Question

A solid sphere rolls without slipping on the roof. The ratio of its rotational kinetic energy and its total kinetic energy is

• A. 44597
• B. 44656
• C. 44599
• D. 44627

Answer 1

The Correct Option is C

Kinetic energy of sphere $K_{r_{0}}=\frac{1}{2} I \omega^{2}$ $\therefore$ Moment of inertia of sphere, $I=\frac{2}{5} M R^{2}$ $\therefore$ Rotational kinetic energy of sphere $K_{r_{0}}=\frac{1}{2} M R^{2} \omega^{2}$ Total energy of sphere $K_{r_{0}} =\frac{1}{2} I \omega^{2}+\frac{1}{2} M v^{2} $ $=\frac{1}{2} \times \frac{2}{5} M R^{2} \omega^{2}+\frac{1}{2} M R^{2} \omega^{2} $ $=M R^{2} \omega^{2}\left(\frac{1}{5}+\frac{1}{2}\right) $ $=\frac{7}{10} M R^{2} \omega^{2}$ Total energy of sphere $K_{t_0}=\frac{7}{10} M R^{2} \omega^{2}$ $\frac{K_{r_{0}}}{K_{t_{0}}}=\frac{\frac{1}{5} M R^{2} \omega^{2}}{\frac{7}{10} M R^{2} \omega^{2}}=\frac{2}{7}$