Question

A massless string is wrapped around a disc of mass $M$ and radius $R$. Another end is tied

to a mass $m$ which is initially at height $h$ from ground level as shown in the figure. If the

mass is released then its velocity while touching the ground level will be :

• A. $\sqrt{2gh}$
• B. $\sqrt{2gh\frac{M}{m}}$
• C. $\sqrt{2gh\frac{m}{M}}$
• D. $\sqrt{\frac{4mgh}{2m+M}}$

Answer 1

The correct option is D $\sqrt{\frac{4mgh}{2m+M}}$

Let the tension in the string be $T$.

Let the acceleration of the mass be $a$.

Let the angular acceleration of the disc be $\alpha$.

Since the string moves without slipping on the disc,

$a=R\alpha$

$a=\frac{mg-T}{m}$

$T=mg-ma$ --- (1)

$\alpha =\frac{\tau }{I}$,

where $\tau =$ torque $=TR$

and $I=\frac{{MR}^2}{2}$

$T=\frac{Ma}{2}$ ----- (2)

From (1) and (2),

$a=\frac{2mg}{m+2M}$

$v^2=u^2+2asu=0s=hv=\sqrt{\frac{4mgh}{2m+M}}$