Question

A circular ring starts rolling down on an inclined plane from its top. Let $V$ be 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. $\frac{V}{2}$
• B. $2V$
• C. $\frac{V}{\sqrt{2}}$
• D. $\sqrt{2}V$

Answer 1

The correct option is D $\sqrt{2}V$

For ring applying directly work-energy theorem,

loss in gravitational potential energy = gain in rotational kinetic energy +gain in translational kinetic energy

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

$mgh=m{V}^2$

$V=\sqrt{gh}$

For the block, Work Energy Theorem,

$mgh=\frac{1}{2}m{V}_1^2$

$V_1=\sqrt{2gh}$

$=\sqrt{2}V$