Question

A cord of negligible mass is wound round the rim of a fly wheel of mass $20$kg and radius $20$cm. A steady pull of $25N$ 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 $2$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).

Answer 1

(a) We use $I\alpha =\tau$
the torque $\tau =FR$
$=25\times 0.20Nm$(as $R=0.20m$)
$=5.0Nm$
$I=M.I.$ of flywheel about its axis $=\frac{M{R}^2}{2}$
$=\frac{20.0\times (0.2{)}^2}{2}=0.4kg$ $m^2$
$\alpha =$ angular acceleration
$=5.0Nm/0.4kg$ $m^2=12.5s^{-2}$
(b) Work done by the pull unwinding $2m$ of the cord $=25N\times 2m=50J$
(c) Let $\omega$ be the final angular velocity. The kinetic energy gained $=\frac{1}{2}I{\omega }^2$,
since the wheel starts from rest. Now,
${\omega }^2={\omega }_0^2+2a\theta ,{\omega }_0=0$
The angular displacement $\theta =$ length of unwound string/ radius of wheel $=2m/0.2m=10$rad
${\omega }^2=2\times 12.5\times 10.0=250(rad/s{)}^2$
$\therefore K.E.$ gained $=\frac{1}{2}\times 0.4\times 250=50J$
(d) The answer are the same, i.e. the kinetic energy gained by the wheel$=$work done by the force. There is no loss of energy due to friction.