CHAPTER SEVEN SYSTES OF PARTICLES AND ROTATIONAL OTION 7.1 Introducton 7.2 Centre of mass 7.3 oton 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 oment of nerta 7.10 Theorems of perpendcular and parallel axes 7.11 Knematcs of rotatonal moton about a fxed axs 7.12 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 represented as a pont mass. It has practcally 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, bend or vbrate and treat them as rgd. 7.1.1 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 block sldng down an nclned plane wthout any sdewse
142 PHYSICS Fg 7.1 Translatonal (sldng) moton of a block down an nclned plane. (Any pont lke P 1 or P 2 of the block moves wth the same velocty at any nstant of tme.) movement. The block s 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. 7.2). The rgd body n ths problem, namely the cylnder, shfts from the top to the bottom of the nclned plane, and thus, has translatonal moton. But as Fg. 7.2 shows, all ts partcles are not movng wth the same velocty at any nstant. The body therefore, s not n pure translaton. Its moton s translaton plus somethng else. 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 along whch the body s fxed s termed as 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. 7.2 Rollng moton of a cylnder It s not pure translatonal moton. Ponts P 1, P 2, 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 most common way to constran a rgd body so (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
SYSTES OF PARTICLES AND ROTATIONAL OTION 143 Fg. 7.5 (a) A spnnng top (The pont of contact of the top wth the ground, ts tp O, s fxed.) Fg. 7.4 A rgd body rotaton about the z-axs (Each pont of the body such as P 1 or P 2 descrbes a crcle wth ts centre (C 1 or C 2 ) on the axs. The radus of the crcle (r 1 or r 2 ) s the perpendcular dstance of the pont (P 1 or P 2 ) from the axs. A pont on the axs lke P 3 remans statonary). axs, 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 2 of the rgd body, P 2 s at a dstance r 2 from the fxed axs. The partcle P 2 moves n a crcle of radus r 2 and wth centre C 2 on the axs. Ths crcle, too, les n a plane perpendcular to the axs. Note that the crcles descrbed by P 1 and P 2 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 s fxed. Fg. 7.5 (b) An oscllatng table fan. The pvot of the fan, pont O, s fxed. 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. You may have observed that the axs of
144 PHYSICS 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 Fg. 7.6(a) oton of a rgd body whch s pure translaton. Fg. 7.6(b) oton 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 2 of the body. The postons O and P at three dfferent nstants of tme are shown by O 1, O 2, and O 3, and P 1, P 2 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 2 = 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 2 and α 3 may all be dfferent. fxed. 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). We shall, n the present chapter, consder rotatonal moton about a fxed axs only. 7.2 CENTRE OF ASS 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 2 respectvely from some orgn O. Let m 1 and m 2 be respectvely the masses of the two
SYSTES OF PARTICLES AND ROTATIONAL OTION 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 1 1 2 2 1 2 (7.1) In Eq. (7.1), X can be regarded as the massweghted mean of x 1 and x 2. If the two partcles have the same mass m 1 = m 2 = m, then mx1 + mx2 x1 + x2 X = = 2m 2 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 2,...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 X m x + m x +... + m x 1 1 2 2 n n = = m + m +... + m m 1 2 n m x (7.2) where x 1, x 2,...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 = 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 2,y 2 ) and (x 3,y 3 ) respectvely. Let the masses of the three partcles be m 1, m 2 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 1 1 2 2 3 3 1 2 3 m y + m y + m y Y = m + m + m 1 1 2 2 3 3 1 2 3 (7.3a) (7.3b) For the partcles of equal mass m = m 1 = m 2 = m 3, m( x + x + x ) x + x + x X = = 3m 3 1 2 3 1 2 3 m( y1 + y2 + y3 ) y1 + y2 + y 3 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) my Y = (7.4b) and Here = m z Z = (7.4c) 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: and r = x ɵ + y ɵ j + z k R = X ɵ + Y j ɵ + Zk m Then R = r (7.4d) 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
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 2... 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 ) x x d m, ( m ) y y d m, and ( m ) z z dm Here s the total mass of the body. The coordnates of the centre of mass now are 1 1 1 X = x d m, Y y dmand Z z dm = = (7.5a) The vector expresson equvalent to these three scalar expressons s 1 R = dm r (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 wdth 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 x dm 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 200g respectvely. Each sde of the equlateral trangle s 0.5m long. Answer Fg. 7.8 Determnng the C of a thn rod. Fg. 7.9
SYSTES OF PARTICLES AND ROTATIONAL OTION 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), (0.25,0.25 3 ). Let the masses 100 g, 150g and 200g be located at O, A and B be respectvely. Then, m x + m x + m x X = m + m + m 1 1 2 2 3 3 ( ) 1 2 3 100 0 + 150(0.5) + 200(0.25) g m = (100 + 150 + 200) g 75 + 50 125 5 = m = m = m 450 450 18 100(0) + 150(0) + 200(0.25 3) g m Y = 450 g 50 3 3 1 = m = m = m 450 9 3 3 The centre of mass C s shown n the fgure. Note that t s not the geometrc centre of the trangle OAB. Why? of 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. 7.11 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 2 and C 3 of the squares are, by symmetry, ther geometrc centres and have coordnates (1/2,1/2), (3/2,1/2), (1/2,3/2) 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 7.2 Fnd the centre of mass of a trangular lamna. Answer The lamna ( LN) may be subdvded nto narrow strps each parallel to the base (N) as shown n Fg. 7.10 Fg. 7.11 Fg. 7.10 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 Q and NR. Ths means the centre of mass les on the pont Hence X = [ + + ] ( 1 + 1 + 1) kg 1(1/2) 1(3/2) 1(1/2) kg m [ + + ] ( + + ) 5 = m 6 1(1/ 2) 1(1/2) 1(3/ 2) kg m 5 Y = = m 1 1 1 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
148 PHYSICS of Fg. 7.11 had dfferent masses. How wll you then determne the centre of mass of the lamna? 7.3 OTION OF CENTRE OF ASS Equpped wth the defnton of the centre of mass, we are now n a poston to dscuss ts physcal mportance for a system of partcles. We may rewrte Eq.(7.4d) as R = m r = m r + m r + + m r (7.7) 1 1 2 2... n n Dfferentatng the two sdes of the equaton wth respect to tme we get dr dr dr drn dt dt dt dt 1 2 = m1 + m2 +... + m n or V = m v + m v + + m v (7.8) where ( = d /dt ) 1 1 2 2... n n v r s the velocty of the frst 1 1 partcle ( = d dt) v r s the velocty of the 2 2 second partcle etc. and V = d R /dt s the velocty of the centre of mass. Note that we assumed the masses m 1, m 2,... 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 dv dv dv dt dt dt dt 1 2 = m1 + m2 +... + m n or A = m a + m a + + m a (7.9) where ( = d /dt) 1 1 2 2... n n a v s the acceleraton of the 1 1 frst partcle, ( = d /dt ) a v s the acceleraton 2 2 of the second partcle etc. and A ( = d V /dt ) 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 by F2 =m2 2 as a and so on. Eq. (7.9) may be wrtten A = F1 + F2 +... + F n (7.10) n 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 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 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
SYSTES OF PARTICLES AND ROTATIONAL OTION 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 (2) a system of partcles wth all knds of nternal moton. Fg. 7.12 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 7.12 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 OENTU OF A SYSTE OF PARTICLES Let us recall that the lnear momentum of a partcle s defned as p = m v (7.12) Let us also recall that Newton s second law wrtten n symbolc form for a sngle partcle s d F = p (7.13) dt where F s the force on the partcle. Let us consder a system of n partcles wth masses m 1, m 2,...m n respectvely and veloctes v1, v 2,... 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 m2v 2 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 + p2 +... + pn = m v + m v + + m v (7.14) 1 1 2 2... n n Comparng ths wth Eq. (7.8) P = 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 = = A (7.16) dt dt Comparng Eq.(7.16) and Eq. (7.11), dp ext dt = F (7.17) Ths s the statement of Newton s second law 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 0 or dt = 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
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 2 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 2 and c 3 are constants. (a) (b) Fg. 7.13 (a) A heavy nucleus (Ra) splts nto a lghter nucleus (Rn) and an alpha partcle (He). The C of the system s n unform moton. (b) The same spltng of the heavy nucleus (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. 7.14 (a) Trajectores of two stars, S 1 (dotted lne) and S 2 (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)
SYSTES OF PARTICLES AND ROTATIONAL OTION 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 sn θ 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. 7.15 (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. 7.15 (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 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 2 sn 0 = 0.
152 PHYSICS From ths follow the results = 0, j j = 0, k k = 0 () j = k Note that the magntude of j s sn90 0 or 1, snce ĩ and j both have unt magntude and the angle between them s 90 0. Thus, j s a unt vector. A unt vector perpendcular to the plane of ĩ and j and related to them by the rght hand screw rule s k. Hence, the above result. You may verfy smlarly, j k = and k = j From the rule for commutaton of the cross product, t follows: j = k, k j =, k = j Note f, j, koccur cyclcally n the above vector product relaton, the vector product s postve. If, j, k do not occur n cyclc order, the vector product s negatve. Now, a b = ( a + a j + a k ) ( b + b j + b k ) x y z x y z = a b k a b j a b k + a b + a b j a b 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 x 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. j k a b = 3 4 5 = 7 j 5k 2 1 3 Note b a = 7 + j + 5k 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. 7.4. 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 = j k 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 = ( 2î + ĵ 3kˆ ) Answer ab = (3 4 j + 5 k ) ( 2 + j 3 k ) = 6 4 15 = 25 Fg. 7.16 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. 7.16 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
SYSTES OF PARTICLES AND ROTATIONAL OTION 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/dt 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 = ωr, 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 ω = dθ dt v = ω r (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 referred as above. Fg. 7.17 (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. 7.17 (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. 7.16 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.