$A$ cord of negligible mass is wound round the rim of a flywheel of mass $20 \; kg$ and radius $20 \; cm$. $A$ steady pull of $25 \; N$ is applied on the cord as shown in the figure. 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 $2 \; m$ 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 $(c)$.

(N/A) We use $I \alpha = \tau$.
The torque $\tau = F R = 25 \times 0.20 \; Nm = 5.0 \; Nm$ (as $R = 0.20 \; m$).
$I$ (Moment of inertia of flywheel about its axis) $= \frac{M R^2}{2} = \frac{20.0 \times (0.2)^2}{2} = 0.4 \; kg \cdot m^2$.
$\alpha$ (angular acceleration) $= \frac{\tau}{I} = \frac{5.0 \; Nm}{0.4 \; kg \cdot m^2} = 12.5 \; rad/s^2$.
$(b)$ Work done by the pull unwinding $2 \; m$ of the cord $= F \times d = 25 \; N \times 2 \; m = 50 \; J$.
$(c)$ Let $\omega$ be the final angular velocity. The kinetic energy gained $= \frac{1}{2} I \omega^2$.
Since the wheel starts from rest,$\omega^2 = \omega_0^2 + 2 \alpha \theta$,where $\omega_0 = 0$.
The angular displacement $\theta = \frac{\text{length of unwound string}}{R} = \frac{2 \; m}{0.2 \; m} = 10 \; rad$.
$\omega^2 = 2 \times 12.5 \times 10.0 = 250 \; (rad/s)^2$.
$\text{K.E. gained} = \frac{1}{2} \times 0.4 \times 250 = 50 \; J$.
$(d)$ The answers are the same,i.e.,the kinetic energy gained by the wheel equals the work done by the force. There is no loss of energy due to friction.