angular displacement d θ is the same for all particles. Since all the torques considered are parallel to the fixed axis, the magnitude τ of the total torque is just the algebraic sum of the magnitudes of the torques, i.e., τ = τ + τ + ..... We, therefore, have W τ θ ( . ) This expression gives the work done by the total (external) torque τ which acts on the body rotating about a fixed axis.
Its similarity with the corresponding expression d W= F d s for linear (translational) motion is obvious. Dividing both sides of Eq. ( . ) by d t gives W P τω or P τω ( .
) This is the instantaneous power. Compare this expression for power in the case of rotational motion about a fixed axis with that of power in the case of linear motion, P = Fv In a perfectly rigid body there is no internal motion. The work done by external torques is Table . Comparison of Translational and Rotational Motion Linear Motion Rotational Motion about a Fixed Axis Displacement x Angular displacement θ Velocity v = d x /d t Angular velocity ω = d θ /d t Acceleration a = d v /d t Angular acceleration α = d ω / d t Mass M Moment of inertia I Force F = Ma Torque τ = I α Work dW = F d s Work W = τ d θ Kinetic energy K = Mv / Kinetic energy K = I ω / Power P = F v Power P = τω Linear momentum p = Mv Angular momentum L = I ω u therefore, not dissipated and goes on to increase the kinetic energy of the body.
The rate at which work is done on the body is given by Eq. ( . ). This is to be equated to the rate at which kinetic energy increases.
The rate of increase of kinetic energy is = We assume that the moment of inertia does not change