Question

The linear speed of a uniform spherical shell after rolling down an inclined plane of vertical height $h$ from rest, is (assume no slipping):

• A. $\sqrt{\frac{10gh}{7}}$
• B. $\sqrt{\frac{4gh}{5}}$
• C. $\sqrt{\frac{6gh}{5}}$
• D. $\sqrt{2gh}$

Answer 1

The correct option is C $\sqrt{\frac{6gh}{5}}$

Using energy conservation,
$mgh=\frac{1}{2}I{w}^2+\frac{1}{2}m{V}^2$
$I_{\text{shell}}=\frac{2}{3}m{R}^2$
$mgh=\frac{1}{2}\times \frac{2}{3}m{R}^2.\frac{V^2}{R^2}+\frac{1}{2}m{V}^2$
(since $V=wR$ in case of pure rolling)
$mgh=\frac{1}{3}m{V}^2+\frac{1}{2}m{V}^2$
$mgh=\frac{5}{6}m{V}^2\Rightarrow V^2=\frac{6gh}{5}$
$\Rightarrow V=\sqrt{\frac{6gh}{5}}$