Question

Solid cylinder of mass $M$ and radius $R$ rolls down an inclined plane of height '$h$' and inclination '$\theta$'. The speed of its centre of mass, when the cylinder reaches the bottom of the inclined plane, is:

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

Answer 1

The correct option is C $\sqrt{\frac{4gh}{3}}$

Using work energy theorem ,

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