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, Reprint 2025-26 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 P1 or P2 describes a circle with its centre (C1 or C2) on the axis of rotation. The radius of the circle (r1or r2) is the perpendicular distance of the point (P1 or P2) from the axis. A point on the axis like P3 remains stationary). Fig. 6.5 (b) An oscillating table fan with rotatingevery particle of the body moves in a circle, 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 P3, 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 Reprint 2025-26 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 P1, P2 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 Reprint 2025-26 96 PHYSICS particles. The centre of mass of the system is m (y1 + y 2 + y 3 ) y1 + y 2 + y 3that point C which is at a distance X from O, Y = = 3m 3where X is given by Thus, for three particles of equal mass, the m 1 x 1 + m 2 x 2 X = centre of mass coincides with the centroid of the (6.1) m 1 + 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 m i x i Thus, for two particles of equal mass the ∑ X = (6.4a) centre of mass lies exactly midway between M them. m i yi ∑ (6.4b) If we have n particles of masses m1, m2, Y = ...mn respectively, along a straight line taken as M the x- axis, then by definition the position of the m i z i ∑ (6.4c)centre of the mass of the system of particles is and Z = given by. M n Here M = m i is the total mass of the ∑ m i x i ∑ ∑ m i x i =1 m 1 x 1 + m 2 x 2 + ... + m n x n = i =X = (6.2) system. The index i runs from 1 to n; mi is the n i mass of the ith particle and the position of the m 1 + m 2 +... + m n ∑ m i ∑ m 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 ir 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: m i = M ri = x i i + yi j + z i k ∑ ɵ ɵ is the total mass of the system. ɵ ɵ and R = X i + Y j + Z k Suppose that we have three particles, not m i rilying in a straight line. We may define x– and y– ∑axes in the plane in which the particles lie and R = 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 i ri = 0 for the m 1 x 1 + m 2 x 2 + m 3 x 3 be the centre of mass then ∑ X = (6.3a) given system of particles. m 1 + m 2 + m 3 A rigid body, such as a metre stick or a m 1y1 + m 2 y 2 + m 3 y 3 flywheel, is a system of closely packed particles; Y = (6.3b) Eqs. (6.4a), (6.4b), (6.4c) and (6.4d) are therefore, m 1 + 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 = m3, so large that it is impossible to carry out the m ( x 1 + x 2 + x 3 ) x 1 + x 2 + x 3 summations over individual particles in these X = = equations. Since the spacing of the particles is 3m 3 Reprint 2025-26 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 ( ∆ m i )z i account of reflection symmetry, for every ∑ along the length of the rod, we can say that on ( ∆ m i )x i ∑ ( ∆m i )y i ∑X = , Y = , Z = m i element dm of the rod at x, there is an element ∑ ∆ m i ∑ ∆ m i ∑ ∆ of the same mass dm located at –x (Fig. 6.8). As we make n bigger and bigger and each ∆mi 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 d m = M , ∑ ∆m i → ∫ is zero. From Eq. (6.6), the point for which the integral itself is zero, is the centre of mass. x d m , ∑ Thus, the centre of mass of a homogenous thin ( ∆m i )x i →∫ rod coincides with its geometric centre. This can y d m , ∑ be understood on the basis of reflection symmetry. ( ∆m i )yi →∫ The same symmetry argument will apply to homogeneous rings, discs, spheres, or even d m and ∑ (∆m i )z i →∫ z 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 X = ∫ x d m , Y = ∫ y d m and Z = ∫ z d m (6.5a) the point (–x, –y, –z). (In other words, the origin M M M 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 their centre of mass coincides with their R = r d m (6.5b) bodies, M ∫ 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 z d m = 0 (6.6) triangle is 0.5m long. ∫ x d m = ∫ y d m = ∫ 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. 6.9 Fig. 6.8 Determining the CM of a thin rod. Reprint 2025-26 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), u Example 6.3 Find the centre of mass of a(0.5,0), (0.25,0.25 3 ). Let the masses 100 g, uniform L-shaped lamina (a thin flat plate)150g and 200g be located at O, A and B be with dimensions as shown. The mass ofrespectively. Then, the lamina is 3 kg. m 1 x 1 + m 2 x 2 + m 3 x 3X = m 1 + 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) g m 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) g m 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 [1(1/2) + 1(3/2) + 1(1/2)] kg m 5 X = = 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 5mass at its midpoint. If we join the midpoint of Y = = m (1 + 1 + 1) kg 6all the strips we get the median LP. The centre 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 Reprint 2025-26 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 m i ri = m 1 r1 + m 2 r2 + ... + m n rn (6.7) MR = ∑ 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 pairs and in the sum of forces of Eq. (6.10), their d R d r1 d r2 d rn M = m 1 + m 2 + ... + m n contribution is zero. Only the external forces d t d t d t dt contribute to the equation. We can then rewrite Eq. (6.10) as or MA = Fext (6.11) M V = m 1 v1 + m 2 v 2 + ... + m n v n (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 m1, m2, ... 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 d V d v1 d v 2 d v n M = m 1 + m 2 + ... + m n may be a collection of particles in which there d t d t d t d t 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 1a 1 + m 2 a 2 + ... + m n a n (6.9) translational and rotational motion. Whatever is the system and the motion of its individualwhere a1 ( = dv1 /dt ) is the acceleration of the particles, the centre of mass moves according first particle, a 2 ( = dv 2 /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 = m 1a1 . of the system, by taking the mass of the whole system to be concentrated at the centre of massThe force acting on the second particle is given 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 Reprint 2025-26 100 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 ,.......v n 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 m 1 v1 , 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 + p 2 + ... + p n = m 1 v1 + m 2 v 2 + ... + m n v n (6.14) Comparing this with Eq. (6.8) P = M V (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, d P d V Fig. 6.12 The centre of mass of the fragments = M = M A (6.16) d t d t 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 d P explosion. = Fext (6.17) d t 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 d P force, namely, the force of gravity acting on the = 0 or P = Constant (6.18a) dtbody, is the same before and after the explosion. 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 momentum of the system is constant. This issame parabolic trajectory as it would have the law of conservation of the total linearfollowed if there were no explosion. momentum of a system of particles. Because of

6.12 Angular momentum in case solved by considering them to be rigid bodies. Ideally a of rotation about a fixed 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 Reprint 2025-26 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 6.3(a) and (b)). an inclined plane. (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.

6.12 ANGULAR MOMENTUM IN CASE OF For computing the total angular momentum ROTATION ABOUT A FIXED AXIS of the whole rigid body, we add up the contribution of each particle of the body. We have studied in section 6.7, the angular momentum of a system of particles. We already Thus know from there that the time rate of total We denote by L ⊥ and L z the components of angular momentum of a system of particles L respectively perpendicular to the z-axis andabout a point is equal to the total external torque along the z-axis;on the system taken about the same point. When OC i × m i v i (6.42a)the total external torque is zero, the total angular L ⊥= ∑ momentum of the system is conserved. where mi and vi are respectively the mass and We now wish to study the angular momentum the velocity of the ith particle and Ci is the centrein the special case of rotation about a fixed axis. of the circle described by the particle; The general expression for the total angular momentum of the system of n particles is N and ˆ (6.42b) L = =∑i 1 ri × p i (6.25b) or L z = Iωk We first consider the angular momentum of The last step follows since the perpendicular a typical particle of the rotating rigid body. We distance of the ith particle from the axis is ri; andthen sum up the contributions of individual by definition the moment of inertia of the body particles to get L of the whole body. m i ri2 . For a typical particle l = r × p. As seen in the about the axis of rotation is I =∑ last section r = OP = OC + CP [Fig. 6.17(b)]. With Note L = L z + L ⊥ (6.42c)p = m v , l = ( OC × m v ) + ( CP × m v ) The rigid bodies which we have mainly considered in this chapter are symmetric about The magnitude of the linear velocity v of the the axis of rotation, i.e. the axis of rotation is particle at P is given by v = ωr⊥ where r⊥ is the one of their symmetry axes. For such bodies, for length of CP or the perpendicular distance of P a given OCi, for every particle which has a from the axis of rotation. Further, v is tangential velocity vi , there is another particle of velocity at P to the circle which the particle describes. –vi located diametrically opposite on the circle Using the right-hand rule one can check that with centre Ci described by the particle. TogetherCP × v is parallel to the fixed axis. The unit vector along the fixed axis (chosen as the z-axis) such pairs will contribute zero to L ⊥ and as a is ˆk . Hence result for symmetric bodies L ⊥ is zero, and CP × m v = r⊥ (mv ) kˆ hence z = Iωkˆ (6.42d) = mr⊥2ω kˆ (since υ = ωr⊥ ) L = L Similarly, we can check that OC × v is For bodies, which are not symmetric about perpendicular to the fixed axis. Let us denote the axis of rotation, L is not equal to Lz and hence the part of l along the fixed axis (i.e. the z-axis) L does not lie along the axis of rotation. by lz, then Referring to Table 6.1, can you tell in which l z = CP × m v = mr⊥2ωkˆ cases L = Lz will not apply? Let us differentiate Eq. (6.42b). Since ˆk is a and l = l z + OC × m v fixed (constant) vector, we get We note that lz is parallel to the fixed axis, ˆbut l is not. In general, for a particle, the angular I ω) k d ( L z ) =  d (  d t momentum l is not along the axis of rotation, d t i.e. for a particle, l and ω are not necessarily Now, Eq. (6.28b) states parallel. Compare this with the corresponding dL fact in translation. For a particle, p and v are = τ dtalways parallel to each other. Reprint 2025-26 122 PHYSICS As we have seen in the last section, only We have already derived this equation using those components of the external torques which the work - kinetic energy route. are along the axis of rotation, need to be taken into account, when we discuss rotation about a 6.12.1 Conservation of angular momentum fixed axis. This means we can take τ = τkˆ . We are now in a position to revisit the principle of conservation of angular momentum in the Since L = L z + L ⊥ and the direction of Lz (vector context of rotation about a fixed axis. From Eq. ˆk ) is fixed, it follows that for rotation about a (6.43c), if the external torque is zero, fixed axis, Lz = Iω = constant (6.44) For symmetric bodies, from Eq. (6.42d), Lz d L z = τkˆ (6.43a) may be replaced by L .(L and Lz are respectively d t the magnitudes of L and Lz.) This then is the required form, for fixed axis d L rotation, of Eq. (6.29a), which expresses theand ⊥= 0 (6.43b) dt general law of conservation of angular momentum Thus, for rotation about a fixed axis, the of a system of particles. Eq. (6.44) applies to many component of angular momentum perpendicular situations that we come across in daily life. You may do this experiment with your friend. Sit on a to the fixed axis is constant. As L z = Iωkˆ , we swivel chair (a chair with a seat, free to rotate get from Eq. (6.43a), about a pivot) with your arms folded and feet not resting on, i.e., away from, the ground. Ask your d ( Iω) = τ (6.43c) friend to rotate the chair rapidly. While the chair d t is rotating with considerable angular speed If the moment of inertia I does not change with stretch your arms horizontally. What happens? time, Your angular speed is reduced. If you bring back d dω your arms closer to your body, the angular speed ( Iω) = I = Iα increases again. This is a situation where thed t d t principle of conservation of angular momentumand we get from Eq. (6.43c), is applicable. If friction in the rotational τ = I α (6.41) Fig 6.32 (a) A demonstration of conservation of Fig 6.32 (b) An acrobat employing the principle of angular momentum. A girl sits on a conservation of angular momentum in swivel chair and stretches her arms/ her performance. brings her arms closer to the body. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 123 mechanism is neglected, there is no external A circus acrobat and a diver take advantage torque about the axis of rotation of the chair and of this principle. Also, skaters and classical, hence Iω is constant. Stretching the arms Indian or western, dancers performing a increases I about the axis of rotation, resulting in pirouette (a spinning about a tip–top) on the toes decreasing the angular speed ω. Bringing the of one foot display ‘mastery’ over this principle. arms closer to the body has the opposite effect. Can you explain? SUMMARY 1. Ideally, a rigid body is one for which the distances between different particles of the body do not change, even though there are forces on them. 2. A rigid body fixed at one point or along a line can have only rotational motion. A rigid body not fixed in some way can have either pure translational motion or a combination of translational and rotational motions. 3. In rotation about a fixed axis, every particle of the rigid body moves in a circle which lies in a plane perpendicular to the axis and has its centre on the axis. Every Point in the rotating rigid body has the same angular velocity at any instant of time. 4. In pure translation, every particle of the body moves with the same velocity at any instant of time. 5. Angular velocity is a vector. Its magnitude is ω = dθ/dt and it is directed along the axis of rotation. For rotation about a fixed axis, this vector ω has a fixed direction. 6. The vector or cross product of two vector a and b is a vector written as a×b. The magnitude of this vector is absinθ and its direction is given by the right handed screw or the right hand rule. 7. The linear velocity of a particle of a rigid body rotating about a fixed axis is given by v = ω × r, where r is the position vector of the particle with respect to an origin along the fixed axis. The relation applies even to more general rotation of a rigid body with one point fixed. In that case r is the position vector of the particle with respect to the fixed point taken as the origin. 8. The centre of mass of a system of n particles is defined as the point whose position vector is ri ∑m i R = M 9. Velocity of the centre of mass of a system of particles is given by V = P/M, where P is the linear momentum of the system. The centre of mass moves as if all the mass of the system is concentrated at this point and all the external forces act at it. If the total external force on the system is zero, then the total linear momentum of the system is constant. 10. The angular momentum of a system of n particles about the origin is n L = ri × pi i =∑1 The torque or moment of force on a system of n particles about the origin is τ = ∑ri × Fi 1 The force Fi acting on the ith particle includes the external as well as internal forces. Assuming Newton’s third law of motion and that forces between any two particles act along the line joining the particles, we can show τint = 0 and Reprint 2025-26 124 PHYSICS dL = τ ext dt 11. A rigid body is in mechanical equilibrium if (1) it is in translational equilibrium, i.e., the total external force on it is zero : Fi = 0 , ∑ and (2) it is in rotational equilibrium, i.e. the total external torque on it is zero : Fi = 0 . ∑ τi = ∑ri × 12. The centre of gravity of an extended body is that point where the total gravitational torque on the body is zero. 13. The moment of intertia of a rigid body about an axis is defined by the formula I m i ri2 =∑ where ri is the perpendicular distance of the ith point of the body from the axis. The 1 2 kinetic energy of rotation is K = Iω . 2 POINTS TO PONDER 1. To determine the motion of the centre of mass of a system no knowledge of internal forces of the system is required. For this purpose we need to know only the external forces on the body. 2. Separating the motion of a system of particles as the motion of the centre of mass, (i.e., the translational motion of the system) and motion about (i.e. relative to) the centre of mass of the system is a useful technique in dynamics of a system of particles. One example of this technique is separating the kinetic energy of a system of particles K as the kinetic energy of the system about its centre of mass K′ and the kinetic energy of the centre of mass MV2/2, K = K′ + MV2/2 3. Newton’s Second Law for finite sized bodies (or systems of particles) is based in Newton’s Second Law and also Newton’s Third Law for particles. 4. To establish that the time rate of change of the total angular momentum of a system of particles is the total external torque in the system, we need not only Newton’s second law for particles, but also Newton’s third law with the provision that the forces between any two particles act along the line joining the particles. 5. The vanishing of the total external force and the vanishing of the total external torque are independent conditions. We can have one without the other. In a couple, total external force is zero, but total torque is non-zero. 6. The total torque on a system is independent of the origin if the total external force is zero. 7. The centre of gravity of a body coincides with its centre of mass only if the gravitational field does not vary from one part of the body to the other. Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 125 8. The angular momentum L and the angular velocity ω are not necessarily parallel vectors. However, for the simpler situations discussed in this chapter when rotation is about a fixed axis which is an axis of symmetry of the rigid body, the relation L = Iω holds good, where I is the moment of the inertia of the body about the rotation axis. EXERCISES 6.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ? 6.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus. 6.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ? 6.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b. 6.5 Show that a.(b × c) is equal in magnitude to the volume of the parallelepiped formed on the three vectors , a, b and c. 6.6 Find the components along the x, y, z axes of the angular momentum l of a particle, whose position vector is r with components x, y, z and momentum is p with components px, py and pz. Show that if the particle moves only in the x-y plane the angular momentum has only a z-component. 6.7 Two particles, each of mass m and speed v, travel in opposite directions along parallel lines separated by a distance d. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken. 6.8 A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are 36.9° and 53.1° respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end. Fig. 6.33 6.9 A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel. Reprint 2025-26 126 PHYSICS 6.10 Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time. 6.11 A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis? 6.12 (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction. (b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy? 6.13 A rope of negligible mass is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? What is the linear acceleration of the rope ? Assume that there is no slipping. 6.14 To maintain a rotor at a uniform angular speed of 200 rad s-1, an engine needs to transmit a torque of 180 N m. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is 100% efficient. 6.15 From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body. 6.16 A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick? 6.17 The oxygen molecule has a mass of 5.30 × 10-26 kg and a moment of inertia of 1.94 ×10-46 kg m2 about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule. Reprint 2025-26 CHAPTER SEVEN GRAVITATION 7.1 INTRODUCTION Early in our lives, we become aware of the tendency of all material objects to be attracted towards the earth. Anything