5.8 THE CONSERVATION OF MECHANICAL a ball of mass m being dropped from a cliff of ENERGY height H. For simplicity we demonstrate this important principle for one-dimensional motion. Suppose that a body undergoes displacement ∆x under the action of a conservative force F. Then from the WE theorem we have, ∆K = F(x) ∆x If the force is conservative, the potential energy function V(x) can be defined such that − ∆V = F(x) ∆x The above equations imply that Fig. 5.5 The conversion of potential energy to kinetic ∆K + ∆V = 0 energy for a ball of mass m dropped from a ∆(K + V ) = 0 (5.10) height H. Reprint 2025-26 WORK, ENERGY AND POWER 79 The total mechanical energies E0, Eh, and EH Answer (i) There are two external forces on of the ball at the indicated heights zero (ground the bob : gravity and the tension (T ) in the level), h and H, are string. The latter does no work since the displacement of the bob is always normal to the EH = mgH (5.11 a) string. The potential energy of the bob is thus 1 2 E h = mgh + mv h (5.11 b) associated with the gravitational force only. The 2 2 total mechanical energy E of the system is E0 = (1/2) mvf (5.11 c) conserved. We take the potential energy of the The constant force is a special case of a spatially system to be zero at the lowest point A. Thus, dependent force F(x). Hence, the mechanical at A : energy is conserved. Thus EH = E0 1 2 1 2 E = mv0 (5.12) or, mgH = mv f 2 2 v f = 2 gH [Newton’s Second Law] a result that was obtained in section 5.7 for a where TA is the tension in the string at A. At thefreely falling body. highest point C, the string slackens, as the Further, tension in the string (TC) becomes zero. EH = Eh Thus, at Cwhich implies, 2 1 2 v h = 2 g(H − h) (5.11 d) E = mv c + 2mgL (5.13) 2 and is a familiar result from kinematics. At the height H, the energy is purely potential. mvc2It is partially converted to kinetic at height h and mg = [Newton’s Second Law] (5.14) L is fully kinetic at ground level. This illustrates the conservation of mechanical energy. where vC is the speed at C. From Eqs. (5.13) ⊳ and (5.14) Example 5.7 A bob of mass m is suspended 5 by a light string of length L . It is imparted a E = mgL horizontal velocity vo at the lowest point A 2 such that it completes a semi-circular Equating this to the energy at A trajectory in the vertical plane with the string 5 m 2 becoming slack only on reaching the topmost mgL = v 0 point, C. This is shown in Fig. 5.6. Obtain an 2 2 expression for (i) vo; (ii) the speeds at points or, v 0 = 5 gL B and C; (iii) the ratio of the kinetic energies (ii) It is clear from Eq. (5.14) (KB/KC) at B and C. Comment on the nature of the trajectory of the bob after it reaches vC = gL the point C. At B, the energy is 1 2 E = mv B + mgL 2 Equating this to the energy at A and employing the result from (i), namely v 02 = 5 gL , 1 2 1 2 mv B + mgL = mv 0 2 2 5 = m g L Fig. 5.6 2 Reprint 2025-26 80 PHYSICS ∴ vB = 3 gL k x m2 W = + (5.16) 2 (iii) The ratio of the kinetic energies at B and C is : 1 2 mv B K B 2 3 = = 2 1 K C 1 mvC 2 At point C, the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise the bob will continue on its circular path and complete the revolution. ⊳

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.

6.11 DYNAMICS OF ROTATIONAL MOTION ABOUT A FIXED AXIS Table 6.2 lists quantities associated with linear motion and their analogues in rotational motion. We have already compared kinematics of the two motions. Also, we know that in rotational motion moment of inertia and torque play the same role as mass and force respectively in linear motion. Given this we should be able to guess what the other analogues indicated in the table are. For Fig. 6.30 Work done by a force F1 acting on a example, we know that in linear motion, work particle of a body rotating about a fixed done is given by F dx, in rotational motion about axis; the particle describes a circular path a fixed axis it should be τdθ , since we already with centre C on the axis; arc P1P′1(ds1) gives the displacement of the particle. know the correspondence d x → dθ and F → τ . Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 119 Table 6.2 Comparison of Translational and Rotational Motion Linear Motion Rotational Motion about a Fixed Axis 1 Displacement x Angular displacement θ 2 Velocity v = dx/dt Angular velocity ω = dθ/dt 3 Acceleration a = dv/dt Angular acceleration α = dω/dt 4 Mass M Moment of inertia I 5 Force F = Ma Torque τ = I α 6 Work dW = F ds Work W = τ dθ 7 Kinetic energy K = Mv2/2 Kinetic energy K = Iω2/2 8 Power P = F v Power P = τω 9 Linear momentum p = Mv Angular momentum L = Iω If there are more than one forces acting on Figure 6.30 shows a cross-section of a rigid the body, the work done by all of them can bebody rotating about a fixed axis, which is taken added to give the total work done on the body. as the z-axis (perpendicular to the plane of the Denoting the magnitudes of the torques due to page; see Fig. 6.29). As said above we need to the different forces as τ1, τ2, … etc,consider only those forces which lie in planes perpendicular to the axis. Let F1 be one such d W = (τ1 + τ2 + ...)dθ typical force acting as shown on a particle of Remember, the forces giving rise to the the body at point P1 with its line of action in a torques act on different particles, but the plane perpendicular to the axis. For convenience angular displacement dθ is the same for all we call this to be the x′–y′ plane (coincident particles. Since all the torques considered are with the plane of the page). The particle at P1 parallel to the fixed axis, the magnitude τ of the describes a circular path of radius r1 with centre total torque is just the algebraic sum of the C on the axis; CP1 = r1. magnitudes of the torques, i.e., τ = τ1 + τ2 + ..... In time ∆t, the point moves to the position We, therefore, have P1′. The displacement of the particle ds1, d W = τdθ (6.39) therefore, has magnitude ds1 = r1dθ and This expression gives the work done by the direction tangential at P1 to the circular path total (external) torque τ which acts on the bodyas shown. Here dθ is the angular displacement rotating about a fixed axis. Its similarity withof the particle, dθ = ∠P1CP1′ .The work done by the corresponding expression the force on the particle is dW= F ds dW1 = F1. ds1= F1ds1 cosφ1= F1(r1 dθ)sinα1 for linear (translational) motion is obvious. where φ1 is the angle between F1 and the tangent Dividing both sides of Eq. (6.39) by dt gives at P1, and α1 is the angle between F1 and the d W dθ P = = τ = τωradius vector OP1; φ1 + α1 = 90°. d t d t The torque due to F1 about the origin is or P = τω (6.40) OP1 × F1. Now OP1 = OC + OP1. [Refer to This is the instantaneous power. Compare Fig. 6.17(b).] Since OC is along the axis, the torque this expression for power in the case of rotational resulting from it is excluded from our motion about a fixed axis with that of power in consideration. The effective torque due to F1 is the case of linear motion, τ1= CP × F1; it is directed along the axis of rotation P = Fv and has a magnitude τ1= r1F1 sinα , Therefore, In a perfectly rigid body there is no internal dW1 = τ1dθ motion. The work done by external torques is Reprint 2025-26 120 PHYSICS therefore, not dissipated and goes on to increase Answer the kinetic energy of the body. The rate at which work is done on the body is given by Eq. (6.40). This is to be equated to the rate at which kinetic energy increases. The rate of increase of kinetic energy is d  I ω2  ( 2ω) d ω I d t  2 = 2 d t We assume that the moment of inertia does not change with time. This means that the mass of the body does not change, the body remains rigid and also the axis does not change its position with respect to the body. Since α= dω/d,t we get d  I ω2  I ω α Fig. 6.31 dt  2 = (a) We use I α = τ Equating rates of work done and of increase the torque τ = F Rin kinetic energy, = 25 × 0.20 Nm (as R = 0.20m) τω= I ωα = 5.0 Nm τ = I α (6.41) I = Moment of inertia of flywheel about its Eq. (6.41) is similar to Newton’s second law 2 for linear motion expressed symbolically as axis = MR F = ma 2 Just as force produces acceleration, torque 2 20.0 × (0.2) produces angular acceleration in a body. The = = 0.4 kg m2 2 angular acceleration is directly proportional to α = angular acceleration the applied torque and is inversely proportional = 5.0 N m/0.4 kg m2 = 12.5 s–2 to the moment of inertia of the body. In this (b) Work done by the pull unwinding 2m of the respect, Eq.(6.41) can be called Newton’s second cord law for rotational motion about a fixed axis. = 25 N × 2m = 50 J u (c) Let ω be the final angular velocity. The Example 6.12 A cord of negligible mass is 1 2 kinetic energy gained = Iω , wound round the rim of a fly wheel of mass 2 20 kg and radius 20 cm. A steady pull of since the wheel starts from rest. Now, 25 N is applied on the cord as shown in 2 2 ω = ω0 + 2αθ, ω0 = 0 Fig. 6.31. The flywheel is mounted on a horizontal axle with frictionless bearings. The angular displacement θ = length of unwound string / radius of wheel (a) Compute the angular acceleration of = 2m/0.2 m = 10 rad the wheel. (b) Find the work done by the pull, when ω2 = 2 × 12 .5 × 10 .0 = 250 (rad/s )2 2m of the cord is unwound. ∴ (c) Find also the kinetic energy of the wheel at this point. Assume that the (d) The answers are the same, i.e. the kinetic energy wheel starts from rest. gained by the wheel = work done by the force. (d) Compare answers to parts (b) and (c). There is no loss of energy due to friction. ⊳ Reprint 2025-26 SYSTEMS OF PARTICLES AND ROTATIONAL MOTION 121