THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA LOS ANGELES GIFT OF John S.Prell 7, V (- 10 /&v / tie The assumed point O is called the pole; OA / ^ is the line of reference; the distance OP is A called the radius vector, and its magnitude is usually denoted by r ; the angle AOP is the direction angle ; its magnitude is denoted by 0, and it is measured around from OA to the left. The polar co-ordinates for a point in a given plane are therefore r and , or a distance and an angle. These are plane polar co- ordinates. Space Polar Co-ordinates. If the point P is not in a given plane, 10 dHAP. II.] TERMS AND DEFINITIONS. 11 we assume as before a pole O, and a reference line OA in space. Through this line we assume any plane, as the plane of this page, OABC, and let OB be the inter- section of this plane with a plane OPB, perpendicular to it and passing through OP. The location of P is then given by the length OP or the radius vector r, the angle A OB or 0, and the angle BOP or 6. The polar co-ordinates for a point not in a given plane are there- fore ?', and 6, or a distance and two angles. These are space polar co-ordinates. If O is a point on the earth's surface, and the reference line OA is a north and south line in the plane of the horizon, the angles 6 and would be the astronomical azimuth and altitude of the point P. Cartesian Co-ordinates. Plane. The data necessary for locating a point by Cartesian co-ordinates, if the point is in a known plane, consists of two distances, parallel to two assumed lines of reference in that plane, passing through the point of reference, which is called the origin. The assumed Lines of reference are usually taken at right angles. Thus if the point P is known to be in the plane of this page, we assume any origin O and draw two reference lines OX and OF through O in this plane and at right angles. These two lines are called the axes of co-ordinates, the horizontal one the axis of a?, or the x axis, the other the axis of y, or the y axis. The distance BP or OA is denoted by x and called the abscissa of the point P. The distance APis denoted by y and called the ordinate of the point P. The abscissa x is positive to the right, negative to the left of the origin, while the ordinate y is positive when laid off above and negative when below the origin. Any point in the plane is thus located with respect to O. If a point is in the first quadrant, its co-ordinates are + a?, + y; if in the second quadrant, x and + y\ if in the third quadrant, x and y; if in the fourth quadrant, -f x and y. When the point is in a known plane, the co-ordi- nates are called plane co-or- dinates. Space Co-ordinates. If the point P is not in a known plane, we take three axes through the origin, all at right angles xisually. Two of these we may denote by X and Y as before ; the third, at right angles to the plane of XY, z we call the axis of z, or the z axis. B __.____ 3 +y -OS +C -y O A ; ' 3 4 -J/ 12 INTRODUCTION. [CHAP. IL Thus the position of the point P is given by the distance OA = x, the distance AC = z, and the distance CP = y. These are the space co ordinates of the point P. The signs prefixed to the co-ordinates indicate the quadrant in which the point is located as before. Thus, + a?, + y and z denote a point in the first quadrant either in front of or behind the plane of XY; x, + y and z, a, point in the second quadrant, either in front of or behind the plane of XY; x, y, z, and + x, y, z, points in the third and fourth quadrants, either in front of or behind the plane of XY. Direction Cosines. If we join the origin O and the point P by a line, and denote the angle of OP with the x axis by a, with the y axis by ft, and with the z axis by y, we have the relations x = OP cos a, y OP cos ft, z = OP cos y. These cosines are called the direction cosines of OP. Since OP is the diagonal of a parallelogram, we have OP 2 = a; 2 + y* + z 1 = OP 2 (cos 2 a + cos 2 ft + cos 2 y). Hence COS 2 a + COS 2 ft + COS 2 7=1 (1) If, therefore, any two of these direction cosines are given, the third can always be fonnd. Since cos 2a = 2 cos 2 a 1, cos 2/3 = 2 cos 2 ft 1, cos 2y 2 cos 2 y 1, we have also cos 2a + cos 2ft + cos 2y = 1 (2) Again, since cos ( + ft) cos a cos ft sin a sin ft, cos ( /?) = cos -4535.89 ft. per sec.; (6) 19.01 miles per sec. nearly. ^f)If the unit of speed is taken at 30 feet per second, and the unit of length at 20 inches, what should be the corresponding unit of time ? Ans. [ F] = YTJT\ or ^ e un i* f speed must always equal the unit of length rrm [-] 20 in. 20 in. Xl sec. 1 per unit of time. Hence \_T\ = rj7-,= n f or [T] = ^ ^ . - = sec. 1 sec. (5) If 1 minute is the unit of time adopted, and 1 decimeter per second is the unit of speed, what is the unit of length ? Ans. \L\ = [F] X [T~\ = 5^' X 60 sec. = 60 decimeters. 1 sec. The distance of a moving point from a fixed point measured in its path is given by s = at + bt 2 , where s is the number of feet passed over in the number of seconds t. (a) What is the unit of a and b f (b) What is the mean speed between the beginning of the Qth and the end of the 12th second from the start f (c) What is the instantaneous speed ? Ans. (a) In order that the equation may be homogeneous, a should be given in ft. per sec. units and 6 in ft.-per-sec. per sec. units. (6) For t = 5 sec. the space passed over is 5a -\- 256. For t = 12 it is 12a -f- 1446. The distance passed over in the interval 12 5 = 7 sec. is 7 -\- 1196. Hence the mean speed is - ' ' a-{- 176. (c) We have for t = 1 1 Si = at\ -f- ~bt-? and hence s Si = a(t t t )-\- b(t* ti*) or - 1 = a -\- b(t -f- ti). When the in- t t\ terval of time t t\ is indefinitely small the instantaneous speed is a -\- 2bt. <" (12) Compare the magnitudes (a) of the foot per second and the mile per hour ; (b) of the mile per hour and the yard per minute. 1 mile x Ihr. 5280 ft. 1 sec. 5280 22 Ans. (a.) - = X Ta = 3600 = 15' 1 sec. 22 is to 1 ft. per sec. as to 1, or as 1.466 to 1. 15 1 mile . ThrT 1760 yds. 1 min. 1760 oni _ (6) -5- = ... . X T r = -577- = 29J. Hence 1 mile per hour 1 yd. 60 mm. 1 yd. 60 1 min. is to 1 yd. per min. as 29J to 1. (13) A point describes 50 feet in 6 minutes and another point de- \C scribes 50 centimeters in 6 seconds. Compare their mean speeds. 50ft. 6 min. 50 X 30.47945 cm. 6 sec. 30.47945 . _ AO Ans. = ft - - X ^ = --57, = 0.508. Hence the 50 cm. 360 sec. 50 cm. 60 6 sec. mean speed in the first case is to that in the second as 0.508 to 1. (14) A man h feet in height walks along a level street at a uniform 20 KINEMATICS GENERAL PRINCIPLES. [CHAP. I. speed of v miles per hour, in a straight line from an electric light I ft. in height. Find the mean speed of the end of his shadow. Ans. I h : vt :: I : x. x Hence = -v = speed required. (15) A passenger in a railroad car moving .\^ with uniform speed counts 50 telegraph poles ^fr at equal intervals of 100 ft. passed in one minute. What is the mean speed of the train ? Ans. 56.8 miles per hour. (16) Three planets describe paths which are to each other as 15, 19, and 12, in times which are as 7, B, and 5. Find their comparative Ans. 225, 665, and 252. (17) Two bodies A and B describe the same path in the same direc- tion, with uniform speeds v and v', and at the start the distance be- tween them is a. Find the time t when they will be at the distance b in the path, and the distance of each from the initial position of A at the end of that time. Ans. Take, as the example requires, the position of A, when t = 0, as the origin, and let distances in front of this origin be (-)-) and behind it be ( ). Then for the distance of A from the origin at the end of any time t we have = vt. For the distance of B from the origin at the end of the same time, if B is initially in front of the origin A, we have s' = a-\- v't; if B is initially behind the origin A, s' = a-}- v't. In general, then, s' = v't a, where the (-(-) sign is taken for a, when B is initially in front, and the ( ) sign when B is initially behind the origin A. We have, then, S s' = vt v't T a. But by the conditions of the problem s s' = b. Hence b = vt v't T a, or t = r . V V Substituting this value of t, we have b a ,ba v'b av s = v ;, s = a-\-v -, = v v v v v v where the (-{-) sign or ( ) sign for a is taken according as B starts ahead of or behind A. (18) In the preceding example, suppose the bodies move in the path in opposite directions. Ans. For the distance of A from the origin at the end of any time t, we have, as before, s = vt. If B is initially in front of A and moves in the oppo- site direction, we have s! = a v't. If B is initially behind A, we Jiave s' = a vft. In both cases, then, s' = v't a, where, as before, the (-f-) sign is taken for a when B is initially in front of A, and the ( ) sign when B is initially behind A. We have, then, s s' = 6 = vt-{-v't T a ', hence _ 6 a b a , _ av v'b t ^^ : . , S V ; 7 , S ^^ ; ; . V-\-V V-\-V V -\- V (19) Required the time when the two bodies are together. Ans. In this case 6 = 0, and a av t = v v" where the (-(-) sign or ( ) sign is to be taken for a according as B starts ahead or behind A, and the (-{-) sign or ( ) sign for v' according as the bodies move in opposite or in the same directions in the path. CHAP. I.] EXAMPLES SPEED. 21 (20) Give finally the general solution. b a b a av v'b Ans. t = v v' v v' v where a has the (-)-) or ( ) sign according as B is in front or behind A, and v' has the (-)-) or ( ) sign according as the bodies move in opposite or in the same directions in the path. (21) Two bodies A and B move in the circumference of a circle of length c, with uniform speeds v and v', the distance apart at the beginning of the time being a. Find the time of the nth meeting, the space described by A and B, and the interval between two successive meetings. Ans. Placing b = in the value for t in example 20, we have - = time of the first meeting. Then a = = time of t lie second meeting, 3 = - 3L = time of the third meeting. f In general, t n - - , - = time of the nth meeting. 9ff Also, s = vt n = v - -^ ~j - = space described by A, V V at)' and s T a = - , - - = space described by B. t> v' The interval between two successive conjunctions is c V V We take (+) or ( ) sign for a according as B is in front of or behind A at start, and (-f ) or ( ) sign for v' according as the bodies move in opposite or same directions in the path. (22) When the earth is in that part of its orbit nearest to Jupiter,, an eclipse of one of Jupiter's satellites is seen 16 min. 30 sec. sooner than when the earth is most remote from Jupiter. The radius of the earths orbit being 92390000 miles, what is the speed of light ? Ans.^86000 miles per sec. i-<23) A train of cars moving with a speed of 20 miles an hour had been gone three hours, when a locomotive ivas dispatched in pursuit, with a speed of twenty -five miles an hour. Find the time of meeting, the speeds being maintained uniform during the time. Ans. t 12 hours. (24) Had the trains in the preceding example started together and moved in opposite directions around the earth, 24840 miles, in what time would they meet ? Ans. 23 days. (See example 21.) (25) It is just one o'clock by a clock. Find tJie time elapsed when, the minute and hour hands will be together. Ans. 5 T 5 T minutes. (See example 21.) (26) The daily motion of Mercury in his orbit is 4. 09239; that of 22 KINEMATICS GENERAL PRINCIPLES. [CHAP. I. Venus 1. 60216; that of the earth .98563. What are the intervals between the epochs at which Mercury and Venus respectively will be in the same direct ion from the sun as the earth ? Ans. 115.876 days and 583.913 days. (See example 21.) (27) A man caught in a shower in which the rain fell vertically, ran ivith a speed of 12 feet per sec. He found that the drops appeared to strike his face at an angle of 10 with the vertical. What was the speed of the drops ? Ans. 12 tan 80 = 68 ft. per sec. (28) When the path of the earth in its orbit is perpendicular to a line drawn from a star to the earth, the direction of the star appears to make an angle of 20". 445 with the perpendicular to the path of the earth. The speed of the earth being 68180 miles per hour, what is the speed of light ? Ans. 191030 miles per sec. (29) Compare the speeds of two locomotives, one of which travels 397f miles in llf hours and the other 262 T % miles in 8f hours. Ans. 91 to 81. - Instantaneous rate of change, = -(- 2b. dv t (11) A steamer leaves Liverpool for New York, and a vessel leaves New York for Liverpool at the same time ; they meet, and when the steamer reaches New York, the vessel has as far to go as the steamer had when they met. If the distance is 3000 miles, hotvfar out from Liverpool did they meet ? Ans. 1854 miles. CHAP. II.] RATE OF CHANGE OF SPEED. 27 (12) A walks at the speed of 3 miles per hour and starts 18 minutes before B. At what speed must B walk to overtake A at the ninth milestone ? Ans. 4.29 rniles per hour. (13) A tourist left behind by his companions, wishes to rejoin them on the following day. He knows they are 5 miles ahead, will start in the morning at 8 o'clock, and will walk at the rate of 3i miles an hour. When must he start in order to overtake them at 1 o'clock p.m., walking at the rate of 4 miles an hour, and resting once for half an hour on the road ? Ans. 7 h. 12 m. A.M. (14) A man walking 4 miles an hour meets 20 street cars in an hour and is overtaken by 4. What is the average speed of the cars, and uihat is the average distance betiveen successive cars ? Ans. 6 miles per hour ; ^ mile. (15) A starts from a railway station, walking 5 miles an hour ; at the end of an hour B starts, walking 4 miles an hour. At the end of another hour a train starts and passes A 25 minutes after it passes B. Find the speed of the train. Ans. 20 miles an hour. A passenger-train going 41 miles an hour and 431 feet long overtakes a freight on a parallel line. The freight-train is 713 feet long and is going 28 miles an hour. Hoiv long does it take the pas- senger-train to pass ? Ans. 1 minute. (17) In a mile race A gives B 50 yards. B passes the line 5 m,inutes after the start. A passes it 5 seconds later. Which would win in an even race, and by ivhat distance ? Ans. A by 21 yards. Equations of Motion of a Point under Different Rates of Change of Speed. The rate of change of speed may be zero, uniform or variable. When variable, it may vary according to any law. (a) Rate of Change of Speed Zero. When the rate of change of speed is zero, the speed is uniform, and the instantaneous speed at any instant is equal to the mean speed for any interval of time. In this case, if Si is the number of units in the initial distance, measured along the path from the origin, and s the number of units in the final distance, we have O _ . Q. v = T , or vt = s Si ....... (1) This equation will be general if we take distances from the origin in one direction as ( + ) and in the other direction as ( ). In such case, if the value of v comes out ( +) it denotes motion in the assumed ( +) direction; if ( ), it denotes motion in the other direction. If t comes out ( + ) it denotes time after, if ( ) time before the start, or beginning of motion. (6) Rate of Change of Speed Uniform. When the rate of change of speed is uniform, the instantaneous rate of change of speed at any instant is equal to the mean rate of change of speed for any interval of time. In this case, if v and v i are the number of units in the initial and 28 KINEMATICS GENERAL PRINCIPLES. [CHAP. II. final instantaneous speeds for any interval of time t units, we have for the uniform rate of change of speed a = V Vi (2) The value of a is (+) when the speed increases and ( ) when it de- creases during the interval of time t. From equation (2) we have v = Vi + at The average speed during the interval t is v + v , 1 mean speed = ~ = Vi + ~-ar. (4) The distance (s Si) betiveen the initial and final positions, de- scribed in the time t, is the mean speed multiplied by the time, or V + Vi. 1 . s s, = g t = vj + -^ar (5> Inserting the value of t from (2) we have 2a Hence (6) a.) (7) In applying these formulas, a is positive (+), when the speed in- creases, and negative ( ), when the speed decreases, without regard to direction of motion. If distances s, Si , from the origin, in one direction are taken as positive (+), distances in the opposite direction are negative ( ). Speeds v, v t are positive ( + ) when motion is in the assumed Positive direction, and negative ( ) when in the other direction, f t is minus ( ), it denotes time before the beginning of motion; if plus ( + ), time after. By means of these equations, if we have given the initial position of a point moving in any path, its initial speed and uniform change of speed, we can determine its final position and speed and the dis- tance described in any given interval of time. [(c) Kate of Change of Speed Variable.] If the rate of change of speed is variable, we have from (1), in Calculus notation, for the instantaneous speed, and from (2), and from (8), dt' dv t s, = / vdt. (8) (9) (10) The preceding equations for constant rate of change of speed can be directly deduced from these three general equations. Thus if we suppose a constant, we have, by integrating (9), v = at -f- C, where C is the constant of integration. When t = 0, we have v = , , and hence C CHAP. II. | EATE OF CHANGE OF SPEED. 29 c, . Hence v r, + at, which is equation (3). Inserting this value of it + $at' 2 -f- 0. But when t = we have s = s, , hence (7 = ,, and, therefore, s s t = u^ -f- laf'. This is equation (5). In any case, if the law of variation of a is given, we can find the relation between v, s and t. Graphic Representation of Rate of Change of Speed. If we rep- resent intervals of time by distances laid off horizontally along the axis of x, and the corresponding speeds by ordinates parallel to the axis of y, we shall have in general a curve for which the change of y with x will show the law of change of speed with the time. (a) Rate of Change of Speed Zero. Lay off from A along AB equal distances, so that the distances from A to 1, 1 to 2, 2 to 3, etc., are all equal and represent each one second of time, and let AB represent the entire time t. Then at A, 1, 2, 3, and B erect the perpendiculars AM, 16, 2c, 3d, BN, and let the length of each represent the speed at the corresponding instant. Since there is no change of speed, these perpendiculars will all be of equal length, we shall have AM = 16 = 2c = Sd = BN = v, and the speed at any interval of time will be given by the ordinate at that instant to the line MN parallel to AB. The space described in any time is given by s s, = vt . This is evidently given by .the area AMNB in the diagram. Therefore, the area corresponding to any time gives the space described in that time. (b) Rate of Change of Speed Constant. Lay off as before the time along AB, and at A, 1, 2, 3, B, the corresponding speeds, so that A M is the initial speed v t and BN the final speed v. Draw MC, be', cd', parallel to AB. Then bb' will be the change of speed in the first sec., cc' the change of speed in the next sec., xand so on. Since these are to be constant, NM is a straight line, the ordinate to which at r ft (1 6 c d N v A 1 2 3 B X sec. sec. sec. any instant will give the speed at that instant. The rate of change of speed is then = - 1 sec. 1 CC dd' sec. 1 sec. -' = a. Hence the rate of change of speed is the tan- But - - = gent of the angle NMC which the line MN makes with the horizontal. Hence a = - or NC = at. L The distance described in the time t is from equation (5) given by - - l t. But this is the area of AMNB. Therefore, the area & corresponding to any time gives the space described in that time. We have then directly from the figure, since NC = at, s _ 8 = - = vd 30 KINEMATICS GENERAL PRINCIPLES. CHAP. II.. If v is greater than v, a will be negative, and the line MN is in- clined below the horizontal MC. [(C) Bate of Change of Speed Variable.] If the rate of change of speed is not constant, we shall have in general a curve MNn. The tangent to this curve at any point N makes an angle with dv ^ the axis of X, whose tangent is = a, equation (9), or the rate of change of speed. The elementary area BNnb = vdt = ds, equation (8), and the total area AMNB = I t/t = = s S,, equation (10). ., dv d*s When- = 0, or - = 0, or a = 0, B b the tangent to the curve is horizontal at the corresponding point, and we have the speed at that point a maximum or minimum, according as the curve is con- cave or convex to the axis of X. EXAMPLES. The speed of a point changes from 50 to BO ft. per sec. in pass- ing over 80 ft. Find the constant rate of change of speed and the time of motion. Ans. a = 900 - 2500 2(s - *,) sign indicates decreasing speed, t = 10 ft.-per-sec. per sec. 9 9, The minus ii ,. 30-50 - = 2 sec. - 10 (2) Draw a figure representing the motion in the preceding exam- ple, and deduce the results directly from it. Ans. Average speed = ~ = 40 ft. per sec. Hence I*- 80 40* = 80, or t = 2 sec. sec. per sec. Also a = 8 5 = - 10 ft.-per- (3) A point starts from rest and moves with a constant rate of change of speed. Shoiv that this rate is numerically equal to twice the number of units of distance described in the first second. Ans. We have t = 1 and v,Q; hence from eq. (5) -!-, which is numerically equal to 2(s s,). 8 - 1 sec. ~=-^a, or a = " 1 sec. 8 *-44) In an air-brake trial, a train running at 40 miles an hour waff stopped in 625.6 ft. If the rate of change of speed was constant during stoppage, what was it ? Ans. From eq. (6), we have for v 0, * s t = 625.6, and , = (40 X 5280)' 60 X 60 ' = 2.75 ft -per-sec. per sec. 2 X 625.6 (60 X 60)" X 2 X625.6 The ( ) sign shows retardation. m^A point starts with a speed v> and has a constant rate of change of speed a. When will it come to rest, and what distance does it describe ? CHAP. II.] RATE OF CHANGE OF SPEED. 31 Ans. From eq. (3), when v 0, we have 0, at = 0, or t ~ . From eq. ) A point describes 150 ft. in the first tnree seconds of its motion and 50 ft. in the next two seconds. If the rate of change of speed is constant, when will it come to rest When will it have a speed of 3Qft. per sec. g Ans. From eq. (5) we have for s s, = 150 and t = 3, 150 = 80, -f- -a ; and 2 26 for s s, = 200 and t = 5, 200 = 50 -}- a. Combine these two equations and * / we have a 10 ft.-per-sec. per sec., and 0, = 65 ft. per sec. From eq. (3), if v = 0, we have 65 Wt = 0, or t = 6.5 sec. From eq. (3) we also have if = 30, 30 = 65 - 10*. or t = 3 5 sec. v^f?) A point whose speed is initially 30 meters per sec. and is decreasing at the rate of 40 centimeters-per-sec. per sec., moves in its path until its speed is 240 meters per minute. Find the distance traversed and the time. Ans. We have v\ = 30 and = 4 meters per sec., and a = 0.4 meters- ia _ QftO per-sec. per sec. From eq. (6) s 81 = ^ = 1105 meters. From (3) we O.o have 4 = 30 - QAt, or t 65 sec. iJ8fA point has an initial speed of v t and a variable rate of change of speed given by + kt, where k is a constant. What is the speed and distance described at the end of a time t ? Ans. From eq. (9) we have a = - = Jet, and integrating, = - -f- C. If, tit & when t = 0, we have = 0, , we obtain = 1)1, and hence = 0, -| -- . From eq. (8) ds = vdt = v^dt -\ -- g . Integrating, = 0, -j- - -j- C. If, kt 3 when t = 0, we have s = 0, we obtain C = 0, and hence * = v,t -J- -^-. b (9) A point has an initial speed of 60 ft. per sec. and a rate of change of speed of + 40 ft.-per-sec. per sec. Find the speed after 8 sec. ; the time required to traverse 300 ft.; the change of speed in traversing that distance ; the final speed. Ans. From eq. (3) we have = 60 + 40x8 = 380 ft. per sec. From eq. /fiQ S (5) we have 300 = 60* + 20<", or -t =f/--- 5 = + 2.6-> sec. or - 5.65 sec. The first value only applies. From eq. (3) we have 0-0, = 40 = 20( V69 3) = -f 106 ft. per sec. or 226 ft. per sec. The first value only applies. We have for the final speed = -f- 166 ft. per sec. or 166 ft. per sec. The first value only applies. That is, the point starts from A with the speed v l = 60 ft. per sec. and de- scribes the path AB = 300 ft in t 2.65 sec., the speed at B being = 166 ft. per sec. __ _j*- In order to interpret the negative values ^- ---- 300 ftT obtained, we observe that = 166 ft. per < A v t + 60 sec. means that the point moves in the opposite t - 2.65 direction. KINEMATICS GENERAL PRINCIPLES. [CHAP. II. t-5.65 Let the point then start from B in the opposite direction with the speed v = 166 ft. per sec. Then from eq. (3) we have v = 166 -j- 40. We see that when t = 4.15 sec., v 0, and the point has passed to some point 7 J , where the speed is zero. This point is the turning- -160 point. For t greater than 4.15 sec. v be- comes positive ; that is, the point moves back towards B, and arrives at a point A where the speed is v = -\- 60 in the time given by 60 = 40t, or t = 1.5 sec. The entire time from B to P and back to A is t = 5.65 sec. This is the time given by the negative value of t in the example ; that is, it is the time before the start, during which the point moves from B to P and back to A. The change of speed v v\ is -(- 60 -j- 166 = + 226, which is the negative value in the example. For the space BA described between the initial and final positions, we have $-1-0! -1-60 166 Q t or 5 X 5.65 = 300 ft., the ( ) sign showing that the dis- tance is on the other side of the origin, from the case of the example. We see then that our equations are general if we have regard to the signs of v, v, , s, s t , and a. (10) If the motion in example 9 is retarded, find (a) the distance described from the starting to the turning point ; (6) the distance de- scribed from the starting-point after 10 sec., the speed acquired and the distance between the final and initial positions ; (c) the distance described during the time in which the speed changes to 90ft. per sec., and this time; (d) the time required by the moving point to return to the starting-point. Ans. The initial speed is v l = -(-60 ft. per sec., the rate of change of speed is a = 40 ft.-per-sec. per sec. Let the time count from the start at A, so that Si = 0, when t = 0, and let dis- tances and motion from A towards P be positive. (a) We have from eq. (6) for the distance from A to the turning-point P, where v = 0, -D--340 2a - 3600 -80 = +45 ft. (5) From eq. (5) we have for the distance between the initial and final posi- tions after 10 sec. AB = * = vJ-^-at* = 60 X 10 20 X 100 = - 1400 ft. A The minus sign shows distance on left of A. The total distance described is then 1490 ft. The speed acquired is given by eq. (5) 1400 = g + 6 X 10 m or v = 340 ft.-per-sec. The minus sign shows motion from A towards B. (c) From eq. (6) the distance between the initial and final positions, when the speed is - 90 ft. per sec., is AC = s = ** ~ V * = 81 ~ 86 ^ = - 56.25 2a 80 ft. The minus sign shows that C is on left of A. The total distance described from the start is then 56.25 + 90 = 146.25 ft. The time, from eq. (5), is _ 56.25 = ~ 9 +6 *. or t = 3.75 sec. 9 (d) The time to reach the turning-point, as we have seen, is 1.5 sec. The time to return is, from eq. (5), 45 = t = 1.5 sec. The entire time to go and, return is then 3 sec. CHAP. II. J RATE OF CHANGE OF SPEED. 33 (11) A railway train runs at a speed of 20 miles an hour, and its speed is increasing uniformly at the rate of 14 feet-per-min. per min. Find its speed after li hours, and the distance traversed in that time. Ans. 14 feet-per-min. per min. = 9.54 miles-per-hour per Lour. From eq. (3) v 20 -(- 9.54 X 1.5 = 34.3 miles per hour. From eq. (6) the distance de- 34 3 2 20 8 scribed is = - -r- = 40.7 miles. 2 X 9.54 (12) A railway train moving with a speed of 50 miles an hour has the brakes put on, and the speed diminishes uniformly for 1 minute, when it is found to be 20 miles per hour. Find the rate of cliange of speed, and the distance, traversed. Also the time in which it ivill come to rest and the distance traversed. Ans. From eq. (3), a = - - = - - - = 1800 miles-per-hour per hour. v J. 60 From eq. (6) the distance traversed is - 5^-7- = -^ mile. In order to come obUU \& to rest, t = -- - = rsKn = 33 hoar = 100 sec. The distance in coming to a 1800 oo - -2500 25 (13) Express a rate of change of speed of 500 centimeters-per- second per second, in terms of the kilometer and minute. Ans. 18 kin.-per-min. per ruin. s (14) The speed and rate of change of speed of a moving point at a certain moment are both measured by 10, the foot and second being the units. Find the number measuring them when the yard and minute are the units. Ans. 200 yards per min. ; 12000 yards-per-min. per min. / (15) What is meant when it is said that the rate of change of speed of a point is + 10, the units being foot and second ? If the point were moving at any instant at the rate of 7| feet per second, after what time would its speed be quadrupled ? and what distance would it de- scribe in that time ? Ans. 2sec.; 42.1875ft. (16) A body describes distances of 120 yards, 228 yards, 336 yards, in successive tenths of a second. Show that this is consistent with constant rate of change of speed, and find the numerical value if the units are a minute and a yard. Ans. 38880000 yards-per-min. per min. .f (17) A point is moving at the rate of 5 feet per sec. , a quarter of a * minute after at the rate of 50 'feet per sec., half a minute after at 95 feet per sec. Show that this is consistent with a constant rate of change of speed, and find its value. Ans. 3 ft.-per-sec. per sec. CHAPTER III. DISPLACEMENT. RESOLUTION AND COMPOSITION OF DISPLACEMENTS. Displacement. The total change of position, measured in units of length, of a point, between its initial and final positions, without reference to the path described, is the linear displacement of the point. Thus if a point moves from the position A l to the position^, the distance A l A* is the linear displacement, no matter what the path may have been from A\ to A*. The term displacement always means linear displacement unless otherwise specified. Line Representative of Displacement. The displacement of a point is therefore completely represented by a straight line. The length of the line gives the magnitude of the displacement, and the direction as denoted by an arrow gives the direction of the dis- placement. Thus the straight line AiA* represents by its length the magnitude of a displacement, and the arrow shows that the displacement is in the direction from Ai to A*. A displacement then has both magnitude and direction, and such a quantity is called a vector quantity. All vector quantities can be represented thus by a straight line.* Relative Displacement. Since the displacement of a point is change of position, it can only be determined by reference to some chosen point of reference. Thus if Ai and A* are the initial and final positions of a moving point A, and Bi is some chosen point of refer- ence, we call AiA* the displacement of A with reference to B\, because to an observer at -Bi the point A would move from Ai to A? in the direction AiA*. But to an observer on the moving point A , the point B^ would appear to move from Bi to Bt , and this is really the displacement of B relative to A. That is, any change in the relative position of two points A and * As we shall see hereafter, linear velocity and acceleration, angular ve- locity and acceleration, moment of linear velocity and acceleration, moment of angular velocity and acceleration, are all vector quantities, and the same prin- ciples apply to all of them as to displacements. 34 J *' CHAP. III.] DISPLACEMENTS. 35 B may be regarded either as a displacement of A relative to B or as an equal and opposite displacement of B relative to A. Relative Displacement Two Points. If then the straight line AB represents the displacement of a point A with reference to a point B, the equal and op- A posite line BA represents the displacement of B relative to A. We shall always denote therefore relative displacement by a line, the length of which gives the magnitude of the displacement, the arrow its direction, and the letters at the end the two points. Thus in the figure one line gives the displacement of A relative to B, the other gives the displacement of B relative to A. Relative Displacement Three Points Triangle of Displace- ments.* Let a moving point A have the dis- placement AB with reference to a point B, and at the same time let B have the displace- ment BC with reference to a point C, , and let B\ , B, be the initial and final positions of the point B. Then it is evident that since the point has the displacement AB with reference to B, and at the same time B moves from B\ to B-, , the point moves from A to C, and AC is the displacement of A with reference to C. Conversely, from the preceding article, CA is the displacement of C relatively to A. Hence, if two sides of a triangle ABC taken the same way round represent the displacements of A relative to B and B relative to C respectively, the third side taken the opposite way round will repre- sent the displacement of A relative to C, and taken the same way/* round, the displacement of C relative to A. At. >H fS/r^ I "' This is called the principle of the triangle of displacements. Tfc-X ' evidently makes no difference whether the displacements are simultaneous or successive. The same principle holds true in. both cases. Polygon of Displacements. Let AB, BC and CD be the given displacement of A relative to B, B relative to (7, and C relative to D. Then if we lay off succes- siyely AB and BC in given direction and magnitude, the line AC gives the displace- ment of A relative to C. If we lay off CD, the line AD gives the displacement of A relative to D. Hence, if any number of displacements in the same plane are represented by the ^ sides of a plane polygon, ABCD, etc., taken D the same way round, the line AD which closes the polygon taken the opposite way round will give the magnitude and direction of the re- sultant displacement of the first point relative to the last, and taken the same way round, of the last point relative to the first. This principle is called the polygon of displacements, and it evi- * We shall see hereafter that all the principles which follow in this chap- ter relating to displacements hold equally good for linear and angular veloci- ties and accelerations, for the moments of linear and angular velocities and ac- celerations, and for forces and moments of forces. B C \/ 36 KINEMATICS GENERAL PRINCIPLES. [CHAP. III. dently holds good whether the displacements are successive or simultaneous. Resolution of Displacements. By the application of these prin- ciples a given displacement or any number of given displacements may be resolved into two components in any two given directions. Thus suppose the displacement of A relative 2* > to B to be given by the line AB. We can re- solve this displacement into components in any two directions given by the arrows a and 6, by drawing lines from A and B parallel to these directions till they intersect at some point C. Then the displacements AC and CB taken the other way round are the component displacements of AB in the required directions. If any number of displacements AB, BC, CD, etc., j are given, we have the resultant displacement AD, and this displacement can be resolved as before into any two directions required. \ *> c Rectangular Components. When a displacement is thus resolved into two directions at right angles, the components are called rectangular components. Unless otherwise specified, when we speak of the components of any displacement, the rectangular components are always under- stood. The Component of the Resultant is equal to the Algebraic Sum of the Components of the Displacements. It is evident that the resultant of any two given displacements is equal to the algebraic sum of their components in the direction of the resultant. For ii AC and C-Bare the given displacements, the resultant AB is equal to the sum of the com- ponents AD and DB. So also for any number of displacements, the resultant AE is equal to the algebraic sum of the components of the displacements in the direction o of AE. \ A The component in any given direction, of this resultant itself, is then equal to the algebraic sum of the components of the displacements in the same direction. Thus the projection of the resultant AE upon the line OP, or the component of AE in the direc- tion OP, is the algebraic sum of the components Y of AB, BC, CD and DE in the direction OP. Components not in the Same Plane. The same principles apply for components not in the same plane. Thus if OA is a given displacement, and we x draw AB perpendicular to the plane of XZ meeting it at B, we have the components OB and BA. Again, we can resolve OB into the compo- nents OC and CB. The components then are OC, CB and BA. Sign of Components of Displacement. A sign of (+) or ( ) pre- fixed to the magnitude of a component displacement indicates direction. Thus for three rectangular axes OX, OF, OZ, a com- CHAP. III.] DISPLACEMENTS. 37 ponent displacement in the directions OX, OY, OZ is positive, and in the opposite directions negative. If polar co-ordinates are used, the com- ponent displacement along the radius vector is positive when away from the pole, nega- tive when towards the pole. Evidently, then, we measure angles in the plane XY from OX around towards OF; in the plane YZ from OF around towards OZ '; in the plane ZX from OZ around towards OX, as shown by the arrows in the figure. Analytical Determination of the Resultant for Concurring Dis- placements. When the line representatives meet in a point they are called concurring. All displacements of a single point must be concurring. The magnitude and direction of any number of such concurring component displacements being given, to find expres- sions for the magnitude and direction of the resultant. (a) When there are Two Given Components. Let OB dj and BC = di be two component displacements making the angle a and in the indicated __ directions. d r Then the resultant is OC = d r and is given at once in magnitude from the tri- angle OBC, dr* = (di + dt cos a)* + (di sin ay = dS + 2d^dy cos a + The resultant d r makes with d, an angle a, given by (1) cos a = + cos 2 (2) (6) When there are Any Number of Components in any Given Direction. Take three rectangular axes, OX, OY, OZ, through the point O, and let the component displacements di, di, d 3 , etc., make the angles ,, /?,, y-\ a a , /?, yt, etc., with the axes of X, Y, Z respect- ively. Then we have for the sum of the components in the direction of X, Y, andZ, d x = di cos a-j + dt cos a* + . . . = 3d cos a ; d y d, cos A + d, cos ft a + . . . = d z = di cosy! + d* cosy* + . . . = 2dcosy. In these summations we must take components with their proper signs as directed in the preceding Article. We have then for the magnitude of the resultant dr , i\AA-> + d' + dz\ . . . ^^r . . ( d r = (4) This resultant makes with the axes angles a, b, c, respectively, given by 38 KINEMATICS GENERAL PRINCIPLES. [CHAP. III. The projections on the planes XY, YZ, ZX make angles with X, F, Z, respectively, whose tangents are dy dz d^ d x * d/ d,' COR. 1. If all the displacements are in one plane, make d z = 0. Then d r = \/d x * + d, (7) ) and, since cos 6 = sin a, cos a = , cos b = -~- ; (8) d r d r ^ (9) d x COR. 2. When there are but two displacements, di and d, take OX corresponding with d, . Then d x = di + d cos a,d v = d sin , and we have at once equations (1) and (2). COR. 3. If the two displacements are in the same direction or in opposite directions, a = o or 180, and d r = di d. That is, the resultant is the algebraic sum of the displacements. --" COR. 4. If the two displacements are at right angles, = 90 and d . , di d d r = 4/d, 2 + d", sin a = - , sin 6 = -- = cos a, tan a = -=- d r d r <*i COR. 5. If the two displacements are equal, di = d and dr 4 = 2d 2 (l + cos a) = 4d* cos' |, hence d r = 2d cos ^. d sin a a a Also, sin a = = sin -=, hence a = 2dcos- 2 2 *** cos 2 COR. 6. If the two displacements are equal and a = 120% then d r = d, and a = 60. EXAMPLES. (1) A wheel whose radius is r rolls on a horizontal plane until it turns through a quarter revolution. Find the displacement of the point of the wheel initially in contact with the plane relative to the point diametric- ally opposite. Ans. Let G represent the initial point of con- tact of the plane, A the point of the wheel in con- tact with G in the plane, B the point of the wheel diametrically opposite A. The displacement of B with reference to G is 30 CHAP. III.] Jo = i r o //3f - ^-3i- JT Xj yy. JO "2- /""O DISPLACEMENTS EXAMPLES. x"J 39 given by BiBy. The displacement of 4 with reference to C is given by A\A t , and therefore the displacement of C with reference to A is given by A t Ai. We have then, by laying oS these displacements, the displacement of A relative to B equal to* 2fj^2 , making an angle of 45 with the the displacement of the end of the fninute-hand with reference to the end of the hour-hand of a clock, between 3 and 3.30 o'clock, the length of each hand being r. Ans. The displacement of the minute-hand with reference to XII is and of the hour-hand with reference to XII, HiH, . Therefore the displacement of XII with refer- ence to NisHyHi. Laying off these displace, ments, we have XI Y 01 MR r 4/6 - 2 cos 15 - 4 sin 15, and for the angle of MR with the vertical 2 sin 5 7i 4/6 2 cos 15 - 4 sin 15 (3) A river flows in a direction N. 33 E., and a boat headed directly across at right angles to the current reaches a point on the other shore from which the starting-point is found to bear S. 3 W. If the distance from the starting-point is 500ft., how far has the boat been carried by the current, and what is the distance across the river if the banks are parallel f Ans. The course of the boat makes an angle of 30 with the bank. The distance across is 250 ft. ; the distance parallel to the bank 433 ft. ' (4) A point A moves 30 ft. in a given direction relative to a fixed point O. Another point B moves relative to O 40 ft. in a direction at right angles to As motion. Find the displacement of A relative to B. Ans. 50 ft. in a direction inclined to A's motion by an angle whose sin . 4 18 5" points move in the circumference of two circles of radius r and 2r respectively. Both start from the point of contact. One moves through an angle of it radians, the other through an angle of radians. Find the displacement of either relative to the other. Ans. If one point moves through the angle it in the small circle and the other through in the large, the displacement is 2r 4/5 , making an angle a with the vertical whose tangent is 2. If one point moves through the angle it in the large circle and the other through -- in the small, the displacement is r^2Q and it makes an angle with Z the vertical whose tangent is 5. (6) In the preceding example let the radius of the small circle be ri and of the large circle TV Find the displacement as before. 40 KINEMATICS GENERAL PRINCIPLES. [CHAP. III.. Ans. 4/4ri !i + 4r 2 n + 2r 1 '' and tangent , 4/4r a + 4?'.,ri 4- 2r? and tangent (7) 7%e displacement of A relative to B is a distance a towards the south, and relative to C, c towards the west. If C is initially a distance b south of B, find the final position of C relative to B. Ans. We have CB for the displacement of C relative to B. Therefore if Ci is the initial po- sition, C? will be the final position, (7,^ being equal to CB. Hence the distance of C's final position from B is BC* = \/(b + a^fc' 2 , and the direction from B to <7 a is east of south an angle whose tangent is - b + a (8) A's displacement relative to B is a to the west. C"s displace- ment relative to B is c in a direction 30 west of south. Find the displacement of A relative to C. Ans. Displacement is 4/a" ac-\- c 1 . Tangent of the angle with the merid- O/T _ ian is - zz -. If this is positive, the angle is west of north; if negative east C4/3 of north. * (9) Two trMns A and B start from the same point and A runs 60 miles northand 50 miles northeast. Find the displacement of B relative to A. Ans. 48.104 miles in a direction 34 52' south of east. * (11) A point undergoes ttvo component displacements, WO feet W. 30 S. and 30 ft. N. What is the resultant ? , Ans. 88.88 ft. 13 south of west. (12) Three component displacements have magnitudes represented by 1,2 and B and directions given by the sides of an equilateral tri- angle, taken the same way around. Find the magnitude of the re- sultant. t Ans. 4/3. acute; a 3 = 120, fis = 100, y* acute. Find the resultant displacement. Ans. We find the angles y 1 (page 12) by the equation cos 2 y = cos (a 4- ft) cos (a ft). Hence r , = 148 2' 31".7, y* = 31 57' 28".3, ^3 = 31 57' 28".3. dx = -f 20 - 8.6824 - 30 = - 18.6824 ft. d y =- 6.946 + 25 - 10.419 = + 7.635 ft. d e =- 33.937 + 42.421 + 50.907 = + 59.391 ft. 1 ft fift24- d r = Vdj? -\- d y * + d g * = 62.73 ft. cos a = g^ or a = 107 19* 36"; = 83 0' 33" ; cos c = "tS , or c = 18 46' 42''- fa) 6**"l^ CHAPTEE IV. VELOCITY. RESOLUTION AND COMPOSITION OF VELOCITIES. RECTANGULAR COMPONENTS OP VELOCITY. SIGN OP COMPONENTS OP VELOCITY. ANALYTIC DETERMINATION OP RESULTANT VELOCITY. Mean Velocity. The distance described by a moving point per unit of time we have called the mean speed of the point (page 15). The displacement (page 34) of a point per unit of time we call the mean linear velocity. Therefore the number of units in the dis- placement of a point in any given time, divided by the number of units in that time, gives the number of units of mean linear velocity or the magnitude of the mean linear velocity while the direction of that velocity is the same as that of the displacement. The term velocity always signifies linear velocity unless other- wise specified. Mean speed, then, is mean time-rate of distance described (page 16). Mean velocity is mean time-rate of displacement. Thus if a point moves in any path PjAP^ from the initial position Pi to the position P 2 in the time t units, the magnitude of the mean speed is given by i- ^ -- ?, while the magnitude of the mean ve- locity is given by and its direc- L tion by the direction of PjP 2 . If a point moves with uniform speed in a circle of radius r units and the 2irr time of revolution is t units, the mean speed is r- units of speed. But the * mean velocity in the time of one revolution is zero, because the displacement TtT in that time is zero. Again, the mean speed in one half a revolution is or 2* units of speed, as before, but the mean velocity for the same time has for I 2r 4r its magnitude T or units of velocity, and the direction is that of a diameter 51 t from the initial to the final position of the point. Instantaneous Velocity. The limiting magnitude of the mean speed when the interval of time is indefinitely small we have called 42 CHAP. IV.] VELOCITY. 43 (page 15) the instantaneous speed. In the same way, the limiting magnitude and direction of the mean velocity when the interval of time is indefinitely small is called the instantaneous velocity. The terms speed and velocity should always be understood to mean instantaneous speed and instantaneous velocity unless other- wise specified. Using- the terms thus, we see that when t is in- definitely small, distance described and displacement coincide and hence the speed at Pi is the magni- tude of the velocity at P, , while the direction of this velocity at any instant is that of the tangent to the p i path at that instant. Velocity is directed speed. Speed is magni- tude of velocity. Unit of Velocity. Since, then, the magnitude of the velocity at any instant is the speed in the direction of the velocity at that in- stant, it follows that the unit of velocity is the same as the unit of speed (page 15), or one unit of length per unit of time, as, for instance, one foot per second. For this reason we have used (page 25) the letter v for the numeric for speed. Uniform Velocity. When the velocity has the same magnitude and direction whatever the interval of time, it is uniform. Uniform velocity, then, is necessarily uniform speed in a straight line in a given direction. The velocity in such case is the same as the mean velocity for any interval of time. Variable Velocity. When either the magnitude or direction of the velocity changes it is variable. When the magnitude alone changes, we have variable speed in a straight line. "When the direction only changes, we have uniform speed in a curved line. When both change, we have variable speed in a curved line. In all these cases the velocity is variable. Thus we can speak of a point moving in a circle with uniform speed, but we cannot speak of a point moving in any curve at all with uniform velocity. If the velocity is uniform, the path must be straight, as well as speed constant. A point can be projected with the same speed in many different directions, but we cannot speak of the same velocity in different directions. A change of direction is a change of velocity whether the speed changes or not. Velocity a Vector Quantity. Since velocity is speed directed, or time-rate of displacement, it has not only sign and magnitude like speed (page 15), but also direction like displacement, and is there- fore a vector quantity. Line Representative of Velocity. Velocity, then, can be repre- sented like displacement (page 34) by a straight line. The length of this line represents the magnitude of the velocity, and the direc- tion as denoted by an arrow gives the direction. Thus the straight line A,A^ represents by its length the magnitude of a velocity, and the arrow shows its direction. Triangle and Polygon of Velocities. The prin- ciples therefore of page 35 hold good for velocities as well as displacements. We have then the triangle and polygon of velocities. Resolution and Composition of Velocities. We can also combine and resolve velocities in the same way precisely as displacements, and the principles of page 36 apply here also. 44 KINEMATICS GENERAL PRINCIPLES. [CHAP. IV. Rectangular Components of Velocity. If a point moves in any path from P to P in the time t, the dis- placement is the chord PP, and the mean '* T 1 T\ f-fc 7 ., . chord PiP velocity is v = t-t, >x If Xi and ?/i are the co-ordinates of P,, and a? and y of P, then the horizontal and vertical components of the mean velocity x v are ,1, i .. T ( ^ ~ r.M If the time is indefinitely short, we have in the notation of the Calculus, for the instantaneous velocities, _ dx _dy dt dt Sign of Components of Velocity. We see that if v x is directed towards the right, we have in the 1st and 4th quadrants x numeric- ally greater than x, and both are positive. In the 2d and 3d quad- rants, if v x is directed towards the right, x, is numerically greater than x and both are negative. In all quadrants, then, v x will be positive when directed towards the right. In the 1st and 2d quad- rants, if Vy is directed upwards, y is numerically greater than y t and both are positive. In the 3d and 4th quadrants, if v a is directed upwards, y t is numerically greater than y and both are negative. In all quadrants, then, v,, will be positive when directed upwards. We have then the following general rule for the signs of the components of the velocity in any quadrant : If the direction of the line representative is toicards the right, v x in positive ; if towards the left, v x is negative. If upwards, v y is positive ; if downwards, negative. The sign then as applied to component velocities indicates direc- tion of motion. For rectangular co-ordinates (+) signifies towards the right or upward, ( ) towards the left or downward. For three rectangular axes OX, OY, OZ, let a point Pbe given by the space co-ordinates x, y, z. Let the velocity v of the point make the angles , ft and Y with the axes of X, Y and Z. Then we have v x v cos a, v y = v cos ft, Vz v cos Y for the components in the direction of the x axes. When the angles a, ft, Y are acute or less than 90, these components are posi- tive ; when the angles are obtuse or more than 90', these components are negative. Therefore, as before, v x towards the right is positive, towards the left negative, v y upwards is positive and downwards negative, and Vz in the direction OZ is positive, in the opposite direction negative. If polar co-ordinates are used, the component velocity along the radius vector must be taken as positive when it acts aicay from the pole, and negative when it acts towards the pole.* * Evidently, then, we measure angles in the plane XY from OX around to- wards OF; in the plane TZ from OF around towards OZ; in the plane ZX from OZ around towards OX, as shown by the arrows in the figure. CHAP. IV.] VELOCITY EXAMPLES. 45 Analytical Determination of the Resultant for Concurring Veloci- ties. When the line representatives meet in a point they are called concurring. We have then the same expressions for the magnitude and direction of the resultant of any number of concur- ring component velocities as for displacements (pages 36 and 37). We have only to substitute v in place of d. EXAMPLES. (1) To a man driving eastward with a speed of 4 miles an hour, th3 wind blows apparently from tlie north, but when he doubles his speed the wind appears to blow from the northeast. Find the real direction and velocity of the wind. Ans. The wind blows from the northwest with a velocity of 4 4/2 miles an hour. *" (2) A point moves in a straight line from A to B, 60 ft. W. 30 S., in 10 sec., and thence in a straight line to C, 3Qft. N., in 20 sec. Find the mean speed and the mean velocity. Ans. The length of path is 90 ft. traversed in 30 sec., or mean speed is 3 ft. per sec. The displacement is 30 V^ ft. W., or the mean velocity is 4/3 ft. per sec. W. *' (3) A ship sails N. 30 E. at 10 miles an hour. Find its easterly velocity and its northerly velocity. Ans. 5 miles an hour; 54/3 miles an hour. (4) A n'verlbswifeirQrn? has a current of 5 miles per hour, and a boat capable of making 10 miles an hour in still water is to go straight across. In what direction must the boat be steered ? Ans. Up stream in a direction inclined 60 with the bank. (5) Find the vertical velocity of a train moving up a 1-per-cent gradient at a speed of 30 miles per hour. Ans. 0.3 mile per hour. (6) A man travelling 4 miles per hour east finds the wind to come from the southeast. When he stands still it shifts 5 to the south. Find its velocity. /t> Ans. 32.52 miles per hour N. 40 "W. U^ff) A point moving with uniform speed in a circle of radius 30 ft. describes a quadrant in 10 sec. Find the mean speed, the mean velocity, the instantaneous speed and the instantaneous velocity. TtT Ans. The length of path described in 10 sec. is -^- = 47.12 ft. The mean speed is then' 4.712 ft. per sec., and since this is uniform it is also the instan- taneous speed. The displacement is r 4/2~= 42.42 ft. at an angle of 45 with the diameter through the starting-point. The mean velocity is then 4.242 ft. per sec. in the same direction. The instantaneous velocity is at any instant tangent to the circle at the point at that instant, and equal in magnitude to the instantaneous speed, or 4.712 ft. per sec. ift A man walks at the rate of 4 miles per hour in a rain-storm, jKVEttd the drops fall vertically with a speed of 200 ft. per sec. In what direction ivill they seem to him to fall ? Ans. Inclined 1 40'. 8 to the vertical. 46 4^ KINEMATICS GENERAL PRINCIPLES. [CHAP. IV. A ship sails east with a speed of 12 miles an hour, and a shot is fired so as to strike an object which bears N.E. If the gun gives the shot a mean horizontal velocity of 90 ft. per sec., towards what point of the compass must it point ? Ans. N. 37 3' E. (10) A. man Qft. tall tvalks at the rate of 4 miles per hour directly away from a lamp-post 10 ft. high. Find the velocity of the ex- tremity of his shadow. Ans. 10 miles per hour in the direction he is walking (see Ex. 14, page I'Jj. (11) Two points moving with uniform speed v start at the same instant in the same direction from the point of contact of their paths. The one moves in a circle of radius r, the other in a tangent to the circle. Find their relative velocity at the end of the time t. Ans. 2v sin %- in a direction inclined to the tangent at an angle - (it -- -I radians. (12) A moves N.E. with a velocity v, and B moves S. 15 E. with the same velocity. Find A's velocity relative to B. Ans. * 4/8, diijaiiiiii w. Mpy* (13) A, ^ailwdy train runs 30 miles per hour north. Another running TO miles per hour appears to a passenger in the first to be running at 25rnites per hour. Find the direction of the velocity of the latter. 3* Ans. N. 56' 15 E. or N. 56' 15' W. (14) Tivo candles A and B, each 1 ft. long and requiring 4 and 6 hours respectively to burn out, stand vertically at a distance of I ft. The shadoiv of B falls on a vertical wall at a distance of 10 ft. from B. Find the speed of the end of the shadow. Ans. 8 inches per hour. (15) A ship has a northeasterly velocity of 12 knots per hour Find the magnitude of her velocity (a) in an easterly direction,' (b) in a direction 15 W. of N. Ans. (a) 6 ^2 ; (b) 6 knots per hour. (16) A boat-crew row 3 miles down a river and back again in 1 hour 40 minutes. If the river has a current of 2 miles per hour, find the rate at which the crew would row in still water. Ans. 5 miles per hour. . (17) A steamer goes 9.6 miles per hour in still water. How long will it take to run 10 miles up a stream and return, the velocity of the stream being 2 miles an hour ? Ans. 2 hours 11 minutes. (18) A steam tug travels 10 miles an hour in still water, but draws a barge 4 miles an hour. It has to take the barge 10 miles up .^ ab ab "(20) A point receives simultaneously three velocities, 60 ft. per sec. N., 88 ft. per sec. W. 30 S., and 60 ft. per sec. E. 30 S. Give the magnitude and direction of the resultant velocity. feet per sec. W. 30 S. A ship sailing due north at the rate of 8 miles per hour is carried to the east by a current of 4 miles per hour. Find the ve- locity with reference to the land. Ans. $.94 miles per hour N. 26 34' E. i^$2) A ship is sailing E. 22$ S. at the rate of 10 miles an hour and the wind seems to blow from the N. W. with a velocity of 6 miles an hour. Find the actual velocity of the wind. Ans. 15.7 miles an hour W. 30 55' N. (23) A point moves in t seconds from A to B, the positions being given by the co-ordinates Xi , y l and x t , y*. What is the mean ve- locity ? Ans . v - '- -yi, making an angle a with the axis of at t y* yi given by tan a = -- . * a a?i (24) A point has four component velocities in the same plane, of 12, 24, 36, 48 ft. per sec., making with the axis of X the angles of 16 D , 29, 33, 75 respectively. What is the resultant velocity ? Ans. v x = 75.14; % = 80.915; v r = 110.434; angle with axis of x, a = 47 7' 10"; angle with axis of y, b = 42 52' 50". '25) A point has three component velocities in the same plane given by v t = 40, v* = 50, Vs = 60 ft. per sec., making with the axes of X and Y angles given by MI = 60, /3i obtuse; /3 3 = 30, a a obtuse; a 3 = 120, fa obtuse. Find the resultant velocity. Ans. We have /?, = 150, a, = 120, fa = 150. Hence v x = - 35 ft. per sec., v v 43.3 ft. per sec. , v r = 55.67 ft. per sec., making the angles with X and F given by a = 128 57' 17", b = 141 2' 43". (26) A point has the component velocities v l = 40, y a = 50, v 3 = 60 ft. per sec., making the angles with X, Y, Z, a 60, /?, = 100, y\ obtuse ; a* = 100, #, = 60, ^ a acute ; a s = 120, 3 = 100, y* acute. Find the resultant velocity. Ans. We find the angles y (page 12) by the equation cos" y =. cos (a + ff) cos (a /?). Hence y, = 148 2' 31" 7, y % = 31 57' 28".3, y* = 31 57' 28".3. v x = 18.6824 ft. per sec., v y = + 7.635 ft. per sec. ,v, = -\- 59.391 ft. per sec., v r = 62.73ft. per sec., making with the axes of X, T, Z, angles given by a = 107 19' 36", b = 38 0' 33", o = 18 46' 42". ^ CHAPTEE V. ACCELERATION. EESOLUTION AND COMPOSITION OF ACCELERATIONS. ANALYTICAL DETERMINATION OP RESULTANT FOR CONCURRING ACCELERA- TIONS. EQUATIONS OP MOTION. THE HODOGRAPH. TANGENTIAL AND NORMAL ACCELERATION. Mean Acceleration. Let P, , P 2 , etc., Fig. (a), be the path of a moving point P, and let the corresponding instantaneous velocities Fig. (a). be v, , v-i , etc. Each velocity is tangent to the path at the corresponding point and is equal in magnitude to the speed at that point. If t is the number of units of time in passing from P, to P 2 , the mean speed for that time (page 15) is P^ thp ^ p ^ V units of speed. The integral change of speed in the time t is v? Vi (page 24), Fig (b) an d the mean rate of change of speed in the time t is a = V * ~ Vl (page 24). V If now in Fig. (6) we draw OQi parallel and equal in magnitude to Vi and OQi. parallel and equal to v* , then the integral change of speed is represented by OQ* OQi and the mean rate of change of , , O#t - O0i speed by - T . The change of velocity, however, in the time t is represented in magnitude and direction by Q> Q* , and this we call the integral ac- celeration. The mean time-rate of change of velocity is given in magnitude by ^'^ 3 , and in direction by QiQ*. We call this the mean linear acceleration. The term acceleration always means linear accelera- tion unless otherwise specified. Mean acceleration is then time-rate of change of velocity, whether that change takes place in the direction of motion or not. Instantaneous Acceleration. The limiting magnitude and direc- tion of the mean acceleration, when the interval of time is indefi nitely small, is the instantaneous acceleration. The limiting direction is not necessarily tangent to the path 48 CHAP. V.] ACCELERATION. 49 except in the case of rectilinear motion, and the limiting magnitude is not the rate of change of speed in the path, except in the case of rectilinear motion. The term acceleration always signifies instantaneous accelera- tion unless otherwise specified. It is the limiting time-rate of change of velocity, whether that change take place in the direction of the motion or not. Acceleration may be zero, uniform or variable. When it is zero the velocity is uniform, and we have uniform speed in a straight line. When it is uniform, it has the same magnitude and the same unchanged direction, whatever the interval of time. In such case the acceleration at any instant is equal to the mean acceleration for any interval of time. If its direction coincides with that of the velocity, we have uniform rate of change of speed in a straight line. If it does not so coincide, we have uniform acceleration and motion in a curved line. When it is variable, either direction or magnitude changes, or both change. Unit of Acceleration. The magnitude of the unit of acceleration is evidently the same as that for rate of change of speed, viz., one unit of length-per-sec. per sec., as for instance one foot-per-sec. per sec. We denote the magnitude of the acceleration thus measured by the letter /, to distinguish it from rate of change of speed, which we have denoted by a (page 25). Line Representative of Acceleration. Since acceleration is time- rate of change of velocity, and is therefore, like velocity and dis- placement, a vector quantity, it can be represented like them by a straight line, whose length and direction give the magnitude and direction of the acceleration (pages 34, 43). Rate of change of speed is given by stating sign and magnitude only. It is a scalar quantity (page 25). Triangle and Polygon of Accelerations. The principles, there- fore, of page 35 hold good for accelerations also, as well as dis- placements. We have then the " triangle and polygon of accelera- tions." Composition and Resolution of Accelerations. For the same reason we can combine and resolve accelerations in the same way as displacements, and the principles of pages 35, 36 apply. Let OB and OD be the initial and final velocities of a point in any given time t. Then BD is the integral acceleration and is the mean acceleration, or the instan- taneous acceleration if it is uniform. If not BD uniform, - , where t is indefinitely small, gives the acceleration. Draw OA and DC at right angles to any line AC through B in any given direction. fiC 1 Then - is the component of the acceleration in this direction. But = AC- AB and AC and AB are the componen t s of the V V velocities in the direction AC. Hence the component in any direction of an acceleration is equal to the acceleration of the component velocities in that direction. 50 KINEMATICS GENERAL PRINCIPLES. [CHAP. V. Sign of Components of Acceleration. We have the same rule for the signs of the horizontal and vertical components f x and f y of any acceleration/, as for the horizontal and vertical components of v x and v y of any velocity v (page 44). If the direction of the linear representative is towards the right or upwards, f x and f y are positive; if towards the left or down- ward, f x and fy are negative. The sign, then, applied to component accelerations indicates di- rection of action. For rectangular co-ordi- nates (+) signifies in the direction OX, OY, OZ, ( ) in the opposite directions. If polar co-ordinates are used, the compo- nent acceleration along the radius vector is positive (+) when it acts away from the pole, and negative ( ) when it acts towards the pole. Evidently, then, we measure angles in the plane XY from OX around towards OF, in the plane YZ from OY around towards OZ, in the plane ZX from OZ around toward OX, as shown by the ar- rows in the figure. Take, for instance, the case of a particle projected vertically away from the earth with the initial velocity , and attaining the final velocity v. As long as the particle ascends, the direction of v is upwards, and v, Vi are both positive. The acceleration due to gravity is always downwards, and hence is negative. When the particle is descending with the initial velocity i) t , then both i and are negative, and the acceleration is negative as before. Analytical Determination of the Resultant for Concurring Accelerations. When the line representatives meet in a point, they are said to be concurring. We have then the same expressions for the magnitude and direction of the resultant of any number of con- curring component accelerations as for displacements (pages 37 and 38). We have only to substitute/ in place of d. Equations of Motion of a Point under Different Accelerations. The magnitude of the acceleration in general is not the same as rate of change of speed, except in the case of rectilinear motion. We have therefore denoted it by /, to distinguish it from rate of change of speed, which we have denoted by a. If, then, we denote by ft the magnitude of the tangential acceleration, or the tangential component of /, we have ft = a, or the magnitude of the tangential acceleration is equal to the rate of change of speed. (a) Acceleration Zero. We have in this case the line representative Q\Qv = 0, and hence the line representative of the velocity does not change either in direction or magni- tude. We have then rectilinear motion with constant velocity, and for any interval of time S Si v =-r- (1) where s, and s are the initial and final distances of the point from any fixed point in the line. This equation is general if we pay attention to the sign of displacement and velocity (pages 37 and 44). (6) When the Direction of the Acceleration coincides with tha Direction of the Velocity. In this case the line representative QiQ CHAP. V.] EQUATIONS OF MOTION. 51 coincides in direction with PQi = v t . We have then rectilinear motion with varying velocity. If / is uniform, the instantaneous acceleration is equal to the mean acceleration for any interval of time. Hence, if v l is the initial and v the final velocity, we have for f constant V Vi f=f~ .......... (2) This value of / is general when we pay attention to the sign for velocity and acceleration (pages 44 and 50). From (2) we have v = v i +ft .......... (3) The average speed is V + Vi 1 The displacement is Inserting the value of t from (2) we have v* - v.* 8-8,= 2f l .......... (6) Hence v* = w, 9 + 2f(s Si) ........ (7) [Iff is variable, we have from (1), in Calculus notation, and from (2), and from (8), /J-t 81 I vdt "i-e vdt. . . (10) The preceding equations can be deduced from these as on page 28.] All these equations are precisely similar to those on page 28, except that we have / in place of a. (c) When the Acceleration is Constant in Magnitude and Direction, but makes an Angle

KINEMATICS GENERAL PKINCIPLES. [CHAP. V. The Hodograph. Let a point P moving in any curve have the positions Pi , P 2 , P 3 , etc., and let the magnitude of the corresponding velocities be v>, Vi, v 3 , etc. These velocities are tangent to the path at Pi, Pi, Ps, and are equal in magni- tude to the speed at these points. If from a point O we draw lines OQ, , OQi, OQs, etc., parallel and equal to v, , t? 2 , v 3 , the extremities of these lines will form a polygon. If, however, the points Pi, P 2 , P 3 are indefinitely near, the polygon be- comes a curve QiQ.Qs, such that when the point P describes the path P,P 2 P 3 , we can conceive another point Q to describe the curve QiQiQs. This curve is called the hodograph.* The point O is called the pole of the hodograph. The points Qi, Q*, Q 3 are called the points cor- responding to P, P 2 , Ps. Any radius vector, as OQi, OQ Z , in the hodograph, represents in magnitude and direction the velocity at the corresponding point Pi, PS, etc., of the path. The magnitude of OQi, OQ?, etc., is the speed u. , v, etc., at P, , P 2 , etc. The direction of OQi, OQi is the direction of v\ , Vi, or that of the tangent to the path at P , P 2 . If t is the interval of time in moving from P l to P 2 , then the chord $1 Q* m the hodograph gives the magnitude and direction of the integral acceleration for that time, and - r ^ ' ^i, which is T the mean velocity in the hodograph, gives the mean acceleration in the path. When the time is indefinitely small, QiQ? becomes tangent to the hodograph and which is now the instantaneous velocity in the hodograph, gives the instantaneous acceleration in the path. The tangent to the hodograph at any point, therefore, gives the direction of the instantaneous acceleration at the corresponding point of the path. The instantaneous speed in the hodograph gives the magnitude of this acceleration. Hence, the velocity at any point of the hodograph is the accelera- tion at the corresponding point of the path. Tangential and Normal Acceleration. The entire resultant ac- celeration / at any point of the path may be resolved into a com- ponent tangential to the path ft and a component normal to the path f a , so that/= Vf? + f n 3 . The magnitude of the tangential ^component ft is the rate of change of speed in the path. Its direction is always tangent to the path, or parallel to the radius vector of the hodograph at the corresponding point of the hodograph. In order to find the normal com- ponent fn , let us first suppose a point to move in a circle with con- stant speed v. * Invented by Sir W. R. Hamilton. CHAP. V.] TANGENTIAL AND NORMAL ACCELERATION. 53 Let the radius of the circle be r, and take any two points Pi and JP on the circle. Then the velocity at Pi is v tangent at Pi or per- pendicular to r at that point, and the velocity at P 2 is v tangent at P 2 or perpendicular to r at P 2 . Now construct the hodograph by making OQi parallel and equal to the velocity at PI and O<^i parallel and equal to the velocity at P 9 , etc. Evidently the hodograph is also a circle of radius u, and the speed in the hodograph is also constant, since the point P moves with constant speed and makes a revolution in the same time as its corresponding point Q in the hodograph. Let t be the time of revo- lution ; then is the speed of Q in the hodograph or the accelera- v tion of P in the path ; and since this speed at any point Q t is at right angles to OQi or t?, it is normal to the path at Pi or parallel to CPi. We have then f n = -~ But the speed in the path is v .. v T Hence t = - , and substituting, we have We can obtain the same result as follows : Since the point P~ moves from Pi to P 2 in the same time that Q moves from Qi to Q* , and the angle PiCP 2 is equal to the angle Q\OQ* , we have v* f n : v::v:r, or f n = . r Since we have supposed v constant in magnitude, the tangential acceleration ft is zero, and therefore f n in this case is the magnitude of the total resultant acceleration /. A normal acceleration, then, has no effect upon the speed, but only changes the direction of motion. Let us now suppose that the speed v is not constant, the point still moving in a circle. Then the hodograph will not be a circle. But if the two points Pi and P 2 are indefinitely near, so that the- arc PI Pa is indefinitely small, the velocities at Pi and P can be taken as equal still, and we shall still have v* fn :v::v:r, or f n = . r Again, if the point P moves in any curve whatever, a circle can always be described whose curvature is the same as that of the curve at any given point. The radius of this circle is the radius of curvature p at the point. In this case, then, we should have v* f n :v::v:p, or / = . Therefore, in general, whatever the path may be and whatever the speed in the path The magnitude at any instant of the normal component of the acceleration is equal to the square of the speed at that instant divided by the radius of curvature. 54 KINEMATICS GENERAL PRINCIPLES. [CHAP. V. ^ " EXAMPLES. (1) A point moves with uniform velocity in a straight line. What is the hodograph f Ans. A point. h ^ (2) A point moves with uniform acceleration either in a straight line or a curve. What is the hodograph ? What is the speed in the hodograph $ Ans. A straight line. Uniform. (3) Show that the direction of motion of any point B on the cir- cumference of a circle rolling with velocity v on a straight line is perpendicular to AB at any instant, when A is the point of contact of the circle with the straight line at the instant considered. (4) AB is a diameter of a circle of which BC is a chord. When is the moment about A of a velocity represented by BC the greatest ? Ans^-When BO and AC are equal. lx(5) A point is moving with uniform speed of a mile in 2 min. 40 sec. round a ring of 100 ft. radius. Find the acceleration. Ans^0.89 ft.-per-sec. per sec. towards the centre. Uf6) A point moves in a horizontal circle with uniform speed v, starting from the north point and moving eastward. Find the integral acceleration when it has moved (a) through a quadrant ; (b) a semicircle ; (c) three quadrants. Ans. (a) v \f%, SW.; (b) 2v, W.; (c) \/2~, NW. (7) If the component velocities of a moving point are represented by the sides of a plane polygon, taken the same way round, the algebraic sum of their moments about any point in their plane is zero. Ans. If the polygon closes, the resultant velocity is zero. If it does not close, the line necessary to make it close taken the other way round is the re- sultant. In either case the algebraic sum of the moments is zero. (8) Show that the hodograph of a point moving with uniform speed in a circle is a circle in which the corresponding point moves also with uniform speed. (9) Show that the locus of the extremity of a straight line repre- senting either of the tivo equal components of a given acceleration is a straight line perpendicular to the straight line representing the given acceleration and through its middle point. (10) Let the velocity of a point moving in a straight line vary as the square root of the product of its distances from two fixed points in the line. Show that its instantaneous acceleration varies as the mean of its distances from the fixed points. Ans. Let be one distance and a -f- s the other. Then from page 51, Chap. V, v k Vs( a + *), where A; is a constant. at _dv _ kads -\- 2ksds _ k*a -+- 2k^s _ k*(a + 2s) i- HGH Cv - , * ~ - - _" ". r" - ^ '-. "" . But the mean of the distances is <}HAP. V.] ACCELERATION EXAMPLES. 55 (11) If the algebraic sum of the moments of the component ve- locities of a moving point about any two points P and Q are each zero, show that the algebraic sum of their moments about any point in the line PQ will be zero. (12) A moving point P has tivo component velocities one of which is double the other. The moment of the smaller about a point O in the plane is double that of the greater. Find the magnitude and di- rection of the resultant velocity. Ans. If is the smaller component and a, ft are the inclinations of the greater and smaller components respectively to PO, the resultant is a 4/5 + 4 cos (ft -f a), and it is inclined to PO at an angle whose sin is 2 sin a -f- sin ft 4/5 + 4 cos (p+a) (13) The velocity v of a point moving in a straight line varies as the square root of its distance s/rom a fixed point in the line. What is its instantaneous acceleration ? ds Ans. We have v = = k 4/s, where A; is a constant. Hence 0V _ dv_ kds kv __ tf _ k 3 ~~~~ ~ ~~" Two railway trains move in directions inclined 60. The one, A, is increasing its speed at the rate of 4 ft. -per-min. per ruin. The other, B, has the brakes on and is losing speed at the rate of 8 ft.-per-min. per min. Find the relative acceleration. Ans. 4 4/7 ft. per min. per min., inclined to the direction of A at an angle whose sin is 4/a and to the direction of B at an angle whose sin is ^ 4/|. ""fTS) The initial and final velocities of a moving point during an interval of tivo hours are 8 miles per hour E. 30 N. and 4 miles per hour N. Find the integral and the mean acceleration. Ans. 4 4/3 miles per hour W., 2 4/3 miles-per-hour per hour W. i^t6T*A point moves in a circle of radius 8 inches and has at a given position a speed of 4 in. per sec., which is changing at the rate of 6 in.-per-sec. per sec. Find (a) the tangential acceleration; (b) the normal acceleration; (c) the resultant acceleration. Ans. (a) 6 in.-per-sec. per sec.; (b) 2 in.-per-sec. per sec.; (c) 2 4/l0^n.-per- sec. per sec. (17) Neivton assumed that the acceleration of gravity varied in- versely as the square of the distance from the earths centre. He then tested this assumption, by applying it to the moon. Assuming the acceleration at tlie earth's surface 32.2 ft. -per -sec. per sec., the radius of the earth 4000 miles, the distance between centres of earth and moon 240,000 miles, and the speed of the moon in its orbit around the earth 3375ft. per sec., shoiv that Newton's assumption is in accord with fact. ifl Ans The acceleration of the moon's centre towards the earth is , or r ft - per - sec " per sec ' But according to Newton's assumption, if g' is the acceleration at the dis- ce of per sec. tance of the moon, -- = , = -. Hence g' = 0.0089 ft.-per-sec. KINEMATICS GENEKAL PRINCIPLES. [CHAP. V (18) .FYwd the resultant of four component accelerations repre- sented by lines drawn from any point P within a parallelogram to the angular points. Ans. If G is the intersection of the diagonals, PC represents the direction of the reetfltant, and 4P(7 its magnitude. 1^(19) A ball is let fall in an elevator which is rising with an ac- celeration of 7.2 kilometers-per-min. per min. The acceleration of the ball relative to the earth is 981 cm.-per-sec. per sec. Find its ac celeration relative to the elevator. Ans. 1181 cin.-per-sec. per sec. towards the floor. (20) Assuming the mean radius of the earth 6370900 meters, the speed of a point on the equator 465.1 m. per sec., acceleration of a falling body 9.81 m.-per-sec. per sec., find ivith what velocity a shot must be fired at the equator with either a westerly or easterly direc- tion in order that, if unresisted, it shall move horizontally round the earth, completing its circuit in 1 or 1| hours respectively. Ans. Westerly velocity, 8370.7 meters per sec. ; easterly velocity, 7440.5 meters^per sec. t^t21) If different points describe different circles with uniform speeds and with accelerations proportional to the radii of their paths, show that their periodic times will be the same. (22) The resultant of two accelerations a and a' at right angles to one another is R. If a is increased by 9 units and a' by 5 units, the magnitude of R becomes three times its former value, and its direc- tion becomes inclined to a at the angle of its former inclination to a'. Find a, a' and R. Ans. 3, 4 and 5 units respectively. (23) If a tangent be drawn at any point of a conic section, the locus of the foot of the perpendicular let fall from a focus on this tangent is a circle in the case of the ellipse and hyperbola, and a straight line in the case of a parabola. Also the locus of the foot of a perpendicular from the vertex of a parabola on a straight line drawn through the focus is a circle. Assuming these properties, show that if a point move in either a circle, ellipse, hyperbola or parabola, so that the moment of its ve- locity about a focus is constant, the hodograph is a circle. (24) Show that if a point moves in an ellipse so that the moment of its velocity about the centre is constant, the hodograph is an ellipse. [Note that the area of the parallelogram formed by drawing tangents to an ellipse at the extremities of a pair of conjugate diameters is constant.] (25) A bullet is fired in a direction towards a second bullet which is let fall at the same instant. Show that the line joining them icill move parallel to itself and that the bullets will meet. (26) Determine whether any of the following equations are pos- sible or not : (1) Wavst + 8v*s = 3gr" 4 ; (2) v*t 4as + Sa = 0; (3) Qv + 2g 3 as*t e = 3a*st\ \L~\z Ans. The first gives us |: in each term and is therefore possible. The second gives us F^, i^ an( i fmL> or eac ^ term re ^ ers to different kinds of CHAP. V.] ACCELERATION EXAMPLES. 57 quantities, and the equation is nonsense on its face. The third gives us ~-J [L] 6 and [L] 3 , and is also nonsense. (27) A point moves in a straight line so that the number of units of distance sfrom the origin at the end of any number of seconds t 53 5 is given by s = 2 + -t + f + -t 3 . Find (a) the number of units of & 4 o distance from the origin at the start ; (b) the velocity v at any in- stant ; (c) the acceleration a at any instant ; (d) the velocity at the start ; (e) the acceleration at the start. Ans. (a) 2 units of distance; (b) v = -jr-|- -^t -f- -^t* ; (c) = H + "7*> a a o & 4 (d) jr units of distance per sec. ; (e) ^ units of distance-per-sec. per sec. a a (28) A point moves in a straight line so that the number of units of acceleration a at the end of any number of seconds t is given by a = 7 t + 2f . If Vi is the number of units of velocity at the o start, and Si the number of units of distance from the origin at the start, find the velocity and the distance from the origin at any in- stant. Ans. v = t>, + It - ^ (29) A point moves in a straight line so that the number of units of velocity v at the end of any number of seconds t is given by o f* v = 5 -t + -t*. Find the acceleration a and the distance s, if Si is & b the initial distance. 3 5 Ans. a = - -t; (30) A point has three component accelerations in the same plane given by f\ = 40, / = 50, fs = 60 ft.-per-sec. per sec., making with the axes of X and Y angles given by , = 60, fti obtuse ; fit = 30% a t obtuse ; a 3 = 120, fa obtuse. Find the resultant acceleration. Ans. We have A = 150, a, = 120, ft 3 = 150. Hence fx = 35 ft.-per-^ec. per sec. ; f y = 43.3 ft.-per-sec. per sec., and f r = 55.67 ft.-per-sec. per sec., making the angles with X and Y given by a = 128 57' 17", b = 141 2' 43". (31) A point has three component accelerations, f, = 40, / = 50, f 3 = 60 ft.-per-sec. per sec., making with the axes of X, Y, Z angles given by a, = 60, /?, = 100', ;', obtuse ; a? = 100, /? = 60, y* acute; a 3 = 120, f} 3 = 100, y* acute. Find the resultant acceleration. Ans. We find the angles y (page 12) by the equation cos 2 Y = cos (a -f- ft) cos (a ft). 58 KINEMATICS GENERAL PRINCIPLES. [CHAP. V. Hence y l = 148 2' 31".7, x - 31 57' 28". 3, y 3 = 31 57' 28".3 ; fx = 18.6824 ft.-per-sec. per sec., fy = -f- 7.635 ft.-per-sec. per sec., fz = -f- 59.391 ft.-per-sec. per sec., / = 62.73 ft.-per-sec. per sec., making witli the axes of X, T, Z angles given by a = 07 19' 36 ', b = 83 0' 33", c 18 46' 42". (32) Investigate the motion of a point whose initial velocity in a horizontal direction is 0, and in a vertical direction 32 ft. per sec. The horizontal acceleration is fx= + 16 ft.-per-sec. per sec., and the vertical acceleration f y = + 4t ft.-per-sec. per sec. Ans. The resultant acceleration is (page 51) /= and it makes an angle A with the horizontal whose cosine is 16 cos A = and an angle /* with the vertical whose cosine is 4t COS JU. = The horizontal velocity at the end of any time is w the same rule as for components of displacement, velocity or acceleration (pages 36, 44, 50). Hence components in the directions OX, OY, OZ are positive ( + ), in the opposite direction negative ( ). If then we look along the line representatives of the compo- nents towards the origin O, the rotation is always seen counter-clockwise. Therefore rotation from X towards Y, Y towards Z, Z towards Xis positive, in the opposite direc- tions negative. For polar co-ordinates, directions away from the. pole are posi- tive, towards the pole negative.* The algebraic sum of the moments of any number of component displacements, velocities or accelerations, about any point in their plane, or about any axis, is equal to the moment of the resultant displacement, velocity or acceleration about that point or axis. Let AB, AC represent two component displace- ments, velocities or acceler- ations of a point A. Then the resultant is AR. Let O be any point in the plane of the components either out- side or inside the angle be- tween the resultant and either component. Then in the first case area OAR = area OAB + area BAR area ROB, and in the second case area OAR = area OAB + area ROB area BAR. In both cases area BAR = area ROB + area OAC, since all three triangles have equal base BR, and the altitude o- BAR is the sum of the altitudes of ROB and OAC. We have then in the first case area OAR = area OAB + area OAC, and in the second case area OAR = area OAB area OAC. * Evidently, then, we measure angles in the plane XY from OX around towards OY ; in the plane YZ f rom OY around to visards OZ; in the plane ZX from OZ around towards OX, as shown by the arrows in the figure. CHAP. VI.] RESOLUTION AND COMPOSITION OF MOMENTS. 63- But twice the area OAR is the moment of -the resultant AR, and twice the areas OAB and OAC are the moments of the component* AB and AC about O. Hence the moment of the resultant is equal to the algebraic sum of the moments of the components. If we have a third displacement, velocity or acceleration at A, the resultant of this and AR would be the resultant for all three. Hence the principle holds for any number of components. Again, let AB, BC, CD represent the components of a point A. Then the resultant is AD. Let OY be an axis and XZ a plane perpendicular to the axis. Let ab, be, cd be the projec- tions on this plane of the component velocities. We have just proved that the mo- ment of ad about O is the algebraic sum of the moments of the compo- nents ab, be, cd. But the moment of each of these about O is the moment of AB, BC, CD about the axis (page 60). Hence the moment of the resultant AD about the axis is equal to the algebraic sum of the moments of the components AB, BC, CD. The moment of acceleration of a moving point relative to any fixed point in the plane of its motion is equal to the time-rate of change of the moment of its velocity about the same point. Let AB Vi be the instantaneous velocity of ,. a point A and f its instantaneous acceleration. Then in any indefinitely small time dt the change of velocity is BC fdt, and the resultant c velocity is AC = v in the plane of Vi and/'. Take a point O in the same plane and drop the per- pendiculars li , I and p upon the directions of Vi , v and /. Then, since the moment of the resultant is equal to the algebraic sum of the moments of the components, we have vl = vdi + fdt . p, or fp = Vl ~dt If the path is a circle of radius r, then Z = li = r, and we have relative to the centre We obtain the same result as follows : Resolve the acceleration / into components f t tangent to the circle and / normal. The latter component passes through the centre, and its moment is zero. We have then the moment of / Y equal to the moment of the other com- ponent, or fp =ftr. General Analytical Determination of Resultant Velocity and Moment for Any Number of Concurring Com- ponent Velocities in the Same Plane. Let the point P be given by the co-ordinates a;, y. Let the component velocities of P be Vi , u a , v 3 , x" a' x \ ..-'-''' y X 64 KINEMATICS GENERAL PRINCIPLES. [ CHAP - VI etc., all in the same plane XY and making with the axes of X and Fthe angles <*, , /tf, ; a. 2 , /7 2 ; a z , /tf a , etc. Let v be the resultant velocity making the angles a and b with the axes of X and Y respectively. For the component velocities parallel to X and Y we have V x Vi COS ai + U 2 COS <* 2 + Vs COS Cf + . . . = .2t> COS ; ) V,/ = t?i COS fti + Vy COS /? 2 + V 3 COS A + . . . = ^V COS /J. } In these summations components towards the right or upwards are positive, towards the left or downwards negative. The resultant velocity is then v r = \/v x * + v v \ ........ (2) making with the axes of X and Y angles a and 6 given by ; (3) V r V r The moment of the resultant velocity with reference to O is the algebraic sum of the moments of the component velocities v x and v y . Up is the lever arm of v r , we have, paying regard to the sign for direction of rotation, for the moment M z about the axis OZ, or the moment in the plane XY, M z = v r p = Vyx v x y ....... (4) Hence the lever-arm is In these equations Vx , V H , x and y are positive in the directions OX, OY and negative in the opposite directions. With these con- ventions the equations are general. If M z comes out positive, the direction of rotation about O is counter-clockwise; if negative, the direction of rotation is clock- wise as shown in the figure. In the first case the line representative is positive and therefore passes through O in the direction OZ. In the second case it has the contrary direction. In both cases, if we look along the line rep- resentative towards the origin, the direction of rotation will be seen as counter-clockwise. Since v,- may be considered, so far as the moment at O is con- cerned, as acting at any point in Its line of direction (page 60), let us take it acting at E, the intersection of the line of direction of v with the axis of Y. Then we have for the distance OE, M v x x OE = M z , or OE -- -. The tangent of the angle which Vx v,- makes with the axis of X is - . Hence the equation of the line of direction of the resultant velocity v is V = ***-*!- ......... (6) Vx Vx If in this equation we make x = 0, we find the ordinate of the point in which the direction of the resultant velocity v r intersects the axis of Y, viz., M OE = y'=- ......... (7) CHAP. VI.] MOMENT OF VELOCITY. 65 If we make y = 0, we find the abscissa of the point in which the direction of the resultant velocity v,- intersects the axis of X, viz., M z W /Q'v - ~y~ (o) General Analytic Determination of Resultant Velocity and Mo- ment for Concurring Component Velocities not in the Same Plane. Let the point P be given by the space co-ordinates x, y, z, and let the component velocities of P be Ui , Vi, Va, etc., making / with the axes of X, Y, and Zthe *M X / angles <*,, fi s , yi; a a , fit, y* ; a s , / pa, ya, etc. Let tv be the resultant velocity making the angles a, b, c with the axes. We have then for the compo- nent velocities parallel to X, Y, z and Z V x = Vi COS ! + Vy COS tt a + Va COS a s = 2v COS <*. \ V y = V t COS /?, + V* COS /? + VaCOS /3 3 = 2vCO8/3; ( . . (1) v z Vi cos^i + V* cos yi + v 3 cos y 3 = 2 v cosy. ; In these summations components in the directions OX, OY, OZ are positive, in the opposite directions negative. The resultant velocity is then v r = VvJ + v^ + v z *, (2) making with the axes of X, Y and Z angles a, b and c given by Vx , V,. V z cos a = , coso^-i'-, cosc= (3) V r V r V r The moment of the resultant velocity v r with reference to O is the algebraic sum of the moments of the component velocities v x , Vy and v z We take the positive direction of rotation in each of the co- ordinate planes in the direction indicated by the arrows in the figure. Thus, rotation about Z from X to Y ) " X " Y " Z\ are positive; " Y " Z " X) in the contrary directions, negative. We have then for the moments about the axes moment about Z parallel to plane XY, M z " X " " " YZ, M x =v z y-v y z; . . . (4) " Y " " " ZX, M y = v x z - v*x. ) In these equations v x , v,, , v z and a*, y, z are positive in the direc- tions OX, OY, OZ, and negative in the opposite directions. With these conventions the equations are general; and if M z M x , My come out positive, we have rotation in each plane counter-clock- wise as indicated by the figure ; if negative, clockwise. In the first case the line representatives pass through O and have the directions OZ, OX, OY. In the second case they have opposite directions through O. In any case the direction of rota- 66 KINEMATICS GENERAL PRINCIPLES. [CHAP. VL. tion is always counter-clockwise when we look along the line repre- sentative towards O. The equations of the projections of the line of direction of the resultant velocity v,- upon the co-ordinate planes are found as in the preceding Article, since each may be considered as acting at any point in its line of direction : -V-17- V V M Z on plane XY, y = * x Vx V x YZ, z = Vy ZX, x =z-, Vz V Z (5) If in these equations we make z = 0, we find the co-ordinates of the point in which the direction of the resultant velocity v r pierces the plane XY, viz., ^ _ M v / _ M* ""' y IP ' ' ' If we make x = 0, we have for the co-ordinates of the point where it pierces the plane YZ, If we make y = 0, we have for the co-ordinates of the point where it pierces the plane ZX, z' = -^, * = *. (8) Vy Vy Combining the line representatives of the moments given by (4) we have Mr = V r p = VM X * + My* + M z * ..... (9) Hence P=^ ........... (10) Vr The line representative for the resultant moment M r passes through O and makes the angles d, e, f with the axes of X, Y, Z given by , MX My j. M Z GQBd = -- r , 0086=^, COS/= - ..... (11) M r M r M r Looking along this line representative towards O, the direction of rotation is always counter-clockwise. The projections of this line representative upon the co-ordinate planes make angles with the axes given as follows: projection on XY tangent of angle with X = ~~; MX " YZ " " " " r=S; . . (12) If, " " ZX " " " " Z = M* ) If we make v z = 0, we obtain the equations of the preceding CHAP. VI.] ACCELERATION EXAMPLES. 67 article. If we make x,y, z zero, we have the equations of page 37 if we put v in place of d. General Analytic Determination of Resultant Acceleration and Moment for Concurring Component Accelerations. The equations of the last two Articles hold good for a point having component accel- erations as well as for component velocities. We have only to sub- stitute / in place of v. General Analytic Determination of Resultant Displacement ancl Moment for Concurring Component Displacements. The same equa- tions hold for a point having component displacements. We have only to substitute d in place of v. If we then make x, y, z = we have the equations of page 37- EXAMPLES. (1) A point P given by the co-ordinates x = + 3ft., y = + 4ft., 2=0 has the component velocities v t = 40, v* = 50, Vs = 60 ft. per sec., making the angles with X, Y and Z, a, = 60, fi t = 150", y\ =. 90; a, = 120, ft, = 30, y, = 90 3 ; a, = 120 \ ft, = 150, y* = 90. Find the resultant velocity and the resultant moment about the origin. Ans. The component velocities are in one plane and v x = + 20 25 30 = 35 ft. per sec. ; v y = 34.64 + 43.3 51.96 = 43.3 ft. per sec. The resultant v r = V^x* + ^ 2 = 55.67 ft. per sec., making with the hori- zontal the angle cos a = = or a = 128 57' 17", and with the vertical angle cos b = ^ = ^y, or b = 141 2' 43". v r 55.67 The moment of the resultant velocity Vr with reference to is Mz 130 + UO -f- 10 sq. ft. per sec. The direction of motion of the radius vector in the plane XT is therefore counter-clockwise. The lever-arm p = r^-^r = about 0. 18 ft. oo. b7 The equation of the line of direction of the resultant velocity is y = 1.237a? + 0.286 ft. The intercepts on the axes are y' + 0.286 ft., y! 0.231 ft. (2) A point P given by co ordinates x = + 3ft., y = + 4ft., z = + 5 ft. has the component velocities v l = 40, v* = 50, Vs = 60 ft. per sec., making the angles with X, Y, Z, a t = 60, ft t = 100, y\ obtuse ; The resultant velocity is Vr = yvy? + Vy* + v,? = 62.73 ft. per sec., making with the axes of X, T, Z angles given by 1 ft fiM24- cos a = gflp, or a = 107 19' 36"; cos b = "tl^f 5 , or b= 83 0' 33" ; or c = 18 46' 42". The moments in the co-ordinate planes are M z + 22.905 + 74.7296 = + 97.6346 sq. ft. per sec. M x = + 237.564 - 38.175 = -f 199.389 " " " " M v = - 93.412 - 178.173 = - 271.585 " " " " The resultant moment is M r = VMx 1 -\- My' + M/ - 350.78 sq. ft. per sec. The line representative makes with the axes of X, T, Z angles given by MX + 199.389 W2 1"W". cosd = = ~' 5 21 37 , Looking along this line towards 0, the motion of the radius vector is counter-clockwise. The equations of the projections of the direction of the resultant velocity v r upon the co-ordinate planes are : on plane XT, y - - 0.408z + 5.226 ft.; on plane TZ, z = + 7.778y - 26 115 ft.; on plane ZX, x= - 0.314z -f- 4.572 ft. The point in which the direction of the resultant velocity pierces the plane XT is given by a?' = + 4.572 ft., y' - -f 3.357 ft. The point in which the direction of the resultant velocity pierces the plane TZ is given by y' = + 5.226 ft., z' = + 14.56 ft. The point in which the direction of the resultant velocity pierces the plane ZX is given by z' = 26.115 ft., x' = + 12.788 ft. (3) A point given by the co-ordinates x = + 3 ft., y = + 4 ft., z = 0, has the component accelerations fi = 40, / 2 = 50, f 3 = 60 ft.-per- sec. per sec., making the angles with X, Y and Z, j = 60, /3, obtuse, y , ^ 90 ; fa = 30, crj obtuse, y* = 90 ; a 3 = 120", /? 3 obtuse, r, = 90. ie resultant acceleration and the resultant moment about the origin. Ans. (page 64). The component accelerations are in one plane and f x = ~\- 20 25 30 = 35 ft.-per-sec. per sec. / = - 34.64 + 43.3 - 51.96 = - 43.3 ft.-per sec. per sec. The resultant f r = I//** +/* = 55.67 ft.-per-sec. per sec., making with CHAP. VI. j ACCELERATION EXAMPLES. 69 f OK the horizontal the angle cos a - '- , or a 128 57' 17", and with J 00. t)7 the vertical the angle cos 6 = ^- = ' , or b 141 2' 43". j 55. b7 The moment of the resultant acceleration with reference to is M z = 130 + 140 = + 10 square feet-per-sec. per sec. The direction of mo- tion of the radius vector in the plane X T is therefore counter-clockwise. The lever-arm p of the resultant is p = ^p-^r about 0.18 ft. oo.bv The equation of the line of direction of the resultant acceleration is y = 1.2372! + 0.286 ft. The intercepts on the axes are y' = + 0.286 ft., ' = 0.231 ft. (4) A point given by the co-ordinates x = + Bft., y= + 4/., = + 5 ft. has the component accelerations fi = 40, / a = 50, f 3 = 60 ft.-per-sec. per sec., making the angles with JT, Y, Z, ai = 60% ft, = 100, yi obtuse; a? = 100, = 60, y* acute; a 3 = 120% /? 3 = 100, y 3 acute. Find the resultant acceleration and the result- ant moment about the origin. Ans. (page 65). We find the angles y (page 12) by cos"?' = cos (a -f- /J) cos (a ft). Hence y l - 148 2' 31",7, y* = 31 57' 28". 3, y* = 31 57' 28".3. f x = + 20 - 8.6824 - 30 = - 18 6824 ft.-per-sec. per sec; /= -6.946 + 25-10.419 =+ 7.635 " " " " " ; fz = - 33.937 + 42. 421 + 50.907 = + 59.391 " " " " " . The resultant acceleration is f r = y/y? +/* +/*" = 62.73 ft.-per-sec. per sec., making with the axes of X, T and Z angles given by cosa= ~ lo'^ 24 ' or a = 107 19 ' 36 "; o.7o cosb= ~^2'^ 5 , or 6 = 83 0' 33"; ' orc= 18 ' 46 ' 42 "- The moments in the co-ordinate planes are Mz = + 22.905 + 74.7296 = + 97.6346 sq. ft.-per-sec. per sec. M x = + 237.564- 38.175 = + 199.389 " " " " " " M y =- 93.412 - 178.173 =- 271.585 " " " " " The resultant moment is M r = tf MX* + MV* + M z * = 350.78 sq. ft.-per-sec. per sec. The line representative makes with the axes of X, T, Z angles given by MX +199.389 = 4<)76 , rd = 70 KINEMATICS GENERAL PRINCIPLES. [CHAP. VI. Looking along this line towards 0, the motion of the radius vector is coun- ter-clockwise. The equations of the projection of the direction of the resultant acceleration f upon the co-ordinate planes are : on plane XT, y = 0.408a; + 5.226 ft.; on plane YZ, 2 = + 7 118y - 26.115 ft.; on plane ZX, x = - 0.3142 + 4.572 ft. The point in which the direction of the resultant acceleration pierces the plane XFis given by x = + 4.572 ft., y' = +3.357 ft. The point in which the direction of the resultant acceleration pierces the plane TZis given by y 1 = + 5.226 ft., z' = - 14.56 ft. The point in which the direction of the resultant acceleration pierces the plane ZX is given by 2' = 26.115 ft., a/ = + 12.788 ft. . (5) A point given by the co-ordinates x = + 3 ft., y= + 4 ft., 2 = has the component displacements d\ = 40 ft., d 3 50 ft., d 3 = 60 ft., making the angles with X, Y and Z, a, == 60, ^, obtuse, y, = 90; ft* = 30, <* 2 obtuse, r* = 90 ; a s = 120, obtuse, y* = 90. find the resultant displacement and moment about the origin. (6) A point given by the co-ordinates x = + 3 ft., y + 4 ft., z = + 5ft. has the component displacements di = 40 ft., d* = 50ft., d s 60/., making the angles with X, Y and Z, i = 60, /Si = 100, yi obtuse; a* = 100, /3 2 = 60, y* acute; a 3 = 120, fi 3 = 100, y 3 acute, find the resultant displacement and moment about the origin. sMr *SV**~ -*- < /e^ AW /- d- ' /^ CHAPTEE VII. ANGULAR REVOLUTION OF A POINT. ANGULAR SPEED. KATE OF CHANGE OP ANGULAR SPEED. EQUATIONS OF MOTION OF A POINT UNDER DIFFERENT RATES OF CHANGE OF ANGULAR SPEED. ANGULAR SPEED IN TERMS OF LINEAR VELOCITY. RATE OF CHANGE OF ANGULAR SPEED IN TERMS OF LINEAR. MOMENT OF LINEAR VELOCITY IN TERM8 OF ANGULAR SPEED. MOMENT OF TANGENTIAL ACCELERATION IN TERMS OF RATE OF CHANGE OF ANGULAR SPEED. NORMAL ACCELERATION IN TERMS OF ANGULAR SPEED. MOTION IN A CIRCLE. Angular Revolution of a Point about a Given Point. When a point moves in any path whatever from the initial position Pi to the final position P 2 in any given time, we have called the distance Pi P 3 the linear displacement (page 34). If we choose any point in space O as a pole and draw the radius vector OPi to the initial and OP* to the final position, we call the angle Pi OP 2 = (I the angular revolution of the point P about O. Since the angle 6 is measured in radians, it ' is independent of the length of the radius vec- tor, or the distance of P, and P 2 from O (page 5). It is also independent of the position of the plane of revolution Pi OPa in space, or of the direction in space of the angular revolu- tion. It has, however, magnitude and sign (+) or ( ), according as the radius vector moves in this plane in one direction or the other. Angular revolution has then magnitude and sign, but not direc- tion. It is therefore a scalar quantity like distance described by a point, and cannot be represented by a straight line. The student must not confound angular revolution with "angular displace- ment," which, as we shall see hereafter (page 170), has like linear displacement, direction as well as sign and magnitude, and is therefore a vector quantity which can be represented by a straight line. Angular Revolution of a Point about a Given Axis. The angular revolution in any given time of a moving point about a given line or axis is the angle between perpendiculars from the initial and final positions of the point to the axis. Thus let OA be a given axis, P and P a the initial and final positions of the mov- ing point, and PiB, P*C perpendiculars to OA. Then the angle between P t B and P 2 C is the angular revolution about OA, whatever the path between P and P 2 . This angle is the same as the angle pCPt, if we complete the rectangle CPi. As the straight line pP* is thus the pro- 71 KINEMATICS GENERAL PRINCIPLES. [CHAP. VII. jection of the line PiP 2 on the plane P*Cp, we see that the angular revolution about the axis is the angular revolution of the projection p of Pi about the point C. Mean Angular Speed of a Point about a Given Point or Axis. The angular revolution per unit of time is the mean angular speed of a point about a given point or axis. Like angular revolution it has then magnitude and sign accord- ing to direction of motion in the plane of revolution, but is inde- pendent of the position in space of that plane. It is therefore a scalar quantity like linear speed (page 16). When the mean angular speed varies with the time it is variable. When it has the same magnitude no matter what the interval of time it is uniform. A point moving with uniform angular speed evi- dently describes equal angles in equal times. Instantaneous Angular Speed of a Point about a Given Point or Axis. The limiting value of the mean angular speed when the interval of time is indefinitely small is the instantaneous angular speed. If the instantaneous angular speed is variable, the mean angular speed has different values for equal intervals of time. The term angular speed always signifies instantaneous angular speed unless otherwise specified. Angular speed like mean angular speed is therefore a scalar quantity, having magnitude and sign according to the direction of motion in the plane of revolution, but independent of the position of this plane in space. The student must not confound angular speed with "angular velocity," which, as we shall see hereafter (page 174), has direction as well as sign and magnitude and is therefore a vector quantity. Numeric Equations of Angular Speed. The unit of angular speed is evidently one radian per second. We denote the magnitude of the angular speed thus measured by the letter GO. If then 6, is the angle measured in the plane of revolution from any fixed line to the initial position of the radius vector and 6 to the final position of the radius vector, we have for the mean angular speed t When the interval of time is indefinitely small, we have in Calculus notation, for the instantaneous angular speed, dO (2) Sign of Angular Speed. These equations are precisely the same as equations (1) and (2), page 16, simply substituting 9 for s. The sign follows the same rule. Thus when the angle is increasing the value of GO is positive ( + ), and when decreasing it is negative ( ). Equation (1) is thus a general if we take angles in any one direc- tion in the plane of revolution measured from a fixed line in that plane as positive, and in the opposite direction as negative. Angular speed, then, whether uniform or variable, mean or in- stantaneous, is independent of direction in space. It is entirely comparable to linear speed (page 15). Change of Angular Speed. When the angular speed of a point varies, the difference between the final and initial instantaneous speeds for any interval of time is the integral change of angular speedy CHAP. VII.] RATE OF CHANGE OF ANGULAR SPEED. 75 Mean Rate of Change of Angular Speed. The integral change of angular speed per unit of time is the mean rate of change of angu- lar speed. When the mean rate of change varies with the time it is vari- able. When it has the same magnitude no matter what the inter- val of time it is uniform. Instantaneous Rate of Change of Angular Speed. The limiting value of the mean rate of change of angular speed when the inter- val of time is indefinitely small is the instantaneous rate of change of angular speed. Rate of change of angular speed should always be understood as meaning instantaneous rate of change unless otherwise specified. Rate of change of angular speed may be zero, uniform or vari- able. When it is zero the angular speed is uniform and the same as the mean speed for any interval of time. When it is uniform the rate of change of angular speed is the same as the mean rate of change for any interval of time. When it is variable the mean rate of change has different values for equal intervals of time. Numeric Equations of Rate of Change of Angular Speed. The unit of rate of change of angular speed is then one radian-per-sec. per sec. We denote its magnitude thus measured by the letter t t But PiN= r sin 9, where r is the radius vector OPi. Hence rsin9 _ chord PiPa . sin t ~~t~ If now the time is indefinitely small, - ^ - - becomes the in- stantaneous velocity v, and becomes the angle AP t v = e between r, 6*^ the radius vector OP\ and the instantaneous velocity v, and sin 9 f\ becomes 9, and becomes the instantaneous angular speed GO. Hence P usin e rca = usm e, or co = . , ..... (1) r In general, then, whatever the path or wherever the pole, The magnitude of the angular speed at any point is equal to the magnitude of the component of the linear velocity at that point per- pendicular to the radius vector, divided by the magnitude of the radius vector. If the pole O is taken at the centre of curvature, so that OPi is equal to the radius of curvature p, then e = 90 and we have/jt = v v or oo = . P Rate of Change of Angular Speed in Terms of Tangential Linear Acceleration. If ft = a is the magnitude of the linear tangential acceleration or rate of change of speed at any point, then we can prove, precisely as in the preceding Article, that (2) where a is the magnitude of the rate of change of angular speed. Hence the magnitude of the rate of change of angular speed at any point is equal to the magnitude of the component of the linear tangential acceleration at that point perpendicular to the radius vector, divided by the magnitude of the radius vector. If the pole O is taken at the centre of curvature, e = 90 and we have pa ft or a = . Moment of Linear Velocity in Terms of Angular Speed. We can resolve the linear velocity v at the point P into a component v cos e along the radius vector and a component v sin e perpendicular to the radius vector. The moment of the first relative to the pole is zero. Since the moment of v is equal to the algebraic sum of the moments of its com- ponents (page 62), if we take moments about the pole, we have vp = v sin e . r. 76 KINEMATICS GENERAL PRINCIPLES. [CHAP. VLU But we have just seen that v sin e = roo. Hence . vp = r*aa (3) That is, the magnitude of the moment of the linear velocity at any point relative to the pole is equal to the magnitude of the angular speed at that point, multiplied by the square of the magnitude of the radius vector. Since v sin e is the normal component of v, v sin e . r is twice the areal velocity of the radius vector (page 61). Moment of Linear Tangential Acceleration in Terms of Rate of Change of Angular Speed. We can resolve the tangential accelera- tion ft into components along and perpendicular to the radius vector and thus obtain, precisely as in the preceding Article, fP=ftPt= 1 * a (4) Hence the magnitude of the moment of the linear tangential ac- celeration a any point relative to the pole is equal to the magnitude of the rate of change of angular speed at that point, multiplied by the square of the magnitude of the radius vector. Since ft sin e is the normal component of ft , ft sin e . r = r*a is twice the areal acceleration of the radius vector (page 61). Normal Linear Acceleration in Terms of Angular Speed. We have seen (page 53) that when a point moves in any path, the v'* magnitude of the normal acceleration f n is given by f n , where p is the radius of curvature. If we take the pole at the centre of curvature, then, we have v = poo, and hence / = V - = P^ = VGO (5) The magnitude of the normal linear acceleration at any point is equal to the magnitude of the radius of curvature at that point, multiplied by the magnitude of the square of the angular speed, or to the velocity vca in the hodograph (page 52). Since for any path roo = v sin e, we have sin e Hence, in general, , 2 tf fn = Vo = P = T = P p Sin' e where r is any radius vector when the pole is not at the centre of curvature, and e is the angle of v with this radius vector. Motion in a Circle. For a point moving in a circle we have e = 90 and r p. Hence from (1), page 75, we have v v = roo, or 00= - (1) r If r is unity, we have the numeric equation K> = v, that is, the number of units of angular speed is equal to the number of units of linear speed at distance unity, From (2), page 75, we have ft f t = ra, or a = - (2} CHAP. VII.] EXAMPLES ANGULAR SPEED. *lU 77 If r is unity, we have the numeric equation a = ft , that is, the number of units of rate of change of angular speed is equal to the number of units of linear tangential acceleration at distance unity. From (5), page 76, we have in any case fn = VK >' (3) or the normal linear acceleration is equal to the velocity in the hodograph (page 52). Inserting the value of v from (1), /="*=. or '=^ (4) If r is unity, we have the numeric equation f n = v 2 = ca 2 , that is, the number of units of the normal linear acceleration is equal to the square of the number of units of linear velocity at distance unity ; or the square of the number of units of angular speed is equal to the number of units of the normal linear acceleration at distance unity. We have also for the total resultant linear acceleration /= W+/ ......... (5) Since the component f n passes through the centre, the moment of / relative to the centre is equal to the moment of f t . Hence vr r"<, ........... (6) fp=f t r = r>a ......... (7) give the moments of v and / with respect to the centre. If the point starts from rest and acquires the velocity v in the U time t, under constant tangential acceleration, we have f t = , 0} a= T Graphic Representation of Rate of Change of Angular Speed. We can represent intervals of time by distances laid off horizontally and the corresponding angular speeds by distances laid off vertically and thus obtain the same diagrams as for linear speed given on page 29. EXAMPLES. (1) The angular speed of a point moving in a plane about some assumed point changes from 50 to 30 radians per sec. in passing through 80 radians. Find the constant rate of change of angular speed and the time of motion. = 10 radians-per-sec. per sec. The minus sign de- notes decreasing speed, t = --- - = 2 sec. (2) Draiv a figure representing the motion in the preceding example, and deduce the results directly from it. , 50 + 30 Ans. Average speed = -- = 40 radians 8 per sec. Hence 40 = 80 or t = 2 seconds. Also ^f 30-50 cc. = - - - = 10 radians-per-sec. per sec. 03-30 78 KINEMATICS GENERAL PBINCIPLES. [CHAP. VII. (3) A point moving in a plane has an initial speed o/60 radians per sec. about an assumed point and a rate of change of speed of + 40 radians-per-sec. per sec. Find the speed after 8 sec.; the time required to describe 300 radians ; the change of speed ivhile describ- ing that angle; the final speed. Ans. See Example (9), page 31. (4) If the motion in the last example is retarded, find (a) the angular revolution from the start to the turning-point ; (b) the angle described from the start after 10 sec. ; the speed acquired and the angle between the final and initial positions ; (c) the angle described during the time in which the speed changes to 90 radians per sec., and this time ; (d) the time required by the moving point to return to the initial position. Ans. See Example 10, page 32. (5) A point moving in a plane describes about a fixed point angles of 120 radians, 228 radians and 336 radians in successive tenths of a second. Show that this is consistent with uniform rate of change of angular speed, and find this rate. Ans. a 10800 radians-per-sec. per sec. (6) Two points A and B move in the circumference of a circle with uniform angular speeds GO and GO'. The angle between them at the start is a. Find the time of the nth meeting, the angles described by A and B, and the interval of time between two successive meet- ings. Ans. See Example (21), page 21. Time of the nth meeting, t n = r^ co K> Angle described by A is = cotn. " " " B is T a. Interval of time between two successive conjunctions is a GO' where we take the (+) or ( ) sign for a according as B is in front of or behind A at start, and (-{-) or ( ) sign for GO' according as the points move in opposite or the same directions. (7) What is the angular speed of a fly-wheel 5 ft. in diameter ivhich makes 30 revolutions per minute, and ivhat is the linear ve- locity of a point on its circumference f Also find its linear normal acceleration and the moment of its velocity with reference to the centre. Ans. n radians per sec.; 2.5?r ft. per sec. , tangent tocirc. ; 2.57T 2 ft.-per- sec. per sec.; 6.257T sq. ft. per sec. (8) Find the linear and angular speed of a point on the eartlCs equator, taking radius 4000 miles ; also the linear normal accelera- tion. Ans. 1535.9 ft. per sec.; - radians, or 15 per hour; 0.112 ft.-per-sec. per 12 sec. q (9; The angular speed of a wheel is it radians per sec. Find the linear speed of points at a distance of 2ft., 4ft. and 10ft. from the centre, also the linear normal acceleration. CHAP. VII.] EXAMPLES ANGULAR SPEED. 79 g Ans. ^7t, 3;r, 7.5ffft. per sec. 9 . 9 , 45 , it', -n l , -$- it 1 ft.-per-sec. per sec. o 4 o (10) If the linear speed of a point at the equator is v, find the speed linear and angular at any latitude X. Ans. v cos A ; - radians per hour, or 15 per hour. 1|W (11) A point moves with uniform velocity v. Find at any instant its angular speed about a fixed point whose distance from the path is a. Ans. -radians per sec., where r is the radius vector. Uniform velocity means uniform speed in a straight line (page 43). Hence the angular speed of a point moving with uniform speed in a straight line is inversely proportional to the square of the distance of the point from a fixed point not in the line. (12) The speed of the periphery of a mill-wheel 12 feet in diame- ter is 6 feet per sec. How many revolutions does the wheel make per sec. ? Ans. ;r revolutions. 2* (13) The time if between 5 and 6 o'clock, and the hour and min- ute hands are together. What is the time f Ans. 5 h. 27 m. 16 sec. (see Example (6) ). (14) Express in degrees and radians the angle made by the hands of a clock at 3.35 o 1 clock. Ans. 102.5 deg. , 1.79 radians. (15) Find the multiplier for changing revolutions per minute into radians per second. Ans. 0.10472 rad. per sec. = 1 rev. per min. (16) The minute and second hands point in the same direction at 12 o'clock. When do they next point in the same direction f Ans. 1 min. lg* y sec. after twelve. (See Example (6)). (17; Two clocks are together at XII. When the first comes to J, it has lost a second; when the second comes to I, it has gained a second. How far are they apart in 12 hours f Ans. 24 sees. (18) Two men start together to walk around a circular course, one taking 75 minutes to the round, the other 90. When will they be to- gether again at the starting-point f Aus. 7.5 hours. (See Example (6)). (19) The hour-hand of a watch is f of an inch long, the minute- hand i of an inch, and the second-hand | of an inch. Compare the lineal speeds of their points and the angular speeds. Ans. 5 : 112 : 2800; 1 : 12 : 720. (20) Deduce the equivalent of longitude for one minute of time and for one second of time. Ans. 15' to 1 min., 15" to 1 sec. 80 KINEMATICS GENERAL PRINCIPLES. [CHAP. VII. (21) The diameter of the earth is nearly 8000 miles. Required the circumference at the equator and the linear speed at latitude 60. Ans. 25000 miles; 521 miles per hour. (22) The wheel of a bicycle is 52 inches in diameter and performs 5040 revolutions in a journey of 65 minutes. Find the speed in miles per hour ; the angular speed of any point about the axle ; the areal velocity of a spoke ; the relative velocity of the highest point with respect to the centre. Ans. 12 miles per hour; 8.12 radians per sec.; 19.06 sq. ft. per sec.; 12 miles per hour. (23) In going 120 yards the front wheel of a carriage makes six revolutions more than the hind wheel. If each circumference were a yard longer, it would make only 4 revolutions more. Find the cir- cumference of each wheel. Ans. 4 yards and 5 yards. (24) If the velocity of a point is resolved into several components in one plane, show that its angular speed about any fixed point in the plane is the sum of the angular speeds due to the several compo- nents. (25) A point moves with uniform speed v in a circle of radius r. Show that its angular speed about any point in the circumfer- . v ence is . 2r (26) Show that the angular speed of the earth about the sun is proportional to the apparent area of the sun's disk. [The radius vector from the sun to the earth sweeps over equal areas in equal times.} (27) A point P moves in a parabola with constant angular speed about the focus S. Show that its linear speed is proportional to (28) A point starting from rest moves in a circle with a constant rate of change of angular speed of 2 radians-per-sec. per sec. Find the angular speed at the end of 20 sec. and the angular displacement of revolution ; also the linear speed and distance described and the number of revolutions ; also the linear tangential acceleration and the normal linear acceleration at the end of 20 sec. Af\f\ Ans. 40 rad. per sec.; 400 radians; 40r ft. per sec.; 400r ft. ; revolu- 35 Jf tions; 2r ft.-per-sec. per sec. tangential acceleration; 1600r ft.-per-sec. per sec. normal acceleration. (29) A point moving with uniform rate of change of angular speed in a circle is found to revolve at the rate of 8i revolutions in the eighth second after starting and 7| revolutions in the thirteenth second after starting. Find its initial angular speed and its uni- form rate of change of angular speed ; also the initial linear speed and rate of change of speed ; also the initial normal acceleration. Ans. 20.2?r radians per sec.; 0.4?r radians-per-sec. per sec.; 20.2?rr ft. pef^sec. ; QAitr ft.-per-sec. per sec.; 408.04?rV ft.-per-sec. per sec. V (30) A point starts from rest and moves in a circle with a uni- form rate of change of angular speed of 18 radians-per-sec. per sec. Find the time in which it makes the first, second and third revolu- tions. - 4/2* 4/67T - 2 4/7T M 1 , _i ' sees. 3 3 .v ' CHAPTER VIII. DIFFERENTIAL EQUATIONS OF MOTION OF A POINT.* Free Motion of a Point Rectangular Co-ordinates. Let a mov- ing point have a position at any instant given by the co-ordinates x t y and 2, and let the distance described in the interval of time dt be ds, and let the direction of motion make the angles or, ft, y with f, T, Z. Then we have dx a dy dz cos a = , cos ft = -/- . cos Y -5-. ds ds ds The magnitude of the velocity is v = ds_ dt 1 (1) (2) and its components in the direction of the axes are ds dx v x =v cos a = cos a = dt dt ds dt ds _ dy f dt ' dz (3) V z = V COS Y = -77 COS Y = -TT- dt dt "We have v x positive towards the right, negative towards the left; ?> posi- tive upwards and negative downwards ; v z positive in the direction OZ, negative in the opposite direction (page 44). Squaring equations (3) and adding, since cos 2 a + cos* ft + cos" y = 1, we have ds dz (5) Let the acceleration be / and its components in the direction of the axes X, F, Z be f x , fy , fz , then we have (6) . _ f _ _ a: = , f y _ , fz- d -f * This Chapter must be omitted by those not familiar with the Calculus. 81 82 KINEMATICS GENERAL PRINCIPLES. [CHAP. VIII. The acceleration f is then -Vf + fi. f*-4 m f. +fy + fz -V y + + - ' ' We have fx positive towards the right, negative towards the left; f y positive upwards, negative downwards; / z positive in the direction OZ r negative in the opposite direction. The tangential acceleration ft = a is the rate of change of speed, or ,. dv d?s Differentiating (4) and substituting (6), we have vdv =f x dx + fydy + f z dz. Hence r t- - (9) v> = 2 / (f x dx + fydy + f z dz) + Const. V ) Dividing by ds, since v = -r-, and dv = , we have from (8) Gilt Ctb f dv _ d*s _ dx , f dy , f dz Jt <* T. 1,0 JKj^ ( & a so (-by a y CL% a % - . _ . ~ ds df ds df ds dt* ' The normal acceleration /, as we have seen (page 76), is/ n = , where P p is the radius of curvature at the point. We have from analytical- Geometry 1_ Hence UJS^A a //7 Q ?A 8 fff*p \ _!_1 4- I " 1 4- f I tiS / \CLS I \CLS i / \ / \ / and the acceleration /is /= V/t 2 + /" ( 11& ) If we denote by $ the angle which the acceleration / makes with the radius of curvature p, and by e the angle which it makes with the tangent to the curve, we have 8 + e 90 and fds\ The moment of the velocity about the origin is the sum of the mo- ments of the components. The moments in planes parallel to JT", YZ, __. dy dx about,, Jfr-- _ dx dz (13a> CHAP. VIII.] DIFFERENTIAL EQUATIONS OF MOTION OF A POINT. 83 The moment about the origin of the resultant velocity v if p is its lever-arm is then vp = M= \/M x * + My* + M7. ..... (136) ve of this moment makes the by MX My Mz - The line representative of this moment makes the angles d, e, f with the axes of X, Y, Z given by (14) jfj at jfj_ Looking along this line representative towards the origin, the direction of rotation is always counter-clockwise. In the same way the moment of the acceleration about the origin is the sum of the moments of the components. We have then precisely the 3^QC d^l/ (J^ same equations as (13), (14), only we put -' , --f , - - in place of at at at* ~r-, -^, in order to find the moments in the co-ordinate planes. dt dt dt Application of the Preceding Formulas. Equations (3) and (6) are the general equations by which the motion of a point is determined. Applications of the use of the equations just deduced will be given here- after. We can only indicate here the general application. If z = 0, we have motion of a point in a plane only. The correspond- ing equations are at once obtained by making z and dz zero wherever they occur in the general equations. If we also make /J = and y = 0, we have motion along the axis of x only. Hence taking x = s, we have from (3) ds d*s *=-; from (6), / = /, = = ; which are equations (8), (9) of page 51. If the velocity v in any case is given, it can be resolved by (3) into its components v x , Vy , v z . Then by differentiating as indicated by (6) the components f x i fy fz of the acceleration f can be found, and the accelera- tion f can then be found by (7). If the component accelerations are given, we find by integration the component velocities and then the resultant velocity. If the path is required, each of equations (6) must be integrated twice, thus introducing two constants of integration for each. The constants of the first integration will depend on the initial velocity, those of the second on the initial position. We thus obtain equations involving x, y, z and t, and by eliminating t we obtain an equation between x and y, or y and 2, or z and x, that is, the equation of the projection of the path ou the co-ordinate planes. Differential Polar Equations for Motion of a Point in a Plane. Let x and y be the rectangular co-ordinates, and ^ r and 9 the polar co-ordinates, of a point P in a plane. Then ' ^^ \M x = rcos 6, y = T sin 9 (15) ' ^ Cr - Differentiating and dividing by dt, we have x dx dr . ^ dB v x = = cose r sine ; (16) dt dt dt dy dr . , , dQ v y = ~- = sin + r cos 6 (17) dt dt dt 84 KINEMATICS GENERAL PRINCIPLES. [CHAP. VIII. Squaring and adding, since sin" 6 + cos" 6 = 1, we have for the magni- tude of the velocity fdsV I drV JdQy *=(lu)=(dt) +r (-dt)' If r is constant, the path is a circle. In this case is zero and t> = at dQ r = ?&?, where oo is the angular speed (page 76). \JLti The velocity along the radius vector is dr dx dy . - The velocity perpendicular to the radius vector is r=-cos e --sin0 ........ (20) dt dt dt d*cc d^y Since by (6) and -~ are the horizontal and vertical components at ctt of the acceleration, we have, by differentiating (16) and (17), d?x . (22) dt dt The acceleration along the radius vector is then f x cos +/# sin9, or (33) If r is constant, the path is a circle. In this case is zero, and the CLv acceleration along the radius vector is^n = rca 2 , where GO is the angular speed (page 76). The ( ) sign denotes direction towards the centre (page 50). The acceleration perpendicular to the radius vector is fy cos 9 fx sin 0, or CHAP. VIII.] DIFFERENTIAL EQUATIONS OF MOTION OF A POINT. 85 dy ^- from (16), (17), for the moment of the velocity with reference to the pole, if p is the lever-arm, vp = M z =1*^ = 1*00, ....... (26) where GO is the angular speed (page 76). From equations (14) and (21) and (22) we have in like manner, for the moment of the acceleration, We see from (25) that this may also be written (28) whore ft is the tangential acceleration and a is the rate of change of angular speed (page 76). Applications of the use of these formulas will be made hereafter. General Polar Equations of Motion of a Point in a Plane Ac- celeration Central. When the acceleration is always directed to or from a fixed point it is called central acceleration, and the fixed point is called the centre of acceleration. Let this fixed point be the pole. Then, since the direction of the acceleration always passes through the pole, its moment with reference to the pole is zero, and we have from (28) (29) where c is a constant of integration. Now -3- = oo = angular speed, and from page 75 we have at r The component along the radius vector is equal to the acceleration itself, if the pole is the centre of acceleration, and we have from equation (23), if the acceleration is towards the centre, where the ( ) sign for /denotes motion towards the centre (page 50). Equations (32) and (33) express all the conditions of central accelera- tion towards the pole, and therefore determine the motion. If the acceleration is away from the pole we have + / instead of f (page 50). From (30) we have and equation (33) becomes But we have seen, page 76, that rco* is the central acceleration for a point moving in a circle of radius r with the speed rco. d?r The rate of change of length of the radius vector ^-, we see from (34), is then the difference between the central acceleration /at any instant and the central acceleration at the same instant of a point moving in a circle of radius r with the same angular velocity. This rate of change of velocity along the radius vector is called the CHAP. VIII.] DIFFERENTIAL EQUATIONS OF MOTION OF A POINT. 87 dr paracentric acceleration. Its integral or -rr is the velocity of approach or recession along the radius vector and is called the paracentric velocity. (a) To find the speed at any point of the path. Central acceleration. If we multiply (32) by rdd and (33) by dr and add, we have --** ...... < 35 > or Integrating, we obtain where Ci is a constant of integration. If the law of variation of /' is given in terms of r for any given case, we can perform the integration denoted by / fdr. From (18) we have ds* dr* + ~ ~ Hence tf = e, - 2 ffdr ......... (37) We have also from (30) v = - ........... (38) p Since in (37) the value of v depends only upon r, we see that the speed for central acceleration at any two points of the path is independent of the path, and is the same for any points equally distant from the centre, the law of acceleration remaining the same. (b) To find the time of describing any portion of the path. Central accelera- tion. Substituting (31) in (36), we have a? + =/' ...... (89) Hence d .= ffl r -S- 3 //^ r 1 i/ We have also from (31) t = - r^dQ, (41) from which r must be eliminated by means of the equation of the path and the integration performed in reference to 9. (c) To find the equation of the path. t*d/f Substitute in (39) for dt* its value from (31), dP = -^-, and we have 88 KINEMATICS GENEKAL PRINCIPLES. [CHAP. VIII. This equation may be simplified by putting r = , and it then becomes (cT) To find the law of the acceleration for any given path. Central accelera- tion. Differentiating (43) with reference to dQ, we have \d&* Substituting in (33) for df its value from (31), dt* = '--, we have d*r \ -^75. (45) which is the same as (44) if we put = u. r r*flfP From (30) we have p* = -^-, and from (18) tfdt* = dr* + rW. C/ Civ Therefore > tAKi (46 > Substitute this in (42) and we have r. ....... (47) Differentiating with respect to p, 7=^-. . . t ...... (48) p*dr The law of acceleration is given by (44) or (45) or (48). From (33) and (35) we have ........ (49) an expression which will often be found useful in reductions. Differential Equations for Constrained Motion of a Point in a Plane. For free motion of a point in a plane we have from (6) for the horizontal and vertical components of the acceleration Under the action of these components, the point, if free to move, de- scribes some curve. But if it is constrained to move in a given curve, these components will be changed by reason of the normal acceleration N due to the given curve. If thus /is the acceleration of a free point P, fx and f y its horizontal and vertical components, N the normal acceleration due to the given V f curve, and 6 the angle of the tangent at P N { v I with the horizontal, we have ==/ -JTsinfi; .. . . (50) . . . (51) CHAP. VIIL] DIFFEKENTIAL EQUATIONS OF MOTION OF A POINT. 89 If JV is zero, the motion is unconstrained and we have (6). These two equations, together with the equation of the given curve, are sufficient to determine the motion completely. In applying them, f x is positive towards right, negative towards left, and the horizontal component of N follows the same rule. We have/y positive upwards and negative downwards, and the vertical component of N follows the same rule. If we multiply (50) by 2dx and (51) by 2dy and add, we have, since dy dx sm 6 = and cos 6 = -, ds ds d? dx* + dy* The first member of this equation is the differential of - y =. = 2 , or is equal to 2vdv. Hence dt* vdv = f x dx +f y dy, } or ! ( 52) t>* = 2 C(f x dx +f y dy) + Constant. J This is precisely the same result as that obtained for free motion, equa- Hence we conclude that if there is no acceleration except that of .ZVdue to the curve alone, or if f x - 0, f y = 0, the speed on the curve is constant and unaffected by the curve. If there is an acceleration besides that due to the curve, the speed will be unaffected by the curve and the same as if the point were free. If the acceleration of the free point is parallel to the axis of y, we have f x and v* = 2 Cfydy + Constant ....... (53) If in this last case/j, is constant, we have tf =2f y y + Constant ........ (54) If the distance of the point from the origin y = Si when v = Vi , we have which is precisely the same as for uniform rate of change of speed for a free point, as given by eq. (7), page 56. Regard must be had to the signs in applying these equations to any special case. Velocity and acceleration upwards are positive, downwards negative ; to the right positive, to the left negative. (a) To find the time of motion of a point on a given curve. In all cases dt = . Hence when the nature of the curve and the v speed at any point of it are known, the value of v may be found from (52) and substituted, and then t may be found by integration If the acceleration of the point is constant and equal to /and paiall< to y, we have from (55) (56) $0 KINEMATICS GENERAL PRINCIPLES. [CHAP. VIII. (b) To find the normal acceleration due to the curve. If we multiply (50) by ~ and (51) by ~ and subtract, we have, since as as d*U (/, i' sin 6 = -A cos = , and dx* + dy* = ds 1 . as as dy dx dxd*y dycPx N = fx ds- ~ fv d*~ + - ds Eliminating dt by the equation v , we have at But if p is the radius of curvature of the constraining curve at the point x, y, _ P == Hence AT_ f d V ffatf " = fa ~5 -- Jv^ > ds ds p * f x sin 6 f y cosQ H -- ...... (57) r' The first two terms give the normal component of / for free motion. The last term is the normal acceleration due to the curve. If f x and f y are zero, the only acceleration is that due to the curve and N = *_ (page 76). KINEMATICS OF A POINT. TRANSLATION. CHAPTEE I. RECTILINEAE MOTION OF TRANSLATION. FALLING BODY. ACCELERATION INVERSELY AS THE SQUARE OP THE DISTANCE. Translation. We have defined translation (page 13) as motion of a rigid system, such that every straight line joining any two points remains always parallel to itself. The paths of all the points are therefore parallel at every instant and equal for any given interval of time, and the velocities of all the points at any instant are equal and parallel. If these velocities are uniform, that is, if all points move in parallel straight lines with equal speed, the translation is uniform. If these velocities change either in magnitude or direction, the translation is variable. When, then, a body has motion of translation only, the motion of the body is the same as that of any one of its points, and the study of the kinematics of a point is therefore the study of the translation of a body. Rectilinear Motion. If the direction of the acceleration of a point does not change and always coincides with the direction of the velocity, then the velocity may change in magnitude but can- not change in direction, and we have motion in a straight line. In such case the magnitude of the velocity is the speed in a straight line, the magnitude of the acceleration is the rate of change of speed and may be either uniform or variable, and the equations of pages 28 or 51 apply. Acceleration is Proportional to Force. Although \ve are now studying change of motion without reference to its cause, it will be well for the student to keep in mind the fact that no material body can change its own motion. Any change of motion is always found to be due to the action of other bodies. This action of external bodies upon the body considered to which change of motion or ac- celeration is due is called force. 91 92 KINEMATICS OF A POINT TRANSLATION. [CHAP. I, The student may figure to himself such a force as the pressure or pull of an imponderable spiral spring upon the body, the axis of the spring having always the direction of the acceleration, and the spring moving with the body, so that its pressure or pull is exerted during the entire time of acceleration and is always proportional to the acceleration. If the acceleration changes in direction, the axis of this spring changes, so that it is always in the same direction as the accelera- tion. If the acceleration changes in magnitude, the pull or push of the spring changes correspondingly. If the acceleration is uniform, that is, does not change either in direction or magnitude, the axis of the spring does not change in direction and its pull or push is constant. The force of gravity upon bodies near the surface of the earth is like the action of such a spring. Its action is practically constant in intensity and direction. The student should note that the direction of the force or accel- eration is not necessarily that of the motion, except in the case of rectilinear motion. Thus in the case of a point moving with uniform speed in a circle, the direction of motion at any instant is tangent to the circle, but the acceleration is always directed towards the centre. Central Acceleration. When the acceleration is thus always directed towards or away from a fixed point, it is called central ac- celeration, and the fixed point is called the centre of acceleration. If the direction of the acceleration is towards the centre, the ac- celeration is negative (page 50) and the force attractive. If away, it is positive and the force repulsive. Uniform Acceleration Motion Rectilinear Force Attractive. When the direction of the uniform acceleration coincides with that of the motion, we have motion in a straight line with uniform rate of change of speed, and equations (2) to (7), page 28 or 51, apply. The most common instance of such motion is that of a body fall- ing freely near the earth's surface.* In this case the acceleration due to gravity is known to be practically constant and is always denoted by g. We have then simply to replace a or / by g in equa- tions (2) to (7), page 28 or 51. We shall take g = 32.2 ft.-per-sec. per sec. or 981 cm.-per-sec. per sec. unless otherwise specified. Value of g. The value of g is usually given in feet-per-sec. per sec. or in centime ters-per-sec. per sec. It has been determined by much careful experiment and found to vary with the latitude A and the height h above sea-level. * Strictly speaking there is no known instance in nature of a uniform ac- celeration (or of a force which does not vary in magnitude and direction). The acceleration g due to gravity (or the force of gravity) varies inversely as the square of the distance from the centre of the earth for a body outside the earth, and directly as the distance for a body inside, i.e., in a shaft or well. But, as we shall see, the variation due to this cause is insensible for all ordinary distances. The decrease of g at a distance of a mile above the earth's surface is only about the 2000th part of its value at the surface. Also two radii of the earth are sensibly parallel when near together. It is therefore customary and practically correct to speak of g as a constant acceleration at any place. It should be borne in mind, however, that even then the resistance of the air very materially modifies the results for falling bodies. We can therefore only assume g as constant for fall in vacuo. CHAP. I.] EECTILINEAK TRANSLATION". 93 The general value is given by g = 32.173 0.0821 cos 2A - 0.000003ft, where h is the height above sea-level in feet, and g is given in feet- per-sec. per sec., or g = 980.0056 2.5028 cos 2A - 0.000003/1, where h is the height above sea-level in centimeters and g is given in centimeters-per-sec. per sec. It will be seen that the value of g increases with the latitude, and is greatest at the poles and least at the equator. It also decreases as the height above sea-level increases. The following table gives the value of g at sea-level in a few localities : Latitude. F. S. Units. C. S. Units. Equator 0' 32.091 978.10 New Haven 41 18 32.162 980.284 Latitude 45 45 32.173 980.61 Paris 48 50 32.183 980.94 London , 5140 32.182 980.889 Greenwich 51 29 32.191 981.17 Berlin 5230 32.194 981.25 Edinburgh 5557 32.203 981.54 Pole 90 32.255 983.11 United States j 49 (25 32.162 32.12 980.26 979.00 For calculations where great accuracy is not required it is cus- tomary to take g = 32 ft.-per-sec. per sec. or g = 981 cm.-per-sec. per sec. For the United States g = 32| is a good average value and is therefore very often used. In exact calculations the value of g for the place must be used. Formulas for a Body Projected Vertically Up or Down. We have then, for a body projected vertically upwards in vacuo, simply to put g in place of / in equations (2) to (7), page 51. We thus obtain gt-, t = Si = Vi V_ 9 V + Vi. (i) (2) (3) hi 2(s - sQ _ t?i VvS - 2g(s - si) . ^ (4) v + vi g V * = v l ! '-2g(s-s 1 ); (5) S Si = tv V" 2g (6) If the starting-point is below the origin, we should change the sign of Si. ,, If the body is projected downwards, we should change the signs of v, vi , s and Si. We see that this is equivalent to simply chang- 94 KINEMATICS OF A POINT RECTILINEAR TRANSLATION. [CHAP. I- ing the sign of g in all equations, leaving the signs of the other quantities unchanged. When the final velocity v is zero, we have from (2), for the time of rising to the highest point or the "turning-point," T = g ' For the time of rising to the highest point and returning to the starting-point we make s Si = in (4) and obtain Hence, the times of rising and returning are equal. For body falling we have T = -- ', the minus sign denoting time before the start necessary to acquire the velocity Vi. The distance from the starting-point to the turning-point is v a found from 16;, by making v = 0, to be ^-. *9 The distance ^- or _ is called the height due to the velocity v\ or v ; that is, the distance a body must fall from rest in order to ac- quire the velocity v t or v. When the distances in rising and falling are equal we have v a v 2 s Si = 0, or ^- = ^ , or Vi = v, that is, the velocity of return is equal to the velocity of projection. If the time of rising is less than T = , the displacement s Si is equal to the distance described. But if the time of rising is 1)l greater than T = , the body reaches the turning-point and then \y falls from rest, and the entire distance described is distance described = ^1 + ^-g(t - T) s = - s = Vl * + *** (7) [Application of Calculus to the preceding Case.] We can deduce the pre- ceding equations from our general equations (8) to (10), page 51. Thus from equation (9) we have, for acceleration directed downwards and therefore ( ), (page 50,) dt dt* Integrating, we have v = = gt -\- Const. ut When t = 0, let v = -f i , the (-(-) sign denoting motion upwards. Then we have Const. = ,, and which is equation (1), page 98. Integrating again, Vit xfft* + Const. CHAP. I.] 1ALLING BODY EXAMPLES. 95- Let s = + s, when t = 0, the (-(-) sign denoting distance upwards. Then we have Const. = Si , and s- Sl =v l t- -gf, (3> which is equation (3), page 93. From equations (1) and (3) we can deduce all the others page 93. The student should note especially that these equations have been deduced for body projected upwards or Vi positive. If we suppose motion towards the centre or downwards, we should have , negative. Also if the starting-point is below the origin, we should change the sign of *i. In all cases we take g minus, as long as the acceleration is directed down- wards. EXAMPLES. Unless otherwise specified g = 32.2 ft.-per-sec. per sec. or981 cm. -per-sec. par sec. All bodies supposed to move in vacuum. .,- v(i) A point moves with a uniform velocity of 2 ft. per sec. Find the distance from the starting-point at the end of one hour. - ^,*Ans. 7200 ft. Motion in a straight line. (2) Two trains have equal and opposite uniform velocities and each consists of 12 cars of 50 ft. They are observed to take 18 sec. to pass. Find their velocities. Ans. 22.73 miles per hour. , Two points move with uniform velocities of 8 and 15 ft. per sec: in directions inclined 90. At a given instant their distance is 10 ft. and their relative velocity is inclined 30 to the line joining them. Find (a) their distance when nearest ; (b) the time after the given instant at which their distance is least. Ans. (a) 5 ft. ; (b) ^3 sec. (4) A body is projected vertically upwards with a velocity of 300 - ft. per sec. Find (a) its velocity after 2 sec.; (b) its velocity after 15 sec.; (c) the time required for it to reach its greatest height ; (d) the greatest height reached ; (e) its displacement at the end of 15 sec.; (/) the space traversed by it in the first 15 sec. ; (g) its dis- placement when its velocity is 200ft. per sec. upwards; (h) the time required for it to attain a displacement of 320 ft. * Ans. (a) 235.6 ft. per sec. upwards; (b) 183 ft. per sec. downwards; (c) 9.3 sec.; (d) 1397.5 ft.; (e) 877.5 ft. upwards; (/) 1917.5 ft.; (y) 776.3 ft. upwards; (h) 1.13 sec. in ascending, 17.5 sec. in descending. (5) A ball is projected upwards from a window half way up a toicer 117.72 meters high, with a velocity of 39.24 m. per sec. Find the time and speed (a) with which it passes the top of the tower ascending ; (b) the same point descending ; (c) reaches the foot of the tower. Ans. () 2 sec.; 19.62 in. per sec.; (b) 6 sec.; 19.62 m. per sec ; (c) (4 _)_ 2 i/f) sec.; 19.62 |/7 m. per sec. * If the student will refer to the Examples, page 114, he will gain an idea of the effect of the air in modifying the motion of falling bodies, and will better appreciate the delusive nature of all problems which ignore it. 96 KINEMATICS OF A POINT RECTILINEAR TRANSLATION. [CHAP. I. * s> yL,(6) A stone is dropped into a well and tlie splash is heard in 3.13 sec. If sound travels in air with a uniform velocity of 332 meters per sec., find the depth of the well. Ans. 44.1 meters. V(7) If in the preceding example the time until the splash is heard isT and the velocity of sound in air is V, find the depth. Ans. Depth = Y( T g + V) - ^V(2Tg+ V)\ ytJ6) Show that a body projected vertically upwards requires twice as long a time to return to its initial position as to reach the highest point of its path, and has on returning to its initial position a speed equal to its initial speed. (9) A stone projected vertically upwards returns to its initial position in 6 sec. Find (a) its height at the end of the first second, and (b) what additional speed would have kept it 1 sec. longer in the air. Ans. (a) 80.5 ft.; (b) 16.1 ft. per sec. (10) A body let fall near the surface of a small planet is found to traverse 204 ft. between the fifth and sixth seconds. Find the ac- celeration. Ans. 20.4 ft.-per-sec. per sec. (11) A particle describes in the nth second of its fall from rest a space equal to p times the space described in the (n l)th second. Find the ivhole space described. Ans. "^7^ J". (12) A body uniformly accelerated, and starting without initial velocity, passes over bfeet in the first p seconds. Find the time of passing over the next b ft. /2~- l) sec. 13) A ball is dropped from the top of an elevator 4.905 meters high. Acceleration of gravity is 9.81 meter s-per -sec. per sec. Find the times in which it will reach the floor (a) when the elevator is at rest; (b) when it is moving with a uniform downward acceleration 0/9.81 m.-per-sec. per sec.; (c) when moving with a uniform down- ivard acceleration of 4.905 m.-per-sec. per sec. ; (d) ivhen moving with a uniform upward acceleration of 4.905 m.-per-sec. per sec. Ans. (a) 1 sec.; (b) oo ; (c) -/2~sec.; (d) i/- * A sec. o (14) If Si , St are the heights to which a body can be projected ivith a given initial vertical velocity at two places on the earth's sur- face at which the accelerations of falling bodies are g\ and g? respec- tively, show that Sigri = Sig?. ) A stone A is let fall from the top of a tower 483 ft. high. At the same instant another stone B is let fall from a window 161 ft. below the top. Hoiv long before A will B reach the ground ? Ans. ( |/6 5 = u, 8oV*f Y , (2) \' / These are the equations (1) and (2) of the preceding Article. They hold, as we see, for motion towards or away from the centre, provided the acceleration is towards the centre. For acceleration away from the centre we change the sign of a'. These equations also hold for any path, straight or curved, if *, 81, / are measured along the path, and a' is the tangential acceleration or rate of change of speed at distance r 7 , and v and Vi the speeds final and initial. If the initial velocity is zero, we have for a body falling from rest L ds Since = , we have from (3) (3) _ ds _ / , J\ 1 ~ dt ~ r \s Sj where we take the ( ) sign for the radical to denote motion towards the centre (page 44). This can be put in the form sds . /2a'r' 3 = dt 4/ . To put this in a form convenient for integration, add and subtract \^ to the numerator of the first term. We then have *, 2* Sids . /2aV 4 === = dt V - -. s - s 3 *' Integrating, we have (4) Let t = when s = Si , then Const. = -- ^. Hence* for the time of falling a i 2s _i/. 2s\ _,/2s \ * We have it versin = it cos 1 -- = cos --- 11. i \ **/ \ Sl I \ 1 2s \ From trigonometry, 2 cos 8 y 1 = cos 2y. Let 2y = cos" M - -- 1 j . Then cos 2w = - - 1 and cosy = -, or y cos~'/i/- Sl Si ' *1 , fs _i2s Hence 2 cos' 1 .i/ - = ?r versin r ! *i and 2 = 102 KINEMATICS OF A POINT RECTILINEAR TRANSLATION. [CHAP. I. from rest we have for acceleration towards the centre If we had taken motion away from the centre, we should have obtained, instead of (4), ,i , . -12* /2ayU. (SiS ") + 5- versm = -- 1 * + Const. & s\ y 81 j Let t = 0, when s = 0, and Const. = 0, and we have Equation (6) applies to a body projected upwards to any height s, Si being the height at which it would come to rest. If we make s = 81 in (6), or s = in (5), we find the time of reaching the turning-point in rising or reaching the centre in Tailing from rest T= * Sl ^ F.S. A. S. 14*. T* r i i - \*> /. CHAPTEE II. SIMPLE HARMONIC MOTION. MOTION IN RESISTING MEDIUM. f Simple Harmonic Motion. The motion of a point moving in any path in such a manner that the tangential acceleration is directly proportional to the distance, along the path, from a fixed point in the path is called simple harmonic motion. Such motion may be rectilinear or curvilinear. The vibrations of such bodies as a tuning-fork or a piano-wire are approximate examples of such motion, and hence the term " harmonic." The vibrations of an elastic body, such as the air, are examples of such motion It is also, as has been stated (note, page 92), the motion of a body under the action of gravitation, within a homogeneous sphere, as it can be shown that in this case the acceleration due to gravity is proportional to the distance from the centre. The motion of the piston of a steam-engine when moved by a crank and connecting-rod approximates the same motion if the rotation of the crank is uniform, the approximation being closer the longer the connecting-rod. This will be evident from the fol- lowing Article. Simple Harmonic Motion in a Straight Line Force Attractive. Let a point M move with uniform speed in a circle of radius CM r. Then the acceleration f n is always direct- ed towards the centre and equal to f n = roa*, where GO is the constant angular velocity (page 76). The projection of f n upon the diameter A 1 CA is tV cos MOP. But r cos MCP is the distance CP=s of the projection P of M upon the diameter CA. Therefore the projection of f n upon the diameter is a = ca j s, or, since GO is constant, a is directly proportional to the distance CP = s. The motion of P is therefore harmonic. If then a point M moves with uniform speed in a circle, its pro- jection P upon any diameter moves with harmonic motion in the meter, the centre of acceleration being the centre of the circle. Let a' be the known acceleration in the line AC of P at a given distance r' from the centre. Then a' = =\ / -^ , and the 103 la! V*" 104 KINEMATICS OF A POINT TRANSLATION. [CHAP. II. speed of M is r time occupied by P in passing from A to A and back to A is But if a is the acceleration at any distance s, and / is the ac- celeration at the extreme distance r, we have for harmonic motion a' f n a <*' Hence the time T of a complete oscillation is The time T of a complete oscillation depends therefore only s 1 upon the constant ratio = , and is independent of the range r or amplitude of the oscillation. For this reason the oscillations are said to be isochronous, or made in equal times, no matter what the range or amplitude. COR. 2. Since the motion of a body under the action of gravity in a homogeneous sphere is harmonic (page 92), if we put g for a' and let r' be the mean radius of the earth, we have from (1) the \motion of a body falling under the action of gravity towards the CHAP. II. J SIMPLE HABMONIC MOTION. 105 \ centre of the earth in a well or shaft, assuming the earth to be a homogeneous sphere and neglecting resistance of the air. In such case (1) becomes u> = ( r + s)(r - s). T If the fall takes place for a short distance compared to r' and near the surface, we have r + s practically equal to 2^ and hence t;" = 2g(r s), which is the same as for uniform acceleration gr, the initial speed being zero. We obtained the same result (page 93) for a body external to the earth. The equations of page 93 hold good, therefore, in all practical cases, whether the fall takes place above the earth or within the earth, neglecting resistance of the ah*. Amplitude Epoch Period Phase. The range r = CA = CA on either side of the centre of acceleration, in harmonic motion, is called the amplitude. A complete oscillation is from A to A and back to A. The time of an oscillation, as we have seen, is independent of the amplitude. From A to A or A to A is a vibration. A vibration is half an oscillation. The time of a vibration is half that of a complete oscillation. If Pi is the initial position from which the time is counted, or the position of P at zero of time, the time of passing from A to Pi is called the epoch. The epoch may also be defined with reference to the auxiliary circle, as the angle ACMi in radians. This is the epoch in angular measure. The epoch in angular measure is then the angle described on the auxiliary circle in the interval of time denned as the epoch. The epoch locates the position of P at zero of time. The entire time which elapses from any instant until the moving point again moves in the same direction through the same position is called the period. The time from Pi to A, then back through Pi to A, and finally back from A to Pi , is a period. It is evidently the time of a complete oscillation from A back to A. That fraction of the period which has elapsed since the moving point P last occupied A is called the phase. Measured on the circle, it is the ratio of the angle ACM radians to 2n radians. The phase locates the position of P at any instant. It therefore varies with the time or with the position of P. The phase at zero of time, then, multiplied by 2* radians gives the epoch in angular measure, and multiplied by the time of an oscilla- v. tion gives the epoch in time. [Application of Calculus to Harmonic Motion. We may deduce the results obtained for simple harmonic motion (page 104), as well as others, from tlie general equations (8) to (10), page 51. We have, as before, a = s (page 104). For acceleration away from the centre we have a positive, for acceleration towards the centre a negative (page 50). 73 106 KINEMATICS OF A POINT TRANSLATION. [CHAP. II. I. Acceleration towards the Centre. In this case we have _ dv _ d*s _ a' = ~dt = W = ~r' 8 ' ds Multiply both sides by ds and then, since = v, we have (tt vdv = -sds. r Oi 8 Integrating, we obtain tf = "^r- -f Const ? . . (a) When s = r let v = Vi. We have then * ' * - * *'. f &> A . , Const. =vS + ( ~, and bence *-!* + ;#*-") (1) If the initial velocity Vi is zero, this becomes r^\ i .... (2) <^ which is the same as equation (1), page 104, already obtained, for initial velocity zero and range r. ds Since v = , we have from (1) ctt where we take the (-(-) sign for motion away from the centre and the (-) sign for motion towards the centre (page 44). This can be written ds If we integrate this between the limits of t and t = when s = r, we have s ./a' r Hence* s = r cos t 4/ Vi \f sin t \f . . (4) T r r a if /a' , B = 7* Then - = sin ( A + B) = sin A cos B -f cos A sin B. i/r + r ',V a fjf , But sin B= , and cos B= \/\ sin 8 B. Substituting these ' *' / r- v-t values and reducing, we, obtain equation (4). "Q/ ^ CHAP. II.] SIMPLE HARMONIC MOTION. 107 If motion is towards the centre, we take the ( ) sign ; if away from the centre, the (-(-) sign. If v\ is zero, or there is no initial velocity, we have /' 1 = rcostf .{/ -.; (5) 8 = R= This reduces to r when v t = 0. If we integrate (3) between the limits of t and t = when s = R, that is, if we count the time from the end of the amplitude where Vi = 0, instead of from = r, we obtain and hence /^ 1 (7) If in (5) .we make = r or in (7) make s = R, we have in both cases /r 7 for the time of a vibration, it A/ , and hence for the time of a complete r a f? oscillation T 2n y . Therefore the time of oscillation or vibration is not affected by the initial velocity. All these equations (1) to (7) hold for motion either towards or away from the centre, provided the acceleration is towards the centre. They also hold for any path, straight or curved, provided r 1 , r and s are measured along the path and a is the rate of change of speed at the distance r'. II. Acceleration Away from the Centre. In this case we have the accelera- tion positive and hence dv , a' a= = -4- *. dt ' r' ds Multiplying by ds, we have, since = v, at a' , vdv = sds. r Integrating this, we have When s = 0, let = d. Then Const. = Vi* and e = + Const. r ' = <"' + >' (8) From (8) we have \ If we make D = in (1), we have for the amplitude 108 KINEMATICS OF A POINT TRANSLATION. [CHAP. II- where we take the (-{-) sign for motion away and the ( ) sign for motion towards the centre. We can put this in the form ds a! If we integrate this between the limits of t and t = when s = 0, we have v = t -f logn , . _ *'A/?I fr* 1 = 2Va' * -- 7^ where e is the base of the ^Taperian system of logarithms. f i - > ' EXAMPLES. g = 32.16 ft.-per-sec. per sec. or 980.23 cm.-per-sec. per sec. (1) If the radius of the earth is 6370900 meters and the accelera- tion of gravity 9.81 meters-per-sec. per sec., what should be the value of a'r' 2 in eq. (2), page 99, */ s and s> are given in kilometers ? Ans. 398171.88 cubic kilometers-per-sec. per sec. (2) A body falls to the earth from a point 1000 miles above the surface. Find its speed on reaching me surface, neglecting resist- ance of the air and taking the earth's radius 4000 miles. Ans. (6) where e = 2.718282 = base of the Naperian system of logarithms. From (5) and (6) we can find the distance for any time or the reverse. From (6) we have cvit = e c ' 1, and substituting this in (3) we have i e* v = (7) or or using common logarithms 2.302585 (8) fr 5 i (9) From (7) and (9) we can find the velocity for any distance or the reverse. From (8) we see that when the velocity becomes zero the space traversed is infinite, and from (2) the time is infinite. Although then the velocity dimin- ishes as the time increases, as we see from (3), it cannot become zero in any finite distance or time. CHAP. II.] RESISTING MEDIUM. HI [Body Falling under the Action of Gravity in a Resisting Medium.] Let fbe the uniform acceleration due to any attractive force. In the case of gravitv we have* M'bfrre-zf-ls the density-or mass of a nnit volume of the medium and $ is the densrtj-of tire body. 'The acceleration / acts away from the starting-point and the retardation erf acts towards the starting-point. If then we take this point as origin, we have dv w=s- c *' ........... a> where c is the coefficient of resistance and has the same value as in the preced- ing Article. Let k be that velocity for which the reta/dation is equal to/, so that/ = c/fe 2 . Then c = ^, and equation (1) becomes w We can write this in the form Integrating, we have k When t = 0, let v = ,. Then Const. = - ~ logn *^i. Hence */ K Vi k , (k + v)(k - ,) or using common logarithms, _ 2.302585ft, (k -f v)(k ,) i) From (4) we have, if e = 2.718282 = base of Naperian system of logarithms, i *H A; i From (5) and (6) we can find the time for any given velocity or the reverse. * As we shall see hereafter, the mass of a body multiplied by g gives the weight of the body, that is, the force of gravity. It is also a well-known fact that a body immersed in a fluid has its weight diminished by the weight of an equal volume of the medium. If then V is the volume of the body, VSg is its weight in vacuo and VAg is its loss of weight due to the medium. Hence VSg VAg is the weight when immersed. Since VS is the mass, Vdf= VSg - VAg, or f = gll- ~\ i ^ , f" ^ ^ T \" ' 112 KINEMATICS OF A POINT TEANSLATION. [CHAP. II. fatr- CW Since = -j- , we have at and therefore from (2) we have dt fc ~W "1 2/ ' or using common logarithms, r 'sJ^V, From (7) we have 2.302585A; 2 # - c,' lo &~ts IF 2/ = A* - (A; 2 - s, (7) (8) (9) where e = 2.718282 = base of the Naperian system of logarithms. From (8) and (9) we can find the distance for any velocity or the reverse. / We see from (9) that when s is great, v approaches k, and k is the limiting value of v. If the initial velocity is zero or less than k, v will continually ap- proach k, but can never exceed k. If the initial velocity is greater than k, v will diminish continually down to k and can never become less than k. [Body Projected Upwards under the Action of Gravity in a Resisting Me- dium. ] In this case we have, taking k as before, since /is negative and the resist- ance is negative, HF-:/-** dv (1) which can be written dt = ^ M i 2- Integrating and determining the constant by the condition that when t = 0, v = i = initial velocity, we have f (2) We have also as before -=> = -^ , and therefore, from (1), ~ vdv ds = 3- .tegrating this, and making s = 0, when v = Vi , we have 2.302585& 2 . * = 57 l 8 (4) The time in which the velocity becomes zero and the body reaches the turn- ing-point is, from (2), - id the corresponding value of s is, from (4), 2.302585& 2 (6) CHAP. II.] RESISTING MEDIUM. 113 At the end of the time T the body begins to return and falls from a state of rest, or , = 0. We have then from the preceding Article, making , = 0, 2.302585& , and 2.302585A 2 , /fc 2 h-s=-- -lo Sw -^ r . ...... (8) Let u be the velocity with which the body returns to the starting-point. Then putting * = in (8) and v = u, we have __ 1 _, * or substituting for h its value, *L Hence We see then that u is less than DI, or the body returns to the point of pro- jection with a velocity less than the velocity of projection. " mist or rain in air " " " ~ = 813.82. " lead in water " " " 3 = H-38- d " " " air ~2 = 9423.61. The coefficient of resistance for a sphere (Vol. Ill, Kinetics, page 61) is 166V where r is the radius of the sphere. For a cone we have c = where r is the radius of the base and h the height. If the cone terminates in a cylinder of length Z, we have c = 26(31 + h)(r* + h') 114 KINEMATICS OF A POINT TRANSLATION. [CHAP. II.. EXAMPLES.* g 32.16 ft.-per-sec. per sec. (1) A lead bullet 1 inch in diameter is projected vertically with a velocity of 2000 ft. per sec. Find (a) the time of ascent with and ivithout resistance of the air ; (b) the distance to which it ascends with and without resistance of the air ; (c) the velocity and time of return with and without resistance of the air. Ans. We have in this case 4 = 9423.61, A* = 67338.1852 and k = 259.49 4 ft. per sec., / = 32.1556 ft.-per-sec. per sec. (a) The time of ascent in vacuo is 62.19 sec.. 259.49 t , 2000 In air T = ...... tan - 1 -- = 12. 60 sec. 32.1o56 2o9.49 (b) The height of ascent in vacuo is 62189 ft. 2.302585 X 67338.1852 / 4000000 \ 32.1556 6-733838-52 (c) The velocity of return in vacuo is 2000 ft. per sec. 4000000 In air ,* = _- t or = 257 ft. per sec. 67338. 1852 (d) The time of return in vacuo is 62.19 sec. 2.302585 X 259.49 . 259.49 + 257 _ In air t ~ T = 2X32.1556 259.49 - 257 (2) A lead bullet 1 inch in diameter is let fall in the air. Find the velocity at the end oft = l sec., 2 sec., 3 sec., 10 sec., 20 sec., with and without the resistance of the air. Ans. We have k = 259.49 ft. per sec.; /= 32.1556 ft.-per-sec. per sec ; tyt v Jc\e k I/ e = 2.718282; and from eq. (6), page 111, making Vi = 0, v = ^ e * + l Let t = 1, 2, 3, 10 and 20, and we have v = 31.98, 63.03, 92.33, 219.3 and 255.86 ft. per sec.; while in vacuo we would have = 32.16, 64.32, 96.48, 321.6 and 643.2 ft. per sec. (3) In the previous example what is the greatest velocity the bullet can attain f Ans. k = 259.49 ft. per sec. As we have seen, this velocity is attained quite early, after which the velocity is uniform. (4) An iron cannon-ball 1 ft. in diameter is projected vertically upwards in the air with a velocity of 2000 ft. per sec. Find (a) the time of ascent ; (b) the distance to which it ascends ; (c) the velocity with which it returns ; (d) the time of return. * Ans. We have in this case -j = 5983.28, & 2 = 513040, k 716.268 ft. per sec.,/ = 32.1546 ft.-per-sec. per sec. * An examination of these problems will give the student an idea of the effect of the air in modifying the motion of a falling body and enable him to> realize the inaccuracy of neglecting it. CHAP. II.] EESISTING MEDIUM EXAMPLES. 115 (a) The time of ascent is T 27.32 sec.; in vacuo, 62.19 sec. (See Ex. (1).) (&) The distance of ascent is h = 34680 ft.; in vacuo, 62189 ft. (c) The velocity of return is u = 6T4.3 ft. per sec.; in vacuo, 2000 ft. per sec. (d) The time of return is t T 38.9 sec.; in vacuo, 62.19 sec. (5) An iron cannon-ball 1 ft. in diameter is let fall in the air. Find the velocity at the end of t = 1 sec., 2 sec., 3 sec., 10 sec., 40 sec., 60 sec. Ans. We have k = 716.268 ft. per sec.,/ = 32.1546 ft.-per-sec. per sec. and i = 0. Hence from eq. (6), page 111 : For t = 1 2 3 10 40 60 sec. = 32.12 64 95.85 301.5 677.8 709.7 ft.-per-sec. In vacuo we would have v = 32.16 64.3296.48 321.6 1286.4 1929.6 ft. per sec. (6) In the previous example, what is the greatest velocity the can- non-ball can attain ? Ans. k = 716.268 ft. per sec. And this is attained in little more than a minute. (7) In Example (5) what are the distances passed through 9 Ans. From eq. (8) we have, making v\ = and taking the values of v already found, For t = 1 2 3 10 40 60 sec. s- 16.03 64.29 144.2 1561 18000 31992ft. In vacuo we would have * = 16.08 64.32 144.72 1608 25728 57888 ft. (8) A lead shot -J- inch in diameter is let fall in the air. Find the greatest velocity it can attain, and the velocity and space traversed in t = 1, 2, 3, 4, 5, 6, 7 and 8 sec. Ans. We have -^- = 9423.61, /= 32.1556, k* = 8417. Greatest velocity = k = 91.7 ft. per sec. For f = 1 2 3 4 5 6 7 8 sec. v = 30.89 55.49 71.76 81.23 86.35 89.01 90.35 91.03ft. per sec. In vacuo, = 32.16 64.32 96.48 128.64 160.8 192.96 225.12 257. 28 ft. per sec. s = 15.67 60.27 123.56 199 283.3 370.7 458.1 545.5ft. In vacuo, 16.08 64.32 145.72 257.3 402 578.8 787.9 1029.12ft. (9) What is the greatest velocity a rain-drop i inch in diameter can acquire, falling in the air f <> Ans. We have ~ = 813 82, / = 32 12, k = 27 ft. per sec. (10) An iron cannon-ball 1 ft. in diameter is let fall in water. Find (a) the greatest velocity it can attain; (b) the final velocity and the space passed through in t = 1, 2, 3 sec. Ans. We have -J- = 7 .2, /= 27.7. Therefore, (a) k = 23.06 ft. per sec. (6) For t = 1 2 3 sec. v = 19.22 22.68 23.02 ft. per sec. s - 11 38 32.8 53.94 ft. We see that the maximum velocity is reached in about 3 sec. 116 KINEMATICS OF A POINT TRANSLATION. [CHAP. II. In the preceding examples it is assumed that the density of the medium is unchanged, and that the acceleration of gravity is con- stant. Near the earth's surface both assumptions are practically true. We have also assumed that the acceleration varies as the square of the velocity. Experiments would seem to indicate that this is not strictly accurate. The effect of resisting media upon the motion of projectiles is therefore best taken account of by means of empirical formulas based upon experiment. We have given the preceding examples in order to call attention to the fact that the influence of the medium, even of the air, is such as to very materially modify the results of the formulas of page 93, which hold good only in vacuo and are not even approximately true except for large and heavy bodies for the first few seconds of fall. The examples of page 95 are therefore devoid of practical value except under such limitations. For very great distances the density of the air and the accelera- tion of gravity are not constant, so that our present assumptions are then no longer in accord with fact. mrv^ -p* <* //7 Ar /3, C/2>, " **** CHAPTER III. TEANSLATION IN A CURVED PATH DIRECTION OF ACCELERATION CONSTANT. PARABOLIC MOTION. PROJECTILE IN A RESISTING MEDIUM. MOTION OF Curved Path. When a point moves in a path such that the di- rection of the acceleration coincides with the direction of motion and does not change, the motion is rectilinear, no matter what the law of variation of the magnitude of the acceleration may be. Such motion we have discussed in the preceding Chapters. If the direction of the acceleration, however, does not coincide with that of the motion, then, whether it is constant in direction \ and magnitude or not, we have motion in a curved path. When a rigid body composed of many points moves so that every straight line through any two of its points remains parallel to itself in all positions of the body, it has a motion of translation only, and we may treat the body as if it were a point. For motion in a curved path the differential equations of page 81 apply. Uniform Acceleration Inclined to Direction of Motion. If the ac- celeration is uniform, that is, constant in magnitude and direction, its component in any given direction is uniform, and the equations for rectilinear motion, page 93, apply to the component motion in that direction. The most common case of curvilinear motion under uniform ac- celeration is that of a body projected with any given velocity in any given direction at the surface of the earth, neglecting the resistance of the air. In such case the acceleration due to gravity is practically uniform and equal to g ft.-per-sec. per sec. Let the initial velocity of projection Vi of the point P make the angle APB ai with the hori- A zontal. Let the co-ordinates of any point M of the path or trajectory \ be PB = x and BM = y. Let the angle MPB = 0. Let \/be the uniform vertical accel- \eration (in the case of gravity/ 'if wo mako Op parallel and equal to the velocity Vi at P, and Om parallel and equal to the ve- locity v at any point M of the trajectory, then pm =ft is the integral acceleration for the time t during which the point passes from P to M, and the straight line pb is the hodograph (page 52) 117 118 KINEMATICS OF A POINT TRANSLATION. [CHAP. III. for motion from P to the point C, where the velocity is horizontal, and Ob is the velocity at this point. The velocity of p in the hodograph is the acceleration in the path (page 52). Therefore the point p moves with uniform velocity / from p to 6, while P moves from P to C. We see at once that the horizontal component of the velocity Vi is Ob v x , or Ob = v x = Vi cos ai ......... (1) The horizontal distance passed over in any time t, while the point P moves from P to M, is then PB = x, or PB x Vit cos ai ........ (2) The vertical component of the velocity v^ is bp = Vi sin ai up- wards. But the acceleration / is downwards Hence the vertical velocity at the end of the time t is bm = bp pm = v y , or bm = Vy = t7isinai ft ....... (3) The vertical velocity at the beginning of the time t is v\ sin <*i . The mean vertical velocity during the time t is then 2i sin a> ft . 1 . - = Vt sin ai - -ft. The vertical distance passed through in the time t is then BM = y, or BA- AM=BM~y = v 1 tsinai -~ft' 2 ..... (4) /* If we combine (2) and (4) by eliminating , we have for the equa- tion of tlie trajectory y = x tan on 2vS cos 1 (5) This is the equation of a parabola. The time of reaching the highest point C is the time of describ- ing the vertical distance DC. Call this time T v . Since at this point the vertical velocity is zero, we have, by making v y = in (3), f J-v - If we substitute this for t in (2) and (4) we obtain the co-ordinates of the vertex C of the parabola, i* sin ai cos ai _ Ui" sin 2cti T~ ~%T~ Vi* sin 2 a\ _ xj P. That is, distance of the directrix above P is the height due to the velocity Vi. If we had taken the origin at the vertex C and let x c and y c be the new co-ordinates, then the horizontal velocity at C would be CHAP. III.] TRANSLATION PROJECTILES. 119 Vi cos a, and the horizontal distance passed over in any time t would be x c = v it COB a i. The mean vertical velocity would be ft and the vertical distance y c = -/f 2 . Combining and eliminating t, we obtain 2vi* cos 2 a i Xc y = j y c , which is the equation of a parabola referred to its diameter CD and the tangent at the vertex C. The parameter is as before v - . To find the velocity at any point of the trajectory. The magnitude of the velocity at any point M is the resultant of the vertical and horizontal velocities, or, from (1) and (3), v j = v x * + Vy* = Vi 2 2vift sin i + ft* (6) The same result is obtained at once from the hodograph from the triangle Opm. Inserting the. value of y from (4), u 2 = vS 2fy (7) If the acceleration is due to gravity, we replace / by g, and have 2 Ui 2 Vl* = y. But we have iust seen that is the distance of the *)n 9/- *^ 9/1 zg zg &y directrix above P. Therefore - y is the distance of the direc- trix above any point M, and ^- is the height due to the velocity v. Hence, the speed at any point is the same as that acquired by a body falling from the directrix to that point. To find the direction of the velocity v at any point M, the mag- nitude of which is given by (6) and (7), let a be the angle which it makes with the horizontal. Then we have directly from the hodo- .graph, since angle mOb = a, v sin a = Vi sin i ft ; v cos a = vi cos ai. Therefore, from (2), tan a = tan i = tan ai E-. ... (8) Vi COS ai X r fx tan a = tan ai ~ ^- (9) ^fe To find the time of flight in a horizontal direction, and the horizontal range. If in (4) we make y = 0, we have for the time Th in which the body reaches the line PX, or the time of flight in a horizontal direction, m 2Vi sin <*i _ J-h = ^ v 1 "/ Inserting this value of t in (2), we have for the horizontal range PX, t . ^/ = -^-^^y-^ ^i = 3l_^ i (11) 120 KINEMATICS OF A POINT TKANSLATION. [CHAP. III.. ii This is twice the distance PD. The greatest value sin 2^, can have is unity, and this occurs when 2i = 90 or i = 45. Therefore the horizontal range, neg- lecting resistance of the air, is greatest for an angle of elevation of 45, and is equal to - To find the greatest height attained, and the corresponding time. t the vertical velocity given by (3) equal to zero, and we have or the time of attaining the greatest height sin ai f (12) just half the whole time of flight as given by (10). Insert the value of the time given by (12) in (4), and we have for e greatest height attained, CD = H, sin" (13) Equations (13) and - Rh given by (11) give the co-ordinates of the a vertex C. To find the displacement in any given direction, and the cor- : esponding time. Let 6 be the angle which any displacement PM = makes with the horizontal, then we have BM = y = x tan f J. Sub- iituting this value of y in (5), we have for the abscissa of the i 2 cos a, sin (an 6) _ Vi 2 [sin (2an 6) sin i ] oint M /yi __ /cose /cose a ad therefore for the displacement or range PM = R, n 2v* cos on sin (on 6) _ ^i 2 [sin (2on 6) sin 6] xv = (14) (15) /cos 2 e /cos s e If in (15) we make 6 = 0, we have the horizontal range Rh as fgiven by (11). If we divide (14) by the horizontal component of the velocity, , we have t = time of flight = ' s * n (a ' ~ (16) This reduces to (10) for 6 = 0. To find the angle of elevation which gives the greatest range in any given direction. The range R given by (15) is a maximum when sin (2ai 6) is a maximum, or when2on 9 = 90 or on = - J90 + 9 j. The direction of projection for the greatest range makes therefore with the vertical an angle 90 on = --(90 6), that is, it bisects the f \ I le between the vertical and the range. To find the elevation necessary to hit a given point. To deter- mine the direction of the velocity Vi in order that the path may pass through a given point given by x and y, we substitute for the equivalent value 1 + tan- , 2 fa? _ * 2f 2vf (21) Equation (21) is then the equation of a curve which passes through all the points in which every two trajectories starting from the same point P at angles of elevation whose difference is indefinitely small cut each other. It is therefore the equation of the envelope or curve which touches all the trajectories or parab- olas described from the same point P with the same initial speed d. Equation (21) is the equation of a parabola AC A', whose axis PC is vertical, whose focus is the point P of projection, and whose vertex C is in the common direction of the trajectories. With the given initial speed Vi, the projectile can reach any point within this envelope by two angles of elevation and two tra- jectories, as proved page 121. It can reach any point in the en- velope by only one elevation and path. It cannot reach with any elevation and with the given velocity Vi any point outside this envelope. The point, therefore, where this envelope cuts the plane of any given range gives the maximum range in that direction for any given Vi. Thus the maximum range on a horizontal plane is found from Vi* (21), by making y = 0, to be -=. The same result is given by (15) J when we make on = 45 and 6 = 0. Questions of maximum range may thus be readily solved by the equation for the envelope. x ii + -J-/ 2 From (2) we have cos a = T-, and from (4) sin ai = - |^ Vit Vit Since cos 2 a t + sin 2 i = 1, we have x 1 + (y + (22) This is the equation of a circle whose radius is Vit and whose centre is situated vertically below P at a distance PD = ft*. / The circumference of this circle is reached in the same time by a point starting from P with the velocity Vi in any direction. "Application of the Calculus.] The same results are obtained by the applica- tion of the differential equations of motion, page 81. Thus in the present case we have for the horizontal component of the acceleration, since /is vertical, CHAP. III.] TRANSLATION PROJECTILES. 123 and for the vertical component cPy * : C, ^ sT* - C, Integrating (a), since f or t = 0, ^ = * cos a lt we have c ' ' , v- <" * o , ^ *//; r ' (1) Integrating again, since for t 0, x = 0, we have x = Hit cos ai . (2) Integrating (b), we have, since when t = 0, = Vi sin a,, ~^fa- ' Mffarf "t - Af *t " T". , *" w dy . C ^. i / t / 7 4 -l^- pC Integrating again, since for t = 0, y = 0, we have y = t> l tsma 1 --ft* (4) Combining (2) and (4) by eliminating , we have for the equation of the trajectory We have also or, from (1) and (2), 2 = a, 2 2/toi sin cti +/ 2 2 (6) Inserting the value of y from (4), a 2 = j 2 2/y (7) If we differentiate (5), we have, for the tangent of the angle which the velocity at any point makes with the horizontal, dv fx tan a = /- = tan a l ^-^-5 , , (8) ax i 2 cos 8 ari These are the same equations as already given, and from them all the others are deduced. EXAMPLES. g = 32.16 ft.-per-sec. per sec. Resistance of air neglected. (1) Show that for parabolic motion the hodograph is a straight line. (2) The sights of a gun are set so that the ball may strike a given object. Show that when the sights are directed to any other object in the same vertical line, the ball will also strike it. (3) Two bodies projected from the same point in directions mak- ing angles ft, ft' with the vertical pass through the same point in 124 KINEMATICS OF A POINT TRANSLATION. [CHAP. III. the horizontal plane through the point of projection. If t and t' are the times of flight, show that sin (ft ft') _ t n t* sin (ft + ft') ~ t" + t* ' what velocity must a projectile be fired at an elevation of 30 so as to strike an object at the distance of 2500 ft. on an ascent of 1 in 40 ? Ans. 311.5 ft. per sec. (5) Find the direction and magnitude of the velocity of projection in order that a projectile may reach its maximum height at a point whose horizontal and vertical distances from the starting-point are b and h respectively. 2h Ans. tan a, = , v^ is fired horizontally at a height of 144.72/tf. above the surface of a lake and the initial speed of the ball is 1000 ft. per sec. Find (a) after what time, and (b) at what horizontal distance, the ball strikes the lake, neglecting resistance of the air. Ans. (a) 3 sec. ; (6) 3000 ft. (7) In the parabola described by a projectile, its speed at any point is that which it would have had had it fallen to that point from the directrix. (8) A particle projected at a given elevation with an initial speed Vi reaches the top of a tower hft. high and 2hft. from the point of projection in t seconds. Find (a) the initial speed of another par- ticle which, being projected at the same elevation from a point dis- tant 4h ft. from the tower, will also reach its summit, and (b) the time it will require. Ans. (a) '* : fl) ball is projected with a velocity of 100 ft. per sec. inclined 75 to the horizon. Find (a) the range on a horizontal plane ; (b) the range on a plane inclined 30 to the horizon ; (c) what other direc- tions of the initial velocity would give the same ranges. Ans. (a) 155.5 ft.; (b) 207.3 (|/3~- 1) ; (c) 15 and 45. (10) Show that with a given initial speed the greatest range on a horizontal plane is just half as great as the greatest range down an incline of 30. (11) Show that if two particles meet which have been projected with the same initial speed, in the same vertical plane, at the same instant, from tivo given points, the sum of their elevations must bz constant. (12) On a small planet a stone projected with a speed of 50 ft. per sec. is found to have a maximum range on a horizontal plane of 400 ft. Find the acceleration of falling bodies at the surface of that planet. Ans. 6 25 ft.-per-sec. per sec. (13) Two stones thrown at the same instant from points 20 yards apart, with initial velocities inclined QO J and 80, respectively, to the horizon, strike a flag-pole at the same point at the same instant. CHAP. III.] TRANSLATION PROJECTILES. 125 Show that the initial speeds are as 1 : V% ; and that the distance of the- pole from the nearer point of projection is 10 yards. \xfl4) At what elevation must a body be projected with a speed of 310.8ft. per sec. that it may hit a balloon 50Qft.from the earth's surface and at a distance of 1000 ft. from the point of projection. or 80 43'. / Wfo) A body is projected with an initial velocity of 30 ft. per sec. inclined 60 to the horizon. Find the velocity after 20 sec. Ans. 617.3 ft. per sec. inclined 148 36'. 6 to the direction of the initial velocity. (16) If from a point A bodies are projected at the same moment and in the same vertical plane at different angles of elevation, with the same initial speed, Vi, the locus of all the positions occupied at the end of a given time t is a circle whose radius is v\t and whose centre is situated vertically below A at a distance ^gf. b i */(' where C is a constant of integration. Let v t cos or, be the component of the initial velocity v lt parallel to x, cti being the angle of elevation at the point of dx projection; then when * = 0, = , cos cti, and C = logn Vi cos a. There- ctt fore, from (3), dx -^ = e i cos a lt (4) where e is the base of the Naperian system of logarithms, 2.718282. Hence dy_ _ dy dx_ _ _< dy_ If we multiply (2) by dx, and (1) by dy, and subtract, we have -v-j = fdx (6) Inserting the value of df 2 from (4), we have d*ydx d*xdy _ f#* _,, d&- "^cos'a. The first member of (7) is equal to d-^-. We have also Substituting these values in (7), we have If we assume the angle of projection i as very small, so that the trajectory is very flat, we have approximately in such case / dif\k ds = dx, and s = x, and (1 + ^i) = ! Therefore (8) becomes , ........ (9) cos 2 w Integrating (9), since when x = 0, -^- tan i , we have dx Integrating (10), we have, since for * = 0, y = 0, y = x tan a, -f :r -/ -- . . / , (e**- iV . .(11) 1 2cVcos 2 tti 4c^i 2 cos 2 a!^ y CHAP. III.] PROJECTILES RESISTING MEDIUM. 129 Equation (11) is the approximate equation of the path. If we expand the last term in a series,. we have . 2, 'cos 2 a, 3fl, 2 cos' 2 a, If the terms containing c are omitted, and/ = g, equation (12) is that of a parabola, which is the path of the projectile in vacuo. The ordinate of the actual curve is therefore less at any distance x than that of the parabola for the same distance. From equation (4) we have d x -ex = e , cos , ; and integrating, since for x = 0, t = 0, we have cv, costtitf = e cx - 1, ........ (13) which gives the time in terms of the abscissa. From (13) we have or, in common logarithms, 2.302585 , log (cvi cos a it -{-!), C which gives the abscissa * in terms of the time. (14) "The general problem of the path of a projectile in a uniform resisting medium, where the resistance varies as the square of the velocity, was pro- posed by Keill as a trial of skill to John Bernoulli, by whom the challenge was received Feb. 1718. Keill, trusting to the complexity of the analysis, which Lad probably deterred Newton from attempting any regular solution of the problem in the second book of the Principia, was in hopes that the exertions of Bernoulli would prove unsuccessful. Bernoulli, however, having expeditiously effected a solution, not only of Keill's problem, but likewise of the more general one where the resistance varies as the nth power of the velocity, expressed a determination not to publish his investigation until he had received intimation that his antagonist had himself been able to solve his own problem. He gave Keill till the following September to exercise his talents, declaring that if he received by that time no satisfactory communication, he should feel himself entitled to question the ability of his adversary. At the request of a common friend, Bernoulli consented to extend the interval to the first of November. It turned out, however, that Keill was unable to obtain a solution. At length Nicholas Bernoulli, Professor of Mathematics at Padua, communicated to John Bernoulli a solution of Keill's problem, which the author afterwards extended to the more general one. Finally, on the 17th of November, information was received by John Bernoulli from Brook Taylor, to the effect that he had ob- tained a solution. John Bernoulli published his own analysis, together with that of his nephew Nicholas, in the Ada, Erudit. Lips. 1719 mai., p. 216." (Walton, Problems in Theoretical Mechanics.) CHAPTER IV. CURVILINEAR MOTION OF TRANSLATION CENTRAL ACCELERATION. HARMONIC AND PLANETARY MOTION. Central Acceleration. If the acceleration of a moving point is always directed towards a fixed point or centre of acceleration, the acceleration is said to be central. The velocity of the moving point at any instant is the resultant of the velocity at the preceding instant and of the integral accelera- tion during the intervening time. But if the acceleration is always directed towards the centre, or fixed point, its moment with reference to that point is zero. Since the moment of the resultant of any two components about any point is equal to the sum of the moments of the components, and since in this case the moment of one of the components, viz., the acceleration, is zero, it follows that the moment of the velocity about the centre, in the case of central acceleration, is constant. Conversely, if the moment of the velocity of a moving point about any fixed point is constant,, the acceleration must always be directed towards that point. If r is the distance of the moving point from the fixed point, p the lever-arm, and GO is the angular speed at any instant, we have for the moment of the velocity pv r*ao = c, equal to twice the areal velocity of the radius vector (page 76). Therefore, in all cases of central acceleration -Vca is constant, or / the area described by the radius vector in a unit of time is constant. (* It follows also that in all cases of central acceleration & = , or r 2 the angular speed is inversely as the square of the radius vector. Cases of Central Acceleration. The two most important cases of central acceleration are those of harmonic motion, where the central acceleration is directly proportional to the distance from the centre, and planetary motion, where it is inversely as the square of the distance. When the velocity is in the same straight line as the central acceleration we have in both these cases rectilinear motion. The first is simple rectilinear harmonic motion, the second is rectilinear planetary motion or that of a body at great distances from the earth. Both these cases have been considered in Chaps. I and II. When the velocity is not in the same straight line with the central acceleration we have compound harmonic motion and planetary motion in general. The first is of great importance in the study of sound, light, heat, etc., as well as in ordinary kinetics. The second is the motion of planets about the sun and of satellites about their primaries. 130 CHAP. IV.] TRANSLATION COMPOUND HARMONIC MOTION. 131 Cases of Harmonic Motion. We have defined simple harmonic motion, page 103, as the motion of a point moving in any path in. such a manner that the tangential component of the acceleration, a is directly proportional to the distance, measured along the path from a fixed point in the path. Such motion may be rectilinear or curvilinear. In the first case it is simple rectilinear, in the second simple curvilinear. If the whole acceleration itself, or/, is central, that is, always directed towards a fixed point not in the path, and is always propor- tional to the distance from this fixed point, the motion is central harmonic, or compound harmonic, so called, because it is the result- ant of two simple rectilinear motions, as will be proved in the next article. Simple rectilinear harmonic motion is also central, because the fixed point is in the path. Any Central Harmonic Motion may be Resolved into Two Simple Rectilinear Harmonic Motions at Right Angles. Let C be the centre of acceleration, and P the position of the moving point at any instant. Let the velocity v of P make an angle a with the axis of X, and let the motion of P be harmonic so that the ac- celeration of P is - r, where a' is the ac- T celeratipn at a known distance r', and r is the distance CP. The velocity v may be resolved into v cos a and v sin a. in the directions CX and CY, and the acceleration may be resolved into r cos PCA or r a' -^- , a' a' rCA, and -j-r cos PCS or rCB, in the same directions. The component accelerations are therefore directly as the dis- tances CA and CB, and the component velocities are in the direc- tions of CA and CB. The central harmonic motion of P, whatever the direction of the velocity v, is therefore the resultant of two simple harmonic motions in the lines CA and CB at right angles. If then any central harmonic motion is resolved into two com- ponents at right angles, the component motions are rectilinear har- monic. Conversely, the resultant of two rectilinear harmonic motions at right angles is a central harmonic motion. Central harmonic mo- tion is therefore called compound harmonic motion. Composition of the Simple Rectilinear Harmonic Motions in Dif- ferent Lines. Let the point M move in a circle AM A' of radius r CA CM with a constant angular velocity a>. Then the motion of the projection P in the line AA is J simple rectilinear harmonic (page 103). Let the point M, move in the circle CBMi of radius r\ = CB = A CMi , with constant angular veloci- ty &7i. Then the motion of the pro- jection Pi in the line CB is simple rectilinear harmonic. Let the angle BCA between the planes of the circles be a. IB FIG. 1. 132 KINEMATICS OF A POINT TRANSLATION. [CHAP. IT. Let the time count from the instant when Mi is at B, so that the epoch of Pi is zero (page 105). At this instant let the epoch of P be . Then e is the difference of epoch, or, in angular measure, the angle of M above or below A at the beginning of the time. In any time t, Mi will have moved from B through the angle oj,t measured from CB, and M through the angle wt e measured from CA. By the preceding Article we can resolve the harmonic motion of Pi into a simple rectilinear harmonic motion at right angles to CA, i, the difference of epoch is constant, and, since the epoch equals the pro- duct of the phase at zero of time by 2n radians (page 105), when the periods are equal the difference of phase is constant. When, then, the periods are equal and e = 0, or the epochs are equal, the phases are also equal at any instant. (For definitions of amplitude, period, epoch and phase, see page 105.) Two Component Simple Rectilinear Harmonic Motions in Differ- ent Lines with the Same Period. In this case GO <*>i and e is con- stant, or the difference of epochs is constant and difference of phase at any instant is constant. We have then, from (1) and (2), x = r cos (out + e) + ri cos a cos (oot), y = r sin a cos (cat). Combining these two equations by eliminating at, we have (ri 2 sin 2 a)x* - 2n sin a(r cos e + ri cos a)xy + (r 2 + 2m cos e cos a+ ri 2 cos 2 )i/ 2 = rVi 2 sin 2 a sin 2 e. (3) This is the equation of an ellipse referred to its centre and rect- angular axes. Hence if a point has two component simple rectilinear harmonic motions in any directions, of any amplitudes, and any difference of epoch, if the periods of the two components are the same, the result- ant motion of the point will be central harmonic in an ellipse, the centre of acceleration at the centra of the ellipse. The areal velocity of the radius vector about the centre is constant (page 130). Such motion is called elliptic harmonic motion. Elliptic har- CEAP. IV.J TRANSLATION COMPOUND HARMONIC MOTION. 133 monic motion, then, is compound harmonic motion when the periods of the components are the same. Equation (3) gives all cases of compound harmonic motion for equal periods of the components. It will be instructive to derive from it special cases. (a) Two Component Simple Rectilinear Motions in Different Lines with, the Same Period and Phase. In this case we make in (3) e = 0, and therefore the phases are equal, and we have at once r + cos a sin a y, This is the equation of a straight line passing through the centre C. The resultant motion is therefore central harmonic in a straight line, or simple rectilinear harmonic. If CA and CB are the amplitudes r and r t inclined at the angle , the result- ant motion has the amplitude CR, in direction and magnitude the diagonal of FIQ. 2. the parallelogram whose adjacent sides are r and n, inclined at the angle a. Conversely, a simple rectilinear harmonic motion whose ampli- tude is CR may be resolved, by completing the parallelogram, into two others in any two directions, of the same period, epoch and phase. /*! If a = 90, we have y = x. Therefore the projection of a simple rectilinear harmonic motion on any straight line is also a simple rectilinear harmonic motion of the same period, epoch and phase. If the component motions are more than two, they may be com- pounded two and two, and therefore any number of component simple rectilinear harmonic motions in any directions, of the same period, epoch and phase, give a single resultant rectilinear har- monic motion of determinate direction and amplitude, which may be resolved into two components in any two directions, of the same period, epoch and phase. (b) Two Component Simple Rectilinear Motions in the Same Line with the Same Period and Different Epochs and Phases. In this case R we make in (3) a = 0, and obtain at once (r 3 + 2rri cos e + r?)y* = 0. But since for a = 0, y = 0, (see Fig. 1,) we have r 3 + 2rri cos e + ri" = constant. In Fig. 3 the points P and Pi move in the line AA' with sim- ple harmonic motion and the diagonal CR = vVT 2rn cos e + r t \ where e is the constant differ- ence of epoch and phase. Since e is constant and CR is constant, its inclination to CM or CJ/i is constant. At any instant the resultant displacement is CPi + CP = CS, and the motion of S is therefore the resultant motion and is simple rectilinear harmonic, with the amplitude CR, the diagonal of the parallelogram on r and n. The epoch and 134 KINEMATICS OF A POINT TRANSLATION. [CHAP. IV. phase are intermediate between the epochs and phases of the com- ponents. If the epochs and phases are the same, e = and the amplitude of the resultant motion is r + n , or the sum of those of the compo- nents. If the difference of epoch or phase is e = TC radians, the am- plitude is 1 ri or the difference of those of the components. By taking CMi and CM of proper lengths we can make MCP and MiCM what we please without changing CR. Therefore any simple rectilinear harmonic motion may be resolved into tivo others in the same line, with any required difference of phase and one of them having any desired epoch, the periods being the same. Three or more component simple rectilinear harmonic motions in the same line and with the same period may be compounded two and two, and the resultant will be rectilinear harmonic with the same period. If the periods are different, the angle MiCM= e will vary and CR will vary. When e = 0, CR will have its maximum value r + rs. When the difference of epoch e is it radians, CR has its minimum value r ri. The angular velocity of CR is also variable. The di- rection of CR will oscillate back and forth about CM, the maximum 1 7*1 inclination being sin " . The resultant motion is therefore not r simple rectilinear harmonic, but a more complex motion. It is, as it were, simple harmonic with periodically increasing and decreas- ing amplitude, and periodical acceleration and retardation of phase, or epoch. (c) Two Component Simple Rectilinear Harmonic Motions at Right Angles with the Same Period and Different Phases or Epochs. The general equation for this case is given by (3). If the directions are at right angles, we have a = 90. Suppose in addition the am- plitudes equal, so that r = n, and the difference of epoch e = 90. We have then, from (3), or 1 + g/ 5 = r*. Since the motion is central harmonic, according to page 130 the areal velocity of the radius vector is constant ; and since the radius is constant, the speed in the circle is constant. We have already seen, page 103, that the projection of the motion of a point moving with uniform speed in a circle, upon a diameter, gives rectilinear harmonic motion. The projection upon two diameters at right angles gives then two component rectilinear harmonic motions of the same period, with a difference of epoch of 90, or of phase of J, since, when one component has its greatest displacement, the other is at the centre with displacement zero. It follows also that two component simple rectilinear harmonic motions at right angles, with the same period and equal amplitudes, differing in epoch by 90 or in phase by one quarter of a period, will give, as a resultant, uniform motion in a circle whose radius is the common amplitude of the components. If the amplitudes are not equal, but and e still 90, and periods the same, we have, from (3), nV + rY* = rVi", which is the equation of an ellipse referred to its centre and axes. The resultant motion is therefore central harmonic in an ellipse, whose semi-diameters are r and r h the centre at the centre of the ellipse. CHAP. IV.] TRANSLATION COMPOUND HARMONIC MOTION. 135 The same result is evidently obtained by projecting the circle in the preceding case upon a plane, so as to obtain the required ampli- tude n, r remaining unchanged. (d) Three or More Component Simple Rectilinear Harmonic Mo- tions in Different Lines with the Same Period but Different Phases or Epochs. We have seen from equation (3) that the resultant of two simple rectilinear component harmonic motions in any two direc- tions, of the same period and different epoch or phase, is elliptic harmonic motion. We have also seen from (a) that any simple rectilinear harmonic motion may be resolved into two others of the same period and phase or epoch in any two given directions. Any number of given simple rectilinear harmonic motions, then, of the same period and different phases or epochs may each be resolved into its own pair in any two given directions. These pairs constitute a number of simple rectilinear harmonic motions in two given lines, all of the same period and different phases or epochs. According to (6), all in one line may be compounded into one re- sultant, and all in the other line into another resultant, these two resultants having the same period and different phases or epochs. The resultant of these two is, according to equation (3), elliptic har- monic motion. Hence the resultant of any number of component simple rectili- near harmonic motions of the same period, whatever their ampli- tudes, directions, phases or epochs, is elliptic harmonic motion, the centre of the ellipse being then centre of acceleration, and the radius vector describing equal areas in equal times. In special cases this becomes, as we have seen, uniform circular motion or simple rectilinear harmonic motion. Since the above holds whatever the inclination of the two result- ants, elliptic harmonic motion may be considered as the resultant of two component simple harmonic motions of the same period and different epochs or phases at right angles. Graphic Representation. We may exhibit graphically simple or compound rectilinear harmonic motion by a curve in which the abscissas represent intervals of time, and the ordinates the corresponding distance of the moving point from its mean position. In the case of a single harmonic motion we have (page 164) jc = rcos(aote). If the distance a; is to be zero when t = 0, we must have the epoch e = radians, or one fourth of the periodic 9 time. This gives x = r sin cot. Since < = ^, where T is the periodic time, we have for t = 0, x = 0; for t = JT, x = r; for t = iT, x = 0; for t = T, x = - r; for t = T, x = 0. The curve is the curve of sines, or the curve which would be de- scribed by the point P (page 103) if, while M maintained its uniform circular motion, the circle itself were to move with uniform speed in a direction perpendicular to CA. 136 KINEMATICS Of A POINT TRANSLATION. [CHAP. IV.- It is the simplest possible form assumed by a vibrating string,, when it is assumed that at each instant the motion of every particle of the string is simple harmonic. If the rectilinear harmonic motion is compound, we have (page 132) in general X = r COS (oot e) + Ti COS (ooit ei). If the displacement of one of the motions is zero when t = 0, we 7t 2 have e = ^-; if ei = e + nit, we have x = r sin oat + r t sin (aoit + nit). [ of one motion is twice that of th< !<, and x = r sin (cot) + r\ sin (2cot + nit). of phase is zero, n = ; i ave X = r sin cot + r sin (2oot). If the period of one motion is twice that of the other, for instance, we have &>i = 2<, and If the difference of phase is zero, n = ; and if the amplitudes are equal also, we have This gives a curve as shown in the figure. ' Periods Unequal. We have in general X = r COS (oat + e), y = TI COS (<*>it + ei) for the two component rectilinear harmonic motions at right angles. The elimination of t in any case gives the curve of resultant com- pound harmonic motion. If the periods of the components are as 1 to 2, and e is the difference of the epochs, we have for equal amplitudes x = r cos (2cat + e), y = r cos out. Eliminating t, . r 2 which is the general equation of the curve for any value of e. Thus for e = 0, or equal epochs, which is the equation of a parabola. For e = ~ , 2 or r r which is also the equation of a parabola. CHAP. IV.] TRANSLATION COMPOUND HARMONIC MOTION. 13T If we make ein succession, 0, 1, 2, etc., eighths of a circumference, we obtain a series of curves as shown. Periodic 2 In the same way we can find the curve for any ratio of periods and difference of epoch. Thus if the periods are as 1 to 3 or 2 to 3, and we make e in succession 0, 1, 2, etc., eighths of a circumference, we obtain the following series of curves : Period 1 : 3 Period 2:3 ^ Blackburn's Pendulum. The motion of a pendulum which swings through a small arc is, as we shall see hereafter (page 154), simple harmonic, and the projection of the bob on c D a horizontal plane moves with simple rectilinear harmonic motion. Curves similar to those just given are therefore traced by Blackburn's pendulum. This consists of two pendulums, CED and EB, arranged so as to swing in two planes at right angles. Any difference of period may be made by ad- justing the lengths of the pendulums, and they may be started with any difference of epoch. If the bob B is made to trace its path on a horizontal plane, we have, approximately, the compound harmonic curve. [Application of the Calculus to Harmonic Motion Let a' be the known acceleration at the known distance ?'. Then the acceleration at any other dis- tance is r, where r is the distance from the centre. r For the acceleration in the direction of the axis of X we have then dt* - and in the direction of the axis of Y, (i) the minus sign denoting direction towards the centre. 138 KINEMATICS OF A. POINT TRANSLATION. [CHAP. IV. Multiply (1) by dx and we have Integrating, d?x , a' . -rrrdx = 7 xdx. dt 2 r 1 dx\* a' , . ) = s-,* 2 4- Const. 2\dt j 2r' Let - = 0. when x = r, then Const. = -r* and dt 2r ' = 4- - \dtj r* where we take the minus sign to indicate that x diminishes as t increases, or motion towards the centre. d'u In the same way we obtain from (2), if -jf~ = 0, when y = TI, 0V where we take the plus sign to indicate that y increases as t increases. Integrating (3) and (4), we have X ( fa' ) ^- -4- C = cos- 1 -, or x = r coslfi/ -, + C t. (5) At' f* 1 ' V \ y = r, sin j ( fi ) where C and C\ are constants of integration. Equations (5) and (6) may be written = r cos C cos 1 1/ % r sin C sin t \/ - = Ai cos 1 4/ "^ -J- A, sin t J /y ' T W /' * y = TI cos (7i sin < { 7 + "i sin (7, cos i = ^ sin where Ai = r cos C, A? = r sin G, Bi = TI cos O t , B* = r, sin d. If we find from these equations the values of sin and cos in terms of x and y and add their squares, we have, by eliminating t, ^ + A^B,) = (A.B, - A*Brf . (7) This is the equation of an ellipse referred to its centre and rectangular axes. If we take one of the principal axes corresponding with the axis of X, and dx count the time from the end of this axis, we have f or t = 0, y = and = dt and x = r. Therefore, from (5) and (6), (7=0 and Ci = 0, and therefore A?. = 0, Ai r, Bi = 0, Bi = TI, and (7) becomes r y -|- r^x* = rVi 2 , or the equation of an ellipse referred to its principal axes. We have also, from (5) and (6), /a' dx ./'. /. Let C be the centre of curvature of the hodograph, so that CQ is perpendic- ular to the tangent at Q and CQ = p = radius of curvature. Then since the linear acceleration / of P is the linear velocity of Q, we have / = po, or p = . 00 c a '/*' 2 a'r* 3 But oo = - and / = 5- . Hence p = . The radius of r ir c curvature p is therefore constant and the hodograph for planetary motion is a circle. The path which, in consequence of aberration, & fixed star seems to de- scribe is the hodograph of the earth's orbit, and is therefore a circle whose plane is parallel to the plane of the ecliptic. The Path for Planetary Motion is a Conic Section. Draw OR perpendicular to CQ and therefore parallel to r. OR is the com- ponent of the velocity v in the direction of the radius vector. Draw QN perpendicular to OCB. Then QN is the component of v per- pendicular to the fixed line CB. But by similar triangles O'R QN O'R O'C ~ ~~ , or - 140 KINEMATICS OF A POINT TRANSLATION. [CHAP. IV, that is, the ratio of the velocity along the radius vector to the- velocity at right angles to any fixed line parallel to O'CB is con- stant and equal to e. If then ri and r 2 are the initial and final values of r for an in- definitely short time, and di , d?. the corresponding distances of P from any given fixed line A'B' parallel to O'CB, we have =e ' or -*-<*- Since this holds whenever we take the fixed line A'B', let us 7*i V\ take the initial distance di such that di = , or e = - T - . e di Then, from (1), d* = , or e = -^-, and we have e a* ri r* -j- = T/~ = e; di at that is, the ratio of the distance r of the moving point P from a fixed point O to its distance d from a fixed line has a constant value. This is the Property of a Point Moving in a Conic Section. If e = 1, then O'R = QN, and the pole O' is on the circumference of the hodograph, and the path of P is a parabola. If e is less than unity, the pole O' is inside the hodograph and the path of P is an ellipse. If e is greater than unity, the pole O' is outside the hodograph and the path of P is an hyperbola. When, therefore, a point has a central acceleration inversely proportional to the square of the distance from the centre, it must move in a conic section with the centre of acceleration as a focus. Conversely, if the path be a conic section and the acceleration is directed towards either focus, it must vary inversely as the square of the distance from the focus. In both cases the radius vector describes equal areas in equal times (page 130). Kepler's Laws. By laborious comparison of the observations which TYCHO BRAHE had made through many years of the planets, especially of Mars, KEPLER discovered the three laws of planetary motion which are known as KEPLER'S LAWS. He gave these laws simply as the expression of facts which seemed warranted by the observations. The three laws are as follows : I. The planets describe ellipses, the sun occupying one of the foci. II. The radius vector of each planet describes equal areas in equal times. III. The "Harmonic Law," so-called. The squares of the periods of the planets are proportional to the cubes of their mean distances, from the sun. The second law, as we have seen (page 130), is a necessary con- sequence of central acceleration. From the first law, as we have just seen, it follows that the acceleration must be inversely as the square of the distance. The third law is a direct consequence of such central accelera- tion, as we shall see in the next Article. Verification by Application to the Moon. Assuming KEPLER'S third law, NEWTON was led directly to the conclusion that the ao- CHAP. IV.] TRANSLATION PLANETARY MOTION. 141 celeration must be inversely as the square of the distance, as follows : The moon and other satellites move around their primaries in sensibly circular orbits, the centre being at the centre of the primary. if T and Ti are the periodic times of two satellites, then accord- ing to KEPLER'S third law, if r and n are the radii of the orbits, we must have If oo is the angular velocity of one satellite, we have (page 76) roo = , or GO = . We have also for the acceleration (page 77) / = - = roo 1 = -=5-. For the other satellite we have in like manner fi = -^. We have then or the acceleration is inversely as the square of the distance. Con- versely, if the acceleration is inversely as the square of the distance, KEPLER'S third law is a necessary consequence. The numerical verification of this conclusion by the moon is given in Example 17 (page 55). [Application of the Calculus to Planetary Motion.] The general formulas for central acceleration have been already given, Chap. VIII, page 85. For any given law of central acceleration we have only to insert the cor- responding value ofjfdr in these general equations. (a) To Determine the Path when the Central Acceleration Varies Inversely as the Square of the Distance from the Pole. When the acceleration is inversely as the square of the distance we have where a' is the acceleration at a known distance r', and therefore a'r' 2 is con- stant. We have in this case / v= Substituting this in equation (42), page 87, we have for the differential equa- tion of the path dr* . 1 r- or, taking dB as always positive, dr , / Cl 8aV 1 V # + c*r r* If we put z = a'r' 2 , and J = a'V 4 + c,c 2 , this becomes 142 KINEMATICS OF A POINT TRANSLATION. [CHAP. IV. Integrating, we have 9 = cos' 1 \- Const. n , ., , 1 a r 2 + n a'r'* . . /a'V 4 Ci When z = n and therefore = r = ^- -4- A/ 1- , let 9 = r G' c* r c 4 c' 2 tp. Then Const. =

see then that the form of the orbit depends solely upon the magnitude of the initial velocity and not upon its direction. Also if a'r"* is negative, that is, when the acceleration is directed away from the pole, we have always an hyperbola. From page 100 we have seen that the speed attained by a body starting from rest at an infinite distance from the centre and moving in a straight line with an acceleration inversely as the square of the distance is A/ - . Heuco the orbit will be an ellipse, parabola or hyperbola according as the velocity of projection is less than, equal to or greater than that acquired from an infinite distance. If 1 is the angle of i with r\ , we have from equation (30), page 85, c = riVi sin 1 (3) The constants are therefore given by (2) and (3) and the orbit is determined when the initial velocity ti\ at the distance r\ and with the direction ei are given. (b) Central Acceleration Inversely as the Square of the Distance from the Focus Path an Ellipse. When the path is an ellipse we have the case of plan- etary motion. Let the point Pmove in the ellipse ABA' with central acceleration always directed towards the sun S in the focus, and varying inversely as the square of the dis- tance or radius vector SP = r. Let the semi-major axis OA = A and the semi-con j ugate axis OB B and the eccen- * - -. os tricity -j- = e. A Let the angle ASX of the major axis with any fixed line SX through the focus be 0. CHAP. IV.] TRANSLATION PLANETARY MOTION. 143 The extremity A of the major axis nearest to the focus S is called in general the lower apsis, or, in the case of a planet, the perihelion. The angle ASX is the longitude of the perihelion. The distance SA is the lower apsidal distance, or the perihelion distance. The other extremity A' is the higher apsis, or, in the case of a planet, the aphelion, and 8 A' is the higher apsidal distance, or aphelion, distance. The angle P8X, or the angle of the radius vector with the fixed line SX, we denote by 6. The angle PSA made by the radius vector with the major axis is then 9 0. This angle is called the true anomaly. The polar equation of an ellipse with reference to the focus 8 as a pole, counting the angle PSA around from the perihelion, is - ecos(S- But equation (1), page 142, just deduced, is the equation of a conic section, which becomes an ellipse, therefore, when Inserting the values of* (2) and (3), we have " where i is the initial velocity at the distance ri making the angle ei with TI. The elliptic orbit is thus determined. From (6) we see that the semi-major axis A depends only on the distance r\ and the velocity of projection Vi and is independent of the direction of projection e,. In whatever direction the body is projected, the major axis will be the same for the same distance and velocity of projection. We have also _ -_, and hence 4 s /1 s rm m n .. -- . "! M + m M + mi ' We see then that Kepler's third law is not, strictly speaking, ex- a' ' T'^ act. The value of , or the acceleration at the same distance, is r 2 not, strictly speaking, constant for all the planets. The more accu- rate expression of the third law is that the squares of the periodic times are directly as the cubes of the semi-major axes and inverse- ly as the sum of the masses of the sun and planet. The error from this source is, however, too slight to be percept- ible, the mass of Jupiter, the largest of the planets, being less than a thousandth part of that of the sun. * Outlines of Astronomy. CHAP. IV.] TRANSLATION PLANETARY MOTION". 145 The motion of translation of the planets is not affected by their rotation on their axes, and we may treat them, therefore, as mate- rial points so far as translation is concerned. The sun is not, strictly speaking, a fixed point in this sense, but both sun and planet move in orbits, so that the pole or focus is not at the sun's centre, and this affects the accuracy of Kepler's first two laws. The sun is also attracted by the other planets, and the plan- ets attract each other. The attraction of the planets for each other sensibly modifies their orbits. The ellipse is therefore only an approximation to the path, and requires correction. Kepler's laws are thus only approximate expressions. If there were but two bodies, one fixed and the other free to move, then Kepler's first two laws would be accurate, and the third law would approach accuracy as the mass of the moving body becomes in- significant with respect to the mass of the other. (d) To Find the Velocity of a Planet at any Point of its Orbit. From equa- tion (37), page 87, Chap. VIII, we have *> 2 = ci 2 jfdv. /a'r'* fdv = -- , where a' is given in the preceding Article, and r' is the mean radius of the earth. Also from equation. (2), page 142, where Vi is the velocity at the distance r\. Therefore or, since (page 143) a'r' 2 = c 2 From page 85, * 2 = 8 , and for an ellipse, from Analytical Geometry, Hence -. 2A - r Equation (2) gives the velocity for any distance r if the semi-major axis A is known. Equation (1) becomes the same as equation (2), page 99, for rectilinear mo- tion. COR. 1. We see that the velocity is greatest where r is least, or at peri- helion, and least at aphelion, where r is greatest. COR. 2. If a point moves in a circle of radius r with a speed Vi, its central acceleration is (page 53). If this acceleration is equal at any instant to r 4,he acceleration of the planet, we have from equation (7), page 143, z>J_ _ a>' 2 _ c*_ _ 1 ~r " ~ 1 r r ~ ~ A(\ e 2 ) r r 146 KINEMATICS OF A POINT TRANSLATION. [CHAP. IV., Therefore, from (2), tf . g ... tfOA-r) . c" 1 " A\\ - e*)r ' A(l - e*)r' or tf : , :: 2A - r : A. That is, the square of the speed in the ellipse is to the square of the speed in. the circle as the distance of the planet from the unoccupied focus is to the semi- major axis. COR. 3. If r\ is the perihelion distance and r a the aphelion distance, we have, from (2), a'r' 2 r a for r TI , ir = -- ; for r = r y , v l -. -- A r,' while f or r = A we have aY* * = . That is, the speed at the extremity of the minor axis is a mean proportional between the speeds at perihelion and aphelion. (e) To Find the Time of Describing any Portion of a Planet's Orbit. From /a'r'* fdr = -- , dr* . "" and ?- = -- Differentiating, we have ^ =(~-^)cos0sin0, and hence CHAP. IV.] TRANSLATION PLANETARY MOTION. 149 Differentiating again, ePr . Mr* drcM - (1 JL\ ( C os 2 - sin 2 0V \ B A I \ J But from equation (49), page 88, we have efO = . Therefore d?r _ dr* _ 2_ J_ J. T~ 4^7C2 ^2 I ^2 ~T~ jg2' Now from equation (45), page 88, we have /= cr ^4 = - C 2 r ri_ rd6* , 1 j _ ffir dr* \ "I + r s< j r 8 ** + r 4 ^ 2 j J The law of acceleration is therefore /ta D cycloid. If then AC and AD are fixed semi -cycloids, symmetrical with refer- ence to the vertical AB, and AB is a simple pendulum, B will describe a cy- cloid, and its oscillations will be isochronous whatever their extent. CHAP. V.] TRANSLATION CONSTEAINED MOTION. 157 and obtain at once, since [Application of the Calculus. To Determine the Motion of a Point Constrained to Move in a Cycloid, the Acceleration being Constant, in the Direction of the Axis and towards the Vertex.] By the application of the general formulas of page 88 we can deduce the results A^ D B already obtained. Let the axis CD = 2r, where r is the radius of the generating circle DP' C. Let the acceleration / act downward. Let CN = y, NP x and the length of arc CP s. Let the initial position be P, at the height CNi = h above C, and the speed at Pi be i = 0. We have thus the case of equation (55), page f is negative and Si = h, v = |/2/(A y) for the speed at any point given by CN = y. When y = 0, we have, at the lowest point C, = 4/2/%, which is the same as that due to the vertical height h. By the property of the cycloid we have 8 - arc CP = 2 tfDC X ON = 2 V2ry = 2 chord OP. Hence IT y We substitute the minus value in equation (56), page 89, because for de- scent the arc decreases as the time increases. We have then dy ds= dy at = - y j Integrating, since for t = 0, y = h, we have (1) For the time of descent to the lowest point where y = 0, or for the time of one quarter of a complete oscillation, '4r /' The periodic time is then T - 2rt 4/ or the same as a simple pendulum (page 154) whose length is 4 times the radius of the generating circle DP' C. The time is independent of h and is the same no matter what the posi- tion from which the point begins to descend. The oscillations are therefore isochronous and hence the cycloid is called the tautochrone. The reason of this remarkable property is easily seen by considering the tangential acceleration. _, , In the cycloid the chord CP' is always parallel to the tangent IP. tangential acceleration or tangential component of /is then =/-8in CDF = The tangential acceleration is therefore directly proportional to the distance from the vertex measured along the path, and the motion of P is simple har- monic (page 103). '8 KINEMATICS OF A POINT TRANSLATION. [CHAP. V- The periodic time is then (page 104) displacement tangential acceleration as already found. If in (1) we make y = = ^CN S , we have t \/ ,oi half the time & A A ' f from Pi to C. The time, therefore, in descending through half the vertical space to C is half the time of passing from P! to C. [To Find a Curve such that a Point Moving on it under the Action of Gravity will Pass from any one Given Position to any Other in Less Time than by any Other Curve through the Same Two Points. ] This is the celebrated problem of the "curve of swiftest descent v first propounded by Bernoulli. The following is founded upon his original solution. If the time of descent through the entire curve is a minimum, that through any portion of the curve is a minimum. It is also obvious that between any two contiguous equal values of a con- tinuously varying quantity, a maximum or minimum must lie. This principle though simple is of very great power, and often enables us to solve problems of maxima and minima, such as require not merely the pro- cesses of the Differential Calculus but those of the Calculus of Variations. The present case is a good example. r p Let, then, PQ, QR and PQ , QR be two pairs of indefinitely small sides of a polygon such that the time of descending through either pair, starting from P, may be equal. Let QQ be horizontal and in- definitely small compared with PQ and QR. The curve of swiftest descent must lie between these paths, / , and must possess any property which they have in common Hence if we draw Qm, Qn perpendicular to RQ, PQ, and let v be the speed down PQ or PQ (supposed uniform) and ' that down QR or Q R, we have for the time from P to R by either path PQ QR PQ QR PQ-PQ _ QR - QR -^ -4 ; _ _ ^ i ; or ^ ^ . 9 ' 9 9 V V V or Qn _ Q'm . v if Now if be the inclination of PQ to the horizon, and & that of QR, we have Qn = QQ cos 0, Q'm = QQ' cos 6'. Hence cos _ cos 0' v v' ' This is true for any two consecutive elements of the required curve, and therefore throughout the curve we have, at any point, v proportional to the cosine of the angle which the tangent to the curve at that point makes with the horizontal. But t? is proportional to the vertical distance h fallen through. Hence the curve required is such that the cosine of the angle it makes with the horizontal line through the point of departure varies as the square root of the distance from that line Now in the figure of page 157 we have, from the property of a cycloid, DP' /J)N cos CP'N = cos TPN = cos GDP' = -^ = y -^ . The curve required is therefore the cycloid. The cycloid has received on account of this property the name of Brachistochrone. CHAP. V.] TRANSLATION CONSTRAINED MOTION". 159> [To Determine the Motion of a Point Constrained to Move in a Circle, the Ac- celeration being Constant and Vertical.] This is the case of equation (55), page 89. A c B Let DN = y, NP = x. Let the speed at P t be \i + -^at 1 . t = 26.9 sec. 2 (5) A point has an initial speed of 7 ft. per sec. and a final speed of 125 ft. per sec., and describes a distance of 3250ft. What is the uniform rate of change of speed f , a = 2.4 ft.-per-sec. per sec. 2a (6) A point has an initial speed of 100 ft. per sec. and a rate of change of speed o/l ft.-per-sec. per sec. Its final speed is 7 ft. per sec. Find the time of motion ana the space described. Ans. v = i at, t = 93 sec. ; *i * , = 4975.5 ft. (7) A locomotive has a speed of 30 miles an hour on a level when the brakes are applied. The loss of speed due to the brakes is 8 ft.- per-sec. per sec. Find (a) the speed at the end of 3 seconds and the distance traversed ; (b) the time of coming to rest; (c) the retarda- tion in order that the locomotive may come to rest in 30 seconds. Ans. v = Vi at. Vi = 44 ft. per sec.; a = 8 ft.-per-sec. per sec.; s Si = Vit -^at*. a (a) 20 ft. per sec., 118 ft.; (b) t = 5.5 sec.; (c) 1.47 ft.-per-sec. per sec. 162 KINEMATICS OF A POINT TRANSLATION. (8) A body falls for 4 seconds in vacuo. Find the final velocity and the space described (g = 32 ft.-per-sec. per sec.). Ans. v = gt, ^-gf; m v = 128 ft. per sec., 8 = 257 ft. (9) A body falling in vacua has a final velocity of 25Qft. per sec. Find the time of falling from rest, and the distance described (g = 32| ft.-per-sec. per sec.). Ans. 7.77 sec.; 971.25ft. (10) A body falling in vacuo from rest describes a distance of 85 feet. Find the time of fall and the final velocity (g = 32^ ft.-per- sec. per sec.). Ans. 2.3 sec. ; 73.9 ft. per sec. (11) A body is projected vertically upwards in vacuo ivith a ve- locity of 40 ft. per sec. Find the height and the time of ascent (g 32 ft.-per-sec. per sec.). Ans. 24.87 ft.; 1.24 sec. (12) A body projected vertically upwards in vacuo returned after 18^ seconds. Find the initial velocity and the height of ascent (g = 32 ft.-per-sec. per sec.). Ans. 297.54 ft. per sec.; 1376 ft. (13) A body falling in vacuo has at a given instant a velocity of 17 ft. per sec., at a later instant a velocity of 90 ft. per sec. Find the time between the two instants and the distance traversed (g = 32^ ft.-per-sec. per sec.). Ans. 2.27 sec.; 121.44ft. (14) A stone is dropped into a well and the splash is heard in 3 seconds. If sound travels in air with a uniform velocity of 1090 ft. per sec., find the depth of the well (g = 32% ft.-per-sec. per sec.). Ans. 130.4 feet. (15) A point has two component velocities (or accelerations), at right angles, of 35 and 87 units. Find the resultant velocity (or ac- C3lerations). Ans. 93.77 units, making an angle with the 35 units of 68 5'. (16) A point has a velocity of 120 ft. per sec. Resolve this into two component velocities at right angles, (a) one of the components being 75 ft. per sec.; (b) one of the components making an angle of 34 7' 3" 'with the resultant. Ans. (a) 93.68 ft. per sec., the resultant making an angle with 75 ft. per sec. of 51 19' 4". (b) 99.343 ft. per sec. adjacent to the given angle and 67.306 ft. per sec. opposite. (17) A point has two accelerations of 115 and 89 ft.-per-sec. per sec. at an angle of 147 8' 3". (a) Find the resultant acceleration ; (b) Find the angle between the given accelerations ichen the resultant is equal to the lesser ; (c) when it is equal to the greater. Ans. (a) 62.865 ft.-per-sec. per sec., making the angle of 83 4' with 89 ft.- per-sec. per sec.; (b) 130 14' 44"; (c) 112 45' 54'. (18) A point has an acceleration of 77.5 ft.-per-sec. per sec. Re- solve it into two components (a) making with the given acceleration the angles 35 7' 11' and 52 9' 8"; (b) when one of the components is 50.5 ft .-per-sec. per sec. and makes an angle of 36 8' 6" with the re- MISCELLANEOUS PEOBLEMS. 163 sultant ; (c) when one of the components is 60 ft.-per-sec. per sec. and the other makes an angle of 47 ' J 10' 11" with the resultant ; (d) when the two components are 46.2 and 35 ft. -per-sec. per sec. Ans. (a) 61.265 and 44.684 ft.-per-sec. per sec. (b) 47.27 ft.-per-sec. per sec., making an angle with the resultant of 39 2' 3". (c) 71.88 or 33.48 ft.-per-sec. per sec., making an angle with the resultant of 61 28' 17" or 24 9' 4". (d) The components make an angle of 35 4', and the resultant makes with the component 46,2 the angle 15 2' 18". (19) A stream flows with a velocity of 1 ft. per sec., and a boat whose speed is 1.3 ft. per sec. is steered up stream at an angle of 30 with the current. Find the resultant velocity. Ana. 0.66 ft. per sec. down stream at an angle of 79 3' with the current. (20) What is the ratio of the speed of light to that of a cannon- ball moving at the rate of 2400 feet per sec., if light passes from the sun to the earth, a distance of 91,500,000 miles, in 8 minutes f Ans. 402600 to 1. (21) ABC is a triangle. Two spheres of radii ri and r 2 start to- gether from A and B, their centres moving along AC and BC with velocities which carry them separately to C in the same time. Find the distances each has gone through when they meet. Ans. If a, b and c are the sides of the triangle, the required distances are -(c n r a ), -(c - fi r a ). C C (22) If a particle is projected vertically in vacuo unth a velocity of 8g, find the time in which it will rise through the height 14gr. Ans. 2 sec. and 14 sec. (23) A body falling in vacuo is observed to describe 144. 9 ft. and 177.1 ft. in two successive seconds. Find g and the time from the beginning of the motion. Ans. g = 32.2 ft.-per-sec. per sec. ; t = 4 sec. to the beginning of the first of the two seconds. (24) A, B, C, D are points in a vertical line, the distances AB, BC, CD being equal. If a particle fall from A, prove that the times of describing AB, BC, CD are as 1 : 1/2"- 1 : 4/3~ V%. (25) A particle describes in successive intervals of 4 seconds each spaces of 24 and 64 feet in the same straight line. Find the accelera- tion and the velocity at the beginning of the first interval. Ans. 2.5 ft.-per-sec. per sec.; one foot per sec. (26) A particle moves 7ft. in the first second, and 11 and 17 ft. in the third and sixth seconds respectively. Show that these facts are consistent with the supposition of a uniform acceleration. (27) A falling particle is observed at one portion of its path to pass through nft. in s seconds. Find the distance described in the next s seconds. Ans. n -j- g&. (28) If s, ms, are the spaces described by a body in times t, nt y respectively, determine the acceleration and the velocity of projection. 2(m n)s (m ri*)s Ans. ; -rr. and - . n(n l)t* n(ln)t 164 KINEMATICS OF A POINT TEANSLATION. (29) If the focus of the path of a projectile be as much below the horizontal plane through the point of projection as the highest point of the path is above it, to find the angle of projection. Ans. If a-j is the angle of projection, a^ = tan- 1 ) ]. \4/2 / (30) Particles are projected from the same point in the same di- rection with different speeds. Find the locus of the foci of their paths. Ans. A straight line through the point of projection making an angle with the horizontal equal to 2ai , where a t is the angle of projection. 9 (31) If a particle is projected in a direction inclined to the hori- zon, show that the time of moving betiveen two points at the extrem- ities of a focal chord of the parabolic path is proportional to the product of the velocities of the particle at the two points. (32) Two particles are projected from two given points in the same vertical line in parallel directions and with equal speeds. Prove that tangents drawn to the path of the lower will cut off from the path of the upper arcs described in equal times. (33) If a particle is projected from a given point so as to strike an inclined plane through that point at right angles, prove that tan (a, e) = _ cot B, where ai is the angle which the direction of 9 projection makes with the horizon, and Q is the inclination of the plane to the horizon. (34) A. particle is projected from a given point with a given velocity. Find the direction of projection in order that its path may touch a given plane. Ans. Let aj be the angle of projection with the horizontal and fi the angle of the given plane with the horizontal, Vi the velocity of projection and d the distance from the point of projection to where the given Diane cuts the hori- zontal through the point of projection. Then 4/W sin 2/3 cos (90 ft ,) = -i* . (35) To find the least velocity with which a body can be projected from a given point so as to hit a given mark, and the direction of projection in this case. Ans. Let d be the horizontal distance from the point of projection to the mark, Vi the velocity of projection, j its inclination to the horizon and ft the angle of elevation of the mark above the horizon. Then (36) If two circles the planes of which are vertical touch each other internally at their highest or lowest points, and if any chord be drawn within the larger circle, terminating respectively at its highest or lowest point, prove that the time of descent down that por- tion of the chord which is exterior to the smaller circle is invariable. (37) AP, PB are chords of a circle, AB being the vertical di- ameter. Particles starting simultaneously from A, P, fall down AP, PB, respectively. Prove that the least distance between them is equal to the distance PB. MISCELLANEOUS PROBLEMS. 165 (38) Two circles lie in the same plane, the lowest point of one being in contact with the highest point of the,other. Prove that the time of descent from any point of the former to a point of the latter along a straight line joining these points and passing through the point of contact, is constant. (39) A particle slides from rest down a smooth sloping roof and then falls to the ground. Find the point where it reaches the ground. Ans. Let I = the length of the slope, a its inclination with the horizontal, h = the height of the lowest point of the slope from the ground. Then the distance of the point where the particle reaches the ground from the foot of the wall is 2 cos a \/l sin a( ^l sin 3 a -f- h \il sin 8 a). (40) Two equal inclined planes are placed back to back, and a ^article projected up one flies over the top and comes to the ground just at the foot of the other. Find the velocity of projection, a being the inclination of each plane and h their common altitude. 1 ^. Ans. -. (41) A particle is projected from a point A with the velocity ac- quired by falling down a height a, up an inclined plane of which the base and height are each equal to b, and after quitting the plane strikes the horizontal plane AB at the point B. Find AB. Ans. AB is equal to a -f- (a 2 6 2 ) . (43) A particle slides down a smooth inclined plane. Determine the point at which the plane is cut by the directrix of the path described by the particle after leaving the plane. Ans. The directrix intersects the plane at the point where the particle began its motion. (43) A particle is projected up a rough plane, inclined to the horizon at an angle of 60, with the velocity which it would have acquired in falling freely through a space of 12 ft., and just reaches the top of the plane. Find the altitude of the plane, its roughness being such that if it were inclined to the horizontal at an angle of 30, tlie particle ivould be on the point of sliding. Ans. 9 ft. (44) A ring slides doum a straight rod whilst the rod is carried uniformly in one plane, at a given angle to the horizon. Find the path described by the ring. Ans. A parabola. (45) A given circle and a given point are in the same vertical plane, the point being within the circle. Find the straight line of quickest descent from the point to the circle. Ans. The required straight line is the distance between the given point and the lower end of that chord of the circle which passes through the given point and terminates at the highest point of the circle. (46) A given point and a given straight line are in the same vertical plane. Determine the straight line of quickest descent from the given line to the point. Ans. From the given point P draw PX horizontally to meet the given line at X. Draw upwards along the line a length .X"P"equal to PX. The straight line joining P and Y is the required straight line. 166 KINEMATICS OF A POINT TKANSLATION. (47) To determine the straight line of slowest descent from a given point to a given circle, the point being without the circle and both in the same vertical plane, the highest point of the circle being lower than the given point. Ans. Let P be the given point and Q the highest point of the circle. Join PQ and produce it to cut the circle at H. Then PR is the line required. (48) Determine the straight line of slowest descent from a given circle to a given point without it, the point and circle being in the same vertical plane, and the point being lower than the lowest point of the circle. Ans. From the given point draw an indefinite straight line cutting the circle at its lowest point, and the second intersection of the indefinite line and the circle is the required line. (49) Find the straight line of slowest descent from one given circle to another, both circles being in the same vertical plane, and each exterior to the other, the highest point of the latter circle being lower than the lowest of the former. Ans. Produce the line which joins the lowest point of the first circle and the highest point of the second, to meet both circles again. The distance be- tween the second intersections is the required line. (50) A smooth tube of uniform bore is bent into the form of a circular arc greater than a semicircle, and placed in a vertical plane with its open ends upwards and in the same horizontal line. Find the velocity with which a ball that fits the tube must be pro- jected along the interior from the lowest point, in order that it may pass out at one end and re-enter at the other. Ans. If r is the radius, h the depth of the centre of the circle below the horizontal through the two ends, Vi the required velocity, , = -|(r 2 + 2hr + 27i 2 ). n (51) A particle slides from rest down a smooth tube in the form of the thread of a screw the axis of which is vertical. Find the time in which it will make a complete revolution about the axis. Ans. If r is the radius of the cylinder on which the helix is described, and a. the angle which the thread makes with the generating line of the cylinder, the required time is \gsin2aj ' (52) A particle falls to the lowest point of a cycloid down any arc of the curve, the axis of the cycloid being vertical and its vertex downwards. Prove that the vertical velocity is greatest when it has completed half its vertical descent. (53) Also prove that it describes half the path in tivo thirds of the whole time. (54) If a clock pendulum lose 5 sec. a day, determine the altera- tion tvhich should be made in its length. Ans. It should be diminished by nearly the ^TT. part of its length. (55) A seconds pendulum was too long on a given day by a quan- tity a. It was then over-corrected so as to be too short by a during MISCELLANEOUS PROBLEMS. 167 the next day. Prove that, I being the length of the seconds pendulum, the number of minutes gained in the two days was nearly (56) A seconds pendulum carried to the top of a mountain is found to lose there 43.2 sec. a day. Find the height of the mountain supposing the radius of the earth to be 4000 miles. Ans. 2 miles. (57) Find the time of vibration of a pendulum 2Qfeet long. Ans. Approximately 2.5 seconds. (58) A body dropped from the top of a wall falls to the ground while a pendulum 6 inches long makes 5 beats. Find the height of the wall. Ans. -i-TT 2 feet. 4 (59) A seconds pendulum is lengthened one hundredth of an inch. Find how many seconds it will lose daily. Ans. About 11 seconds. (60) If the length of a seconds pendulum is 39.1386 inches, what will be the length of one which vibrates 40 times a minute. Ans. 88.06185 inches. (61) If the length of a seconds pendulum is 39.1393 inches, find the value of g. Ans. g = 32.190 feet-per-sec. per sec. (62) A pendulum which beats seconds at the equator gains 5 minutes a day at the pole. Compare polar and equatorial gravity. Ans. 144 to 145, approximately. (63) Two pendulums the lengths of which are li and h vibrate at different points on the earth's surface. The number of vibrations which they make in the same time are in the ratio nil to m 3 . Find the ratio of g at the tivo places. li-mS Ans. ~. ly.niy (64) A seconds pendulum is carried to the top of a mountain of tvhich the height s 1 mile. Find the number of seconds it will lose daily, gravity being supposed to vary inversely as the square of the distance from the centre of the earth, and the radius of the earth to be 4000 miles. Ans. About 21.6 seconds. (65) A body revolves uniformly in a circle of 4 ft. radius in 10 seconds. Find the angular velocity and the velocity at the circum- ference. Ans. 0.628 radian per sec.; 2.5 ft. per sec. (66) A body revolves uniformly in a circle of 4 ft. radius with a velocity of 8 ft. per sec. Find the normal acceleration. Ans. 16 ft.-per-sec. per sec. (67) A body revolves uniformly in a circle of 18 ft. radius. The normal acceleration is 5 ft.-per-sec. per sec. Find the velocity. Ans. 9.5 ft. per sec. 168 KINEMATICS OF A POINT TRANSLATION. (68) A body whose velocity is 10 ft. per sec. is made to move uni- formly in a circle by a normal acceleration of 2 ft.-per-sec. per sec* Find the radius. Ans. 50 ft. (69) Two points have velocities Vi and Vt and are made to move tn a circle by reason of central accelerations inversely proportional to the square of the distance from the centre. The distance of one point is ri. Find the distance r 2 of the other. Ans. : = r a 2 : n 2 ; r a =( (70) Find the relation between the distances r\ and r* and the times of revolution ti and t-t. .1 n KINEMATICS OF A EIGLD SYSTEM. CHAPTEE I. RIGID SYSTEM WITH ONE POINT FIXED. BOTATION. ANGULAR DISPLACEMENT. LINEAR DISPLACEMENT IN TERMS OF ANGULAR. LINE REPRESENTATIVE OF ANGULAR DISPLACEMENT. RESO- LUTION AND COMPOSITION OF ANGULAR DISPLACEMENTS. ANGULAR VELOCITY. INSTANTANEOUS AXIS OF ROTATION. ANGULAR ACCELERA- TION. RESOLUTION AND COMPOSITION OF ANGULAR VELOCITIES AND ACCELERATIONS. EQUATIONS OF MOTION OF A POINT OF A ROTATING SYSTEM. MOMENT OF ANGULAR VELOCITY AND ACCELERATION. GENERAL ANALYTICAL DETERMINATION OF RESULTANT FOR CONCURRING ANGULAR VELOCITIES AND ACCELERATIONS. Rotation. When a rigid system moves so that all its points describe circles in parallel planes about a common straight line or axis passing through the centres of the circles and perpendicular to their planes, the system is said to rotate or have a motion of rotation about that axis. Any plane parallel to the planes of the circles is the plane of rotation. Since the system is rigid, every point must describe its circle in the same time, or the angular speed (page 72) of every point is the same. If the angular speed does not change and the plane of rotation does not change, the rotation of the system is uniform. If either the angular speed changes or the plane of rotation changes, the rotation is variable. Motion of a Rigid System with One Point Fixed. We have de- nned translation (page 13) as motion of a system such that every straight line joining every two points remains always parallel to itself during the motion. In such case the motion of the system is the same as that of any one of its points, and the study of the trans- * The advanced student should read this portion of the work in connection with the analogous portions of Statics referred to in the text. The student new to the subject would do well to omit this portion of the work and take it in. connection with Statics later. 169 170 RIGID SYSTEM WITH OXE I'OTNT FIXED KOTATION. [CHAP. I. lation of a system is the same as the study of the translation of a point. In the preceding Chapters we have treated of the kinematics of a point or translation. If one point of a system is fixed, the motion of which it is capa- ble will be more or less complex according as its points can or can not move relatively to each other. We restrict our discussion to rigid systems, that is, systems whose points can not move relatively to each other. . If one point of such a system is fixed, there can be no translation and the only motion of which it is capable is one of rotation as just defined. In such motion it is evident that all straight lines in the system must remain straight lines of unchanged length and mutual inclina- tion, and all planes must remain planes of unchanged form, area and mutual inclination. Also the motions of any two points indefi- nitely near must be indefinitely nearly the same. Angular Displacement of a Rigid System. Let AB be the axis of rotation of a rigid system. Then, since every point must com- plete its circle in the same time, the angle de- scribed by all points in any given time must be the same as the angle described by any one point P. The angle 6 between the initial and final positions, in any given time, of the perpendic- ular PO from any point Pto the axis, is called the angular displacement of the system. Since the angle 6 is measured in radians, it is independent of the length PO (page 5). Linear Displacement in Terms of Angular. Let OPi = OP 2 = r be the radius for any point P which moves in a circle perpendicular to the axis at O, through the angular displace- ment PiOP 2 = 0, from the initial position Pi to the final position P 2 . Then the triangle PiOPz is isosceles, and if we draw ON per- pendicular to PiP we have the linear dis- placement PiP 3 =2rsin4- a 9 Line Representative of Angular Displacement of a Rigid System. An angular displacement of a rigid system is given when we know not only its magnitude and the direction of rotation in the plane of rotation, but also the direction of that plane in space. It is therefore a vector quantity having not only magnitude and sign, but also direction, and it can be completely represented by a straight line, like linear displace- ment (page 34). Thus the length of the straight line AB denotes the magnitude of the angular displacement PiOP = 6. The plane of rotation is at right angles to the line AB, which is therefore coincident ivith the axis of rotation. The direction of rotation is always clockwise when we look along this line in the direction indicated by the arrow. By direction of an angular displacement we mean always the direction of its line representative as denoted by the arrow. CHAP. I.] ANGULAR DISPLACEMENTS. 171 Composition and Resolution of Successive Angular Displacements. Let a rigid system with one point fixed undergo successive angular displacements. It is required to determine the resultant angular displacement. Evidently the successive angular displacements may be about the same or about different axes, and in either case may be finite or indefinitely small. (a) About the Same Axis Finite or Indefinitely Small. If the axis of all the angular displacements is the same, the plane of rotation does not change, and the magnitude and sign or direction of the resultant displacement in that plane is given by the algebraic sum of the magnitudes of the successive displacements, whether they are finite or indefinitely small Inversely, any angular displacement about a given axis may be resolved into any number of successive displacements about the same axis, whether finite or indefinitely small, provided the alge- braic sum of their magnitudes is equal to the magnitude of the given displacement and the same in sign. (b) About Different Axes Displacements Finite. The axes must pass through the fixed point of the system. First let the displace- ments be finite. Let O be the fixed point of the system, and ORi , OR* the initial positions of the two given axes. Take ORi OR*, and let us sup- pose first a displacement O t of the system about ORi and then a dis- placement f J* about the new position of the other axis. During this motion R* and Ri will move on the surface of a sphere. When the system is rotated an angle 0i about ORi, the axis OR* moves from OR* to OR?'. Now join RiRi, RiR*' and R*R*' by great circles of the sphere. Then the angle R~FZiR' = G>. Bisect this angle by a great circle meeting R*'R* at D. Draw a great circle through R*' inclined to R*'Ri at the angle ^ and meeting RiD at R. fy Then draw R*'C*, making the same r\ angle ^ with R*'Ri on the other side, and make R*'C* = R*'R. Then R& will equal RiR, and the angle R-2 .ZviG-2 = ^". When then the system is rotated about ORi , and the axis OR, moves to OR,' through the angle RiRtR*' = 9,, the line OR will move to OC-2 through the angle RRiC* = 9i. If now the system is rotated about OR* through the angle CiR-'R = i and OB K> are the line representatives of the initial and final angular velocities of a rigid system with one point fixed, during any time t, then AB is the line represent- ative of the integral angular acceleration of the system during the time t, and r- gives the < v magnitude of the mean angular acceleration whose direction is AB. (Compare page 48.) Mean angular acceleration then is time-rate of change of angular velocity, whether that change takes place in the direction of the an- gular velocity or not. Instantaneous Angular Acceleration of a Rigid System. The limiting magnitude and direction of the mean angular acceleration when the interval of time is indefinitely small is the instantaneous angular acceleration. It is the limiting time-rate of change of an- gular velocity whether that change takes place in the direction of the angular velocity or not. Angular acceleration always signifies instantaneous angular acceleration unless otherwise specified. It may be represented by a straight line just like angular dig placement (page 170). By direction of an angular acceleration we mean the direction of its line representative. Instantaneous Axis of Angular Acceleration. The instantaneous angular acceleration of a rigid system is then given by its line rep- resentative. This line representative coincides in position with the axis of angular acceleration at the instant. This axis is then the instantaneous axis of angular acceleration. Angular acceleration may be zero, uniform or variable. W n< it is zero, the angular velocity is uniform and we have uniform angular speed and an unchanging plane of rotation. When it is uniform, it has the same magnitude and the san direction whatever the interval of time. In such case the accelera- tion is equal to the mean acceleration for any .^terval of time the direction coincides with that of the b^^Q** uniform rate of change of angular speed and an unchanged plane of rotation. If it makes an angle with the velocity, we have a changing plane of rotation and variable velocity. 176 RIGID SYSTEM WITH ONE POINT FIXED ROTATION. [CHAP. I. When it is variable, either direction or magnitude changes or both change. If the angular acceleration is always at right angles to the an- gular velocity, it only changes the direction but not the magnitude of that velocity. Hence, just as on page 53 a normal linear acceleration has no effect upon the linear speed, but only changes the direction of mo- tion, so, if a rigid system rotating with given angular speed about an axis has an angular acceleration about an axis always perpendic- ular to the first, there is change of direction of this axis but no change of angular speed about it. The gyroscope is an illustration of this principle. Resolution and Composition of Angular Velocity and Accelera- tion. Since for an indefinitely small time the angular displace- ment is indefinitely small, we see from page 171 that we can com- bine angular velocities and accelerations, whether simultaneous or successive, by means of their line representatives just like linear velocities and accelerations (page 43). Sign of Components of Angular Velocity and Acceleration. The sign of the line representatives of the components along the axes X, Y, Z of an angular velocity or acceleration follows the same rule as for linear velocities and accelerations (pages 44, 50). Unit of Angular Acceleration. Angular acceleration is meas- ured in terms of the same unit as rate of change of angular speed (page 73), or one radian-per-sec. per sec. We denote its magni- tude then by the same letter, . Equations of Motion of a Rotating Rigid System under Different Angular Accelerations. Since angular velocities and accelerations are represented by straight lines, just like linear velocities and ac- celerations, we have the same equations for motion of a rotating rigid system as on page 50. We have only to substitute GO for v, Q for s, a for /. With these substitutions equations (1) to (14), page 50, hold good and it is unnecessary to repeat them here. Moment of Angular Displacement. Just as we called the prod- uct of the magnitude of a linear displacement by the magnitude of the perpendicular let fall from any given point upon its direc- tion the moment of the linear displacement (page 60), so for angular displacement we call the product of its magnitude by the magnitude of the perpendicular from any point upon the direction of the line representative the moment of the angular displacement We take its sign just as for moment of linear displacement, page 62. -Since the line representative is coincident with the axis, the perpendicular is the distance of the point from the axis. Thus if AB = d is the line representative of an angular dis- placement OiOOa = 6 of a rigid system, the axis has the position CHAP. I.] CONCURRING ANGULAR DISPLACEMENTS, ETC. 177 AOB. If then Oi is the initial position of any point of the sys- tem and OOi = p is the perpendic- ular from Oi upon the axis or direc- tion of the line representative AB, the moment is p r i according to direction, just as for moment of linear displace- ment (page 62;. But pfj is the length A of the arc OiO* described by the point Oi in a plane perpendicular to the plane of AB and OOi. Hence, the moment pB of the angular 'isplacement of a rigid system rela- d m tive to any point of the system gives the length of the arc OiO 2 described by that point in a plane perpendicular to the plane of the axis AB and the radius vector p. The corresponding linear displacement of Oi is evidently T = 2p sin - (1) Since the angle OOiO 2 equals the angle OOaOi, we have for the direction of the linear displacement relative to OOi , angle OOiO, = 90 - (2) We have also, just as on page 62, the algebraic sum of the moments of any number of component angular displacements, rel- ative to any point, equal to the moment of the resultant. Also, just as on page 60, the line representative of an angular displacement may be laid off from any point in its line of direction without affecting its moment. Moment of Angular Velocity or Acceleration. Just as we called the product of the magnitude of a linear velocity or accelera- tion by the magnitude of the perpendicular from any given point upon its direction the moment of the linear velocity or acceleration, so for angular velocity or acceleration we call the product of the magnitude by the magnitude of the perpendicular from any point upon the direction of the line representative the moment of the angular velocity or acceleration. We take its sign just as for moment of linear velocity or accele- ration (page 60). Since the line representative is coincident with the axis, the perpendicular is the distance of the point from the axis. Thus if velocity of is the line representative of an angular system, the instantaneous axis has the position AOB. If then O\ is any point of the system and OOi = p is the perpendicu- lar from Oi on the axis or direction of the line representative, the moment is pw according to direction, just as for moment of linear velocity (page 60). But this is B the linear velocity v of Oi at the instant, in a direction perpendicular to the plane of AB and OOi. Hence, the moment poo of the angular velocity GO of a rigid system relative to any point of the system gives the linear velocity v of that point in a direction perpendicu- lar to the plane of the instantaneous axis of rotation AB and the instantaneous radius vector p. 178 RIGID SYSTEM WITH ONE POINT FIXED ROTATION. [CHAP. L. In the same way, the moment pa of the angular acceleration a of a rigid system relative to any point of the system gives the linear tangential acceleration ft of that point in a direction perpendicular to the plane of the instantaneous axis of angular acceleration and the instantaneous radius vector. We have also, just as on page 62, the algebraic sum of the mo- ments of any number of component angular velocities or accelera- tions, relative to any point, equal to the moment of the resultant. Also, just as on page 60, the line representative of an angular velocity or acceleration may be laid off from any point in its line of direction without affecting its moment. Concurring Angular Displacements, Velocities or Accelerations. We see then that angular displacements, velocities or accelera- tions are represented by straight lines, called line representatives, which coincide with the axis of rotation. We deal with them entirely by means of these line representatives. When we speak of their "direction," we mean the direction of the line representa- tives. We resolve and combine them by means of their line repre- sentatives, and in the same way we have their moments just as for linear displacements, velocities or accelerations. Following the same analogy, we can speak of them as "applied " or "acting" at. certain points. When they all intersect at the same point, we may call them concurring, just as if they were linear. When they do not intersect at the same point they are non-concurring. When they act in the same direction in the same line they are conspiring. When in the same or opposite directions in parallel lines they are parallel. When in opposite directions in the same line or in paral- lel lines they are opposite. When they lie in the same plane they are co-planar. Condition for Rotation only. If a rigid system has one point fixed, it can have no translation but only rotation, and therefore all the component angular displacements, velocities or accelerations must reduce to a concurring system, so that we have a single result- ant angular displacement, velocity or acceleration about an axis through this point, which is therefore at rest. General Analytical Determination of Resultant Angular Dis- placement, Velocity or Acceleration for any Number of Concurring Components. We see then that all the equations of pages 63 to 65 hold good for angular displacements, velocities or accelerations, as well as for linear. For angular displacements we have only to substitute 6 in place of v. The moments M x , My , M z then give the arcs of displacement about the axes of X, Y, Z of the origin, considered as a point of the rigid system, rotating about the resultant axis. For angular velocities we have only to substitute a> for v. The moments M* , My, Mz then give the component linear velocities v x , V H , v z along the axes of X, Y, Z of the origin, considered as a point of the rigid system, rotating about the resultant axis. For angular accelerations we have only to substitute for v. The moments M x , M v , M z then give the component linear tangential accelerations ftx, fty, ftz along the axes of X, Y, Z of the origin, considered as a point of the rigid system, rotating about the re- sultant axis. To make our notation consistent we should also replace cos a r cos &, cos c, page 65, by cos d, cos e, cos /, and replace cos d, cos e cos f, page 66, by cos a, cos 6, cos c. We have then from page 65, equation (4), for the component linear velocities v x , v y , v z along the axes of X, Y, Z of the origin* CHAP. I.] TWO CONCURRING ANGULAR VELOCITIES. 179 considered as a point of the rigid system, rotating about the re- sultant axis, Vx = cozy oayZ ; Vy = COxZ GOzX; Vz = <&yX coxy- We have also in the same way fix = a z y a y z ; ftz = a y x a x y. (1) (2) Equations (2) give the component linear tangential accelerations along the axes of X, F, Z of the origin, considered as a point of the rigid system, rotating about the resultant axis. If we multiply the first of equations (1) by w x , the second by co v , the third by oo z and add, we obtain Vx<*>X + VyCOy + = 0. Equation (3) is the condition for rotation only. When it is ful- filled, we know that the motion of the system is that of rotation only about the instantaneous axis. Resultant of Two Concurring Component Angular Displacements, Velocities or Accelerations.* It will be of profit to specially discuss the case of two concurring component angular displacements, velocities or accelerations. Let the two angular velocities on, oj. 2 be in the same plane and pass through the points A and B of a rigid system, so that they intersect at O. Then the resultant GO,- must pass through O and be in the plane Of ftJi, 003. Take any point P in this plane and draw the perpendiculars Pni = pi t Pn*=pi, Pn=p,. Then, since the moment of the resultant is equal to the algebraic sum of the moments of the components, w,-p r = cojp, + &> 2 p, (1) where regard must be paid to the Thus we have in the figure COifl 71 GOffJi' OOi^Jl ut/2^/3. // Draw the line AB, intersecting the /'GO resultant w r at the point C. Let ai be the angle of a>i, and 2 the angle of ca 2 with AB. If we take moments < / f F10.2.X /w / \ <0 r about (7, we have N^\ GO I . ACsinai = G0 2 .5(7 sin a 3 . "*C * Compare Statics Non-concurring Forces. 180 KIGID SYSTEM WITH ONE POINT FIXED ROTATION. [(3HAP. But AC + BC = AB. Hence cot . AS sin -t y Dr> GL>I . AB sin ori . o)i Sin ai + ao* sma a ooi sin tr 1 + a> 2 sin a* We thus know the position of the resultant a?,- in the plane of . Then we have GO X ooi COS a i + G0 2 COS a ; I f ...... (o) GOy = ooi COS /Ji + ca 2 COS /?; ) where we must pay regard to signs. Thus components in the direction OX, O Y are positive, in the opposite directions negative. If the resultant o> r makes the angles d and e with the axes of X and F, we have , oox oo cos d = cos e = -=- ........ (4) 00 r Gar Squaring and adding, G0 r = \/K>x* + GOy* ......... (5) The magnitude and direction of the resultant are thus deter- mined. Also if Qi is the angle of 001 with the resultant, and 9 2 the angle of caa with the resultant, and 6 the angle between 001 and GO* , we have directly from Fig. 2 sin 0! = sin 6, sin 0, = ' sin 6, ..... (6) 00 r Mr and ___ (Or = i/GO* + GO'S 2G01GO* COS 0, ..... (7) where the ( + ) sign is used when 6 is less than 90, and the ( ) sign when 9 is greater than 90. The tangent of the angle d which the resultant makes with AB or OX is tand = " .......... (8) oox From (6) and (7) we can find the magnitude and direction of the resultant directly if is known. If 2 are known, we can locate the resultant by describing a circle with centre P and radius p r and drawing oo r tangent to this circle in the direction given by (6). * Compare Statics. CHAP. I.] TWO PARALLEL ANGULAR DISPLACEMENTS, ETC. 181 The same formulas hold good for two concurring angular ac- celerations. We have only to replace GO \>y a. [The student will of course not confuse this a, which stands for angular acceleration, with a i , a a in the formulas, which stand for angles.] The same formulas hold good also for two concurring component angular displacements. We have only to replace &> by d. When the Angular Displacements, Velocities or Accelerations are Parallel. In this case i and 2 are equal, 6 = 0, the intersection O is at an infinite distance, oo r = &>, + &> 2 , and we have from (2) AC=-.AB, BC=-.AB; (1) and hence, multiplying the first by &n and the second by a> 2 , GJ, . AC = (2 . BC, or ^ = (2 To prove this independently, take C as the point of moments Then whether the line representatives act in the same or in opposite directions, we have oo,pi aovp* = 0, or coipi = co*p*. But from similar triangles : AC i and e 2 always pass through the points A and B no matter what their common direction, the resultant oo r always passes through C. The point C is then the point of application of the resultant <*> r for all directions. , , . Hence, the resultant of two parallel component angular dis- placements, velocities or accelerations is in their plane and equal in magnitude to their algebraic sum. It acts parallel to the com- ponents in the direction of the greater. If the components always pass through two given points A and 5, the resultant always passes through a point C no matter ivhat the common direction. IMS point C is then the point of application of the resultant. It is on the straight line AB or this line produced, and divides it into seg- ments inversely as the components. Or the products of the com- ponents into their adjacent segments are equal. (Compare Stati Parallel Forces.) 182 RIGID SYSTEM WITH ONE POINT FIXED ROTATION. [CHAP. I. COR. 1. When the components act in the same direction, the resultant lies within the components and nearest the larger. When the components act in opposite directions, the resultant lies without the components and on the side of the larger. COR. 2. When the components are opposite and equal in magni- tude, oj r = 0. Also from (1), AC =00, BC = oo, or the resultant is zero and acts at an infinite distance. That is, equal and opposite parallel components cannot have a single resultant. Such a system is called a couple. (Compare Statics Parallel Forces.) EXAMPLES. '(I) A rigid system has two component rotations of 2 and 4 radians about axes inclined 60. Find the resultant rotation. Ans. Component rotations are understood to be simultaneous unless other- wise specified (page 172). Hence magnitude of resultant rotation is 2 |/7 radians; axis inclined at an angle with the greater component whose sine is /3 (2) A sphere with one of its superficial points fixed has two com- ponent rotations one of 8 radians about a tangent line and one of 15 radians about a diameter. Find the axis of the resultant dis- placement and the number of complete revolutions made about it. Ans. Inclination of axis to greater component at an angle whose tang is . Resultant displacement 17 radians, number of complete revolutions . 15 Ait (3) A sphere is rotating uniformly about a diameter at the rate of 10 radians per min. Find (a) the component angular velocity about another diameter inclined 30 to the former, and (b) the com- ponent rotation produced in 2 min. about a diameter inclined 45 to the first. Ans. (a) 5 |/3 radians per min. ; (b) 10 |/2 radians. (4) A pendulum suspended at a point in the polar axis of the earth oscillates in a vertical plane. Find the motion of this plane relative to the earth. Ans. The plane of the pendulum is fixed in space, and the motion of the earth with reference to this plane is a rotation from west to east at the rate of one revolution per day. The motion of the plane relative to the earth is theu from east to west at the same rate of one revolution per day. (5) A pendulum is hung at a place of latitude A and oscillates in a vertical plane. Find (a) the angular velocity of the plane of the pendulums motion relative to the earth, and (b) the time in which this plane will make one complete revolution at a place in latitude 60 AT Ans. The angular velocity of the earth about its axis is 2?r radians per day. The component of this in the direction of an axis through the centre of the earth and the point of suspension of the pendulum is 2n sin A radians per day from west to east. This is the motion of the earth relative to the plane of the pendulum. Hence (a) The motion of the plane of the pendulum relative to the earth is 2n sin \ radians per day from east to west; -CHAP. I.] EXAMPLES. 183 (b) The time of revolution is - 1 - 2 2n sin A sin 60 " ^ Qays ' (6) A cube rotates about a vertically upward axis through one of its edges. At a given instant at which the diagonal of the upper surf ace passing through the axis points north the cube has an an- gular velocity o/40 radians per sec., and begins to have a uniform angular acceleration about an axis vertically downwards through the same edge of 6 rad.-per-sec. per sec. Find (a) the direction in ichich the diagonal will point after 20 sec. ; (b) the number of revolu- tions made by the cube. Aiis. From the equations of motion page 73 we have GO = , at, B 6, = oo L t ft-at*- We have M, = 40 radians per sec., a = 6 rad.-per-sec. per sec., t = 20 sec., 9, = 0. In the time t t = ^~- sec. the cube comes to rest and has the angular 400 displacement 9j = -5- radians towards the east. o It then moves in the opposite direction towards the west during the time 20 40 t = 20 - = sec - an 3 = 60 radians per sec., the line representatives making the angles i = 60, ft\. = 150, y, = 90; 2 = 120% fa = 30, y? = 90; a 3 = 120% fa = 150% y* = 90% Find the resultant angular velocity. Ans. (See Example (1), page 67.) The component angular velocities are in one plane and co x = 35 radians per sec., <*>y = 43.3 radians per sec. The resultant is oo r = 55.67 radians per sec., its line representative or the instantaneous axis of rotation making with the horizontal the angle d 128 57' 17", and with the vertical the angle e = 141 2' 43". If we look along this 184 KINEMATICS OF A RIGID SYSTEM ROTATION. [CHAP. I, line representative which passes through the fixed point, towards the origin, the rotation will be seen as counter-clockwise. The moment of the resultant angular velocity oo r with reference to the point gives us the linear velocity of rotation at about the instantaneous axis, V* = -f- 10 ft. per sec. in a direction through perpendicular to the plane XT, or along Z, from towards Z. The distance of from the axi is p = about 0.18 ft. The equation of the axis is y = 1.237x -|- 0.286. Its intercepts on the axis are y' = + 0.286 ft., x 0.232 ft. (9) Express and solve the same example for component angular accelerations and displacements. (10) A rigid system has one point fixed. The co-ordinates of this point with reference to any point of the system taken as an origin are at any instant x = + % ft., y = + 4/., z +5ft. The compo- nent angular velocities at this instant are 001 = 40, eaa = 50, &3 3 60 radians per sec., the line representatives making the angles with the axed o-j = 60, 0i = 100, y\ obtiise ; X = - 18.6824, coy - + 7.635, co z = + 59.391 radians per sec. The resultant angular velocity is co r = 62.73 radians per sec., its line rep- resentative making with the axes the angles d = 118 17' 33", e = 85 6' 12", / = 12 30' 24". This line representative passes through the fixed point and gives the in- stantaneous axis of rotation. If we look along this line towards the origin, the rotation will be seen as counter-clockwise. The velocities of rotation at along the axes are v z = + 97.6346, v x = + 199.389, vy = - 271.585 ft. per sec. The resultant velocity of rotation at is t>r = 407.6 ft. per sec., making with the axes angles a = 60 42' 57", b = 131 46' 24", c = 76 8' 31". The equations of the projections of the axis upon the co-ordinate planes are: on plane -XT, y = 0.408z + 5.226 ; " " TZ, z = + 7.778^-26.115; " " ZX, x = - 0.314s -f 4.572. The axis pierces the plane XY at a;' = + 4.572 ft., & = + 3.357 ft.; " " " " " TZ " y 1 = + 5.226ft., z' = - 14.56 ft.; " " " " " ZX " z' =- 26.115 ft., *' =-f- 12. 809 ft. (11) Express and solve the same example for component angular accelerations and displacements. (12) Let the axes of two concurring angular velocities of a rigid system, aoi = 20, o> 2 = 30 radians per sec., pass through the points A, B of the system, the distance AB = 2 ft., and the angles a, = 60, a 2 = 30. Find the point C on the line AB through which the resultant axis passes, and the magnitude and direction of the resultant angu- lar velocity. CHAP. I.] EXAMPLES. 185 Ans. AC = 0.928 ft., oo x + 15.98 radians per sec., ao y = 32.32 radians per sec. The angle of the resultant with AB is given by 82 32 tan d = - rf = = - 2.022, or d = 63 41' = BCcor. lo. yo The resultant is cor = 36.05 radians per sec. We have also for the angle of the resultant with coi , since = 90, Q = 0.832, or 6, = 56 19'. OO.UO (13) Express and solve the same example for component angular accelerations and displacements. CHAPTEE IL MOMENT OF A COUPLE. DISPLACEMENT OP A RIGID SYSTEM. RIGID PLANE SYSTEM. COMPOSITION AND RESOLUTION OP TRANSLATION AND ANGULAR DISPLACEMENT. COM- POSITION AND RESOLUTION OP TRANSLATION AND ANGULAR VELOCITY. CENTRAL AXIS. SCREW MOTION. ROTATION AND RECTILINEAR TRANS- LATION. COMBINED PARALLEL ROTATIONS, ONE AXIS FIXED. INTER- SECTING AXES, ONE AXIS FIXED. ANALYTIC DETERMINATION OF RE- SULTANT ANGULAR VELOCITY AND VELOCITY OF TRANSLATION FOR NON- CONCURRING ANGULAR VELOCITIES. Moment of a Couple.* We have just seen in the preceding Chapter, page 181, that two parallel equal and opposite components acting at different points of a rigid system constitute a couple. We may have then an angular-displace- ment couple or angular- velocity cou- ple or angular-acceleration couple. Let + ?, co, acting at the points A, B of a rigid system, constitute an c 2 angular-velocity couple. If we take any point C between the components, or any point C\ , C 2 on either side, in the plane of the components, we have in the first case, denoting the distance AB by p, for the moment about C, just as for linear velocities (page 60), - oo . AC oo.BC = <*>(AC + BC) = poo. In the second case, for the moment about C\ we have oo . dA - B = - Go(C>B - CiA) = -pas. In the third case, for the moment about C 2 we have oo . CiA + GO . CiB = co(C 2 A CiB) = pan. Hence the moment about every point in the plane of the couple is constant and equal to poo, the ( + ) or ( ) sign denoting direction just as for moment of linear velocity (page 60). For an angular-acceleration couple we have in the same way pa, for an angular-displacement couple pQ. We see then that the moment of a couple is the same for every point in its plane and equal to the product of either of the compo- nents by the distance between them. * Compare Statics Parallel Forces. 186 (CHAP. II.] DISPLACEMENT OF A RIGID SYSTEM. 187 Composition and Resolution of Translation and Angular Displace- ment. Let a rigid system have a rotation of OiOO 2 = 9 radians aJbout an axis AOB through the point O, and AB = B be the line representative. If we take any other point of tiie system, as Oi , and at this point apply the two equal and opposite angular displacements OM = 6 and Oib = + 6, both parallel to AB, it is evident that the motion of the system is not affected. We have then the angular displace- ment about the axis AOB reduced to an equal angular displace- ment Oib about a parallel axis through Oi and a couple AB and OM. The moment of this couple is the same for every point in its plane and equal to p6, where p is the perpendicular distance be- tween the components AB and OM of the couple. But we have seen (page 177) that the moment pO corresponds to a linear displacement in a plane perpendicular to the plane of the couple of T = 2psin-, making an angle OOiO-, withp given by 7C-B OOiO, = (1) (2) Hence, an angular displacement Q about any given axis can be resolved into an equal angular displacement about a parallel axis through any point of the system and a linear translation in the plane of rotation of the system whose magnitude and direction are given by (1) and (2). Conversely, the resultant of the rotation of a rigid system about a given axis and a translation in any given direction, is an equal rotation about a parallel axis, whose position with reference to the first can be determined by (I) and (2). COR. 1. Two non- concurring angular displacements can be re- duced to a resultant angular displacement about a resultant axis at any point and a couple which causes translation. COR. 2. Hence if we have any number of component angular rotations about any axes, whether these axes intersect or not, we can reduce each to an equal rotation about a parallel axis through some one point of the system and a translation of the system. The resultant translation can then be found as on page 35, and i lie resultant rotation as on page 173, for simultaneous angular dis- placements. COR. 3. Therefore any number of component translations and rotations can all be reduced to a single translation and a single rotation about any given point. It is evident that this single rota- tion is not affected by the position of the point, which affects the translation only. Displacement of a Rigid System. Any displacement of a rigid system may be produced by a translation and an angular displace- ment. 188 RIGID SYSTEM TRANSLATION AND ROTATION. [CHAP. II.. Let Ai , J3i, Ci be the positions of any three points which determine the initial position of the rigid system. Let A? , Bi , C-i be the final position of these points after any displacement. First let the system be translated, so that Ai comes to its final position At. Then B\ and Ci will take the positions b and c, the lines Bib and C\c being equal and parallel to AiA*. We see then that Ai is a fixed point in the c. system so far as the two positions AiBiCi , AiBiCi are concerned. But we have seen (page 173) that in every possible displacement of a rigid system with one point fixed there is an axis of rotation fixed in the system which remains unchanged. Hence A*cb can be brought to the position AiBiCi by rotation about that axis. COR. 1. It follows that the displacement of a rigid system is known if the magnitude and direction of the linear displacement of any point is known, and also the magnitude and direction of the angular displacement of the system about that point. COR. 2. Also, the displacement of a rigid system is known if the magnitude and direction of the component linear displacements of any point parallel to three rectangular axes and of the component angular displacements of the system about axes parallel to the first through the point are known. Rigid Plane System. Any displacement of a rigid plane system in its own plane may be produced by rotation about some point in the plane. Let AiBi and A 2 -B 2 be the initial and final positions in the plane, of the same line of the system, o , so that AiBi and A 2 _B 2 are of vvss. ^. equal length. Join AiA^ and \V^v ^--^ BiBi by lines and bisect these lines at C and D. Erect per- pendiculars at the points of \ \ \ ^x^ 7D bisection C and D and produce them to intersection at O. Then by construction OAi = OAv, and OB, - OB,, and AiB, = AsBz. Hence the two triangles OAiBi and OA-tB* are in all respects equal and the line AiBi may be brought to coincide with AzB? by rotation about the point O. If AiBi is parallel to AiBi , we have translation only and the point O is at an infinite distance. CHAP. II.] DISPLACEMENT OF A RIGID SYSTEM. 189 Since the angle AiOBi = the angle A*OB* , if we take the angle A~OBi from both we have AtOA* = BiOB>. If then the displace- ment is such that Ai falls on OBi or on OB, produced, A* must be on OB* or OB* produced. In both cases OC and OD coincide and do not intersect, but it is evident that in such case the point O in which AiBi and A*B inter- sect is the point about which rotation would produce the given dis- placement. If in any case AiA* and BiB* are indefinitely small, the point O is called the instantaneous centre of motion. Any Displacement of a Rigid System. Any displacement of a rigid system may be produced by rotation about an axis and a trans- lation in the direction of that axis. Let AB and BC represent the resultant translation and rotation to which the compo- nent translations and rotations of the system can be reduced (page 187, Cor. 3). Draw AD and DB parallel and perpendic- ular to BC. Then the translation AB is re- solved into the two components AD and DB. But the resultant of DB and BC (page 187) is an equal rotation about an axis parallel to BC. Hence the translation AD and the rotation BC are reducible to the translation AD and a rotation about an axis parallel to AD. Composition and Resolution of Translation and Angular Ve- locity or Angular Acceleration. Let a rigid system have an f angular velocity f I radians per sec. t about an \ angular acceleration ) ( a radians-per-sec. per sec. ( axis AOB through the point O, and let AB = j * I be the linear rep- resentative. If we take any other point of the system, as Oi , and at this point apply two equal and opposite I angular velocities ) a \ angular accelerations \ ' J- "land 0,6= j J"|,itiBevi- \.vi f ( a i ' ~*~ ' rf dent that the motion of the system ,, j angular velocity [ is not affected. We have then the j angl f lar acceleration f the axis AOB reduced to an equal { ang^ar^acceleratfon [ Olb about a parallel axis through Oi and a couple represented by AB and OM. The moment of this couple is the same for every point m its plane (page 186) and equal to j ^ \ , where p is the perpendicular distance between the two axes. But we have seen (page 177) that the moment j jj" [ gives the linear j acceleration f \ in a direction perpendicular to the plane of the couple. Since the moment of the couple is the same for every point in its plane, we have then translation of the entire system in a direction perpendicular to the plane of the couple, as well as simultaneous rotation about the axis through Oi. The direction 190 RIGID SYSTEM TRANSLATION AND ROTATION. [CHAP. 11^ and magnitude of this translation will depend upon the point Oi , but the rotation will be the same wherever the point Oi may be taken. Hence,* an j ^g^^acceleration \ f a rigid s y stem about any axis can be resolved into an equal j ^guUr'acceUratl \ about a parallel axis at any distance p, and a j a cceleratio ( f t rans ^ a ^ on ) V f ^ ? ( in a direction at right angles to the plane of the axes. Conversely, the resultant of an j angular acceleration a\- fa rigid system about a given axis and a < accelvrat'on 1 f ^ rans ^ a ^ on < ^ > in any direction is an equal j anguiai velocity i about a parallel axis distant { \ in a direction perpendicular to the \P= \ I a } plane of \ ^ > and the given axis. This parallel axis is the instantaneous axis. COR. 1. Hence if we have any number of component j accelerations [ a ^ ou ^ an y ax i s > each can be reduced to an equal about point of the system, and a j acceleration [ ^ t rans l a tion of the system. We can then find the resultant j acceleration 1 of trans ' lation as on page { g \ and the resultant { Jjj*^^ \ as on page 176. COR. 2. Therefore any number of component j anf ular Ic- celerltns } and | acSSSns \ trandation, can all be reduced to a single resultant j acceleration f f trans ^ at i n and a single re- sultant 1 a^St^S^^n | about an axis throu g h an ^ one P int of the system. The j acceleration I of transla ti n w ^ vary in di- rection and magnitude with the point chosen. The j acceie C at^o f w ^ ^ e *^ e sanie no ma tter what point is chosen. Central Axis. Any number of component angular velocities of a rigid system can be reduced to a single angular velocity about a determinate axis and a simultaneous velocity of translation of the system along that axis. Such an axis is called the central axis, and such motion is called screw motion. * Compare Statics Non-concurring Forces. CHAP. II.] CENTKAL AXIS. 191 Thus let OA and OC represent the resultant linear velocity of translation V r and the resultant angu- lar velocity ca r , to which, as we have just seen, all the angular velocities can be reduced. Draw AD and OD parallel and per- A pendicular to OC. Then the velocity of translation OA = V r is resolved into the two components AD and OD. But the resultant of OC and OD is an equal angular velocity about an axis parallel to OC (page 190). Hence the velocity of translation OA = V r and the angular ve- locity OC = oo r are reducible to an equal angular velocity about an axis parallel to OC and a linear velocity of translation AD along that axis This axis is called the central axis, and may be located by the following geometric construction. At any point O of the system taken arbitrarily let the velocity of translation be V, and the ro- tation axis through O be &?,-, making the angle (f>. Through O draw a line OD =p perpen- dicular to the plane of V r and oa r , so that poor = V r sin 0, or Vr sin 4> p = tor Then a line through D parallel to the rotation axis at O will be the central axis. (Compare Statics Non-concurring Forces. Screw Motion. Let u,- denote the resultant velocity of trans- lation along the central axis. This is called the velocity of advance. The distance d advanced during one complete rotation of the sys- tem is called the pitch of the screw, and the distance advanced dur- ing a rotation of one radian, or , we call the unit pitch of the iiTt screw. If oa r is the resultant angular velocity of rotation, the time ot a complete rotation is t = 00 r We have then for the value of the pitch hence u r = card and for the unit pitch d Ur_ oar (2) If r is the radius vector of any point of the system, then the linear velocity of that point due to rotation about the axis is (3) v = roa r in a direction perpendicular to the plane of the central axis and the radius vector. The resultant velocity at that point is then Vr = 192 RIGID SYSTEM TRANSLATION AND ROTATION. [CHAP. IT. The inclination of the path at that point to the plane of rotation is given by tan* = ?*= * ....... (5) or the tangent of the angle of inclination at any point is equal to the ratio of the pitch to the circumference of the circle described by the point relatively to the axis ; or it is equal to the ratio of the unit pitch to the radius vector of the point. Centre of Parallel Angular Velocities.* Let &?,, <, oo s , etc., be any number of parallel angular velocities passing through the points Ai , A-t v As , etc. , of a rigid system. Then the resultant oo r must be parallel to the components and equal in magnitude to their algebraic sum, or co r = KOI + + 033 + ...= Take any two components G>I and GO?, and produce the line A^ to intersection K with the plane Z2T. Drop perpendiculars Ai A-tB* to this plane and draw the line KBiB* in this plane. Now, from page 181, the resultant of aoi and ca 2 is &?i = coi + and its point of application is at A on the line AiAa , so that Drop the perpendicular AB to the plane similar triangles Then we have by Denote the distance AiJS, , A 2 B 2 by yi , ?/ 2 , respectively, and the distance AB, or the ordinat of the point of application of the re- sultant oo i of coi and (a, by 2/1. Then we have by similar triangles Hence or ?/i = In the same way for three angular velocities, coi , o> 2 , co s , we can combine the resultant i + OOt + GOa In general, then, for any number of parallel angular velocities we have for the ordinate y ot the point of application of the result- ant u-^M y ~ 2oo' In precisely similar manner, if we denote the distances AC and AD of the point of_application of the resultant from the planes YZ and XY by x and y, we have Equations (1), (2) and (3) give the co-ordinates of the point of application of the resultant for any number of parallel angular velocities. This point is called the centre of parallel angular veloc- ities. The same equations evidently hold for parallel angular ac- celerations, by replacing GO by a. We have then the centre of parallel angular accelerations, i The position of this centre depends only upon the magnitude and position of the line representatives and is independent of their com- mon direction. If z is zero, z\ , Zi, etc., are zero, and the line representatives all lie in the plane XY. The centre is then given by (1) and (2). If z and y are zero, the centre is in the axis of .STand is given by (2). Rotation and Rectilinear Translation Combined. Let a rigid system have an angular velocity Ob = GO about an axis through O, perpendicular to the plane of the paper, and at the same time a velocity of H . translation v in a straight line. Then, as we have seen, page 177, v can be re- placed by the couple Oa and IB, and we have at any instant a resultant rotation IB = GO about a parallel axis though I at a distance 01 = p = in a direction 00 perpendicular to that of v. This axis is the instantaneous axis; that is, the point I at any distant has the velocity v in one direction due to translation, and the velocity v = poo in the other direction due to the couple, and is therefore at rest. It is evident that every straight line in the system paralle the moving axis at O and at a constant distance from O of p = - becomes in turn the instantaneous axis when it arrives at (a the position I with reference to O. Hence when a rigid system has a velocity of translation m a straight line and at the same time an angular velocity m about a given axis 06, the resultant motion of the system is the same as tf 194 RIGID SYSTEM TRANSLATION AND ROTATION. [CHAP. II.. Curtate Cycloid If the radius is , we have poo = v and hence = p. CO Cycloid The in- a cylindrical surface fixed in the system, of radius p, = , rolled on GO a plane HIH parallel to the plane of Ob and v. The path described by any point in the axis Ob is a straight line. The path described by any point not in this axis is called a trochoid. The special form of trochoid described by any point in the cylindrical surface is called a cycloid. Any internal point describes a prolate cycloid ; any external point, a curtate ) )\/ cycloid. / X The general form of these curves is shown in the accompanying figures. A common illustration of such motion is a wheel rolling hi a straight line on a plane. stantaneous axis is at right angles to the plane of the wheel and passes through the point of contact with the plane. The velocity at the centre is v, and at the opposite extremity of the diameter through the point of contact 2 in the direction of translation. The velocity of any point at a distance d from the instantaneous axis is doo in a direction per- pendicular to the plane of the instantaneous Prolate Cycloid ax j s an( j the instantaneous radius vector d. Combined Parallel Rotations One Axis Fixed. Let a rigid system rotate with the angular velocity &>i about a moving axis at Oi, and at the same time let this axis revolve with the angular velocity t 2 about a parallel fixed axis at Oa. Then, as we have seen, page 181, the resultant oo r is in the plane of the components ci and i is the greater ; Fig. 3, that in which ci and GO? are in opposite directions and wa is the greater. The resultant angular velocity is in all cases then given by GO = CO +00 (I) where we take ooi and oo y with their proper signs. This resultant rotation oo r takes place about the instantaneous axis through 7, so situated that (page 181) FIG. l. EiGi-2. E1&.3. (2) so that at any instant 1 has two opposite and equal linear velocities and is therefore at that instant at rest. Since then cai . 10 1 = oo* . /Oa, CHAP. II.] ROTATION ABOUT INTERSECTING AXES. 195 we see at once that in Fig. 2, an is greater than oo 2 , and in Fig. 3, aot is less than 003. We have also as on page 181, taking moments about Oa and Oi , i COS (ai + aa) GJj tan ai= sin (ai + - + COS (ai + aa) (1) (2) From (1) and (2) we can find ai and aa , when the angle between the axes (ai +a a) and the angular velocity ratio are given. We have also &) r y = cOi a + fi5a a 2c3i(a COS (ai + aa), .... (3) and GOT _ sin (ai + aa) ao r _ sin (ai + a a ) oo\ _ sin a 2 , A ^ 09] Sin Sin Sin All lines which come successively into the position of the in- stantaneous axis are in the surface of a cone described by the revo- lution of CI about COi ; and all the positions of the instantaneous axis lie in the surface of a cone described by the revolution of CI about COa. Hence the motion of the rigid system is such as would be pro- duced by the rolling of the cone CIOi , fixed in the system, about the fixed cone CIO?. If r 2 is the radius O 2 1 of the fixed cone, and ri the radius Oil of the rolling cone, we have For the height COa = hot the fixed cone we have + r 2 cos (ai + aa r 2 cotang 2 = sin a a ) (5) (6) CHAP.II.] ANALYTICAL DETERMINATION OF RESULTANT VELOCITY. 197 and for the height COi = hi of the rolling cone h> = n cotang a, = *", + r. cos (a. + a,) Sin (i + ) The plane through the instantaneous axis and the axis of the fixed cone passes through the axis of the rolling cone and turns about the axis of the fixed cone with the angular velocity *_/ O fji \J U. path described by a point relatively to ed axis is called a spherical epitrochoid 0,0, 198 RIGID SYSTEM TRANSLATION AND ROTATION. [CHAP. II. Resultant Angular Velocity. We have, just as on page 65, re- placing v by oj, for the component angular velocities parallel to X, Y, Z, oox = 001 COS ai + tt> 2 COS as + . . . = .Sc^COS a ; ] K>y GOi COS fti + i, <*>*, etc. (page 190;, into an equal angular velocity about a parallel axis through the origin O, and a velocity of translation of the system due to a couple, given by the moment of z = 0, that is, when oo r = or there is no rotation. In this case all the angular velocities must reduce to two equal and opposite resultant angular velocities. The first condition is fulfilled when equations (4) are zero. That is, the two equal and opposite resultant angular velocities must pass through the same point, so that their moment is zero at any point. We have then for the equations of condition of rest, from (1), (10) = 0; = 0; = 0; and, from (4) Scoy cos y 2& >z cos /? = 0; ") SooZCOSa 2 oax COS ^ = 0; I ...... (H) 2a>x cos /3 2aay cos a = 0. J If equations (11) only are fulfilled, then the two opposite resultant angular velocities pass through the origin, which is therefore at rest; but unless (10) is also fulfilled they are not equal, and we have rotation about an axis through O, but no translation. If equations (10) only are fulfilled, there is no rotation, the two resultant angular velocities are opposite and equal, but unless (11) is also fulfilled they do not pass through the same point. Hence they form a couple, and we have translation and no rotation. Condition that the Angular Velocities shall Reduce to a Single Angular Velocity. If the angular velocities, then, all intersect in one point of the system, the moment at that point is zero. It has therefore no translation and the system rotates about an . axis through that point. If the angular velocities do not intersect in a single point, we have in general translation and rotation. There is, however, one case in which the angular velocities may not all intersect in one point, and yet we may have rotation only without translation. In this case the angular velocities must re- duce to three, any tivo of which intersect, while the other, although it does not pass through their point of intersection, yet intersects their resultant. * Compare Statics Xon-concurring Forces. 200 RIGID SYSTEM TRANSLATION AND ROTATION. [CHAP. II. Thus let the resultant angular velocities GO X , <% intersect in a point A. We can then take them as acting at any point in, their resultant AC. Let ooz intersect AC at B. Then we can take all three acting at J5, and we thus have rotation only, about an axis (12) Let x, y, z be the co-ordi- nates of B. Then, since we can take co x , a^, oa z at -B, we have for the components of the velocity of the origin O My = Vy 00 X Z OOzX; MZ =z Vz == GOyX OOxy* If we multiply the first of these by oo x , the second by co y , and the third by w z and add, we have (compare Statics Non-concurring Forces) Vx co* + V !/ ooy+Vz(a z = (13) We should obtain the same result for any other two components intersecting and a third passing through a point on their resultant. Equation (13) then gives the condition that all the angular ve- locities acting upon the system reduce to a single angular velocity at a point whose co-ordinates are x, y and z, and we have rotation only (page 179). We have evidently for the equations of the projection of the line of the resultant on the co-ordinate planes Cdy Vz Gdx Vy GOz I^X y = x , x = z , z = y ~ GOX <&x ooz (*>z G&y G^y Parallel Velocities. (Compare Statics Non-concurring Forces.) If the axes of all the angular velocities are parallel, we have <*, ft, y constant and the same for all. Hence from (3) and (1) cos d = cos = oo r Wy=K)r COS e = COS K)z = G0 r COS / = COS (14) The resultant ao f must have the common direction of the paral- lel components, or d = a, e ft, f = r, and .v - ?/'2z - z'2w] - cos ) [2wj; - *'2w] ; V . (16> V K = COB 02wte - x 1 ) - cos a2co(2/ - y') = cos (3[2wa: - a-'2z of the angular velocity along the axes are given. The motion of the system at any instant is then known when these six quantities, V x , V y , V z , GO*, eo y , ooz are given. These six quantities are called the components of motion of the system. Equivalent Screw. (Compare Statics Non-concurring Forces.) The motion of a rigid system being thus known, it is required to find the screw motion to which it is equivalent. That is, to find the central axis, the linear velocity along the central axis, and the angular velocity about it. Since V x , Vy, V z , <*>x , w y , oo z are .given, we have : (1) The angular velocity about the central axis cor = Vx* + * ........ (1) (2) The direction cosines of the central axis cos/ = ^ ..... (2) (3) The linear velocity of every point resolved in a direction parallel to the central axis must be the same and equal to that along the central axis. Let u r be the resultant linear velocity of every point of the system along the central axis, and let its compo- nents along the co-ordinates axes be Ux, u v , u z . Take the point for which V x , V y , Vz are given, as the origin, and let the co-ordinates of any point of the central axis be x", y", z". Then the components v x , tty, v z of the velocity of the origin due to rotation about the central axis are, from equations (1), page 179, V y =co x z" G> Z X"; \ ....... (3) v z = ooyx" <*> x y We have then Vx=U x +Vx, Vy = U,, + Vy, V z = U z + Vz , or Ux = Vx V X , Uy=VyVy, Uz = V z - V z . . 202 RIGID SYSTEM TRANSLATION AND ROTATION. [CHA?. II. Hence U r (V x Vx) COS d + (Vy Vy} COS 6 + (V z V z ) COS/. (5) Inserting the values of the direction-cosines of the central axis from (2), we obtain Ur C0 r = (Vx Vx)x + Vy oo y + Vz <*>z ....... (6) We also have from (4) tt,- COS d = Ux = Vx V x , Ur COS 6 = Vy Vy , U r COS / = Vz V z . (7) Hence from (2) and (3) Ur_ _ Vx + <*>yZ" oozy' _ Vy + K> z x" K> X Z" _ V z + <*> x y" ' <30 r ~ K> x G> y K) z Equations (8) give the equation of the central axis. From ( 6 1 and (1) we have Wr_ _ Vx<*>x + Vy&y + Vz<*>z oo r ~ ca x * + co* + coz* This we have called the unit pitch (page 191), or the distance of advance during a rotation of one radian about the central axis. If we substitute (9) in (8) and reduce, we have for the equation of the central axis 1 / Vz*>y VyQQz\ _ 1 / _ Vx<*>z VzOQx \ X - y Therefore the central axis passes through a point whose co- ordinates are * _ Vy&z n _ Vx<*>z VzGQx ^, _ Vyx Vx^y n * , ' If we substitute these values of x", y", z", in (3) and (7), we have from (2) V x = u>- cos d a>r(z" cos e y" cos/), r cos d; 1 V y = it,- cos e oo,{x" cos / z" cos d), oo y = oo r cos e; j- . (12) Vz = Ur COS / G>r(y" COS d X" COS 6), &> Z = r COS/, j When, therefore, the components of motion V x , V y , V z , a> x , z is the component in the direction of Xof the normal linear acceleration of the origin due to rotation about the central axis. The normal linear acceleration is pao^. Hence g y z - is the projection of p on the axis of X. CHAP. II.] COMPOSITION AND RESOLUTION OF SCREWS. 203 On the other hand, if the position of the central axis 'jc" v" z") is known together with the linear velocity u r along it and the 'an- gular velocity tor round it, the components of the motion for the origin are given by (12). The Invariant. (Compare Statics Non-concurring Forces ) From (6) we see that the quantity VzK>z is always equal to u r y , oo z , that is, what- ever the direction of the axes. This quantity is therefore 'called the invariant of the components. Since oo r is also invariable what- ever point is taken and whatever the direction of the axes, it may be called the invariant of the rotation. If the motion is such that the invariant is zero, it follows that either UT = or oo r = 0. The condition VxGOx + VyGOy + V Z GO Z = is therefore the condition that the motion is equivalent to either a simple translation or a simple rotation. If &>,- is not zero and this condition is fulfilled, we have rotation only (see pages 179, 200). Composition and Resolution of Screws. (Compare Statics Non-concurring Forces.) If two screws are given, then by equa- tion (12) we can find the six components of motion of each screw. Adding these two and two, we have the six components of the resultant screw. Then by (1), (2), (6) and (11) the central axis together with the linear and angular velocities of the screw may be found. Conversely, we may resolve any given screw motion into two screws in an infinite number of ways. Since a screw motion is represented by six components at any point, we have in the two screws twelve quantities at our disposal. Six of these are required to make the two screws equivalent to the given screw. We may therefore in general satisfy six other conditions at pleasure. Thus we may choose the axis of one screw to be any given straight line we please with any given linear velocity along it and any angu- lar velocity round it. The other screw may then be found by re- versing this assumed screw and joining it thus changed to the given motion. The screw equivalent to this compound motion is the second screw, and it may be found in the manner just explained. Or again, we may represent the motion by two screws whose unit pitches are both zero, the axis of one being arbitrary. We can thus represent any motion by two angular velocities, one, GO, about an axis which we may choose at pleasure, and the other, &/, about some axis which does not in general cut the first axis. These are called conjugate axes. These angular velocities are such that &?,- would be their resultant if their axes were placed parallel to their actual posi- tions, so as to intersect the central axis. If then d is the shortest distance between the axes, we have V r = da>; and if rf> is the angle between GO and r , and the angle between GO and &/, we have Ur "/$( V r sim(> = Ur, or BW.I(> = ==-. Vr Also oo sin 9 oo r sin ^ = K> sin 9, or sm^>= 204 RIGID SYSTEM TRANSLATION AND ROTATION. [CHAP. II_ Hence (1) V r <*>' sin = Ur co r , or dooao' sin 6 = hence Ur ODr d = 0000' sinfl' EXAMPLES. (1) A line DE moves, keeping its extremities in two fixed lines ADB, AEC. Find the instantaneous centre and the direction of motion of any point G at any instant. Ans. From D and E draw DF and EF perpendicular to AB and A C, meet- ing at F. Then F is the instantaneous centre (page 189). Join OF. The di- rection of motion of G is perpendicular to GF. (2) A line DE moves, keeping its extremities in two fixed lines, one, ADB, vertical and the other, AEC, horizontal, and makes at a given instant an angle of 30 with the horizontal. Find (a) the di- rection of motion of the middle point of DE at the instant, and (b) the point whose motion is inclined at that instant 30 to AC. Ans. (a) Inclined 60 to AC ; (b) -DE from E. (3) A line moves so that its extremities remain in a given circle. Find the instantaneous centre of motion for any instant. Ans. The centre of the circle. (4) Find the ratio of the velocity of any point of a screw to its velocity of advance. Ans. where d is the pitch, r the radius of the screw (page 191). (5) Let Ci and Ct be fixed axes about which turn the cranks CAi, CiB, tvhose free ends are connected by the link AB, jointed at A and B. The axes are perpendicular and the plane of motion parallel to- the paper. If the linear and angular velocities of A are Vi, 001, find the linear and angular velocities v* , cat of B. Ans. Let ACi = r t and BG* = r 9 . Produce dA and C*B to meet in I. Then at any instant the linear velocities of A and B are perpendicular to A d c, and BCi respectively. Hence at that instant AB is rotating about the instan- taneous axis at /. Let GO be the angular velocity of AB about /. From Ci , <7 a , /let fall the perpendiculars C\D, C?F, IE on the line of the link AB or CHAP. II.] EXAMPLES. 205 its prolongation. Also draw the line of centres C\ C, cutting the link, pro- longed, if necessary, in the point K. Then a = Bl . &, = BI _ IE GO T? Ctf Since TiGOi = Vi and r,ao, = v, , we have BI GO, CtD C,K V, KL GO, Ui-U - = ^rr ; als . = 77-^. = Hence 1. The linear velocities of B and A are to each other as their distances from the instantaneous axis. 2. The angular velocities of the cranks are to each other inversely as the perpendiculars from their centres of motion upon the line of the link ; or in- versely as the segments into which the line of the link cuts the line of centres. (6) In the case of the crank and connecting rod, since B moves in a straight line CiB, we have BI always perpen- ,,1 dicular to GiB, and hence v, BI =Al' or AT 1 Let the distance CiB = 8, the length of the connecting-rod = I, and the angle of the crank r t with CiB = 0i. Then / we have A /= tan 0,, AI = COS 0i s = r, cos 0, + 4/f - ri 2 sin 2 0,; or, if is very long compared to r\ , approximately r, 2 sin 2 0, Hence = S = Ti COS 0j stan 0i . riGO 21 s sin 0i 8 Ti COS 0i C080! When 0i = 90, we have v t =r 1 co 1 =v 1 , or the velocity of A and B are equal, and BI and Al&re infinite. When 0, = or 180, we have v, = and g = i _j_ 7>1 or i _ ri . These are the " dead points " of the crank, or the eiids of the stroke. (7) A rod (length = I) hangs by a small ring at its upper end from a fixed horizontal rod. To the former an angular velocity ao is given in a vertical plane through the fixed rod, so that the centre of the movable rod moves vertically. Find the linear velocity of its centre when its inclination to the vertical is B. Ans. caZsinS. (8) A disk (radius = r) rolls ivithout sliding on a plane. Find the relation between its angular velocity eo and the linear velocity v of its centre. Ans. The point of contact with the plane is at rest at any instant, 01 TOO = V. 206 RIGID SYSTEM TRANSLATION AND ROTATION. [CHAP. II. (9) A rod AB (length = I) rotates about a hinge at A and rests with its end B on the surface of a wedge BCD. The wedge advances towards A with velocity v. The angles BAG = 9, BCD = 0. Find the angular velocity a> of the rod. F5 xW _ _v sin ~ I cos (0 0)' (10) Two bevel-gear wheels have the angle between their axes 70. The rolling wheel is required to make 3i revolutions about its axis while going around the axis of the fixed ivheel once. Find the angles of the bevel. If the inner radius of the fixed ivheel is 50 inches, find that of the rolling wheel and the length of the axes. Ans. (See page 196.) The angular-velocity ratio - = - . Hence 7 sin 70 and hence : = 13 45'. We have also r\ = =- 100 - -f 50 cos 70 inches, and A a = - : ^ - =33.4 inches, sm70 100 50 + cos 70 hi - ^-=KB - = 58.4 inches. sin 70 (11) The angle between the plane of the earth's equator and the plane of the ecliptic is 23 27' 28". The earth rotates about its polar axis in one sidereal day and makes a revolution about the axis per- pendicular to the plane of the ecliptic in 25868 years. Find the instantaneous axis. Ans. (Page 196.) We have 2it oo i = 2rt radians per day, and co 2 = -- radians per day. ~5obo X Also the angle 0(70, = ! + ,, = 23 27' 28". Therefore OJ is r sin (o-i + a 2 ) Oil '= r tan cti = + cos (a, + a 2 ) GJo where r is the polar radius of the earth, or 3950 miles. The radius of the roll- ing cone is then Oil 5.52 ft. and the angle aj = 0". 00867. (13) A rigid system has an angular velocity of 001 = 40 radians per sec. about an axis parallel to the axis of X, passing through a point whose co-ordinates are x\ = 2 ft., y\ = 3 ft., z\ = 0. and 2 = 30 radians per sec. about au axis coinciding with the axis of Y. Find the resultant angular velocity and the instantaneous axis. Ans. (Page 198.) We have co x -\- 40, ca y = + 30, co z = 0, x + Uyooy + uzoo z = is sat- isfied. Therefore there is rotation only about the instantaneous axis which passes through the intersection I of cox and coy. The velocity of any point whose co-ordinates are x' = 2, y = 2, z' = 3 is given by Vx = ooyz' = + 90 ft. per sec. ; Vy = <*>xz' 120 ft. per sec. ; Yz = Uz <*>yx' + ooxy" = 100 ft. per sec. z The resultant velocity of this point is then Vr = 180 ft. per sec. and its direction cosines are _ 90 7 1 _ 120 100 COS d ^Trt7\9 COS . rtrt* COS C ^^ . - - I 180 180 180 or a = 60, b =131 48', c = 123 45'. (13) A rigid system has the angular velocities a?! = 50, ao* = 30, o0 3 = 70, co t = 90, and GO* = 120 radians per sec. about axes passing through points of the rigid system given by != + 5ft., #1 =+10ft. ; a? a = + 9ft., y, = + 12 ft.; x 3 = + 17 ft., y 9 = + 14 ft. ; = + 20 ft., y = + 13 ft.; x& - _|_ 15 ft., y b - + 8ft.; and making with the co-ordinate axes the angles a, = 70, /?i = 20; a, = 60, fa = 150; a 3 = 120, /?, = 0; 4 = 150, 04 = 120; 6 = 90, 5 = 0. Find the resultant, etc. (Compare Vol. II, Statics.) Ans. (Page 198.) We have for the components of the angular velocities parallel to the axes a>x = 50 cos 70 + 30 cos 60 - 70 cos 60 - 90 cos 30 = - 80.842 rad. per sec. ; coy = 50 cos 20 - 30 cos 30 + 120 + 70 cos 30 - 90 cos 60 =+ 156.626 rad. [per sec.; GOz = 0. The resultant angular velocity is given in magnitude by oo r = i/oj-g' + ttjj/ 2 = + 176.259 radians per sec., and its direction-cosines by GOx ^ oU. o4^ , . . mQ j o/ j /. d = -^= 176^59-' r * = 117181 = +156-626 or e = 37 18 > r . car 176.259 ' We have from equation (4), page 198, 2 cox cos ft = + 50 cos 20 X 5 - 30 cos 30 X 9 + 70 cos 80 X 17 - 90 cos 60 X 20 + 120 X 15 = + 1931.67 ft. per sec.; 208 EIGID SYSTEM TRANSLATION AND ROTATION. [CHAP. II. We have then for the components of the linear velocity of the origin x = 0, Vy = 0, Vz 2oox cos ft 2aoy cos a -(-3083.915 ft. per sec. Since then equation (13), page 200, VxOOx + VyGOy + VzOOz = is satisfied, the angular velocities reduce to a single resultant angular velocity and we have rotation only. The moment of this resultant angular velocity relative to the origin gives us v r = i/vx* + V + flz* = v z = + 3083.915 ft. per sec. Its lever-arm is r 3083.915 r= = -. n - A = 17.5 ft. cor 176. 2o9 The equation of the line of direction of the resultant angular velocity is o^ _ ^_ = _ 05 38. 14 GO X GOx The co-ordinates of the point through which this resultant angular velocity passes are given from equations (17), page 201 : (14) .Find ffee resultant angular velocity for a number of parallel angular velocities given by co, = + 33 rad. per sec.; *, = + 25 ft.; y l = + 13 ft.; 00,, = + 20 " " " x, = 10 " y t = 15 " a> 3 = - 35 " " " s = + 15 " y 3 = - 27 " oo 4 = _ 72 " " " x< = - 31 " y 4 = + 17 " co 5 = + 120 " " " x, = + 23 " y 6 = - 19 " Ans. oor = + 66 radians per sec.; x + 77.15 ft.; y = 36.82 ft. (15) Find the resultant, etc., for the angular velocities given by GDI 50, co a = 70, oo 3 = 90, oo 4 = 120 radians per sec. j = 60; fi t 40; ^, acute; a;, =0; y l = 0; z^ = 0; - G0 r <*>r or d = 62 9' 48", e = 30 39* 20", / = 101 49'. CHAP. H.] EXAMPLES. 209 We also Lave for the components of the velocity of the origin, from equa- tions (4), page 198, ex = - 1838.604, vy = + 928.947, = - 86.903 ft. per sec. The resultant linear velocity of the origin is then v r = VW* + V + z* = + 2061.789 ft. per sec., and its direction-cosines are given by Vx Vy Vz cos a = , cos o = , cos c = ; Vr Vr Vr or a = 153 5' 40", b = 63 14' 15", c = 92 24' 56". The equations of the projections of the resultant angular velocity on the co-ordinate planes are y = 1.885x + 0.746, a? = - 2.28a + 18.19, z=- 0.23% - 8.57. We see that Vxc&x + Vy&y + Vz&z does not, in this case, equal zero. Hence {page 179) the angular velocities do not reduce to a single resultant, but to a resultant angular velocity about the central axis and translation along that axis due to an angular- velocity couple. The resultant angular velocity about the central axis is, as already found, co r = -f- 249.325 rad. per sec., and its angles d, e, /with the axes are already found. The co-ordinates of the central axis are given by equations (11), page 202 : of' = -^ = + 0.463 ft; y" = " = + 1.678 ft.; = GOxVy - flOyte = , 8>og ft COr* The resultant linear velocity -WT along the central axis is given by equation (6), page 202: = Its direction-cosines are the same as for oo r . The components of Ur are given by equations (7), page 202 : ux = Ur cos d 19.481, uy = u r cos e = 35.806, u z u r cos/ = + 8.5238 ft. per sec. (16) In the preceding example find what the co-ordinates Xi,y*, z* of the angular velocity y , <*>z , 4 has no effect on the magnitude or direction of the resultant oa r . We have then vx=- 659.571 - 93.2622/4 - 68.8292 4 ; \ v y = + 369.629 + 31.0592* + 93.262*4; > ..... (1) Vz= - 107.036 + 68.829*4 - 31.05% 4 . ) We have as the equation of condition for a single resultant Vx<*>x + VyGOy -\- Vz<*>Z = 0, 116.4230* + 214.48% - 51.057tfe = 0. ........ (2) 210 RIGID SYSTEM TRANSLATION AND ROTATION. [CHAP. II., From (1) we obtain (Ox + 659.571) 31.059 + (v y - 369.629) 68.829 = (v z + 107.036) 93.262, or Dx +2.216% -3. 003% = +481.034 (3) From (2) and (3) we obtain 0.3740J, - 2.564%; = + 481.034. If we retain for vy its value in the preceding example, + 928.947 ft. per sec., we shall have v z = 52.108, v x = 1733.975 ft. per sec. If we substitute these values in (1), we obtain 93.262#4 + 68.829z 4 = + 1074.4; 31.0592* + 93.262*4 = + 559.308; 68.829*4 - 31.059^4 = + 54.934. Hence *4 = - 0.33324 + 5.997, y t = - 0.738z 4 + 11.520. If then we assume z 4 = 0, we have * 4 = + 5.997, y t = + 11.52 ft. (17) Using the values of the preceding example, find the point through which the resultant angular velocity passes. (Compare Vol. II, Statics.) Ans. We have aa x = + 116.423, coy = + 214.480, ca z = - 51.057, co r = + 249.325 radians per sec. ; d = 62 9' 48", e = 30 39' 20'', /= 101 49'; v x = - 1733.975, % = + 928.947, v z = - 52.108, v r = + 1967.823 ft. per sec.; a = 151 47', b = 61 49' 53", c = 91 31' 3". The co-ordinates *, y, z are given (page 200) by - 1733.975 = cozy - ooyz = - 51.057y - 214.480?; 4- 928.947 = (Oxz - oozx + 116.4233 + 51.057*; - 52.108 = ooyx - oo x y = + 214.480* - 116. 423^. Hence we obtain "= - 2.2802^ + 18.194; y = -4.2008a + 33.961. If we assume z = 0, we have x = + 18.194, y = + 33.961 ft. If we should introduce then a fifth angular velocity, o> 5 = + 249.325, whose direction makes with the axes the angles 5 = 117 50' 12", ft t = 149 20' 40", y b = 78 11', passing through a point whose co-ordinates are * 6 = + 18.194, y 6 = + 33.961 and 2 5 = 0, the conditions for rest (page 199) would be satisfied, and we should have aor = 0, v r = 0. (18) A point of a rigid system rotates about an axis at a distance of 5 feet. The linear displacement of the point is 8ft. Find the angular displacement ana the direction of the linear displacement. Ans. sin-- = = ^. = angular displacement = 106 14' = 1.858 ra- 2 2r 5 dians. The linear displacement makes an angle of 36 53' with the radius of rotation. CHAP. II.] EXAMPLES. 211 (19) A point of a rigid system has at any instant the component linear velocities V x = + 6, V y = - 18, V z = + 40 ft. per sec., and the system at the same instant rotates about an axis perpendicular to the plarte of XY with an angular velocity of 6 radians per sec. in the direction from X towards Y. Find the equivalent screw motion of the system. Ans. (Page 201.) We take the given point as the origin. Since the axis is perpendicular to the plane of XT, we have oo x =. 0, ca y = 0, oo z = ao r = + 6 radians per sec. Since the condition Vxx-\- V y z = 0; y x - 4, 30 ft. per sec., Vy = 0, Vz = + 40 ft. per sec, If the velocities Vx and Vz of the point P, do not change in direction or magnitude, we have the case of a system translated in the direction 03 and rotating about an axis through P, , while at the same time this axis has a ve- locity of translation in a straight line (page 193). 212 EIGID SYSTEM TRANSLATION AND ROTATION. [CHAP. II. The motion of the system would then be the same as if a cylindrical surface of radius P 3 I= 5 ft., fixed to the system with its axis passing through P 2 at right angles to the plane of XT, rolled on the plane HIH parallel to the plane JP"with the angular velocity co z + 6 radians per sec., while at- the same time the cylinder is translated parallel to OZ with the velocity Vz = -\- 40 ft. per sec. (20) A base-ball rotates about an axis through its centre in a horizontal plane with an angular velocity oo z = 60 radians per sec., and its centre has a horizontal velocity of translation of V 50 ft. per sec. in a direction making an angle of 36 52' with the axis of rotation. Find the motion of the ball. Ans. Let the plane of ZX be horizontal and take the centre as origin. Then, since V is in this plane, we have Fa; =+30 ft. per sec., Vy = 0, Vz = -\- 40 ft. per sec. Also, oo x = 0, ooy = 0, G>Z = 60 rad. per sec. The rotation is then clockwise, or from Y towards X, as shown in the figure. Then, just as in the preceding example, the central axis is parallel to the axis of Z, and the position of the central axis is, from equation (11), page 202, given by x" = 0, y' = 01= - 1 ft., z" = 0. 8 If then we neglect the acceleration due to the attraction of the earth, the motion of the ball is a screw motion consisting of a velocity u z + 40 ft. per sec. along the axis of rotation OZ through the centre of the bull, and a rota- tion of ooz = 60 radians per sec. about this axis, together with a translation of this axis of Vx + 40 ft. per sec. Or, neglecting the acceleration due to gravity, the motion is the same (page 193) as if the ball were part of a cylinder of radius 01= -^ ft. whose axis OZ a is the axis of rotation of the ball, and this cylinder rolls on the horizontal plane HIH with angular velocity co z = 60 radians per sec., while at the same time the cylinder is translated along OZ with the velocity -j- 40 ft. per sec. The centre of the ball moves then in the resultant of Vx and Vz , or along the straight line V in the horizontal plane XZ, with a velocity F = 50 ft. per sec. , at an angle of 36 52' with the axis of rotation OZ. If now, owing to gravity, the ball falls vertically while the centre moves along 0V, then we must consider the plane HIH as falling vertically with the ball. The centre moves then in a curve Odb, the projection of which upon the plane XZ is a straight line Oc. (21) Ball-players assert that the projection of this curve Oab (preceding example) upon the plane XZ is not a straight line but a curve. Explain hoiv this can be. -z Ans. We have seen in the preceding example that if the centre of the ball has a velocity V and at the same time the ball has an angular velocity caz CHAP. II.] RELATIVE MOTION OF A BODY. 213 around an axis OZ, the centre moves with the velocity V z along the axis and at the same time the axis itself moves with the velocity V x , V z and V x being the components of V along and perpendicular to the axis. At the same time the ball rotates about the axis OZ. The motion of is then in the straight line 0V. But no account has been taken of the resistance of the air. The air acts to cause a retardation of Vz and V x . If the retardation in each case were proportional to the velocity, we should still have motion of in the straight line V. But the retardation in each case is not proportional to the velocities but more nearly proportional to the squares of the velocities. Hence the greater component is retarded propor- tionally more than the less. If then the rotation axis OZ makes an angle less than 45 with the direc- tion of V, Vz is greater than Vx and is therefore retarded proportionally more than Vx. The centre moves then in an " out-curve " OB. If, however, the axis of rotation OZ makes an angle greater than 45 with the direction of V, Vx is greater than V z and is therefore retarded proportionally more than Vz. The centre moves then in an "in-curve." In either case the velocity is retarded least in the direction of least resistance and the centre swerves in the direction of the smallest component of V. Thus by "twisting " the ball the pitcher is able to make it curve slightly by either to right or to left according as the axis of rotation makes an angle with the velocity of projection greater or less than 45. If the axis of rotation makes an angle of 45 with the velocity of projection, there should be no curve. If it is at right angles to the velocity of projection, there should be no curve. The cause of curvature is thus due to the resistance of the air, but it is not, as is generally supposed, due to the ball rolling upon a cushion of compressed air in front of it, since in that case we should always have curvature in one direction for one direction of rotation. In the first of our figures preceding, such action tends to increase the "out- curve." But in the second it tends to decrease the "in-curve." The "in-curve" would not be possible if this action were the only cause of curvature. It ought to be less than the out-curve, so far as this action is effective, in the figures given. If we have rotation in the opposite direction from that in the figure, or if the line representative of t. disk The corresponding CHAP. II.] ACCELERATION OF RELATIVE MOTION. relative position P of the particle is then the point which N would occupy if the line AB were turned -counter-clockwise through the angle cot. Repeating the construction for successive values of t we obtain the relative path( APCD. The end D corresponds to the\ rotation angle aoT, where T is the time of the actual motion from A to B. If GO T = 7t, the point D would coincide with A. Let r be the radius of the disk. Then from the two equations cT = 2r, and K)T=n, we have for the condition of this coincidence of D and A, 00 It c~ = 2r' If the actual path AB makes an acute angle with the axis of rotation through C, the relative path lies on the surface of a cone. Acceleration of Relative Motion. Let a particle describe the path MN with any motion, and at the same time let this path have a motion of translation. Then we can regard the first motion as relative with reference to the second, and its acceleration fi is the relative accelera- tion. Besides this relative acceleration at any instant, the particle has the acceleration /a of the motion of trans- lation at that instant. The actual acceleration of the particle is then the resultant of the two accelerations /i and / a . It is, however, different when the path MN has any motion in general, because such motion may be resolved into a motion of translation of any point of the path and a motion of rotation about an axis through that point. Take for this point the point of space occupied by the particle at any instant. Then we have besides the accelera- tion /i of the particle in its path, and the acceleration / 2 of the point of space occupied by the particle, a third acceleration, /a, due to the rotation, which we can determine as Let v be the relative velocity of the particle. Then in an in- definitely small time dt, vdt will be the element MN of the relative path This element in the time dt is translated to PQ and at the same time has the angular velocity a? about the point of space oc- UP Let the axis OP through this point niake the angle 6 with PQ. Then at the end of the time dt, Q will be at R. If .A is the accelera- tion in the direction QR, then the distance QR will be 216 KIGID SYSTEM TRANSLATION AND EOTATION. [CHAP. II. The radius of rotation is OQ = vdt sin 6. Hence the distance QR is also given by QR = OQ . a>dt vdt sin B . is the angular velocity of the earth, the acceleration / 2 of the point of space occupied by the particle is/ 2 = rcosA . < 2 . We have also fa 2voo sin A acting at right angles to the plane of the element of the relative path and the axis through P parallel to the earth's axis, that is, tangent to the latitude circle at P. It acts towards the east. The actual acceleration is the resultant of these three accelerations. CHAPTEE III. GENERAL ANALYTICAL RELATIONS FOR A POINT OF A RIGID ROTATING SYSTEM. EULER'S GEOMETRIC EQUATIONS. General Analytical Relations for a Point of a Rigid Rotating System. Let a rigid system rotate at any instant about the axis 1C with the angular velocity GO and the angular acceleration a. Take any point O of the system as origin, and let the direction- cosines of co be cos a, cos ft, cos y. Then we have for the components of DO and a 03 X = 03 COS a, G0y= CO COS /?, GOz = GO COS Y \ a x = a COS a, a y = a COS /?, a z = a COS y ; GOx COS a = = oo ax a COB /,= .=:&., cosr = ^ = -; oo a ao a (1) and since cos 2 a + cos 2 + cos 2 y = 1, Let (x v z) be the co-ordinates of any point on the axis 1C, and equations (6) : v'x = aoyZ co z y 'i 1 V'y = tozX' - ooxZ ; I ........ (4) f'z = oo x y' GOyOC. J We have thus the total components of the velocity P, just as on. page 198, equation (7): Vx = v x + v'x ; I Vy = Vy + V'y J J- ......... (5) Vz = V Z + V'z. The components of the linear tangential acceleration of P due to rotation about the parallel axis through O, we see from (4), are given by ftx = a y z' a z y' ; fty a-zX' otxZ ; f ..... ... (6) ftz = axy 1 a y x . J Since poo = v and pW = v 2 = v/ + *V + Vz*, we have from (3) , , y - x . 2 00 s *>' Let/'n be the normal linear acceleration of the point P due to rotation about the parallel axis through the origin O. Then f' n = v'co ; and since velocity in the hodograph is the normal acceleration in the path (page 52), we have directly from the figure, for the com- ponents of/ n, fnx = f'afij,/ V'i,eo z ; fny = v'iK>z V'^oox ; fnz = V'y0) x V'xoo^. We have then for the components of the acceleration /' of the point P, from (8) and (6), fx =fnx +ftx = (V'zaoy - v' y co z ) + (a y z' a z y')\ "j f y =fny +fty = (v'xGDzv'z&x) + (cCgX a x z') ; I (9) / * =f'nz + ftz = (V'ytox ~ v'xOOy) + (o. x tf - y y' + oozZ')ao x v?x! + (a y z' a z y) ; ~\ fy (GOxrf + OOyy' + OOzZ')GOy K?y' + (ctzX' OixZ') } I (10) fg = (osxX 1 + ooyy' + Go z z)a>z ccfz + (a x y a y z"). j Equations (10) give the values of the components of the linear acceleration / of any point P of the system whose co-ordinates are (X, y', z'), in terms of these co-ordinates and the components of a> and a. The moments of the component linear accelerations with refer- ence to the origin O are fx Vy* + *\ f y Vx'* + z", fz Vx'* + y\ . . . (11) For the moments about the axes of the components of t/ and /' we have: about X parallel to plane FZ, M x = v' z y' Y " " " ZX, M y = vf x z'- Z " " " XT, Mz^v'ytf-v' The resultant moment in both cases is given by Mr = VM X * + My 11 + Mz* ....... (13) Its line representative has the direction-cosines Mx_ My_ Mz_ (U} Mr ' Mr ' Mr" Looking along the line representative towards the origin the rota- tion is counter-clockwise. [We can deduce equations (9) directly by the Calculus. Thus if we dif- ferentiate the values of v'x, v' y , v'z given by (4), then, since ='- f ='> ='" and doo z we have at once _ at f y = Euler's Geometrical Equations. To determine the geometrical equations between the motion of a rigid system in space and the angular velocity of the system about an axis in the system. 222 EULER'S GEOMETRICAL EQUATIONS. [CHAP. in. Let OX, OFi, OZi be rectangular co-ordinate axes, fixed in the system and therefore rotating with it, and let the system rotate about some axis fixed in the system, and therefore making in- variable angles with these axes, so that the component angular velocities in the co-ordinate planes are K> XI , <&y l , a> Z] . We take direction of rotation as al- ways about X l from F to Z, ] " F " Z, "X [posite di- ll 7 < v K v \ rection ^1 -A I - 1 . J negative. Let now OX, OF, OZbe rect- angular co-ordinate axes whose directions in space are invariable. For instance, the axis OZ may be always directed towards the North Pole, then XY is the plane of the celestial equator. Let the point O be taken as the centre of a sphere of radius r. Let X, F, Z and Xi , Yi , Z\ be the points in which this sphere is pierced by the fixed and moving axes. Let the axes OXi , OF , OZi have the initial positions OX, OF, OZ. First turn the system about OZ as an axis through the angle XZP = #, so that OX moves to OP, and OF to OD. Then turn the system about OD as an axis through the angle ZOZi 6, so that OP moves to OE, and OZ to OZi. Finally turn the system about OZi as an axis through the angle EZXi = 0, so that OE moves to OXi , and OD to OF.. It is required to find the geometric relations between 6, 0, ty and oo x , , GO a*, GO Z} as the system rotates. These geometric relations are called Euler's Geometric Equations. Let the angular velocity of Z\ perpendicular to the plane ZOZ\ at any instant be denoted by --. This is called the angular velocity Cl/t/ of precession. Let the angular velocity of Zi along ZZi at the same dfi instant be denoted by - - . This is called the angular velocity of nu- CvV tation. Let the angular velocity of Xi with reference to E at that instant be denoted by -~r. Gut Draw ZiN perpendicular to OZ. Then Z\N r sin 0, and the linear velocity at any instant of Zi perpendicular to the plane ZOZ^ is r sin . -, and along ZZ\ at the same instant it is r- . The linear dt at velocity at the same instant of Z\ along YiZi is rao Xl , and along it is roo,,,. We have then directly from the figure r = rK) yi cos + roo Xl sin ; i> r sin 6 . - = rcoy, sin rco Xj cos 0. at CHAP. III.] KINEMATICS OF A RIGID SYSTEM. Since the radius r cancels out, dQ fit = w vi cos + d$ sm dT ~ l>l sm ^ ~~ Combining these two equations, cos 225 d> , sm 6 cos ty, = -VT cos + -=r- sin 6 sin (2) In the same way by drawing a perpendicular from E to OZOE we have the linear velocity of E perpendicular to ZOE equal to r cos 6 * and of X relative to # along * uc> The whole velocity of Xi in space along Xi Yi is ro,. Hence _ d*f> , d Equations (1), (2) and (3) are Euler's Geometric Equations. EXAMPLES. (3), (1) Deduce the angular velocities co*, coy, &> z about the fixed axis, in terms of 6, 0, if>. Ans. Let oo r be tlie resultant angular velocity about the fixed azes. If we impress on space and also on the system, in addition to its existing motion, an angular velocity equal to cor about the resultant axis of rotation, the axes OXi, OYi, OZi will become fixed and OX, OT, OZ will move with angular velocities cox , ooj/ , ooz. Hence in the equations already found we have only to replace

by Xl , otyj , K>Z I will become <&x , cay , ooz , and we have ; at 'if Go y = cos if? + -JT sm 6 sin ^ ; at ut __ -. (4) (2) Refer the axes fixed in space to the axes fixed in the system. Ans We have simply to interchange in the figure Xi, Yi, Z* with X, Y, Z each with each. If then the angles 0, , if> are still measured as indicated in the figure, the relations connecting them with the angular velocities are ob- tained by changing co x ,, ooy,, coz^ into GO X , ooy , ooz. If we measure 6 in the direction opposite to that indicated m the figure, th& expressions for sin -|- sin ^ cos cos 0; V ... (6) cos ZXi = sin cos 0. J cos XYi = sin if> cos cos tf> sin cos 6; 1 cos FFi = cos T)) cos sin ^ sin cos 0; V . . . (7) cos ZY\ = sin sin 0. cos JTZj = sin cos ij>\ j cos YZi = sin sin $;>.... (8) cos ZZi = cos 0. J Ans. We have from the figure tLe following spherical triangles for which we know two sides and the included angle : Triangle. Sides. Angle. Triangle. Sides. Angle. = 9+^ ^#=0 =90-0 -1^1=0 DXYi DX =90 =90-0 YDX i= 6 = 6 1^,= 180 -0^27, Fi2,=90 F,^Z=90-0 QA 7 *7 A Triangle. Sides. Angle. PZi =90 "& = 90 P^, =90-0 ^ p PF = 90 - ^ ^^^ = 9< Solving these triangles we have at once equations (6), (7), (8). (5) Prove in the same, way the following : cos XiX = sin if> sin -)- cos rf> cos cos 0; } cos Fi-3T= sin tf> cos cos if) sin cos 0; V . . . (9) cos ZiX = sin cos tb. J cos .Xi F = cos if) sin + sin ^ cos cos ; } cos Fj F = cos if> cos sin ^ sin cos 9; V . . . (10) cos Zi Y = sin sin ib. \ CHAP. III.] KINEMATICS OF A KIGID SYSTEM ROTATION. 225 cos XiZ = sin cos 0; cos TiZ = sin sin 0; > (11) cos ZiZ = cos 6. j (6) Find the relations between the co-ordinates x, y, z, of the fixed system of axes and the co-ordinates Xi , y\ , z\ of the moving system. Ans. If we multiply the first of equations (6) by*, the second by y, the third by z and add, and do the same for (7) and (8), we have at once, as we see from the figure, x\ = ( sin if} sin -j- cos if> cos cos 6)x -J- (cos ip sin -\- sin $ cos cos Q)y sin cos 0.2; y l = ( sin i/} cos cos i[> sin cos 0)# f . . (12) -f- (cos if) cos sin if} sin cos 0)y + sm 9 8m 0-s g! = sin cos ip.x + sin sin ip.y -\-cos 9 . z. In the same way we have from equations (9), (10), (11), x = ( sin if> sin -f- cos ip cos cos Q)XI -\- ( sin ^ cos cos ^ sin cos 0)^i -j- sin 9 cos i y = (cos V sin + sin V cos cos 0)xi } . (18) -|- (cos if) cos sin V sin cos 0)y t -j- sin sin ip.Zi; g = sin 9 cos Hayford's Text-book of Geodetic Astronomy Svo. * Michie and Harlow's Practical Astronomy 8vo, * White's Theoretical and Descriptive Astronomy 12mo, 2 00 3 BOTANY. GARDENING FOR LADIKS, ETC. Baldwin's Orchids of New England Small 8vo, $1 50 Thome's Structural Botany 16mo, 2 25 Westermaier's General Botany. 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