SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

Transcription

1 CHAPTER SEVEN SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 7.1 Introducton 7. Centre of mass 7.3 Moton of centre of mass 7.4 Lnear momentum of a system of partcles 7.5 Vector product of two vectors 7.6 Angular velocty and ts relaton wth lnear velocty 7.7 Torque and angular momentum 7.8 Equlbrum of a rgd body 7.9 Moment of nerta 7.10 Theorems of perpendcular and parallel axes 7.11 Knematcs of rotatonal moton about a fxed axs 7.1 Dynamcs of rotatonal moton about a fxed axs 7.13 Angular momentum n case of rotaton about a fxed axs 7.14 Rollng moton Summary Ponts to Ponder Exercses Addtonal exercses 7.1 INTRODUCTION In the earler chapters we prmarly consdered the moton of a sngle partcle. (A partcle s deally represented as a pont mass havng no sze.) We appled the results of our study even to the moton of bodes of fnte sze, assumng that moton of such bodes can be descrbed n terms of the moton of a partcle. Any real body whch we encounter n daly lfe has a fnte sze. In dealng wth the moton of extended bodes (bodes of fnte sze) often the dealsed model of a partcle s nadequate. In ths chapter we shall try to go beyond ths nadequacy. We shall attempt to buld an understandng of the moton of extended bodes. An extended body, n the frst place, s a system of partcles. We shall begn wth the consderaton of moton of the system as a whole. The centre of mass of a system of partcles wll be a key concept here. We shall dscuss the moton of the centre of mass of a system of partcles and usefulness of ths concept n understandng the moton of extended bodes. A large class of problems wth extended bodes can be solved by consderng them to be rgd bodes. Ideally a rgd body s a body wth a perfectly defnte and unchangng shape. The dstances between all pars of partcles of such a body do not change. It s evdent from ths defnton of a rgd body that no real body s truly rgd, snce real bodes deform under the nfluence of forces. But n many stuatons the deformatons are neglgble. In a number of stuatons nvolvng bodes such as wheels, tops, steel beams, molecules and planets on the other hand, we can gnore that they warp (twst out of shape), bend or vbrate and treat them as rgd What knd of moton can a rgd body have? Let us try to explore ths queston by takng some examples of the moton of rgd bodes. Let us begn wth a rectangular

2 14 PHYSICS Fg 7.1 Translatonal (sldng) moton of a block down an nclned plane. (Any pont lke P 1 or P of the block moves wth the same velocty at any nstant of tme.) block sldng down an nclned plane wthout any sdewse movement. The block s taken as a rgd body. Its moton down the plane s such that all the partcles of the body are movng together,.e. they have the same velocty at any nstant of tme. The rgd body here s n pure translatonal moton (Fg. 7.1). In pure translatonal moton at any nstant of tme, all partcles of the body have the same velocty. Consder now the rollng moton of a sold metallc or wooden cylnder down the same nclned plane (Fg. 7.). The rgd body n ths problem, namely the cylnder, shfts from the top to the bottom of the nclned plane, and thus, seems to have translatonal moton. But as Fg. 7. shows, all ts partcles are not movng wth the same velocty at any nstant. The body, therefore, s not n pure translatonal moton. Its moton s translatonal plus somethng else. most common way to constran a rgd body so that t does not have translatonal moton s to fx t along a straght lne. The only possble moton of such a rgd body s rotaton. The lne or fxed axs about whch the body s rotatng s ts axs of rotaton. If you look around, you wll come across many examples of rotaton about an axs, a celng fan, a potter s wheel, a gant wheel n a far, a merry-go-round and so on (Fg 7.3(a) and (b)). (a) Fg. 7. Rollng moton of a cylnder. It s not pure translatonal moton. Ponts P 1, P, P 3 and P 4 have dfferent veloctes (shown by arrows) at any nstant of tme. In fact, the velocty of the pont of contact P 3 s zero at any nstant, f the cylnder rolls wthout slppng. In order to understand what ths somethng else s, let us take a rgd body so constraned that t cannot have translatonal moton. The (b) Fg. 7.3 Rotaton about a fxed axs (a) A celng fan (b) A potter s wheel. Let us try to understand what rotaton s, what characterses rotaton. You may notce that n rotaton of a rgd body about a fxed axs,

3 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 143 Fg. 7.5 (a) A spnnng top (The pont of contact of the top wth the ground, ts tp O, s fxed.) Axs of oscllaton Fg. 7.4 A rgd body rotaton about the z-axs (Each pont of the body such as P 1 or P descrbes a crcle wth ts centre (C 1 or C ) on the axs of rotaton. The radus of the crcle (r 1 or r ) s the perpendcular dstance of the pont (P 1 or P ) from the axs. A pont on the axs lke P 3 remans statonary). Axs of rotaton from blades every partcle of the body moves n a crcle, whch les n a plane perpendcular to the axs and has ts centre on the axs. Fg. 7.4 shows the rotatonal moton of a rgd body about a fxed axs (the z-axs of the frame of reference). Let P 1 be a partcle of the rgd body, arbtrarly chosen and at a dstance r 1 from fxed axs. The partcle P 1 descrbes a crcle of radus r 1 wth ts centre C 1 on the fxed axs. The crcle les n a plane perpendcular to the axs. The fgure also shows another partcle P of the rgd body, P s at a dstance r from the fxed axs. The partcle P moves n a crcle of radus r and wth centre C on the axs. Ths crcle, too, les n a plane perpendcular to the axs. Note that the crcles descrbed by P 1 and P may le n dfferent planes; both these planes, however, are perpendcular to the fxed axs. For any partcle on the axs lke P 3, r = 0. Any such partcle remans statonary whle the body rotates. Ths s expected snce the axs of rotaton s fxed. Fg. 7.5 (b) An oscllatng table fan wth rotatng blades. The pvot of the fan, pont O, s fxed. The blades of the fan are under rotatonal moton, whereas, the axs of rotaton of the fan blades s oscllatng. In some examples of rotaton, however, the axs may not be fxed. A promnent example of ths knd of rotaton s a top spnnng n place [Fg. 7.5(a)]. (We assume that the top does not slp from place to place and so does not have translatonal moton.) We know from experence that the axs of such a spnnng top moves around the vertcal through ts pont of contact wth the ground, sweepng out a cone as shown n Fg. 7.5(a). (Ths movement of the axs of the top around the vertcal s termed precesson.) Note, the pont of contact of the top wth ground s fxed. The axs of rotaton of the top at any nstant passes through the pont of contact. Another smple example of ths knd of rotaton s the oscllatng table fan or a pedestal fan [Fg.7.5(b)]. You may have observed that the

4 144 PHYSICS axs of rotaton of such a fan has an oscllatng (sdewse) movement n a horzontal plane about the vertcal through the pont at whch the axs s pvoted (pont O n Fg. 7.5(b)). Whle the fan rotates and ts axs moves sdewse, ths pont s fxed. Thus, n more general cases of rotaton, such as the rotaton of a top or a pedestal fan, one pont and not one lne, of the rgd body s fxed. In ths case the axs s not fxed, though t always passes through the fxed pont. In our study, however, we mostly deal wth the smpler and specal case of rotaton n whch one lne (.e. the axs) s fxed. Fg. 7.6(a) Moton of a rgd body whch s pure translaton. Fg. 7.6(b) Moton of a rgd body whch s a combnaton of translaton and rotaton. Fg 7.6 (a) and 7.6 (b) llustrate dfferent motons of the same body. Note P s an arbtrary pont of the body; O s the centre of mass of the body, whch s defned n the next secton. Suffce to say here that the trajectores of O are the translatonal trajectores Tr 1 and Tr of the body. The postons O and P at three dfferent nstants of tme are shown by O 1, O, and O 3, and P 1, P and P 3, respectvely, n both Fgs. 7.6 (a) and (b). As seen from Fg. 7.6(a), at any nstant the veloctes of any partcles lke O and P of the body are the same n pure translaton. Notce, n ths case the orentaton of OP,.e. the angle OP makes wth a fxed drecton, say the horzontal, remans the same,.e. α 1 = α = α 3. Fg. 7.6 (b) llustrates a case of combnaton of translaton and rotaton. In ths case, at any nstants the veloctes of O and P dffer. Also, α 1, α and α 3 may all be dfferent. Thus, for us rotaton wll be about a fxed axs only unless stated otherwse. The rollng moton of a cylnder down an nclned plane s a combnaton of rotaton about a fxed axs and translaton. Thus, the somethng else n the case of rollng moton whch we referred to earler s rotatonal moton. You wll fnd Fg. 7.6(a) and (b) nstructve from ths pont of vew. Both these fgures show moton of the same body along dentcal translatonal trajectory. In one case, Fg. 7.6(a), the moton s a pure translaton; n the other case [Fg. 7.6(b)] t s a combnaton of translaton and rotaton. (You may try to reproduce the two types of moton shown, usng a rgd object lke a heavy book.) We now recaptulate the most mportant observatons of the present secton: The moton of a rgd body whch s not pvoted or fxed n some way s ether a pure translaton or a combnaton of translaton and rotaton. The moton of a rgd body whch s pvoted or fxed n some way s rotaton. The rotaton may be about an axs that s fxed (e.g. a celng fan) or movng (e.g. an oscllatng table fan [Fg.7.5(b)]). We shall, n the present chapter, consder rotatonal moton about a fxed axs only. 7. CENTRE OF MASS We shall frst see what the centre of mass of a system of partcles s and then dscuss ts sgnfcance. For smplcty we shall start wth a two partcle system. We shall take the lne jonng the two partcles to be the x- axs. Fg. 7.7 Let the dstances of the two partcles be x 1 and x respectvely from some orgn O. Let m 1 and m be respectvely the masses of the two

5 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 145 partcles. The centre of mass of the system s that pont C whch s at a dstance X from O, where X s gven by m x X = m + m x + m (7.1) In Eq. (7.1), X can be regarded as the massweghted mean of x 1 and x. If the two partcles have the same mass m 1 = m = m, then mx1 + mx x1 + x X = = m Thus, for two partcles of equal mass the centre of mass les exactly mdway between them. If we have n partcles of masses m 1, m,...m n respectvely, along a straght lne taken as the x- axs, then by defnton the poston of the centre of the mass of the system of partcles s gven by. mx m1x1+ mx mnxn = 1 X = = = n m1+ m mn m n = 1 m x m (7.) where x 1, x,...x n are the dstances of the partcles from the orgn; X s also measured from the same orgn. The symbol (the Greek letter sgma) denotes summaton, n ths case over n partcles. The sum m = M s the total mass of the system. Suppose that we have three partcles, not lyng n a straght lne. We may defne x and y axes n the plane n whch the partcles le and represent the postons of the three partcles by coordnates (x 1,y 1 ), (x,y ) and (x 3,y 3 ) respectvely. Let the masses of the three partcles be m 1, m and m 3 respectvely. The centre of mass C of the system of the three partcles s defned and located by the coordnates (X, Y) gven by m x + m x + m x X = m + m + m m y + m y + m y Y = m + m + m (7.3a) (7.3b) For the partcles of equal mass m = m 1 = m = m 3, m( x + x + x ) x + x + x X = = 3m m( y1 + y + y3) y1 + y + y3 Y = = 3m 3 Thus, for three partcles of equal mass, the centre of mass concdes wth the centrod of the trangle formed by the partcles. Results of Eqs. (7.3a) and (7.3b) are generalsed easly to a system of n partcles, not necessarly lyng n a plane, but dstrbuted n space. The centre of mass of such a system s at (X, Y, Z ), where mx X = (7.4a) M my Y = (7.4b) M mz and Z = (7.4c) M Here M = m s the total mass of the system. The ndex runs from 1 to n; m s the mass of the th partcle and the poston of the th partcle s gven by (x, y, z ). Eqs. (7.4a), (7.4b) and (7.4c) can be combned nto one equaton usng the notaton of poston vectors. Let r be the poston vector of the th partcle and R be the poston vector of the centre of mass: r = x + y j + z k and R = X + Y j + Zk m r Then R = (7.4d) M The sum on the rght hand sde s a vector sum. Note the economy of expressons we acheve by use of vectors. If the orgn of the frame of reference (the coordnate system) s chosen to be the centre of mass then m r = 0 for the gven system of partcles. A rgd body, such as a metre stck or a flywheel, s a system of closely packed partcles; Eqs. (7.4a), (7.4b), (7.4c) and (7.4d) are therefore, applcable to a rgd body. The number of partcles (atoms or molecules) n such a body s so large that t s mpossble to carry out the summatons over ndvdual partcles n these equatons. Snce the spacng of the partcles s

6 146 PHYSICS small, we can treat the body as a contnuous dstrbuton of mass. We subdvde the body nto n small elements of mass; m 1, m... m n ; the th element m s taken to be located about the pont (x, y, z ). The coordnates of the centre of mass are then approxmately gven by ( m ) x ( m ) y ( m ) z X =, Y =, Z = m m m As we make n bgger and bgger and each m smaller and smaller, these expressons become exact. In that case, we denote the sums over by ntegrals. Thus, m d m = M, ( m ) x x dm, ( m ) y y dm, and ( m ) z z dm Here M s the total mass of the body. The coordnates of the centre of mass now are X = x dm, Y = y dm and Z = z dm (7.5a) M M M The vector expresson equvalent to these three scalar expressons s R 1 = M r d m (7.5b) If we choose, the centre of mass as the orgn of our coordnate system, R = 0.e., r dm = 0 or x dm = y dm = z dm = 0 (7.6) Often we have to calculate the centre of mass of homogeneous bodes of regular shapes lke rngs, dscs, spheres, rods etc. (By a homogeneous body we mean a body wth unformly dstrbuted mass.) By usng symmetry consderaton, we can easly show that the centres of mass of these bodes le at ther geometrc centres. Let us consder a thn rod, whose wh and breath (n case the cross secton of the rod s rectangular) or radus (n case the cross secton of the rod s cylndrcal) s much smaller than ts length. Takng the orgn to be at the geometrc centre of the rod and x-axs to be along the length of the rod, we can say that on account of reflecton symmetry, for every element dm of the rod at x, there s an element of the same mass dm located at x (Fg. 7.8). The net contrbuton of every such par to the ntegral and hence the ntegral tself s zero. From Eq. (7.6), the pont for whch the ntegral tself s zero, s the centre of mass. Thus, the centre of mass of a homogenous thn rod concdes wth ts geometrc centre. Ths can be understood on the bass of reflecton symmetry. The same symmetry argument wll apply to homogeneous rngs, dscs, spheres, or even thck rods of crcular or rectangular cross secton. For all such bodes you wll realse that for every element dm at a pont (x, y, z) one can always take an element of the same mass at the pont ( x, y, z). (In other words, the orgn s a pont of reflecton symmetry for these bodes.) As a result, the ntegrals n Eq. (7.5 a) all are zero. Ths means that for all the above bodes, ther centre of mass concdes wth ther geometrc centre. Example 7.1 Fnd the centre of mass of three partcles at the vertces of an equlateral trangle. The masses of the partcles are 100g, 150g, and 00g respectvely. Each sde of the equlateral trangle s 0.5m long. Answer Fg. 7.8 Determnng the CM of a thn rod. Fg. 7.9

7 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 147 Wth the x and y axes chosen as shown n Fg. 7.9, the coordnates of ponts O, A and B formng the equlateral trangle are respectvely (0,0), (0.5,0), (0.5,0.5 3 ). Let the masses 100 g, 150g and 00g be located at O, A and B be respectvely. Then, m x + m x + m x X = m + m + m = Y ( ) (0.5) + 00(0.5) g m ( ) g = = m = m = m (0) + 150(0) + 00(0.5 3) g m 450 g = m = m = m The centre of mass C s shown n the fgure. Note that t s not the geometrc centre of the trangle OAB. Why? concurrence of the medans,.e. on the centrod G of the trangle. Example 7.3 Fnd the centre of mass of a unform L-shaped lamna (a thn flat plate) wth dmensons as shown. The mass of the lamna s 3 kg. Answer Choosng the X and Y axes as shown n Fg we have the coordnates of the vertces of the L-shaped lamna as gven n the fgure. We can thnk of the L-shape to consst of 3 squares each of length 1m. The mass of each square s 1kg, snce the lamna s unform. The centres of mass C 1, C and C 3 of the squares are, by symmetry, ther geometrc centres and have coordnates (1/,1/), (3/,1/), (1/,3/) respectvely. We take the masses of the squares to be concentrated at these ponts. The centre of mass of the whole L shape (X, Y) s the centre of mass of these mass ponts. Example 7. Fnd the centre of mass of a trangular lamna. Answer The lamna ( LMN) may be subdvded nto narrow strps each parallel to the base (MN) as shown n Fg Fg Fg Hence X = [ + + ] ( ) kg 1(1/) 1(3/) 1(1/) kg m 5 = m 6 By symmetry each strp has ts centre of mass at ts mdpont. If we jon the mdpont of all the strps we get the medan LP. The centre of mass of the trangle as a whole therefore, has to le on the medan LP. Smlarly, we can argue that t les on the medan MQ and NR. Ths means the centre of mass les on the pont of [ + + ] ( + + ) 1(1/) 1(1/) 1(3/) kg m 5 Y = = m kg 6 The centre of mass of the L-shape les on the lne OD. We could have guessed ths wthout calculatons. Can you tell why? Suppose, the three squares that make up the L shaped lamna

8 148 PHYSICS of Fg had dfferent masses. How wll you then determne the centre of mass of the lamna? 7.3 MOTION OF CENTRE OF MASS Equpped wth the defnton of the centre of mass, we are now n a poston to dscuss ts physcal mportance for a system of n partcles. We may rewrte Eq.(7.4d) as MR = m r = m r + m r + + m r (7.7) n n Dfferentatng the two sdes of the equaton wth respect to tme we get dr dr1 dr drn M = m1 + m m n or M V = m v + m v + + m v (7.8) where ( = d / ) n n v r s the velocty of the frst 1 1 partcle ( = d ) v r s the velocty of the second partcle etc. and V = d R / s the velocty of the centre of mass. Note that we assumed the masses m 1, m,... etc. do not change n tme. We have therefore, treated them as constants n dfferentatng the equatons wth respect to tme. Dfferentatng Eq.(7.8) wth respect to tme, we obtan dv dv1 dv dv M = m1 + m m n or MA = m a + m a + + m a (7.9) where ( = d / ) n n a v s the acceleraton of the 1 1 frst partcle, ( = d /) a v s the acceleraton of the second partcle etc. and ( = d / ) n A V s the acceleraton of the centre of mass of the system of partcles. Now, from Newton s second law, the force actng on the frst partcle s gven by F1 = m1a 1. The force actng on the second partcle s gven F as = m by a and so on. Eq. (7.9) may be wrtten M A = F1 + F F n (7.10) Thus, the total mass of a system of partcles tmes the acceleraton of ts centre of mass s the vector sum of all the forces actng on the system of partcles. Note when we talk of the force F1 on the frst partcle, t s not a sngle force, but the vector sum of all the forces on the frst partcle; lkewse for the second partcle etc. Among these forces on each partcle there wll be external forces exerted by bodes outsde the system and also nternal forces exerted by the partcles on one another. We know from Newton s thrd law that these nternal forces occur n equal and opposte pars and n the sum of forces of Eq. (7.10), ther contrbuton s zero. Only the external forces contrbute to the equaton. We can then rewrte Eq. (7.10) as M A = F (7.11) ext where F ext represents the sum of all external forces actng on the partcles of the system. Eq. (7.11) states that the centre of mass of a system of partcles moves as f all the mass of the system was concentrated at the centre of mass and all the external forces were appled at that pont. Notce, to determne the moton of the centre of mass no knowledge of nternal forces of the system of partcles s requred; for ths purpose we need to know only the external forces. To obtan Eq. (7.11) we dd not need to specfy the nature of the system of partcles. The system may be a collecton of partcles n whch there may be all knds of nternal motons, or t may be a rgd body whch has ether pure translatonal moton or a combnaton of translatonal and rotatonal moton. Whatever s the system and the moton of ts ndvdual partcles, the centre of mass moves accordng to Eq. (7.11). Instead of treatng extended bodes as sngle partcles as we have done n earler chapters, we can now treat them as systems of partcles. We can obtan the translatonal component of ther moton,.e. the moton of the centre of mass of the system, by takng the mass of the whole system to be concentrated at the centre of mass and all the external forces on the system to be actng at the centre of mass. Ths s the procedure that we followed earler n analysng forces on bodes and solvng

9 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 149 problems wthout explctly outlnng and justfyng the procedure. We now realse that n earler studes we assumed, wthout sayng so, that rotatonal moton and/or nternal moton of the partcles were ether absent or neglgble. We no longer need to do ths. We have not only found the justfcaton of the procedure we followed earler; but we also have found how to descrbe and separate the translatonal moton of (1) a rgd body whch may be rotatng as well, or () a system of partcles wth all knds of nternal moton. Fg. 7.1 The centre of mass of the fragments of the projectle contnues along the same parabolc path whch t would have followed f there were no exploson. Fgure 7.1 s a good llustraton of Eq. (7.11). A projectle, followng the usual parabolc trajectory, explodes nto fragments mdway n ar. The forces leadng to the exploson are nternal forces. They contrbute nothng to the moton of the centre of mass. The total external force, namely, the force of gravty actng on the body, s the same before and after the exploson. The centre of mass under the nfluence of the external force contnues, therefore, along the same parabolc trajectory as t would have followed f there were no exploson. 7.4 LINEAR MOMENTUM OF A SYSTEM OF PARTICLES Let us recall that the lnear momentum of a partcle s defned as p = m v (7.1) Let us also recall that Newton s second law wrtten n symbolc form for a sngle partcle s d F = p (7.13) where F s the force on the partcle. Let us consder a system of n partcles wth masses m 1, m,...m n respectvely and veloctes v1, v,... v n respectvely. The partcles may be nteractng and have external forces actng on them. The lnear momentum of the frst partcle s m1v 1, of the second partcle s mv and so on. For the system of n partcles, the lnear momentum of the system s defned to be the vector sum of all ndvdual partcles of the system, P = p1 + p pn = m1v1 + mv mnv n (7.14) Comparng ths wth Eq. (7.8) P = M V (7.15) Thus, the total momentum of a system of partcles s equal to the product of the total mass of the system and the velocty of ts centre of mass. Dfferentatng Eq. (7.15) wth respect to tme, dp dv = M = MA (7.16) Comparng Eq.(7.16) and Eq. (7.11), dp ext = F (7.17) Ths s the statement of Newton s second law of moton extended to a system of partcles. Suppose now, that the sum of external forces actng on a system of partcles s zero. Then from Eq.(7.17) dp or = 0 P = Constant (7.18a) Thus, when the total external force actng on a system of partcles s zero, the total lnear momentum of the system s constant. Ths s the law of conservaton of the total lnear momentum of a system of partcles. Because of Eq. (7.15), ths also means that when the total external force on the system s zero the velocty of the centre of mass remans constant. (We assume throughout the dscusson on systems of partcles n ths chapter that the total mass of the system remans constant.) Note that on account of the nternal forces,.e. the forces exerted by the partcles on one another, the ndvdual partcles may have

10 150 PHYSICS complcated trajectores. Yet, f the total external force actng on the system s zero, the centre of mass moves wth a constant velocty,.e., moves unformly n a straght lne lke a free partcle. The vector Eq. (7.18a) s equvalent to three scalar equatons, P x = c 1, P y = c and P z = c 3 (7.18 b) Here P x, P y and P z are the components of the total lnear momentum vector P along the x, y and z axes respectvely; c 1, c and c 3 are constants. (a) (b) Fg (a) A heavy nucleus radum (Ra) splts nto a lghter nucleus radon (Rn) and an alpha partcle (nucleus of helum atom). The CM of the system s n unform moton. (b) The same spltng of the heavy nucleus radum (Ra) wth the centre of mass at rest. The two product partcles fly back to back. As an example, let us consder the radoactve decay of a movng unstable partcle, lke the nucleus of radum. A radum nucleus dsntegrates nto a nucleus of radon and an alpha partcle. The forces leadng to the decay are nternal to the system and the external forces on the system are neglgble. So the total lnear momentum of the system s the same before and after decay. The two partcles produced n the decay, the radon nucleus and the alpha partcle, move n dfferent drectons n such a way that ther centre of mass moves along the same path along whch the orgnal decayng radum nucleus was movng [Fg. 7.13(a)]. If we observe the decay from the frame of reference n whch the centre of mass s at rest, the moton of the partcles nvolved n the decay looks partcularly smple; the product partcles (a) Fg (a) Trajectores of two stars, S 1 (dotted lne) and S (sold lne) formng a bnary system wth ther centre of mass C n unform moton. (b) The same bnary system, wth the centre of mass C at rest. move back to back wth ther centre of mass remanng at rest as shown n Fg.7.13 (b). In many problems on the system of partcles, as n the above radoactve decay problem, t s convenent to work n the centre of mass frame rather than n the laboratory frame of reference. In astronomy, bnary (double) stars s a common occurrence. If there are no external forces, the centre of mass of a double star moves lke a free partcle, as shown n Fg.7.14 (a). The trajectores of the two stars of equal mass are also shown n the fgure; they look complcated. If we go to the centre of mass frame, then we fnd that there the two stars are movng n a crcle, about the centre of mass, whch s at rest. Note that the poston of the stars have to be dametrcally opposte to each other [Fg. 7.14(b)]. Thus n our frame of reference, the trajectores of the stars are a combnaton of () unform moton n a straght lne of the centre of mass and () crcular orbts of the stars about the centre of mass. As can be seen from the two examples, separatng the moton of dfferent parts of a system nto moton of the centre of mass and moton about the centre of mass s a very useful technque that helps n understandng the moton of the system. 7.5 VECTOR PRODUCT OF TWO VECTORS We are already famlar wth vectors and ther use n physcs. In chapter 6 (Work, Energy, Power) we defned the scalar product of two vectors. An mportant physcal quantty, work, s defned as a scalar product of two vector quanttes, force and dsplacement. (b)

11 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 151 We shall now defne another product of two vectors. Ths product s a vector. Two mportant quanttes n the study of rotatonal moton, namely, moment of a force and angular momentum, are defned as vector products. Defnton of Vector Product A vector product of two vectors a and b s a vector c such that () magntude of c = c = ab snθ where a and b are magntudes of a and b and θ s the angle between the two vectors. () c s perpendcular to the plane contanng a and b. () f we take a rght handed screw wth ts head lyng n the plane of a and b and the screw perpendcular to ths plane, and f we turn the head n the drecton from a to b, then the tp of the screw advances n the drecton of c. Ths rght handed screw rule s llustrated n Fg. 7.15a. Alternately, f one curls up the fngers of rght hand around a lne perpendcular to the plane of the vectors a and b and f the fngers are curled up n the drecton from a to b, then the stretched thumb ponts n the drecton of c, as shown n Fg. 7.15b. (a) (b) Fg (a) Rule of the rght handed screw for defnng the drecton of the vector product of two vectors. (b) Rule of the rght hand for defnng the drecton of the vector product. A smpler verson of the rght hand rule s the followng : Open up your rght hand palm and curl the fngers pontng from a to b. Your stretched thumb ponts n the drecton of c. It should be remembered that there are two angles between any two vectors a and b. In Fg (a) or (b) they correspond to θ (as shown) and (360 0 θ). Whle applyng ether of the above rules, the rotaton should be taken through the smaller angle (<180 0 ) between a and b. It s θ here. Because of the cross ( ) used to denote the vector product, t s also referred to as cross product. Note that scalar product of two vectors s commutatve as sad earler, a.b = b.a The vector product, however, s not commutatve,.e. a b b a The magntude of both a b and b a s the same ( ab sn θ ); also, both of them are perpendcular to the plane of a and b. But the rotaton of the rght-handed screw n case of a b s from a to b, whereas n case of b a t s from b to a. Ths means the two vectors are n opposte drectons. We have a b = b a Another nterestng property of a vector product s ts behavour under reflecton. Under reflecton (.e. on takng the plane mrror mage) we have x x, y y and z z. As a result all the components of a vector change sgn and thus a a, b b. What happens to a b under reflecton? a b ( a) ( b) = a b Thus, a b does not change sgn under reflecton. Both scalar and vector products are dstrbutve wth respect to vector addton. Thus, a.( b + c) = a. b + a. c a ( b + c) = a b + a c We may wrte c = a b n the component form. For ths we frst need to obtan some elementary cross products: () a a = 0 (0 s a null vector,.e. a vector wth zero magntude) Ths follows snce magntude of a a s a sn0 = 0.

12 15 PHYSICS From ths follow the results () ˆ ˆ = 0, ˆj ˆj = 0, kˆ kˆ = 0 () ˆ ˆj = kˆ Note that the magntude of ˆ ˆj s sn90 0 or 1, snce î and ĵ both have unt magntude and the angle between them s Thus, ˆ ˆj s a unt vector. A unt vector perpendcular to the plane of î and ĵ and related to them by the rght hand screw rule s ˆk. Hence, the above result. You may verfy smlarly, ˆj kˆ = ˆ and kˆ ˆ = ˆj From the rule for commutaton of the cross product, t follows: ˆj ˆ = kˆ, kˆ ˆj = ˆ, ˆ kˆ = ˆj Note f ˆ, ˆj, kˆ occur cyclcally n the above vector product relaton, the vector product s postve. If ˆ, ˆj, k ˆ do not occur n cyclc order, the vector product s negatve. Now, a b = ( a ˆ + a ˆj + a kˆ ) ( b ˆ + b ˆj + b kˆ ) x y z x y z = a b kˆ a b ˆj a b kˆ + a b ˆ + a b ˆj a b ˆ x y x z y x y z z x z y = ( a b a b ) + ( a b a b ) j + ( a b a b ) k y z z y z x x z x y y x We have used the elementary cross products n obtanng the above relaton. The expresson for a b can be put n a determnant form whch s easy to remember. ˆ ˆj kˆ a b = = 7ˆ ˆj 5kˆ 1 3 Note b a = 7ˆ + ˆj + 5k ˆ 7.6 ANGULAR VELOCITY AND ITS RELATION WITH LINEAR VELOCITY In ths secton we shall study what s angular velocty and ts role n rotatonal moton. We have seen that every partcle of a rotatng body moves n a crcle. The lnear velocty of the partcle s related to the angular velocty. The relaton between these two quanttes nvolves a vector product whch we learnt about n the last secton. Let us go back to Fg As sad above, n rotatonal moton of a rgd body about a fxed axs, every partcle of the body moves n a crcle, a b = ˆ ˆj kˆ a a a x y z b b b x y z Example 7.4 Fnd the scalar and vector products of two vectors. a = (3î 4ĵ + 5kˆ ) and b = ( î + ĵ 3kˆ ) Answer ab = (3ˆ 4ˆj + 5 kˆ ) ( ˆ + ˆj 3 kˆ ) = = 5 Fg Rotaton about a fxed axs. (A partcle (P) of the rgd body rotatng about the fxed (z-) axs moves n a crcle wth centre (C) on the axs.) whch les n a plane perpendcular to the axs and has ts centre on the axs. In Fg we redraw Fg. 7.4, showng a typcal partcle (at a pont P) of the rgd body rotatng about a fxed axs (taken as the z-axs). The partcle descrbes

13 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 153 a crcle wth a centre C on the axs. The radus of the crcle s r, the perpendcular dstance of the pont P from the axs. We also show the lnear velocty vector v of the partcle at P. It s along the tangent at P to the crcle. Let P be the poston of the partcle after an nterval of tme t (Fg. 7.16). The angle PCP descrbes the angular dsplacement θ of the partcle n tme t. The average angular velocty of the partcle over the nterval t s θ/ t. As t tends to zero (.e. takes smaller and smaller values), the rato θ/ t approaches a lmt whch s the nstantaneous angular velocty dθ/ of the partcle at the poston P. We denote the nstantaneous angular velocty by ω (the Greek letter omega). We know from our study of crcular moton that the magntude of lnear velocty v of a partcle movng n a crcle s related to the angular velocty of the partcle ω by the smple relaton υ = ω r, where r s the radus of the crcle. We observe that at any gven nstant the relaton v = ω r apples to all partcles of the rgd body. Thus for a partcle at a perpendcular dstance r from the fxed axs, the lnear velocty at a gven nstant v s gven by v = ωr (7.19) The ndex runs from 1 to n, where n s the total number of partcles of the body. For partcles on the axs, r = 0, and hence v = ω r = 0. Thus, partcles on the axs are statonary. Ths verfes that the axs s fxed. Note that we use the same angular velocty ω for all the partcles. We therefore, refer to ω as the angular velocty of the whole body. We have charactersed pure translaton of a body by all parts of the body havng the same velocty at any nstant of tme. Smlarly, we may characterse pure rotaton by all parts of the body havng the same angular velocty at any nstant of tme. Note that ths charactersaton of the rotaton of a rgd body about a fxed axs s just another way of sayng as n Sec. 7.1 that each partcle of the body moves n a crcle, whch les n a plane perpendcular to the axs and has the centre on the axs. In our dscusson so far the angular velocty appears to be a scalar. In fact, t s a vector. We shall not justfy ths fact, but we shall accept t. For rotaton about a fxed axs, the angular velocty vector les along the axs of rotaton, and ponts out n the drecton n whch a rght handed screw would advance, f the head of the screw s rotated wth the body. (See Fg. 7.17a). The magntude of ths vector s ω = dθ referred as above. Fg (a) If the head of a rght handed screw rotates wth the body, the screw advances n the drecton of the angular velocty ω. If the sense (clockwse or antclockwse) of rotaton of the body changes, so does the drecton of ω. Fg (b) The angular velocty vector ω s drected along the fxed axs as shown. The lnear velocty of the partcle at P s v = ω r. It s perpendcular to both ω and r and s drected along the tangent to the crcle descrbed by the partcle. We shall now look at what the vector product ω r corresponds to. Refer to Fg. 7.17(b) whch s a part of Fg reproduced to show the path of the partcle P. The fgure shows the vector ω drected along the fxed (z ) axs and also the poston vector r = OP of the partcle at P of the rgd body wth respect to the orgn O. Note that the orgn s chosen to be on the axs of rotaton.

14 154 PHYSICS Now But Hence ω r = ω OP = ω (OC + CP) ω OC = 0 as ω s along OC ω r = ω CP The vector ω CP s perpendcular to ω,.e. to the z-axs and also to CP, the radus of the crcle descrbed by the partcle at P. It s therefore, along the tangent to the crcle at P. Also, the magntude of ω CP s ω (CP) snce ω and CP are perpendcular to each other. We shall denote CP by r and not by r, as we dd earler. Thus, ω r s a vector of magntude ωr and s along the tangent to the crcle descrbed by the partcle at P. The lnear velocty vector v at P has the same magntude and drecton. Thus, v = ω r (7.0) In fact, the relaton, Eq. (7.0), holds good even for rotaton of a rgd body wth one pont fxed, such as the rotaton of the top [Fg. 7.6(a)]. In ths case r represents the poston vector of the partcle wth respect to the fxed pont taken as the orgn. We note that for rotaton about a fxed axs, the drecton of the vector ω does not change wth tme. Its magntude may, however, change from nstant to nstant. For the more general rotaton, both the magntude and the drecton of ω may change from nstant to nstant Angular acceleraton You may have notced that we are developng the study of rotatonal moton along the lnes of the study of translatonal moton wth whch we are already famlar. Analogous to the knetc varables of lnear dsplacement (s) and velocty (v) n translatonal moton, we have angular dsplacement (θ) and angular velocty (ω) n rotatonal moton. It s then natural to defne n rotatonal moton the concept of angular acceleraton n analogy wth lnear acceleraton defned as the tme rate of change of velocty n translatonal moton. We defne angular acceleraton α as the tme rate of change of angular velocty; Thus, d α = ω (7.1) If the axs of rotaton s fxed, the drecton of ω and hence, that of α s fxed. In ths case the vector equaton reduces to a scalar equaton dω α = (7.) 7.7 TORQUE AND ANGULAR MOMENTUM In ths secton, we shall acquant ourselves wth two physcal quanttes (torque and angular momentum) whch are defned as vector products of two vectors. These as we shall see, are especally mportant n the dscusson of moton of systems of partcles, partcularly rgd bodes Moment of force (Torque) We have learnt that the moton of a rgd body, n general, s a combnaton of rotaton and translaton. If the body s fxed at a pont or along a lne, t has only rotatonal moton. We know that force s needed to change the translatonal state of a body,.e. to produce lnear acceleraton. We may then ask, what s the analogue of force n the case of rotatonal moton? To look nto the queston n a concrete stuaton let us take the example of openng or closng of a door. A door s a rgd body whch can rotate about a fxed vertcal axs passng through the hnges. What makes the door rotate? It s clear that unless a force s appled the door does not rotate. But any force does not do the job. A force appled to the hnge lne cannot produce any rotaton at all, whereas a force of gven magntude appled at rght angles to the door at ts outer edge s most effectve n producng rotaton. It s not the force alone, but how and where the force s appled s mportant n rotatonal moton. The rotatonal analogue of force n lnear moton s moment of force. It s also referred to as torque or couple. (We shall use the words moment of force and torque nterchangeably.) We shall frst defne the moment of force for the specal case of a sngle partcle. Later on we shall extend the concept to systems of partcles ncludng rgd bodes. We shall also relate t to a change n the state of rotatonal moton,.e. s angular acceleraton of a rgd body.

15 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 155 of the lne of acton of F from the orgn and F ( = F sn θ) s the component of F n the drecton perpendcular to r. Note that τ = 0 f r = 0, F = 0 or θ = 0 0 or Thus, the moment of a force vanshes f ether the magntude of the force s zero, or f the lne of acton of the force passes through the orgn. One may note that snce r F s a vector product, propertes of a vector product of two vectors apply to t. If the drecton of F s reversed, the drecton of the moment of force s reversed. If drectons of both r and F are reversed, the drecton of the moment of force remans the same. Fg τ = r F, τ s perpendcular to the plane contanng r and F, and ts drecton s gven by the rght handed screw rule. If a force acts on a sngle partcle at a pont P whose poston wth respect to the orgn O s gven by the poston vector r (Fg. 7.18), the moment of the force actng on the partcle wth respect to the orgn O s defned as the vector product τ = r F (7.3) The moment of force (or torque) s a vector quantty. The symbol τ stands for the Greek letter tau. The magntude of τ s τ = r F snθ (7.4a) where r s the magntude of the poston vector r,.e. the length OP, F s the magntude of force F and θ s the angle between r and F as shown. Moment of force has dmensons M L T -. Its dmensons are the same as those of work or energy. It s, however, a very dfferent physcal quantty than work. Moment of a force s a vector, whle work s a scalar. The SI unt of moment of force s newton metre (N m). The magntude of the moment of force may be wrtten τ = ( r sn θ) F = r F (7.4b) or τ = r F sn θ = rf (7.4c) where r = r snθ s the perpendcular dstance 7.7. Angular momentum of a partcle Just as the moment of a force s the rotatonal analogue of force n lnear moton, the quantty angular momentum s the rotatonal analogue of lnear momentum. We shall frst defne angular momentum for the specal case of a sngle partcle and look at ts usefulness n the context of sngle partcle moton. We shall then extend the defnton of angular momentum to systems of partcles ncludng rgd bodes. Lke moment of a force, angular momentum s also a vector product. It could also be referred to as moment of (lnear) momentum. From ths term one could guess how angular momentum s defned. Consder a partcle of mass m and lnear momentum p at a poston r relatve to the orgn O. The angular momentum l of the partcle wth respect to the orgn O s defned to be l = r p (7.5a) The magntude of the angular momentum vector s l = r p sn θ (7.6a) where p s the magntude of p and θ s the angle between r and p. We may wrte l = r p or r p (7.6b) where r (= r snθ) s the perpendcular dstance of the drectonal lne of p from the orgn and p ( = p sn θ) s the component of p n a drecton perpendcular to r. We expect the angular momentum to be zero (l = 0), f the lnear momentum vanshes (p = 0), f the partcle s at the orgn (r = 0), or f the drectonal lne of p passes through the orgn θ = 0 0 or

16 156 PHYSICS The physcal quanttes, moment of a force and angular momentum, have an mportant relaton between them. It s the rotatonal analogue of the relaton between force and lnear momentum. For dervng the relaton n the context of a sngle partcle, we dfferentate l = r p wth respect to tme, dl d ( ) = r p Applyng the product rule for dfferentaton to the rght hand sde, d dr dp ( r p) = p + r Now, the velocty of the partcle s v = dr/ and p = m v Because of ths d r m 0, p = v v = as the vector product of two parallel vectors vanshes. Further, snce dp / = F, dp r = r F = τ Hence d ( r p) = τ or d l = τ (7.7) Thus, the tme rate of change of the angular momentum of a partcle s equal to the torque actng on t. Ths s the rotatonal analogue of the equaton F = dp/, whch expresses Newton s second law for the translatonal moton of a sngle partcle. Torque and angular momentum for a system of partcles To get the total angular momentum of a system of partcles about a gven pont we need to add vectorally the angular momenta of ndvdual partcles. Thus, for a system of n partcles, L = l1 + l ln = l n = 1 The angular momentum of the th partcle s gven by l = r p where r s the poston vector of the th partcle wth respect to a gven orgn and p = (m v ) s the lnear momentum of the partcle. (The An experment wth the bcycle rm Take a bcycle rm and extend ts axle on both sdes. Te two strngs at both ends A and B, as shown n the adjonng fgure. Hold both the Intally After strngs together n one hand such that the rm s vertcal. If you leave one strng, the rm wll tlt. Now keepng the rm n vertcal poston wth both the strngs n one hand, put the wheel n fast rotaton around the axle wth the other hand. Then leave one strng, say B, from your hand, and observe what happens. The rm keeps rotatng n a vertcal plane and the plane of rotaton turns around the strng A whch you are holdng. We say that the axs of rotaton of the rm or equvalently ts angular momentum precesses about the strng A. The rotatng rm gves rse to an angular momentum. Determne the drecton of ths angular momentum. When you are holdng the rotatng rm wth strng A, a torque s generated. (We leave t to you to fnd out how the torque s generated and what ts drecton s.) The effect of the torque on the angular momentum s to make t precess around an axs perpendcular to both the angular momentum and the torque. Verfy all these statements. partcle has mass m and velocty v ) We may wrte the total angular momentum of a system of partcles as L = l = r p (7.5b) Ths s a generalsaton of the defnton of angular momentum (Eq. 7.5a) for a sngle partcle to a system of partcles. Usng Eqs. (7.3) and (7.5b), we get dl d dl l τ (7.8a) = ( ) = =

17 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 157 where τ s the torque actng on the th partcle; τ = r F The force F on the th partcle s the vector sum of external forces F ext actng on the partcle nt and the nternal forces F exerted on t by the other partcles of the system. We may therefore separate the contrbuton of the external and the nternal forces to the total torque τ = τ = r F as where τ = τext + τ, nt = r ext τ ext F nt and τ nt = r F We shall assume not only Newton s thrd law of moton,.e. the forces between any two partcles of the system are equal and opposte, but also that these forces are drected along the lne jonng the two partcles. In ths case the contrbuton of the nternal forces to the total torque on the system s zero, snce the torque resultng from each actonreacton par of forces s zero. We thus have, τ nt = 0 and therefore τ = τ ext. that Snce τ = τ, t follows from Eq. (7.8a) dl = τ ext (7.8 b) Thus, the tme rate of the total angular momentum of a system of partcles about a pont (taken as the orgn of our frame of reference) s equal to the sum of the external torques (.e. the torques due to external forces) actng on the system taken about the same pont. Eq. (7.8 b) s the generalsaton of the sngle partcle case of Eq. (7.3) to a system of partcles. Note that when we have only one partcle, there are no nternal forces or torques. Eq.(7.8 b) s the rotatonal analogue of dp = F ext (7.17) Note that lke Eq.(7.17), Eq.(7.8b) holds good for any system of partcles, whether t s a rgd body or ts ndvdual partcles have all knds of nternal moton. Conservaton of angular momentum If τ ext = 0, Eq. (7.8b) reduces to dl 0 = or L = constant. (7.9a) Thus, f the total external torque on a system of partcles s zero, then the total angular momentum of the system s conserved,.e. remans constant. Eq. (7.9a) s equvalent to three scalar equatons, L x = K 1, L y = K and L z = K 3 (7.9 b) Here K 1, K and K 3 are constants; L x, L y and L z are the components of the total angular momentum vector L along the x,y and z axes respectvely. The statement that the total angular momentum s conserved means that each of these three components s conserved. Eq. (7.9a) s the rotatonal analogue of Eq. (7.18a),.e. the conservaton law of the total lnear momentum for a system of partcles. Lke Eq. (7.18a), t has applcatons n many practcal stuatons. We shall look at a few of the nterestng applcatons later on n ths chapter. Example 7.5 Fnd the torque of a force 7î + 3ĵ 5kˆ about the orgn. The force acts on a partcle whose poston vector s î ĵ + kˆ. Answer Here r = ˆ ˆj + kˆ and F = 7ˆ + 3ˆj 5k ˆ. We shall use the determnant rule to fnd the torque τ = r F or τ = ˆ + 1ˆj + 10kˆ Example 7.6 Show that the angular momentum about any pont of a sngle partcle movng wth constant velocty remans constant throughout the moton.

18 158 PHYSICS Answer Let the partcle wth velocty v be at pont P at some nstant t. We want to calculate the angular momentum of the partcle about an arbtrary pont O. acceleraton nor angular acceleraton. Ths means (1) the total force,.e. the vector sum of the forces, on the rgd body s zero; n F1 + F Fn = F = 0 (7.30a) = 1 If the total force on the body s zero, then the total lnear momentum of the body does not change wth tme. Eq. (7.30a) gves the condton for the translatonal equlbrum of the body. () The total torque,.e. the vector sum of the torques on the rgd body s zero, Fg 7.19 The angular momentum s l = r mv. Its magntude s mvr snθ, where θ s the angle between r and v as shown n Fg Although the partcle changes poston wth tme, the lne of drecton of v remans the same and hence OM = r sn θ. s a constant. Further, the drecton of l s perpendcular to the plane of r and v. It s nto the page of the fgure.ths drecton does not change wth tme. Thus, l remans the same n magntude and drecton and s therefore conserved. Is there any external torque on the partcle? 7.8 EQUILIBRIUM OF A RIGID BODY We are now gong to concentrate on the moton of rgd bodes rather than on the moton of general systems of partcles. We shall recaptulate what effect the external forces have on a rgd body. (Henceforth we shall omt the adjectve external because unless stated otherwse, we shall deal wth only external forces and torques.) The forces change the translatonal state of the moton of the rgd body,.e. they change ts total lnear momentum n accordance wth Eq. (7.17). But ths s not the only effect the forces have. The total torque on the body may not vansh. Such a torque changes the rotatonal state of moton of the rgd body,.e. t changes the total angular momentum of the body n accordance wth Eq. (7.8 b). A rgd body s sad to be n mechancal equlbrum, f both ts lnear momentum and angular momentum are not changng wth tme, or equvalently, the body has nether lnear n τ1 + τ τn = τ = 0 (7.30b) = 1 If the total torque on the rgd body s zero, the total angular momentum of the body does not change wth tme. Eq. (7.30 b) gves the condton for the rotatonal equlbrum of the body. One may rase a queston, whether the rotatonal equlbrum condton [Eq. 7.30(b)] remans vald, f the orgn wth respect to whch the torques are taken s shfted. One can show that f the translatonal equlbrum condton [Eq. 7.30(a)] holds for a rgd body, then such a shft of orgn does not matter,.e. the rotatonal equlbrum condton s ndependent of the locaton of the orgn about whch the torques are taken. Example 7.7 gves a proof of ths result n a specal case of a couple,.e. two forces actng on a rgd body n translatonal equlbrum. The generalsaton of ths result to n forces s left as an exercse. Eq. (7.30a) and Eq. (7.30b), both, are vector equatons. They are equvalent to three scalar equatons each. Eq. (7.30a) corresponds to n F x = 0 = 1 n, F y = 0 and = 1 n F z = 0 = 1 (7.31a) where F x, F y and F z are respectvely the x, y and z components of the forces F. Smlarly, Eq. (7.30b) s equvalent to three scalar equatons n τ x = 0 = 1 n, τ y = 0 and τ z = 0 = 1 n = 1 (7.31b) where τ x, τ y and τ z are respectvely the x, y and z components of the torque τ.