A solid cylinder of mass $M$ and radius $R$ rolls without slipping on an inclined plane of length $L$ and…
A solid cylinder of mass $M$ and radius $R$ rolls without slipping on an inclined plane of length $L$ and height $h$. What is the speed of its centre of mass when the cylinder reaches its bottom?
$\sqrt{2 g h}$
$\sqrt{\frac{3}{4} g h}$
$\sqrt{\frac{4}{3} g h}$
$\sqrt{4 g h}$
Solution
Here K.E. of centre of mass when cylinder reached the bottom $=\frac{1}{2} m v^2+\frac{1}{2} \mathrm{I} \omega^2 \ldots(\mathrm{i})$
Here $\mathrm{I}=\frac{1}{2} M K^2$ and $\omega=\frac{V}{R}$
$\begin{aligned}
& \therefore \mathrm{KE}=\frac{1}{2} m v^2+\frac{1}{2} M K^2 \cdot \frac{V^2}{r^2} \\
& =\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right]
\end{aligned}$
But for solid cylinder $K=\frac{R}{\sqrt{2}}$ or $\frac{K^2}{R^2}=\frac{1}{2}$
$\therefore K E=\frac{3}{4} m v^2$
PE of the solid cylinder at height $h$ is
$\mathrm{PE}=m g h$
From eq. (ii) and (iii)
$\begin{aligned}
& m g h=\frac{3}{4} m v^2 \\
& \Rightarrow v=\sqrt{\frac{4}{3} g h}
\end{aligned}$