Question

A disc rolls down a plane of length $L$ and inclined at angle $\theta$, without slipping. Its velocity on reaching the bottom will be

• A. $\sqrt{\frac{2gL\sin \theta }{3}}$
• B. $\sqrt{\frac{4gL\sin \theta }{3}}$
• C. $\sqrt{4gL\sin \theta }$
• D. $\sqrt{\frac{10gL\sin \theta }{3}}$

Answer 1

The correct option is B $\sqrt{\frac{4gL\sin \theta }{3}}$

Let velocity of the disc at bottom $=v$
For the situation show in diagram,
Initial height of the disc
$h=L\sin \theta$
using energy conservation

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$
here, $I=\frac{m{R}^2}{2},\omega =\frac{v}{R}$
$\Rightarrow mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{m{R}^2}{2}{\left(\frac{v}{R}\right)}^2$

$\Rightarrow g\left(L\sin \theta \right)=\frac{1}{2}{v}^2+\frac{1}{4}{v}^2$

Therefore, $v=\sqrt{\frac{4gL\sin \theta }{3}}$

Hence, option $\left(A\right)$ is correct.