Question

A block of mass $m$ is suspended from a pulley in the form of a circular disc of mass $m$ and radius $R$. The system is released from rest. Find the angular velocity of the disc when the block has dropped by height $h$. [Assume there is no slipping between string & pulley]

• A. $\frac{1}{R}\sqrt{\frac{4gh}{3}}$
• B. $\frac{1}{R}\sqrt{\frac{2gh}{3}}$
• C. $R\sqrt{\frac{2gh}{3}}$
• D. $R\sqrt{\frac{4gh}{3}}$

Answer 1

The correct option is A $\frac{1}{R}\sqrt{\frac{4gh}{3}}$

By conservation of mechanical energy:-
$T{E}_i=T{E}_f$
$\Rightarrow \text{Initial P.E of block = Final P.E of block + Final K.E of block and pulley}$
Take final P.E of block as zero (reference point). Let the final velocity of block be $v$ and final angular velocity of disc be $\omega$.
Then, $mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
As we know,
$v=\omega R$ (for no slipping condition)
and, $I_{disc}=\frac{m{R}^2}{2}$

$\Rightarrow mgh=\frac{1}{2}m{\omega }^2{R}^2+\frac{1}{2}\frac{m{R}^2}{2}{\omega }^2$
$\Rightarrow mgh=\frac{3}{4}m{\omega }^2{R}^2$
$\therefore \omega =\sqrt{\frac{4gh}{3R^2}}=\frac{1}{R}\sqrt{\frac{4gh}{3}}$