A sphere rolls down an inclined plane through a height h. Its velocity at the bottom would be
\[ mgh=\frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}= \frac{1}{2}mv^{2}+ \frac{1}{2}(\frac{2}{5}mR^{2})\frac{v^{2}}{R^{2}} \]
Solving we get,
\[ v= \sqrt[]{\frac{10}{7}gh} \]
A body rolls down an inclined plane. If its kinetic energy of rotation is 40% of its kinetic energy of translation, then the body is
Given, rotational kinetic energy is 40% of total energy. so,
\[ \frac{1}{2}I\omega^{2}=\frac{40}{100}\left( \frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2} \right) \]
Solving ,
\[ I = \frac{2}{5}mR^{2} \]
Object is Solid Sphere.