Physics - Rotation Question with Solution | TestHub

Kinetic energy of rotation is $K_{\text{rot }}=\frac{1}{2}l{\omega }^2=\frac{1}{2}M{K}^2\frac{v^2}{R^2}$ where, $k$ is radius of gyration. Kinetic energy of translation is $K_{\text{trans }}=\frac{1}{2}M{v}^2$ Thus total energy, $\begin{array}{rl}E& =K_{\text{rot }}+K_{\text{trans }}\\ & =\frac{1}{2}M{K}^2\frac{v^2}{R^2}+\frac{1}{2}M{v}^2\\ & =\frac{1}{2}M{v}^2\left(\frac{K^2}{R^2}+1\right)\\ & =\frac{1}{2}\frac{M{v}^2}{R^2}\left(K^2+R^2\right)\end{array}$ Hence, $\begin{array}{rl}\frac{K_{\text{rot }}}{\text{ Total energy, }E}& =\frac{\frac{1}{2}M{K}^2\frac{v^2}{R^2}}{\frac{1}{2}\frac{M{v}^2}{R^2}\left(K^2+R^2\right)}\\ & =\frac{K^2}{K^2+R^2}\end{array}$