Physics - Rotation Question with Solution | TestHub

By energy conservation $$\begin{array}{l}(\mathrm{K}.\mathrm{E}{)}_{\mathrm{i}}+(\mathrm{P}.\mathrm{E}{)}_{\mathrm{i}}=(\mathrm{K}.\mathrm{E}{)}_{\mathrm{f}}+(\mathrm{P}.\mathrm{E}{)}_{\mathrm{f}}\\ (\mathrm{K}.\mathrm{E}{)}_{\mathrm{i}}=0,(\mathrm{P}.\mathrm{E}{)}_{\mathrm{i}}=\mathrm{mgh},(\mathrm{P}.\mathrm{E}{)}_{\mathrm{f}}=0\\ (\mathrm{K}.\mathrm{E}{)}_{\mathrm{f}}=1/2\mathrm{I}{\omega }^2+1/2{\mathrm{mv}}_{\mathrm{cm}}^2\end{array}$$ For solid cylinder, moment of inertia, $\mathrm{I}=1/2{\mathrm{mR}}^2$ $(\mathrm{K}.\mathrm{E}{)}_{\mathrm{i}}+(\mathrm{P}.\mathrm{E}{)}_{\mathrm{i}}=(\mathrm{K}.\mathrm{E}{)}_{\mathrm{f}}+(\mathrm{P}.\mathrm{E}{)}_{\mathrm{f}}$ $$\begin{array}{l}\text{ so, }\mathrm{mgh}=1/2\left(1/2{\mathrm{mR}}^2\right)\left(\frac{{\mathrm{v}}_{\mathrm{cm}}^2}{{\mathrm{R}}^2}\right)+1/2{\mathrm{mv}}_{\mathrm{cm}}^2\\ \Rightarrow {\mathrm{v}}_{\mathrm{cm}}=\sqrt{4\mathrm{gh}/3}\end{array}$$