Question

A solid cylinder of mass ' $M$ ' and radius ' $R$ ' rolls down an inclined plane of length ' $L$ ' and height ' $h$ ' without slipping. Find the speed of its centre of mass when the cylinder reaches its bottom?

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

Answer 1

Answer: (C)

The given situation is shown in the following figure.

When cylinder reaches at bottom, then its whole potential energy is converted into its rotational kinetic energy and linear kinetic energy of its centre of mass.
Hence, $Mgh=\frac{1}{2}I{\omega }^2+\frac{1}{2}M{v}_{COM}^2$
$=\frac{1}{2}\times \frac{M{R}^2}{2}\times {\left(\frac{v_{COM}}{R}\right)}^2+\frac{1}{2}M{v}_{COM}^2$
$[\because I=\frac{M{R}^2}{2}$ and $\omega =\frac{v_{COM}}{R}]$
$=\frac{M{v}_{COM}^2}{4}+\frac{M{v}_{COM}^2}{2}$
$\Rightarrow Mgh=\frac{3}{4}M{v}_{COM}^2$
$\Rightarrow v_{COM}^2=\frac{4gh}{3}$
$\Rightarrow v_{COM}=\sqrt{\frac{4gh}{3}}$