Question

A disc of mass $m$ and radius $r$ is free to rotate about its center as shown in the figure. A string is wrapped over its rim and a block of mass $m$ is attached to the free end of the string. The system is released from rest. The speed of the block as it descends through a height $h,$ is?

Answer 1

From the law of energy conservation,the potential energy of the block gets converted to kinetic energy of block and pulley:

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$ where $I$ is the moment of inertia of pulley$=\frac{1}{2}m{R}^2$

Also, $v=R\omega$

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{1}{2}m{R}^2\right)(\frac{v}{R}{)}^2$

$mgh=\frac{1}{2}m{v}^2+\frac{1}{4}m{v}^2$

$mgh=\frac{3}{4}m{v}^2$

$v=\sqrt{\frac{4gh}{3}}$