Question

A solid cylinder starts rolling down on an inclined plane from its top and $V$ is the velocity of its centre of mass on reaching the bottom of inclined plane. If a block starts sliding down on an identical inclined plane but smooth, from its top, then the velocity of block on reaching the bottom of inclined plane is :

• A. $\frac{V}{\sqrt{2}}$
• B. $\sqrt{2}V$
• C. $\sqrt{\frac{3}{2}}V$
• D. $\sqrt{\frac{2}{3}}V$

Answer 1

The correct option is C $\sqrt{\frac{3}{2}}V$

Apply conservation of Total mechanical energy:

Loss in gravitational potential energy = gain in rotational kinetic energy +gain in translational kinetic energy
$mgh=\frac{1}{2}m{V}^2+\frac{1}{2}\left(\frac{m}{2}{R}^2\right){\left(\frac{V}{R}\right)}^2$
$mgh=\frac{3}{4}m{V}^2$
$V=\sqrt{\frac{4}{3}gh}$

for block
Again apply conservation of Total mechanical energy:
$mgh=\frac{1}{2}m{V}_0^2$
$V_0=\sqrt{2gh}$
$=\sqrt{\frac{2}{3}}V$