Physics - Rotation Question with Solution | TestHub

Here K.E. of centre of mass when cylinder reached the bottom $=\frac{1}{2}m{v}^2+\frac{1}{2}\mathrm{I}{\omega }^2\dots (\mathrm{i})$ Here $\mathrm{I}=\frac{1}{2}M{K}^2$ and $\omega =\frac{V}{R}$ $\begin{array}{rl}& \therefore \mathrm{KE}=\frac{1}{2}m{v}^2+\frac{1}{2}M{K}^2\cdot \frac{V^2}{r^2}\\ & =\frac{1}{2}m{v}^2\left[1+\frac{K^2}{R^2}\right]\end{array}$ But for solid cylinder $K=\frac{R}{\sqrt{2}}$ or $\frac{K^2}{R^2}=\frac{1}{2}$ $\therefore KE=\frac{3}{4}m{v}^2$ PE of the solid cylinder at height $h$ is $\mathrm{PE}=mgh$ From eq. (ii) and (iii) $\begin{array}{rl}& mgh=\frac{3}{4}m{v}^2\\ & \Rightarrow v=\sqrt{\frac{4}{3}gh}\end{array}$