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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= d ω/dt 4 Mass M Moment of inertia I 5 Force F = Ma Torque τ = Iα 6 Work dW = F ds Work W = τ dθ 7 Kinetic energy K = Mv /2 Kinetic energy K = I ω2/2 8 Power P = F v Power P = τω 9 Linear momentum p = Mv Angular momentum L = I ω Figure 6.30 shows a cross-section of a rigid If there are more than one forces acting on the body, the work done by all of them can be body 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 page; see Fig. 6.29). As said above we need to Denoting the magnitudes of the torques due to the different forces asτ1,τ2, … etc, consider only those forces which lie in planes perpendicular to the axis. Let F b1 one such dW = ( τ1+ τ2 + ...θd typical force acting as shown on a particle of Remember, the forces giving rise to the the body at point P 1ith its line of action in a torques act on different particles, but the plane perpendicular to the axis. For convenience angular displacement d θ 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 P 1 parallel to the fixed axis, the magnitudeτ of the describes a circular path of radius r with centre total torque is just the algebraic sum of the 1 C on the axis; CP 1 r 1 magnitudes of the torques, i.e., τ = τ1+ τ2+ ..... In time ∆t, the point moves to the position We, therefore, have P ′. The displacement of the particle ds , 1 1 dW = τd θ (6.39) therefore, has magnitude ds = r d 1 1 θ and This expression gives the work done by the direction tangential at P 1o the circular path as shown. Here d θ is the angular displacement total (external) torque τ which acts on the body rotating about a fixed axis. Its similarity with of the particle, θ = ∠P 1P′1 .The work done by the corresponding expression the force on the particle is dW = F . ds = F ds cos φ = F (r d θ)sinα dW= F ds 1 1 1 1 1 1 1 1 1 for linear (translational) motion is obvious. where φ1is the angle between F1and the tangent Dividing both sides of Eq. (6.39) by dt gives at P1,nd α 1is the angle between F an1 the dW dθ radius vector OP ; φ + α = 90°. P = =τ = τω 1 1 1 dt d t The torque due to F ab1ut the origin is or P = τω (6.40) OP × F . Now OP = OC + OP . [Refer to 1 1 1 1 This is the instantaneous power. Compare Fig.6.17(b).] SinceOCisalongtheaxis,thetorque 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 F is the case of linear motion, τ = CP×F ; it is directed along the axis of rotation 1 1 P = Fv and has a magnitude τ1= r1F1sinα , Therefore, In a perfectly rigid body there is no internal dW = τ dθ motion. The work done by external torques is 1 1 Reprint 2026-27