Question

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

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

Answer 1

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

Potential energy of the solid cylinder of mass $M$ at height($h)=Mgh$
$K.E$ of center of mass when reached at bottom
$=\frac{1}{2}M{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}M{v}^2+\frac{\mathrm{M}}{2}{k}^2{v}^2/R^2=\frac{1}{2}M{v}^2(1+\frac{k^2}{R^2})$
For a solid cylinder
$\frac{k^2}{R^2}=\frac{1}{2}\therefore K.E.=\frac{3}{4}M{v}^2\therefore Mgh=\frac{3}{4}M{v}^{2}v=\sqrt{\frac{4}{3}gh}$