Systems of Particles and Rotational Motion PDF

CHAPTER SIX SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 6.1 INTRODUCTION In the earlier chapters we primarily considered the motion of a single...

CHAPTER SIX SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 6.1 INTRODUCTION In the earlier chapters we primarily considered the motion of a single particle. (A particle is ideally represented as a 6.1 Introduction point mass having no size.) We applied the results of our 6.2 Centre of mass study even to the motion of bodies of finite size, assuming 6.3 Motion of centre of mass that motion of such bodies can be described in terms of the 6.4 Linear momentum of a motion of a particle. system of particles Any real body which we encounter in daily life has a 6.5 Vector product of two vectors finite size. In dealing with the motion of extended bodies 6.6 Angular velocity and its (bodies of finite size) often the idealised model of a particle is relation with linear velocity inadequate. In this chapter we shall try to go beyond this 6.7 Torque and angular inadequacy. We shall attempt to build an understanding of momentum the motion of extended bodies. An extended body, in the 6.8 Equilibrium of a rigid body first place, is a system of particles. We shall begin with the 6.9 Moment of inertia consideration of motion of the system as a whole. The centre 6.10 Kinematics of rotational of mass of a system of particles will be a key concept here. motion about a fixed axis We shall discuss the motion of the centre of mass of a system of particles and usefulness of this concept in understanding 6.11 Dynamics of rotational motion about a fixed axis the motion of extended bodies. A large class of problems with extended bodies can be 6.12 Angular momentum in case of rotation about a fixed solved by considering them to be rigid bodies. Ideally a axis rigid body is a body with a perfectly definite and unchanging shape. The distances between all pairs of Summary particles of such a body do not change. It is evident from Points to Ponder this definition of a rigid body that no real body is truly rigid, Exercises since real bodies deform under the influence of forces. But in many situations the deformations are negligible. In a number of situations involving bodies such as wheels, tops, steel beams, molecules and planets on the other hand, we can ignore that they warp (twist out of shape), bend or vibrate and treat them as rigid. 6.1.1 What kind of motion can a rigid body have? Let us try to explore this question by taking some examples of the motion of rigid bodies. Let us begin with a rectangular 2024-25 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 93 most common way to constrain a rigid body so that it does not have translational motion is to fix it along a straight line. The only possible motion of such a rigid body is rotation. The line or fixed axis about which the body is rotating is its axis of rotation. If you look around, you will come across many examples of rotation about an axis, a ceiling fan, a potter’s wheel, a giant wheel in a fair, a merry-go-round and so on (Fig Fig 6.1 Translational (sliding) motion of a block down an inclined plane. 6.3(a) and (b)). (Any point like P1 or P2 of the block moves with the same velocity at any instant of time.) block sliding down an inclined plane without any sidewise movement. The block is taken as a rigid body. Its motion down the plane is such that all the particles of the body are moving together, i.e. they have the same velocity at any instant of time. The rigid body here is in pure translational motion (Fig. 6.1). In pure translational motion at any instant of time, all particles of the body have the same velocity. Consider now the rolling motion of a solid metallic or wooden cylinder down the same (a) inclined plane (Fig. 6.2). The rigid body in this problem, namely the cylinder, shifts from the top to the bottom of the inclined plane, and thus, seems to have translational motion. But as Fig. 6.2 shows, all its particles are not moving with the same velocity at any instant. The body, therefore, is not in pure translational motion. Its motion is translational plus ‘something else.’ Fig. 6.2 Rolling motion of a cylinder. It is not pure translational motion. Points P1, P2, P3 and P4 have different velocities (shown by arrows) (b) at any instant of time. In fact, the velocity of Fig. 6.3 Rotation about a fixed axis the point of contact P3 is zero at any instant, (a) A ceiling fan if the cylinder rolls without slipping. (b) A potter’s wheel. In order to understand what this ‘something Let us try to understand what rotation is, else’ is, let us take a rigid body so constrained what characterises rotation. You may notice that that it cannot have translational motion. The in rotation of a rigid body about a fixed axis, 2024-25 94 PHYSICS Fig. 6.5 (a) A spinning top (The point of contact of the top with the ground, its tip O, is fixed.) Axis of oscillation Axis of rotation from blades Fig. 6.4 A rigid body rotation about the z-axis (Each point of the body such as P 1 or P2 describes a circle with its centre (C1 or C2) on the axis of rotation. The radius of the circle (r1or r 2 ) is the perpendicular distance of the point (P1 or P2 ) from the axis. A point on the axis like P3 remains stationary). every particle of the body moves in a circle, Fig. 6.5 (b) An oscillating table fan with rotating blades. The pivot of the fan, point O, is which lies in a plane perpendicular to the axis fixed. The blades of the fan are under and has its centre on the axis. Fig. 6.4 shows rotational motion, whereas, the axis of the rotational motion of a rigid body about a fixed rotation of the fan blades is oscillating. axis (the z-axis of the frame of reference). Let P1 In some examples of rotation, however, the be a particle of the rigid body, arbitrarily chosen axis may not be fixed. A prominent example of and at a distance r1 from fixed axis. The particle this kind of rotation is a top spinning in place P1 describes a circle of radius r1 with its centre [Fig. 6.5(a)]. (We assume that the top does not C1 on the fixed axis. The circle lies in a plane slip from place to place and so does not have perpendicular to the axis. The figure also shows translational motion.) We know from experience another particle P2 of the rigid body, P2 is at a that the axis of such a spinning top moves distance r2 from the fixed axis. The particle P2 around the vertical through its point of contact moves in a circle of radius r2 and with centre C2 with the ground, sweeping out a cone as shown on the axis. This circle, too, lies in a plane in Fig. 6.5(a). (This movement of the axis of the perpendicular to the axis. Note that the circles top around the vertical is termed precession.) described by P1 and P2 may lie in different planes; Note, the point of contact of the top with both these planes, however, are perpendicular ground is fixed. The axis of rotation of the top to the fixed axis. For any particle on the axis at any instant passes through the point of like P 3 , r = 0. Any such particle remains contact. Another simple example of this kind of stationary while the body rotates. This is rotation is the oscillating table fan or a pedestal expected since the axis of rotation is fixed. fan [Fig.6.5(b)]. You may have observed that the 2024-25 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 95 axis of rotation of such a fan has an oscillating Thus, for us rotation will be about a fixed axis (sidewise) movement in a horizontal plane about only unless stated otherwise. the vertical through the point at which the axis The rolling motion of a cylinder down an is pivoted (point O in Fig. 6.5(b)). inclined plane is a combination of rotation about While the fan rotates and its axis moves a fixed axis and translation. Thus, the sidewise, this point is fixed. Thus, in more ‘something else’ in the case of rolling motion general cases of rotation, such as the rotation which we referred to earlier is rotational motion. of a top or a pedestal fan, one point and not You will find Fig. 6.6(a) and (b) instructive from one line, of the rigid body is fixed. In this case this point of view. Both these figures show the axis is not fixed, though it always passes motion of the same body along identical through the fixed point. In our study, however, translational trajectory. In one case, Fig. 6.6(a), we mostly deal with the simpler and special case the motion is a pure translation; in the other of rotation in which one line (i.e. the axis) is fixed. case [Fig. 6.6(b)] it is a combination of translation and rotation. (You may try to reproduce the two types of motion shown, using a rigid object like a heavy book.) We now recapitulate the most important observations of the present section: The motion of a rigid body which is not pivoted or fixed in some way is either a pure translation or a combination of translation and rotation. The motion of a rigid body which is pivoted or fixed Fig. 6.6(a) Motion of a rigid body which is pure in some way is rotation. The rotation may be translation. about an axis that is fixed (e.g. a ceiling fan) or moving (e.g. an oscillating table fan [Fig.6.5(b)]). We shall, in the present chapter, consider rotational motion about a fixed axis only. 6.2 CENTRE OF MASS We shall first see what the centre of mass of a system of particles is and then discuss its significance. For simplicity we shall start with a two particle system. We shall take the line Fig. 6.6(b) Motion of a rigid body which is a joining the two particles to be the x- axis. combination of translation and rotation. Fig 6.6 (a) and 6.6 (b) illustrate different motions of the same body. Note P is an arbitrary point of the body; O is the centre of mass of the body, which is defined in the next section. Suffice to say here that the trajectories of O are the translational trajectories Tr1 and Tr2 of the body. The positions O and P at three different instants of time are shown by O1, O2, and O3, and P 1, P 2 and P3, respectively, in both Figs. 6.6 (a) and (b). As seen from Fig. 6.6(a), at any instant the velocities of any particles like O and P of the body are the same in pure translation. Notice, in this case the orientation of OP, i.e. the angle OP makes Fig. 6.7 with a fixed direction, say the horizontal, remains the same, i.e. α1 = α2 = α3. Fig. 6.6 (b) illustrates a case of combination of translation and rotation. In this case, Let the distances of the two particles be x1 at any instants the velocities of O and P differ. Also, and x2 respectively from some origin O. Let m1 α1, α2 and α3 may all be different. and m2 be respectively the masses of the two 2024-25 96 PHYSICS particles. The centre of mass of the system is m (y1 + y 2 + y 3 ) y1 + y 2 + y 3 that point C which is at a distance X from O, Y = = where X is given by 3m 3 Thus, for three particles of equal mass, the m 1 x1 + m 2 x 2 X = (6.1) centre of mass coincides with the centroid of the m1 + m 2 triangle formed by the particles. In Eq. (6.1), X can be regarded as the mass- Results of Eqs. (6.3a) and (6.3b) are weighted mean of x1 and x2. If the two particles generalised easily to a system of n particles, not have the same mass m1 = m2 = m, then necessarily lying in a plane, but distributed in space. The centre of mass of such a system is mx 1 + mx 2 x 1 + x 2 X = = at (X, Y, Z ), where 2m 2 Thus, for two particles of equal mass the X = ∑mi xi (6.4a) centre of mass lies exactly midway between M them. If we have n particles of masses m1, m2, Y = ∑ m i yi (6.4b)...mn respectively, along a straight line taken as M the x- axis, then by definition the position of the centre of the mass of the system of particles is and Z = ∑ m i zi (6.4c) given by. M n ∑ mi xi Here M = ∑mi is the total mass of the m x + m 2 x 2 +... + m n x n X= 1 1 = i =1 = ∑ mx i i (6.2) system. The index i runs from 1 to n; mi is the n m1 + m 2 +... + m n ∑ mi ∑ m i mass of the ith particle and the position of the i =1 ith particle is given by (xi, yi, zi). where x1, x2,...xn are the distances of the Eqs. (6.4a), (6.4b) and (6.4c) can be particles from the origin; X is also measured from combined into one equation using the notation the same origin. The symbol ∑ (the Greek letter of position vectors. Let ri be the position vector sigma) denotes summation, in this case over n of the ith particle and R be the position vector of particles. The sum the centre of mass: ∑ mi =M r = x ɵi + y ɵj + z k i i i i is the total mass of the system. and R = X ɵi + Y ɵj + Z k Suppose that we have three particles, not lying in a straight line. We may define x– and y– axes in the plane in which the particles lie and R= ∑ m i ri represent the positions of the three particles by Then M (6.4d) coordinates (x1,y1), (x2,y2) and (x3,y3) respectively. The sum on the right hand side is a vector Let the masses of the three particles be m1, m2 sum. and m3 respectively. The centre of mass C of Note the economy of expressions we achieve the system of the three particles is defined and by use of vectors. If the origin of the frame of located by the coordinates (X, Y) given by reference (the coordinate system) is chosen to m 1x 1 + m 2 x 2 + m 3 x 3 be the centre of mass then ∑ mi ri = 0 for the X = (6.3a) given system of particles. m1 + m 2 + m 3 A rigid body, such as a metre stick or a m 1y1 + m 2y2 + m 3y3 flywheel, is a system of closely packed particles; Y = (6.3b) Eqs. (6.4a), (6.4b), (6.4c) and (6.4d) are therefore, m1 + m 2 + m 3 applicable to a rigid body. The number of For the particles of equal mass m = m1 = m2 particles (atoms or molecules) in such a body is = m 3, so large that it is impossible to carry out the m (x1 + x 2 + x 3 ) x1 + x 2 + x 3 summations over individual particles in these X = = equations. Since the spacing of the particles is 3m 3 2024-25 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 97 small, we can treat the body as a continuous Let us consider a thin rod, whose width and distribution of mass. We subdivide the body into breath (in case the cross section of the rod is n small elements of mass; ∆m1, ∆m2... ∆mn; the rectangular) or radius (in case the cross section ith element ∆mi is taken to be located about the of the rod is cylindrical) is much smaller than point (xi, yi, zi). The coordinates of the centre of its length. Taking the origin to be at the mass are then approximately given by geometric centre of the rod and x-axis to be along the length of the rod, we can say that on X = ∑ (∆m i )x i , Y = ∑ (∆m i )yi , Z = ∑ (∆m i )zi account of reflection symmetry, for every ∑ ∆m i ∑ ∆m i ∑ ∆m i element dm of the rod at x, there is an element of the same mass dm located at –x (Fig. 6.8). As we make n bigger and bigger and each ∆m i smaller and smaller, these expressions The net contribution of every such pair to become exact. In that case, we denote the sums over i by integrals. Thus, the integral and hence the integral itself ∑ ∆m i → ∫ dm = M , is zero. From Eq. (6.6), the point for which the integral itself is zero, is the centre of mass. ∑ ( ∆m i )x i → ∫ x dm , Thus, the centre of mass of a homogenous thin rod coincides with its geometric centre. This can be understood on the basis of reflection symmetry. ∑ ( ∆m i )yi → ∫ y dm , The same symmetry argument will apply to homogeneous rings, discs, spheres, or even and ∑ (∆m i )zi → ∫ z dm thick rods of circular or rectangular cross Here M is the total mass of the body. The section. For all such bodies you will realise that coordinates of the centre of mass now are for every element dm at a point (x, y, z ) one can 1 1 1 always take an element of the same mass at M∫ ∫ y dm and Z = M ∫ z dm X= x dm , Y = (6.5a) M the point (–x, –y, –z ). (In other words, the origin The vector expression equivalent to these is a point of reflection symmetry for these three scalar expressions is bodies.) As a result, the integrals in Eq. (6.5 a) all are zero. This means that for all the above 1 M∫ R= r dm (6.5b) bodies, their centre of mass coincides with their geometric centre. If we choose, the centre of mass as the origin of our coordinate system, u Example 6.1 Find the centre of mass of R= 0 three particles at the vertices of an equilateral triangle. The masses of the i.e., ∫ r dm = 0 particles are 100g, 150g, and 200g respectively. Each side of the equilateral or ∫ x dm = ∫ y dm = ∫ z dm = 0 (6.6) triangle is 0.5m long. Often we have to calculate the centre of mass of Answer homogeneous bodies of regular shapes like rings, discs, spheres, rods etc. (By a homogeneous body we mean a body with uniformly distributed mass.) By using symmetry consideration, we can easily show that the centres of mass of these bodies lie at their geometric centres. Fig. 6.8 Determining the CM of a thin rod. Fig. 6.9 2024-25 98 PHYSICS With the x – and y–axes chosen as shown in Fig. concurrence of the medians, i.e. on the centroid 6.9, the coordinates of points O, A and B forming G of the triangle. ⊳ the equilateral triangle are respectively (0,0), (0.5,0), (0.25,0.25 3 ). Let the masses 100 g, u Example 6.3 Find the centre of mass of a 150g and 200g be located at O, A and B be uniform L-shaped lamina (a thin flat plate) respectively. Then, with dimensions as shown. The mass of the lamina is 3 kg. m 1x 1 + m 2 x 2 + m 3 x 3 X = m1 + m 2 + m 3 Answer Choosing the X and Y axes as shown in Fig. 6.11 we have the coordinates of the 100 (0 ) + 150(0.5) + 200(0.25) gm vertices of the L-shaped lamina as given in the = (100 + 150 + 200) g figure. We can think of the L-shape to consist of 3 squares each of length 75 + 50 125 5 1m. The mass of each square is 1kg, since the = m= m= m 450 450 18 lamina is uniform. The centres of mass C1, C2 and C3 of the squares are, by symmetry, their 100(0) + 150(0) + 200(0.25 3) gm geometric centres and have coordinates (1/2,1/2), Y = (3/2,1/2), (1/2,3/2) respectively. We take the 450 g masses of the squares to be concentrated at these points. The centre of mass of the whole 50 3 3 1 = m= m= m L shape (X, Y) is the centre of mass of these 450 9 3 3 mass points. The centre of mass C is shown in the figure. Note that it is not the geometric centre of the triangle OAB. Why? ⊳ u Example 6.2 Find the centre of mass of a triangular lamina. Answer The lamina (∆LMN ) may be subdivided into narrow strips each parallel to the base (MN) as shown in Fig. 6.10 Fig. 6.11 Hence X = [1(1/ 2) + 1(3 / 2) + 1(1/2)] kg m 5 = m Fig. 6.10 (1 + 1 + 1) kg 6 By symmetry each strip has its centre of [1(1/2) + 1(1/ 2) + 1(3 /2)] kg m 5 mass at its midpoint. If we join the midpoint of Y = = m all the strips we get the median LP. The centre (1 + 1 + 1) kg 6 of mass of the triangle as a whole therefore, has The centre of mass of the L-shape lies on to lie on the median LP. Similarly, we can argue the line OD. We could have guessed this without that it lies on the median MQ and NR. This calculations. Can you tell why? Suppose, the means the centre of mass lies on the point of three squares that make up the L shaped lamina 2024-25 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 99 of Fig. 6.11 had different masses. How will you Thus, the total mass of a system of particles then determine the centre of mass of the lamina? times the acceleration of its centre of mass is ⊳ the vector sum of all the forces acting on the system of particles. 6.3 MOTION OF CENTRE OF MASS Note when we talk of the force F1 on the first Equipped with the definition of the centre of particle, it is not a single force, but the vector mass, we are now in a position to discuss its sum of all the forces on the first particle; likewise physical importance for a system of n particles. for the second particle etc. Among these forces We may rewrite Eq.(6.4d) as on each particle there will be external forces exerted by bodies outside the system and also MR = ∑ m i ri = m1r1 + m 2 r2 +... + m n rn (6.7) internal forces exerted by the particles on one Differentiating the two sides of the equation another. We know from Newton’s third law that with respect to time we get these internal forces occur in equal and opposite dR dr dr dr pairs and in the sum of forces of Eq. (6.10), their M = m1 1 + m 2 2 +... + m n n contribution is zero. Only the external forces dt dt dt dt contribute to the equation. We can then rewrite Eq. (6.10) as or MA = Fext (6.11) M V = m1 v1 + m 2 v 2 +... + m n vn (6.8) where Fext represents the sum of all external where v1 ( = dr1 /dt ) is the velocity of the first forces acting on the particles of the system. particle v 2 ( = dr 2 dt ) is the velocity of the Eq. (6.11) states that the centre of mass of a system of particles moves as if all the mass second particle etc. and V = dR / dt is the of the system was concentrated at the centre velocity of the centre of mass. Note that we of mass and all the external forces were assumed the masses m 1, m 2,... etc. do not applied at that point. change in time. We have therefore, treated them Notice, to determine the motion of the centre as constants in differentiating the equations of mass no knowledge of internal forces of the with respect to time. system of particles is required; for this purpose Differentiating Eq.(6.8) with respect to time, we need to know only the external forces. we obtain To obtain Eq. (6.11) we did not need to specify the nature of the system of particles. The system dV dv1 dv 2 dv n M = m1 + m2 +... + m n may be a collection of particles in which there dt dt dt dt may be all kinds of internal motions, or it may or be a rigid body which has either pure translational motion or a combination of MA = m 1a1 + m 2 a 2 +... + m n an (6.9) translational and rotational motion. Whatever where a1 ( = dv1 /dt ) is the acceleration of the is the system and the motion of its individual particles, the centre of mass moves according first particle, a 2 ( = dv2 /dt ) is the acceleration to Eq. (6.11). Instead of treating extended bodies as single of the second particle etc. and A (= d V / dt ) is particles as we have done in earlier chapters, the acceleration of the centre of mass of the we can now treat them as systems of particles. system of particles. We can obtain the translational component of Now, from Newton’s second law, the force their motion, i.e. the motion of the centre of mass acting on the first particle is given by F1 = m1a1. of the system, by taking the mass of the whole The force acting on the second particle is given system to be concentrated at the centre of mass and all the external forces on the system to be by F2 = m 2 a 2 and so on. Eq. (6.9) may be written acting at the centre of mass. as This is the procedure that we followed earlier MA = F1 + F2 +... + Fn (6.10) in analysing forces on bodies and solving 2024-25 100 PHYSICS problems without explicitly outlining and where F is the force on the particle. Let us justifying the procedure. We now realise that in consider a system of n particles with masses m1, earlier studies we assumed, without saying so, m2,...mn respectively and velocities v1 , v 2 ,.......vn that rotational motion and/or internal motion respectively. The particles may be interacting of the particles were either absent or negligible. and have external forces acting on them. The We no longer need to do this. We have not only found the justification of the procedure we linear momentum of the first particle is m1v1 , followed earlier; but we also have found how to of the second particle is m 2 v 2 and so on. describe and separate the translational motion For the system of n particles, the linear of (1) a rigid body which may be rotating as momentum of the system is defined to be the well, or (2) a system of particles with all kinds vector sum of all individual particles of the of internal motion. system, P = p1 + p2 +... + pn = m1 v1 + m 2 v 2 +... + m n vn (6.14) Comparing this with Eq. (6.8) P=MV (6.15) Thus, the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass. Differentiating Eq. (6.15) with respect to time, dP dV Fig. 6.12 The centre of mass of the fragments =M = MA (6.16) dt dt of the projectile continues along the same parabolic path which it would Comparing Eq.(6.16) and Eq. (6.11), have followed if there were no dP explosion. = Fext (6.17) dt Figure 6.12 is a good illustration of Eq. (6.11). This is the statement of Newton’s second law A projectile, following the usual parabolic of motion extended to a system of particles. trajectory, explodes into fragments midway in Suppose now, that the sum of external air. The forces leading to the explosion are forces acting on a system of particles is zero. internal forces. They contribute nothing to the Then from Eq.(6.17) motion of the centre of mass. The total external dP force, namely, the force of gravity acting on the = 0 or P = Constant (6.18a) body, is the same before and after the explosion. dt The centre of mass under the influence of the Thus, when the total external force acting external force continues, therefore, along the on a system of particles is zero, the total linear same parabolic trajectory as it would have momentum of the system is constant. This is followed if there were no explosion. the law of conservation of the total linear momentum of a system of particles. Because of 6.4 LINEAR MOMENTUM OF A SYSTEM OF Eq. (6.15), this also means that when the PARTICLES total external force on the system is zero Let us recall that the linear momentum of a the velocity of the centre of mass remains constant. (We assume throughout the particle is defined as discussion on systems of particles in this p=mv (6.12) chapter that the total mass of the system Let us also recall that Newton’s second law remains constant.) written in symbolic form for a single particle is Note that on account of the internal forces, dp i.e. the forces exerted by the particles on one F= (6.13) another, the individual particles may have dt 2024-25 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 101 complicated trajectories. Yet, if the total external force acting on the system is zero, the centre of mass moves with a constant velocity, i.e., moves uniformly in a straight line like a free particle. The vector Eq. (6.18a) is equivalent to three scalar equations, Px = c1, Py = c2 and Pz = c3 (6.18 b) (a) (b) Here Px, Py and Pz are the components of the total linear momentum vector P along the x–, y– Fig. 6.14 (a) Trajectories of two stars, S1 (dotted line) and z–axes respectively; c 1 , c 2 and c 3 are and S2 (solid line) forming a binary constants. system with their centre of mass C in uniform motion. (b) The same binary system, with the centre of mass C at rest. move back to back with their centre of mass remaining at rest as shown in Fig.6.13 (b). In many problems on the system of particles, as in the above radioactive decay problem, it is convenient to work in the centre of mass frame rather than in the laboratory frame of reference. (a) (b) In astronomy, binary (double) stars is a common occurrence. If there are no external forces, the centre of mass of a double star Fig. 6.13 (a) A heavy nucleus radium (Ra) splits into moves like a free particle, as shown in Fig.6.14 a lighter nucleus radon (Rn) and an alpha particle (nucleus of helium atom). The CM (a). The trajectories of the two stars of equal of the system is in uniform motion. mass are also shown in the figure; they look (b) The same spliting of the heavy nucleus complicated. If we go to the centre of mass radium (Ra) with the centre of mass at frame, then we find that there the two stars rest. The two product particles fly back are moving in a circle, about the centre of to back. mass, which is at rest. Note that the position of the stars have to be diametrically opposite As an example, let us consider the to each other [Fig. 6.14(b)]. Thus in our frame radioactive decay of a moving unstable particle, of reference, the trajectories of the stars are a like the nucleus of radium. A radium nucleus combination of (i) uniform motion in a straight disintegrates into a nucleus of radon and an line of the centre of mass and (ii) circular alpha particle. The forces leading to the decay orbits of the stars about the centre of mass. are internal to the system and the external As can be seen from the two examples, forces on the system are negligible. So the total separating the motion of different parts of a linear momentum of the system is the same system into motion of the centre of mass and before and after decay. The two particles motion about the centre of mass is a very useful technique that helps in understanding produced in the decay, the radon nucleus and the motion of the system. the alpha particle, move in different directions in such a way that their centre of mass moves 6.5 VECTOR PRODUCT OF TWO VECTORS along the same path along which the original decaying radium nucleus was moving We are already familiar with vectors and their [Fig. 6.13(a)]. use in physics. In chapter 5 (Work, Energy, Power) If we observe the decay from the frame of we defined the scalar product of two vectors. An reference in which the centre of mass is at rest, important physical quantity, work, is defined as the motion of the particles involved in the decay a scalar product of two vector quantities, force looks particularly simple; the product particles and displacement. 2024-25 102 PHYSICS We shall now define another product of two A simpler version of the right hand rule is vectors. This product is a vector. Two important the following : Open up your right hand palm quantities in the study of rotational motion, and curl the fingers pointing from a to b. Your namely, moment of a force and angular stretched thumb points in the direction of c. momentum, are defined as vector products. It should be remembered that there are two angles between any two vectors a and b. In Definition of Vector Product Fig. 6.15 (a) or (b) they correspond to θ (as shown) A vector product of two vectors a and b is a and (3600– θ). While applying either of the above vector c such that rules, the rotation should be taken through the (i) magnitude of c = c = ab sin θ where a and b smaller angle (