A solid cylinder of mass $m$ and radius $\mathrm{R}$ rolls down an inclined plane of height $\mathrm{h}$…
A solid cylinder of mass $m$ and radius $\mathrm{R}$ rolls down an inclined plane of height $\mathrm{h}$ without slipping. The speed of its centre of mass when it reaches the bottom is
$\sqrt{(2 \mathrm{gh})}$
$\sqrt{\frac{4 \mathrm{gh}}{3}}$
$\sqrt{\frac{3 \mathrm{gh}}{4}}$
$\sqrt{\frac{4 \mathrm{~g}}{\mathrm{~h}}}$
Solution
$\begin{aligned} & \text { K. E. }=\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} \mathrm{mv}^2 \\ & \text { K.E. }=\frac{1}{2}\left(\frac{1}{2} \mathrm{mr}^2\right) \omega^2+\frac{1}{2} \mathrm{mv}^2 \\ & =\frac{1}{4} \mathrm{mv}^2+\frac{1}{2} \mathrm{mv}^2=\frac{3}{4} \mathrm{mv}^2\end{aligned}$
Now, gain in K.E. $=$ Loss in P.E.
$\frac{3}{4} \mathrm{mv}^2=\mathrm{mgh} \Rightarrow \mathrm{v}=\sqrt{\left(\frac{4}{3}\right) \mathrm{gh}}$