Physics - Rotation Question with Solution | TestHub

When body rolls dawn on inclined plane with velocity ${\mathrm{V}}_0$ at bottom then body has both rotational and translational kinetic energy. Therefore, by law of conservation of energy, $$\begin{array}{rl}& P.E.=K.E_{\text{trans }}+K.E_{\text{rotational }}\\ & =\frac{1}{2}m{V}_0^2+\frac{1}{2}I{\omega }^2\\ & =\frac{1}{2}m{V}_0^2+\frac{1}{2}m{k}^2\frac{V_0^2}{R_0^2}\dots \text{ (i) }\\ & \left[\because \mathrm{I}=m{k}^2,\omega =\frac{V}{R_0}\right]\end{array}$$ When body is sliding down then body has only translatory motion. $$\begin{array}{rl}\therefore P.E.& =K.E_{\text{trans }}\\ & =\frac{1}{2}m{\left(\frac{5}{4}{v}_0\right)}^2\end{array}$$ Dividing (i) by (ii) we get $$\begin{array}{rl}& \frac{P.E.}{P.E.}=\frac{\frac{1}{2}m{v}_0^2\left[1+\frac{K^2}{R_0^2}\right]}{\frac{1}{2}\times \frac{25}{16}\times m{V}_0^2}\\ & =\frac{25}{16}=1+\frac{K^2}{R_0^2}\Rightarrow \frac{K^2}{R_0^2}=\frac{9}{16}\\ & \text{ or, }K=\frac{3}{4}{R}_0.\end{array}$$