SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

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 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 /d , t we get d t ω α      = Equating rates of work done and of increase in kinetic energy, τω ω α ( . ) Eq.

( . ) is similar to Newton’s second law for linear motion expressed symbolically as F = ma Just as force produces acceleration, torque produces angular acceleration in a body. The angular acceleration is directly proportional to the applied torque and is inversely proportional to the moment of inertia of the body. In this respect, Eq.( .

) can be called Newton’s second law for rotational motion about a fixed axis. Example . A cord of negligible mass is wound round the rim of a fly wheel of mass kg and radius cm. A steady pull of N is applied on the cord as shown in Fig.

. . The flywheel is mounted on a horizontal axle with frictionless bearings. (a) Compute the angular acceleration of the wheel.

(b) Find the work done by the pull, when 2m of the cord is unwound. (c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest. (d) Compare answers to parts (b) and (c).

Answer Fig. . (a) We use