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

AIIMSAIIMS 2012

Solution:

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 $K_{t_0}=\frac{7}{10} M R^{2} \omega^{2}$
$\frac{K_{r_{0}}}{K_{t_{0}}}=\frac{\frac{1}{5} M R^{2} \omega^{2}}{\frac{7}{10} M R^{2} \omega^{2}}=\frac{2}{7}$