Question

As shown in the figure, a solid cylinder of mass $m$ and radius $R$ starts rolling down an inclined plane of inclination $\theta$. Sufficient friction is present for the pure rolling of the cylinder. Determine the speed of its centre of mass when the cylinder has fallen a height $H=1m$.
(Take $g=10m/s^2$)

• A. $\sqrt{\frac{20}{3}}m/s$
• B. $\sqrt{10}m/s$
• C. $\sqrt{\frac{10}{3}}m/s$
• D. $\sqrt{\frac{40}{3}}m/s$

Answer 1

Answer: (D)

$\sqrt{\frac{40}{3}}m/s$

The situation is shown in the figure:-

For no slipping there will be no loss of mechanical energy.

Energy at position $1$ is $E_1=mgH$

Energy at position $2$ is $E_2=\frac{1}{2}m{V}_{cm}^2+\frac{1}{2}{I}_{cm}{\omega }^2$

$\because \omega =\frac{V_{cm}}{R}$ (For pure rolling)

$I_{cm}=\frac{m{R}^2}{2}$
$\Rightarrow E_2=\frac{3}{4}m{V}_{cm}^2$

From the law of conservation of energy:-

$E_1=E_2\Rightarrow mgH=\frac{3}{4}m{V}_{cm}^2\Rightarrow V_{cm}=\sqrt{\frac{4}{3}gH}=\sqrt{\frac{40}{3}}m/s$

Correct Option - (D)