MOTION RELATIVE TO THE EARTH. [CHAP. VIII. When the moving point P has ascended to a height h, the attraction of the earth is altered and is nearly equal tog (I 2/t/a), where a is the radius of the earth. Since h is usually not more than a few hundred feet and a is roughly 4000 miles, it is obvious that the change in the value of gravity is so small that, for a first approximation at least, we may regard gravity as a force constant in direction and magnitude. Since 27r/o> is 24 hours, we find that o> 2 a is nearly equal to #/289. Hence if we neglect gh/a we miist also neglect co 2 h at all points near 0. The applied force co-b is not neglected because at points near the equator b is nearly as large as the radius of the earth. 614. The equations of motion of a particle referred to axes moving with the earth have been already formed in Art. 499. We have here merely to express the components it 2 , &z in terms of the angular velocity co of the earth. We then substitute the values of the space velocities u, v, w in the equations of the second order and neglect all terms of the form o> 2 #, o> 2 y, co*z. We thus find dx .. d*x _ dy ~ dz v where X, Y, Z are the impressed forces other than gravity, the mass being unity. 615. It will clearly be convenient to choose as the axis of z the vertical at 0. If the axis of x be directed along the meridian towards the south and the axis of y towards the west, we have 6 l = a> cos X, 2 = 0, 8 3 = co sin X, since X is the latitude of the place. It is sometimes necessary to take the axis of x inclined to the meridian at some angle yS, the angle /3 being measured from the south towards the west. We then have 6 l = co cos X cos /3, 2 = a> cos X sin ft, 3 = co sin X. 616. If we wish the axes to move round the vertical with an angular velocity p, we have @ = pt + e, where e is some constant. ART. 618.] FALLING BODIES. 379 We then have L a> cos X, cos ft, # 2 = &> cos X sin ft, 3 = ta sin X + p. The components 1} 2 , 3 are not now constants, and in making the substitutions for u, v, w in the equations of motion their differential coefficients will not disappear. But if p be any small quantity of the same order as w, these differential coefficients are of the order w 2 . The equations of motion will then be still represented by the forms given in Art. 614. 617. As in some few cases it is necessary to examine the terms which contain w 2 , we give the results of the substitution when the axis of z is vertical, while those of x, y point respec- tively southward and westward : CbOc dii 'i |- 2w sin \ - ' ,- &>* sin 2 \x cos X -77 2&) sin X -r- &> ?/ = J, eft 2 cfa or rt" 2 .2 d/v -TTT -f 2w cos X -^ to 2 cos 2 \z eo 2 sin X cos \x = a + Z. ac 2 at 618. Ex. A particle P is attached to a point A at the summit of a high tower and when in relative rest the particle is allowed to fall freely. The point A being at a height h vertically above 0, it is required to find the point at which the particle strikes the horizontal plane at 0. Taking the axes of x, y to point due south and west, the equations of motion are x" - 2y'6 3 = 0, y" - 2z'6 1 + 2x'0 s = 0, z" + 2y'0 1 = - g, where l = u cos X, 3 = - w sin X, and the accents denote d/dt (Art. 614). We solve these by successive approximation. As a first approximation, we neglect the terms which contain w. Eemembering that initially x, y, x', y', z' are each zero and z = h, we arrive at x=Q, y = 0, z = h-$gt 2 . As a second approximation we substitute these values of x, y, z in the terms of the differential equations which contain or w. We obtain after an easy integration x=At + B, y = Ct + D-> 3 1 , z = Et + F-$gt' 2 . The particle being initially in relative rest we have x'=0, t/' = 0, z'=0, hence .4 = 0, (7=0, E=0. The initial velocities in space are not required here, but (after has been reduced to rest) these are given by w=0, v= -hO l} w = Q. To the value of v we may add the velocity of 0, viz. - ub. Also when t = 0, we have x = 0, We see from the value of z that the vertical motion is unaffected by the rotation of the earth. The time of falling is given by h = ^gt y . Since x = throughout the motion, the particle strikes the horizontal plane on the axis of y, and there is 380 MOTION RELATIVE TO THE EARTH. [CHAP. VIII. no southerly deviation. Since 1 = w cos X we have y= - Jflwcos Xt 3 ; there is there- fore a deviation towards the east which is proportional to the cube of the time of descent. This deviation is greatest at the equator. 618. Ex. 1. Show that the path of a particle falling from relative rest is nearly the curve 325a?/ 2 = cos 2 \z 3 . Ex. 2. A particle is projected vertically upwards in vacuo with a velocity F. Prove that when the particle reaches the ground there is no deviation to the south, and that the deviation to the west is 4wcosXF 3 /3<7 2 . [Laplace iv., p. 341.] Ex. 3. A particle falls from relative rest at a point A situated at a height h above the point 0. Supposing the resistance of the air to be represented by KV where v is the velocity and K a small quantity, find the effect on the easterly deviation. Measuring z upwards and neglecting the terms x'6 3 , y'0 lt as we now know that they are of the order w 2 (Art. 618), the equations of motion become y" - 20^' = - Ky', z" = - g - KZ'. The vertical motion is sensibly the same as if the earth were at rest. Substi- tuting z'= - gt in the first equation, / 3/c \ where d=djdt. This leads at once to y= - Jf/^t 3 ( l--j- M The easterly deviation is therefore slightly diminished by the resistance of the air. Ex. 4. Prove that, if the attraction of the earth on the falling particle were represented by X= -gxja, Y= -gyla, Z= -g (1-22/a), the time of falling from rest at a height h, as deduced from the equations of Art. 614, would be increased by the inappreciable fraction 5hj6a of itself. Thence show that the easterly deviation is not perceptibly altered. Ex. 5. Tlie southern deviation. A particle falls from relative rest at a point A situated on the vertical at a point on the surface of the earth. Let the southern horizontal component of the attraction of the earth be represented by .X=sinXcosX (Ax+Cz), where A and C are very small functions of the ellipticity and the angular velocity of the earth, the point having been reduced to rest. Prove that the southern deviation measured on the tangent plane at O is sin X cos X<7 4 (f 2 + 7^(7). This result is obtained by substituting the approximate values of y and z obtained in Art. 618 in the small terms given in Art. 617. Expressions for the components of the attraction of the earth are to be found in treatises on the "figure of the earth" (see Stokes' Mathematical and Physical Papers, vol. n. p. 142). These give approximately (after some reduction) C=(2m-e)20/a, where m=u a a/g and e= 1/300, hence (7 = 2w 2 nearly. 620. Two cases of motion. Two special cases of the motion of a particle deserve attention ; (1) when the particle in its motion does not deviate far from the vertical and (2) when the motion is nearly horizontal. ART. 621.] TWO CASES OF MOTION. 381 Supposing the axis of z to be vertical, the horizontal velocities dx/dt and dy/dt are small compared with the vertical velocity dz/dt in the first case. The products of the horizontal velocities by co are therefore of a higher order of small quantities than the product of the vertical velocity by co and should be neglected in a first approximation. In the second case, on the contrary, dz/dt is small and we neglect its product by co. The two sets of equations are therefore as follows (Art. 614) : d 2 x dz d 2 x ^dy ? X \ d? dt 3 , dt 2 "* dx Z dt { 9 3 =Y, d 2 z dt 2 " 9 dx dt 9 +2^(9 =-g + Z. dt ' i "" y g dt 2 dt d 2 z _ dt 2 ~ We notice that when the motion is nearly vertical the com- ponents 6 l , #2 enter into the equations, while 3 does not appear until we proceed to higher approximations. It is therefore the component of the angular velocity about a tangent to the earth which affects the motion. On the other hand when the motion of the particle is nearly horizontal it is the component of the earth's rotation about the vertical, viz. 3 , which plays the principal part. If we compare the x and y equations for the case in which the motion is nearly horizontal with those given in Art. 614, when the square of co is neglected we see that they express the motion of a particle moving freely in space but referred to axes which turn round the vertical with an angular velocity 3 . If, as is generally the case, the forces X, Y are either zero or in- dependent of the changes of the nearly constant quantity z, we can thus obtain these equations in an elementary way. The particle moves freely in space, unaffected by the rotation of the earth, but the axes of reference move round the vertical and leave the particle behind. This geometrical interpretation of the equations may be made more evident by considering some simple cases. 621. As an example consider the case of a pendulum. When the bob makes small oscillations the motion is nearly horizontal. To construct the motion we suppose the pendulum to oscillate freely in space (with the proper initial conditions). This oscillation is left behiud by the earth, and the effect is that the plane of 382 MOTION RELATIVE TO THE EARTH. [CHAP. VIII. oscillation appears to revolve about the vertical with an angular velocity equal and opposite to the vertical component of the earth's angular velocity. The plane of oscillation therefore turns from west to south with an angular velocity w sin X. This problem is more fully considered in Art. 624. 622. Flat trajectories. A bullet is projected from a gun, situated at the point O, with a great velocity V, in a direction making a small angle a with the horizon so that the trajectory is nearly flat. It is required to find the motion. The initial velocity of the bullet in space (after has been reduced to rest) is V. After leaving the gun the bullet describes a parabolic path in space, while the axes of reference turn with the earth round the vertical at 0, and the bullet is left behind by the axes (Art. 620). Supposing that the initial plane of xz contains the direction of projection, the coordinates of the bullet at the time t are evidently x = Vt cos a, y = - x6 3 t where 6 3 = - u sin X. The deviation y is therefore always to the right of the plane of firing in the northern hemisphere, and to the left in the southern hemisphere. If R be the range the whole deviation is Etu sin X. We notice also that the deviation ?/ is independent of the azimuth of the plane of firing, and that the time of describing a given distance x is independent of the rotation of the earth. The third equation of motion (Arts. 614, 615) gives cfiz dx -T-;J= -0 + 20 2 , /. z = Vt sina- \gtf- Fwt 2 cos a cos X sin /3, where 2 = -wcosXsin/3 and is the angle the plane of firing makes with the meridian. The vertical deviation of the bullet from its parabolic path at the moment of reaching a target distant x from the gun is therefore - xtu cos X sin p. 623. Deviation of a projectile. Ex. A particle is projected with a velocity V in a direction making an angle a with the horizontal plane, and the vertical plane through the direction of projection makes an angle /3 with the plane of the meridian, the angle /3 being measured from the south towards the west. If x is measured horizontally in the plane of projection, y horizontally in a direction making an angle /3 + ^ir with the meridian, and z vertically upwards from the point of projection, prove that x= V cos at + (V sin at 2 - gt 3 ) u cos X sin /3, y = (V sin at 2 - gt 3 ) u cos X cos /3 + Fcos at 2 u sin X, z=Vsinat-^gt z - V cos at 2 u cos X sin , where X is the latitude of the place, and w the angular velocity of the earth. Prove also (1) that the increase of range on the horizontal plane through the point of projection is 4w sin /3 cos X sin a ( sin 2 a - cos 2 a) F 3 /sinX+p becomes zero, as explained in Art. 616. The equations of motion are now ml' ~dt z where T is the tension of the string, and lt 2 have the values given in Art. 616. The third equation proves that the tension T differs from mg by quantities of the order Iwo. at least. Since x/l and y\l are of the order a, and we have agreed to reject terms of the order wet 2 , we must put T = mg in the two first equations. Since the two first equations are independent of &>, the motion of a real pendulum when affected by the rotation of the earth is the same as that of an ideal pendulum, unaffected by the rotation, but whose path, viewed by a spectator moving with the earth, appears to turn round the vertical with an angular velocity p = G> sin X in a direction south to west. If In? = g } the solutions of the equation are clearly x = A cos(nt + C), y = Bsm(nt + D) ......... (2). It appears that the time of oscillation, viz. 27T/W, is unaffected by the rotation of the earth. To determine the constants of inte- gration, we notice that when the particle is drawn aside from the vertical and not yet liberated, it partakes of the velocity of the earth and has therefore a small velocity relative to the axes. This is equal to law sin X and is transverse 'to the plane of displacement. Taking the plane of displacement as the plane of xz at the time t = 0, the initial conditions are x = lct, y = Q, dxjdt = 0, dyjdt = law sin X. 384 MOTION RELATIVE TO THE EARTH. [CHAP. VIII. It is then easy to see that A = la, En-- law sin \, (7=0, D = 0. The particle therefore describes an ellipse whose semi-axes are A and B. Since the ratio of the axes, viz. a> sin X \l(ljg) is very small, the ellipse is very elongated and the particle appears to oscillate in a vertical plane. The effect of the rotation of the earth is to make this plane appear to turn round the vertical with an angular velocity &> sin X. 635. It is known that, independently of all considerations of the rotation of the earth, the path of the bob of a pendulum is approximately an ellipse whose axes have a small nearly uniform motion round the vertical. This progression of the apses vanishes when the angle subtended at the point of suspension by either axis of the ellipse is zero ; see Art. 566. As the presence of this progression will complicate the experiment, it is important (1) that the angle of displacement should be small, (2) that the pendulum when drawn aside should be liberated without giving the bob more transverse velocity than is necessary. This is usually effected by fastening the bob when displaced to some point fixed in earth by a thread, and when the mass has come to apparent rest it is set free by burning the thread. The progression of the apses due to the angular magnitude of the displacement is in the opposite direction to that caused by the rotation of the earth. The advantage of using a long pendulum is that the linear displacement of the bob may be considerable though the angular displacement of the wire is very small. The bob should also be of some weight, for otherwise its motion would be soon destroyed by the resistance of the air; Art. 113. 626. As we have rejected some small terms it is interesting to examine if these could rise into importance on proceeding to solve the equations (1) to a second approximation. To determine this we substitute the first approximation of Art. 624 (2) in the differential equations. The third equation shows that T/m - g has two sets of terms. First, there are terms independent of w which lead to the solution already obtained in Art. 555, and need not be again considered here. Next, there are terms which contain w as a factor and have the form sin(n) where J3=pt, Art. 616. These when multiplied by x\l or yjl give no terms of the form sin nt or cos nt. None of the terms which contain u can rise into importance (Art. 303). 627. The idea of proving the rotation of the earth by making experiments on falling bodies originated with Newton. But more than a hundred years elapsed before any observations of value were made. In 1791 Guglielmini of Bologna made some experiments in a tower 300 feet high. The liberation of the balls was effected by burning the thread by which they were suspended, and this was not done until they had entirely ceased to vibrate as observed by a microscope. The vertical was determined by a plumb line, but he had to wait several months before it came to rest. The results were disappointing for they showed a deviation towards the south nearly as great as that towards the east. This discrepancy was due to two causes, (1) the numerous apertures in the walls of the tower caused slight winds, (2) the vertical was not ascertained until a change in the seasons had ART. 628.] INVERSION OF THE PATH. 385 altered its position. Other experiments were made by Benzenberg about 1802 in Hamburg, but Eeich's experiments in 1831 3 in the mines of Freiberg are generally considered to be the most important. The height of the fall was 158^ metres and the mean of 106 experiments gave a deviation to the east of 28 millimetres, the deviation to the south being about a twentieth of that towards the east. These were the experiments that Poisson selected to test the theory; he showed that the observed easterly deviation was within a thirtieth of that given by calculation. Poisson also investigates the general equations of motion of a particle relative to the earth and obtains equations equivalent to those given in Art. 617. He then applies them to a variety of problems. Journal de I'ecole poly technique, 1838. The defect of experiments on falling bodies is the smallness of the quantities to be measured. In 1851 Foucault invented a new method; he showed that the plane of oscillation of a simple pendulum appeared to rotate round the vertical with an angular velocity equal and opposite to the component of the earth's angular velocity. The advantage of this method is that the experiment can be continued through several hours, so that the slow deviation of the pendulum can be (as it were) integrated through a time long enough to make the whole displace- ment very large. Foucault's experiment was widely repeated with many improve- ments. Among English experiments we may mention those by Worms in 1859 at King's College, London, in Dublin by Galbraith and Haughton, at Bristol, at Aberdeen, at Waterford in 1895. The accuracy of the method is such that it is possible to deduce the time of rotation of the earth. Foucault's observations gave 23 h , 33 m , 57 s , while the repetition of the experiment at Waterford led to 24 h , 7 m , 30 s , the true time lying between the two (see Engineering, July 5, 1895). Though the experiment can be easily tried when only the general result is required, yet many difficulties arise when the deviation has to be found with accuracy. Indeed Foucault admitted that it was only after a long series of trials that he made the experiment succeed (see Bulletin de la Societe Astronomique de France, Dec. 1896). Inversion and Conjugate functions. 628. Inversion*. Let a point P of unit mass move under the action of forces whose potential in polar coordinates is U =f(r, 0, <). Produce any radius vector OP of the path to Q, where OP . OQ = k 2 ; the locus of Q is called the inverse path of that of P and any two points thus related are called inverse points. Let OP = r, OQ = p. Let P', Q' be two other inverse points near the former, then since OP . OQ = OP' . OQ', a circle can be described about the quadrilateral PQP'Q'. The elementary arcs PP', QQ' are there- fore ultimately in the ratio r: p. If the points P, Q move so as * The reader may consult a paper by Larmor in The Proceedings of the London Mathematical Society, vol. xv. 1884. The principle of least action is there applied to both the method of Inversion and that of Conjugate functions. R. D. 25 386 INVERSION. [CHAP. vin. to be always inverse points, their velocities u, u lt are connected by the equation M/MI = r/p. The position of the point P in space is determined either by the quantities (p, 6, $) or (r, 0, ). Choosing the former as the coordinates, the Lagrangian equations of the motion of P are deduced from T = * u* = } - 4 (p'* + pW* + ? sin 2 0f ), p These equations contain only the polar coordinates of Q. They primarily give the motion of a point Q describing the inverse path in such a manner that P and Q are always at inverse points. Let us now transpose the factor &*/p 4 from T to U. We then have (Art. 524) The Lagrangian equations derived from these give the motion of a particle which describes the same path as that of Q, but in a different time. Let the particle be called II. The form of T shows that II moves as a free particle, acted on by forces whose potential is U z . We see also that the masses of the particles P and II are equal. See also Art. 650, Ex. 2. The path of either particle may be inferred from that of the other. If the path of the particle P described with a work function f(r, 0, <) + (7 is known, then the other particle II, if properly pro- jected, will describe the inverse path, with a work function 629. To find the relation between the velocities u, v of the particles P, II, when passing through any inverse points P, Q, we notice that by the principle of vis viva ^u* = U+C, |t; 2 = i7 2 . It follows immediately that v = uk?/p 2 , and therefore that ur = vp. Since the planes of motion OPP', OQQ 7 coincide, the angular momenta of the particles, when at inverse points of their paths, about every axis through the centre of inversion are equal. The constant G is determined by the consideration that the ART. 631.] THE PRESSURES AND THE FORCES. 387 known velocity u in the given path must satisfy the equation The particles P, IT do not necessarily pass through inverse points of their respective paths at the same instant. Let t, r be the times at which they pass through any pair P, Q, of inverse points; t + dt, r + dr the times at which they pass through a neighbouring pair P', Q' of inverse points. Since the elementary arcs PP', QQ' are in the ratio r : p while the velocities of P, II are in the ratio l/r : l/p, it follows by division that the elementary times dt, dr are in the ratio r* : p*. The relation between t and r dt T 2 is found by integration from -7- = . This agrees with the ratio given in Art. 524. Supposing that the particles P, II are projected from inverse points on their respective paths, their initial velocities must be inversely as their distances from the centre of inversion. The initial directions of motion must be in the same plane and make supplementary angles with the radius vector which passes through both the initial positions. 630. If the particle P is constrained to move on a surface the argument needs but a slight alteration. The inverse point Q describes a curve which lies on the inverse surface. Let (p, 0, (p) be the polar coordinates of Q ; then these may also be taken as the Lagrangian coordinates of P. Using the equation of the inverse surface, we have p' -A 0' + -r $' Substituting the values of p, p' in the do u

1 dA 1 dU 2 - = - n - * -T- p sm 6 dr p sin d

. [This follows easily from the expression for F given in Art. 310. When h = h', Art. 629, this agrees with Art. 631, Ex.] Ex. 4. A particle P moves on a sphere under the action of a centre of attractive force situated at a point on the surface, and the velocity v at any point is JB/r 8 where r=OP. Prove that the path is a circle whose plane passes through 0. Inverting the sphere, we find that the stereographic projection is a straight line. The result follows at once, see Art. 609. 633. Conjugate functions. Let the Cartesian coordinates {x, y), (, 77) of two corresponding points P, Q be so related that ^') 00' ART. 634.] CONJUGATE FUNCTIONS. 389 where /is any real function and i = V( 1). Expanding the right- hand side we have a> + y*' = 0(f i7) + ^(f 17)1 .................. (2), where and -fy are real functions. The transformation is therefore effected by using the equations * = *('/), y = *(,'?) .................. (3), the motion of P following geometrically from that of Q. Differ- entiating (1) we find .-. x '* + y'* = ^.{?* + v' 2 } .................. (4), where /x, 2 is a real positive quantity given by A*' =/'( + *)/'( -17*') .................. (5). Let U=F(x,y) be the work function of the forces which act on the particle P. The motions of P and Q may be deduced by the Lagrangian rule from the constant of U being included in F for the sake of brevity. Transposing the factor yu, 2 to the work function, the equations give by the same rule the motion of a particle II, whose mass is equal to that of P, which (when properly projected) will describe the same path as the point Q, but in a different time, Art. 524. To find the relation between the velocities u, v of the particles P, IT at corresponding points of their paths, we observe that since ^u- = U, \tf = U 2 , the velocities are such that v = fiu. To find the ratio of the times dt, dr we notice that, by (4), the corresponding arcs ds, da- are such at ds = fjuda; while pu = v. It follows by division that dt = p?dr. 634. Ex. It is known that a particle can describe the ellipse x 2 /a 2 + ?/ 2 /6 2 =l, with a force tending to the centre equal to KT. It is required to find the conjugate path and law of force when we use the transformation xyi = (t}i) n lc n ~ 1 . Let x = rcos6, y = rsin0; =pcos#, ?7 = psin0; the equation of transforma- tion then gives r=p n /c"- 1 , e = n. The equation of the path is therefore cos 2 n sin 2 n0 _ c 2 " ~ 2 ~ = ~ ' 390 GROUPING OF TRAJECTORIES. [CHAP. VIII. Also, fjf =/'( + 7,0 /'(f-T^n'^ + ^-'/c 2 "- 2 ; .: /x = np n ~ 1 / cn ~ 1 - Again in the elliptic orbit, w 2 = 2 ( U+ C) = K (a 2 + 6 2 - r 2 ). Hence since r=/xu, The ratio of the angular momenta, viz. vpjur, is easily seen to be equal to n. When n= -1, this transformation becomes r = c 2 /p, 6= - is measured positively in the opposite direction to 6. 635. Ex. If the particle P is constrained to move on any given curve with a work function U, while the equal particle II is constrained to move on the conjugate curve, with a work function U%=fj?U, the pressures R lt R 2 on- the two curves are in the ratio of the cubes of the velocities, i.e. R^ii^-R^v 3 . This gives also R 2 =fi 3 R l , The grouping of trajectories and Jacobi's solution. 636. The Cartesian equations of the motion of a free particle of unit mass are ,,- y - and to these we join the equation of energy v* = x' 2 + y' 2 + z' 2 = 2J7+2C ............... (2). When the equations (1) have been integrated we have x, y, z expressed by three functions of t with six constants whose values become known when the initial values a, b, c of the coordinates and the initial velocities a', b', c' are given. Since t enters into the equations (1) only in the form dt, the differential equations are not altered by writing t + e for t. One of the constants of integration therefore enters into the solution as a mere addition to the time. When we eliminate the time we arrive at two equations which are the equations of all the possible trajectories in space. The constant e disappears with t, and the equations of the possible trajectories contain five constants, of which the energy C may be regarded as one. To understand the ART. 637.] ACTION. 391 relations of these trajectories to each other it becomes necessary to group them into systems. We first group the trajectories according to the values of the energy 0. Taking any one group, having any given energy, the four remaining constants are determined for any special trajectory when the coordinates of some two points A, B arbitrarily chosen on it are given. 637. Action. If ds be an element of the arc of the trajectory, the integral V=Jmvds is called the action as the particle passes from A to B. If mv z be the vis viva of the particle in any position we also have V=fmv 2 dt, the limits being the times ti and t of passing through A and B. When we are only concerned with the motion of a single particle, it is convenient to suppose its mass to be taken as unity. Considering a single particle, let s be measured from A to B along the trajectory of least action and let the length AB be L Let A'B' be a neighbouring trajectory (Art. 590) from some point A' near A to a point B' near B. Proceeding as in Art. 591, writing v for <, we find dx o dv d / dx\ ' dv 1 , [[(dv d / . + \j---r- [ J J [_{dx ds\ , , t> J-B + C. + j---r- v - j - ds J J [_{dx ds\ ds where the part outside the integral is to be taken between the limits A and B and the energy C has been varied for the sake of generality. It is easy to deduce from the equations of motion (as in Art. 599) that the coefficients of Bx, By, Bz inside the integral are zero. Also since t> 2 = U+C, we have vdv/dC = l. Since vdx/ds is the x component of the velocity we thus have B V = x'Bx + y'8y + z'Bz - a'Ba - b'Sb - c'Bc + (t- *0 BC. . .(4). When we consider the motion of a system of particles, either constrained or free, and all taking different paths, it is more convenient to take t as the independent variable. Let us imagine the system to be moving in some manner which we will call the actual course. Let the work function of the field be U and let L be the Lagrangian function, then L = T+U (Art. 506). Let ff 1 , 6 2 , &c. be any indepen- dent coordinates of the system, a lt a. 2 , &c. their values in some position A occupied by the system at a time ^. Then 0,, 2 , &c. are functions of t, whose forms it is our object to discover. Let us next suppose the system to move in some varied manner, i.e. let the coordinates be functions of t slightly different from those in the actual course. By 392 JACOBl'S SOLUTION. [CHAP. VIII. the fundamental theorem* in the calculus of variations, we have d dL where u = 80 - 6'St, 2 implies summation for all the coordinates 9 l , 6. 2 , &c. and the limits of integration are t l and t. Since each separate term inside the integral vanishes by Lagrange's equations (Art. 506), we have If the geometrical conditions do not contain the time explicitly T will be a /77 1 homogeneous function of t ', 2 ', &c. (Art. .510) and therefore 2 ,0' = 2T. We also suppose that for each varied course the velocities are so arranged that the principle of energy holds, i.e. T - U=C, though C may be different for each course. Hence L=2T-C, and s]cdt = S{C (t-tj}. We now have the two equations .................. (A) S 50 -2 , Sa ..................... (B). The action F of the system is the sum of the actions of the several particles. We therefore have V \1Tdt. When the system reduces to a single particle of unit mass 2T=x' 2 + y" 2 + z' 2 , and the equation (B) becomes the same as (4). 638. Let us consider the motion of a single free particle and let the energy C be given, therefore 8C = 0. Let v x , v 2 be the velocities at A, B; 8a- 1} S}*>J(2C) ........................ (17). If we choose the first integral the surfaces V=K are planes and the trajectories are grouped into systems of parallel lines, the lines taking all directions. If we choose the second integral, the surfaces V = K are spheres having their centres on the plane of xy. The trajectories are grouped into systems of straight lines diverging from points on that plane. To illustrate the use of equations (9) let us substitute in them the second integral. We have at once where r a =(x-a)* + (y-p)- + z*. These evidently give a system of straight lines diverging from the point x = a, y=fi, 2=0, described with a velocity V(2C). 644. When the coordinates chosen are not Cartesian the expression for the kinetic energy does not take the simple form given in (2). Let the kinetic energy T be given by 2T = P6' 2 + Q' 2 + RV .................. (19), where P, Q, R are functions of the coordinates 0, , i/r. Let us now take as the Hamiltonian equation * 2 ART. 645.] JACOBl'S SOLUTION. 397 Proceeding exactly in the same way as before, we prove that if F=/(0,4>,t,tf,,/3) + 7 ............... (21), be an integral of (20), the first integrals of the Lagrangian equations of motion (Art. 506), are The trajectories, &c. are given by where a 1} /3 1} and e are new constants. This enunciation includes the most useful cases of Jacobi's rule. But his method applies also to any dynamical system, in which T is a quadratic function of the velocities. For these generalizations we refer the reader to treatises on Rigid Dynamics. 645. Ex. 1. Apply Jacobi's rule to find the path of a projectile. The Hamiltonian equation is Separating the variables, we find that one complete integral is V = V(2a) x - ^- (2(7 - 2a - 2gy)% + y. Ex. 2. Apply Jacobi's method to find the path of a particle in three dimensions about a fixed centre of force which attracts according to the Newtonian law. Taking polar coordinates we have 2 00' !! , U=-. The Hamiltonian equation (Art. 644) may be put into the form {*(%)'-* } +(%)'+,(%)'=* If we equate these three expressions respectively to a, -a + /3cosec 2 aud - j8 cosec 2 6, we obtain three differential equations in which the variables are separated and whose solutions satisfy the Hamiltonian equation. Let the inte- grals of these be V=f 1 (r, a), F=/ 2 (0, a, /3), V=f a (tf>, /3). It is obvious that F=/ 1 +/ 2 +/3 + 7 is a complete integral from which all the trajectories may be deduced. Ex. 3. Apply Jacobi's method to find the motion of a particle in elliptic co- ordinates (X, /j., v) when the work function is (M 2 - v 2 ) /, (\) + (v* - x 2 ) /, (AC) + (\ 2 - M 2 ) / 3 00 (X'-MW-^M" 2 -^) Taking the expression for T given in Art. 577, the Hamiltonian equation (Art. 644) after a slight reduction becomes 398 JACOBI'S SOLUTION. [CHAP. VIII. (X 2 _ 7,2) (X 2 _ **) ( M 2 _ ) _ + (^ _ 7,1) (^ _ k *) (* _ x 2 ) + ( - ft*) (, 2 - fc 2 ) (X 2 - M 2 ) = - 2 { ( M 2 - ,")/! ( X) + (^ _ x 2 )/ 2 ( M ) + (X 2 - M 2 ) / () } - 2CD, where D = (X 2 - M 2 ) (M 2 - " 2 ) (" 2 - X 2 ) Since (M 2 - " 2 ) + (" 2 - X 2 ) + (X 2 - (**) = 0, X 2 (M* - * 2 ) + M 2 (* a - X 2 ) + " 2 (X 2 - M 2 ) = 0, X 4 ( M 2 - * 2 ) + M 4 (" 2 - X 2 ) + " 4 (X 2 - M 2 ) = - D, the differential equation is satisfied by assuming //iF\ 2 (X 2 - ft 2 ) (X 2 - fc 2 ) f ^ J = - 2/, (X) + a + /3X 2 + 2CX 4 , with similar expressions for dVjdfj. and dV/dv. In these trial solutions the variables X, fj., v have been separated, the first containing X, the second /t, and the third v. Supposing the integrals to be V=F 1 (X, o, /3, C), V = F Z (/a, 2 dt or JW is positive and therefore, whatever path from A to B may be taken, we can increase the whole action by conducting the particle along a sufficiently circuitous but neighbouring path. Thus, if C be any point on the free course AB we can conduct the particle along that course to C, then compel it to make a circuit, and after returning to the neighbourhood of C conduct it along the remainder CB of the free path. Additional positive terms are thus given to the integral and the action is increased. The energy of the motion is unaltered, but the time of transit is longer. Since every element of the integral is positive, there must be some path joining A and B which makes the action a true minimum. If the theory of max-min in the Calculus of Variations gives only one path, that path must be a minimum. 648. It may be that there are several free paths by which the particle could travel from A to B. Selecting one of these, say ADB, we may ask if the action along it is a true minimum. Let a neighbouring free path starting from A (the energy being the same) intersect ADB in C. To simplify matters let no other free path intersect ADB nearer to A than C. If B lie between A and C there is only one free path from A to B which is in accordance with the principles of mechanics, and that path makes the action a true minimum ; Art. 647. If B is beyond C, there are two neighbouring free paths from A to C. It may be proved that the action from A to B is not in general a true minimum, the action for some neighbouring courses being greater and for others less than for the free path AB (Art. 653). 649. It may be that there is no free path from A to B, yet there must be a path of minimum action. For example, a heavy particle projected from A with a given velocity can by a free path arrive only at such points as lie within a certain paraboloid whose focus is at A, Art. 159. The path of minimum action from A to a point B beyond the paraboloidal boundary is not a free path. When deduced from the Calculus of Variations it falls under the case mentioned in Art. 646. Its position is such that it cannot be varied arbitrarily on all sides, i.e. the signs of the variations dx, Sy, Sz are not arbitrary along the whole length of the course. Such limitations exist when the path runs along the boundary of the field of motion (Art. 299). We therefore draw verticals from A and B to intersect the level of zero velocity (which in this case is the directrix) in C and D. Let us conduct the particle from A along AC to a point as near C as we please, and thence along a course coinciding indefinitely nearly with the directrix to a point as near D as we please. The particle is finally conducted along the vertical DB to the given point B. Throughout this course the velocity is always supposed to be ^/(2gz) where z is the depth below the directrix. The velocity being ultimately zero along the directrix the whole action from A to B is reduced to the sum of the actions along the vertical paths AC, DB. The path close to the directrix cannot be varied arbitrarily, because the particle cannot be conducted above that level without making the velocity imaginary. This minimum path is therefore not given by the ordinary rules of the Calculus of Variations. A similar anomaly occurs in the case of brachistochrones. The parabola is a brachistochrone when the force acts parallel to the axis and is such that the velocity is inversely proportional to the square root of the distance from the ART. 651.] EXAMPLES. 401 directrix; Art. 605. The directrix being given in position, the initial and final points A, B of the course may be so far apart that no such parabola can be drawn. In this case the brachistochrone is found by conducting the particle along the vertical straight line A C in accordance with the given law of velocity, thence with an infinite velocity along the directrix CD, and finally along the vertical line DB to B. The further discussion of these points is a part of the Calculus of Variations. Some remarks on the dynamics of the problem may be found in the author's Rigid Dynamics, vol. n. chap. x. 650. Ex. 1. Prove that the same path is a brachistochrone for v 2 =/ (x, y, z) and a path of least action for v' 2 =Ajf(x, y, z); Art. 599. The brachistochrone is deduced from the calculus of variations by making Jds/v a minimum ; the path of least action by making fa'ds a minimum. These must give the same curve if v'=k 2 jv ; (Jellett and Tait). Ex. 2. Prove that, if a path be described by a particle P with such a work function that v z =f(r, 6, ), the inverse path can be described by a particle II with fc 4 / Jk 2 \ a velocity t/, such that v' 2 =-^f( , 0,

2 taken between the limits is zero, since both paths begin at A and end at B. Let us choose the function X so that (5), (6). Since this integral is essentially positive it follows from (3) that the action along every varied path from A to E is greater than that along the circle. This argument requires that X should not be infinite within the limits of integration. By taking pa = ^tr e where e is a quantity as small as we please the values of X given by (5) can be made finite from = to 0=7r/p-e' where e' is a quantity as small as we please. The argument therefore requires that the point B should not make the angle AOB>7rjp. When the angle AOB is greater than irjp we can prove that tJie action along some varied curves extending from A to B is less, and along others is greater, than that in the circle. To prove this let us conduct the particle from A to B along the varied path whose equation is p=Lsmg6. Let ft be the angle AOB, then since p vanishes at each end, g is arbitrary except that gfi is a multiple of TT. Since pfi>ir one value at least of g is less than p and the others are greater than p. Substituting in (3), we find that the integral is the limits being = to 6= ft. The smaller values of g make I negative, while the greater values (which correspond to the more circuitous routes) make I positive. The conclusion is that when the angle AOB>irlp, the action along the circle is not a true minimum. 654. Ex. A particle moves in a plane with a velocity v = (x, y) beginning at a given point A and ending at B. The path taken being that of minimum action, it is required to find in Cartesian coordinates the equation of the path and the change of action when the path is varied in an arbitrary manner. Let the elementary action vds = f> J(l + y' z ) dx be represented by f(x, y, p) dx, where p has been written for y'=dyjdx. Then writing y + 8y, p + Sp for y and p, * Lagrange Th^orie des fonctions Analytiques 1797. He refers to Legendre, Memoirs of the Academy of Sciences 1786, and adds that it must be shown that X does not become infinite between the limits of integration. Not being able to settle this question, he just missed Jacobi's discovery. See also Todhunter's History of the Calculus of Variations, page 4. 26-2 404 PRINCIPLE OF LEAST ACTION. [CHAP. V11I. (but not varying x) the whole increase of action on the varied curve is by Taylor's theorem, SA = \[f v Sy +f p 5p + } {/ (5 +/-( + Ac.] dx, where suffixes as usual represent partial differential coefficients. Integrating the second term by parts, as in Art. 591, we have 8 A = [f p Sy] + /{(/- f p ') &y + Ac. } dx, where the part outside the integral, being taken between fixed limits, is zero, and accents denote total differentiation with regard to x. The path of minimum action is found by equating the coefficient of 8y to zero, Art. 591. This path is therefore given by f v -f?' = ....................................... (1), and the change of action in any varied path by 8A=^[f in/ (8y)*+2f yp 8ySp+f pp (Sp)*]dx ..................... (2). To find the path in Cartesian coordinates we integrate the equation (1). This can only be effected when the form of the function

3w 2 . In an ellipsoid A > B if the axis in the direction of is less than that in the direction of 17. It follows that if the axis of is the least axis, A is greater for an ellipsoid than for a sphere. The swarm is therefore more stable for an ellip- soidal than for a spherical swarm provided the least axis of the ellipsoid is placed along the radius vector from the sun. Let us suppose that all the particles are describing the same principal oscillation. The projections of their paths on the plane 77 are therefore given by = acos0, r) = b sin 6, where B =pt + a. These paths are coaxial ellipses described in the same periodic time 2ir/p, the semi-axes of any ellipse being a, b. By substituting these values of , t\ in the second of equations (I), we find r ^ ; it follows that all o znp the ellipses are similar to each other. There will therefore be no collisions between the particles. ELLIPSOIDAL SWARM. 407 The ratio of the axes of the ellipses is not altogether arbitrary. By using (III) we find where A, B and therefore^ 3 are known functions of the ratios of the axes of the ellipsoid. We may deduce from the values of A, B given in the theory of Attrac- tions that Aa? is less or greater than Bb- according as a 2 is greater or less than 6 2 . It then follows from this equation that in both the principal oscillations the axis of the ellipsoid in the direction of the radius vector from the sun is less than the axis of the ellipsoid in the direction of motion of the centre. If P, Q, R be any three particles describing similar co-axial ellipses in the same time with an acceleration tending to their common centre, it is not difficult to prove that the area of the triangle PQR is constant throughout the motion. Let us apply this theorem to the motion of the projections of the particles on the plane of ;. Joining adjacent triads of particles, we divide the whole area into elementary triangles. If the swarm is homogeneous, the areas of these triangles are initially equal and we see that they will remain equal throughout the motion. The swarm will therefore remain homogeneous. Consider next the motions of the particles perpendicular to the plane of ij. These are harmonic oscillations and are all described in the same time lirjq. The amplitude of each oscillation is the ordinate of the ellipsoid corresponding to the ellipse described by the projection and this is constant for the same particle. The distance between two adjacent particles moving in the same ordinate in the same direction is increasing or decreasing according as they are approaching or receding from the plane of 17. As there are as many particles approaching as receding, the uniformity of the density is not affected by this motion. When both the principal oscillations are being described simultaneously the state of the motion becomes more complicated. The outer boundary is not strictly ellipsoidal, being dependent on both the states of motion. Since also the rotations in the principal oscillations are in opposite directions, we can no longer neglect the collisions between the particles. To take account of the collisions we must have recourse to a statistical theory analogous to the kinetic theory of gases. But this would lead us too far from the methods of this treatise. For an example of the application of the kinetic theory the reader is referred to a memoir by G. H. Darwin, On the mechanical conditions of a swarm of meteorites, dfcc., Phil. Trans. 1889. He supposes a number of meteorites to be falling together from a condition of wide dispersion and to have not yet coalesced into a system of a sun and planets. No account is taken of the rotation of the system. Callandreau has discussed the case in which a comet, regarded as a spherical swarm of particles, is heterogeneous, the density being a function of the distance from the centre. The effect of a passage near Jupiter has also been taken into account. See his Etude sur la theorie des cometes periodiques. He considers it probable that the periodic comets are undergoing a gradual disintegration and he points out that according to this hypothesis a few comets captured by the action of Jupiter could by repeated subdivisions produce all those known to exist. See The Observatory, Feb. 1898. LAGRANGE'S EQUATIONS. Note on Art. 524. THIS rule may be put into another form. We know that ifL = T+C7+<7be the Lagrangian function and 0, , &c. the coordinates, the equations of motion are d_d_L_dL d_dL^_dL dt dO'~d0' dtd$'~d4>' ......................... ( We now see that we may use the same equations, if we substitute (2), where M is any arbitrary function of the coordinates 0, , &c. which we may find suitable when solving the equations. The expression for T 2 differs from T only in the fact that the differential co- efficients are taken with regard to a different independent variable, which has been represented by T. Thus When the equations luive been solved the paths of the particles are found by eliminating T without enquiry into its meaning. The equation of energy is supposed to be T- U= C the constant C is therefore known when the initial values of 0, , &c., 0', tf>', &c. are given. We notice that one solution must be analogous to that given by the principle T of vis viva. We therefore have -^=M(U+C). Since this must agree with the equation T= U+ C, it immediately follows that T=2' 2 fj , T 2 =3f 2 T. The relation between T and t is therefore Mdr=dt. When the paths of the particles are alone required, we may eliminate the time from the Lagrangian equations by using a new function instead of the Lagrangian function. In this method we choose some one coordinate to be the independent variable and regard the others , \j/, &c. as unknown functions of whose forms are to be determined by the altered equations of motion. Let where accents denote differential coefficients with regard to the time. Let also where the suffixes of , \f>, &c. here denote differentiations with regard to the new independent variable 0. **=**> ^ = ^V (6) " d' rift d d LAGRANGE'S EQUATIONS. 409 The equation of energy gives T'0*=U+C, .: 0'=^ ........................ (7). z . d dT dT dU, The Lagrangian equation __-_ = _ becomes C\* d l(U+C\ldT'\ _dT' ^) de\\ T' ) dfal ~ d U+C dU where all the differential coefficients are partial except the d/dO. Bemembering that U is not a function of fa , this becomes If then we use Q = {(U+C)T'}^ as if it were the Lagrangian function and regard 6 as the independent variable, we have the equations d_dQ_dQ d_dQ^_dQ, d6 dfa d ' dO dfi fy ' ' from which the paths may be found. This result follows easily from the theorem of Art. 524 by putting dr = dd, and we have here reproduced so much of that article as is required for our present purpose. If dr dO, we have MdO = dt and therefore by (7) of this note / T' \^ H= I TT -- } Substituting in (2) the Lagrangian function becomes We notice that however the expressions for the vis viva and the work function may be different in different problems, yet so long as the product (U+C)T r remains unchanged, the paths are determined by the same relations between the coordinates 6, , dbc. Since in the Lagrangian equations, the letters 0, , &c. represent arbitrary functions of the quantities or coordinates which determine the position of the system, it is evident that we have here taken as the independent variable any arbitrary function of the coordinates. If some one coordinate, say , is absent from the product (U+C) T' (though T contains the differential coefficients of . But when C is given by the initial conditions this limitation is not necessary. If we substitute for dT'jdfa and T' the values given by (6) and (7) this integral becomes dT/d0' = 2a, which is the same as that obtained in Art. 521. We may deduce this extension directly from the Lagrangian equations. Suppose where M is a function of 0, , &c. while A llt . In this case the product T(U+C) is not a function of gives d dT dM ., 1 dM dt d? - ^ ( __ dtd'~ d M If then the initial circumstances are such that the equation of energy is dT T=U+C, we have ^-, = 0. dtp As a simple example, consider the case of a projectile moving under the action of gravity. We have T=$(x' 2 + y' 2 ), U=-gy. Since the product of these is independent of x we choose some other coordinate as the independent variable. Writing x 1 = dxjdy we have This by an easy integration leads to the parabola (x - /3) 2 = 4', , &c. but not of t. We then find as in Art. 512, Ex. 3, that the equation of energy becomes L 2 -L = C ....................................... (13). Proceeding as in Art. 524, we change dt into dr and write C) .............................. (14). We may now use this as the Lagrangian function. INDEX. The numbers refer to the articles. ACCELERATION. Components in two dimensions, 38. Moving axes, 223. Three dimensions, 490, &c. Moving axes, 498. Hyper acceleration, 233. Accele- rating force, 68. ADAMS, J. C. Motion of a heavy projectile, 178. The true and mean anomalies, 347. Proof of Lambert's theorem, 352. Eesistance to comets, 386. ALGOL. Two problems, 405. ALLEGBET. Problem on the resistance to a projectile, 176, Ex. 4. AMBIGUOUS SIGNS. In rectilinear motion