The figure shows a frictionless horizontal track which smoothly turns into a frictionless inclined s

Question

The figure shows a frictionless horizontal track which smoothly turns into a frictionless inclined surface of inclination $	heta$. A sphere of radius $R$ rolling on the horizontal surface with angular speed $\omega_0$ moves towards the inclined surface. The maximum distance moved up the inclined surface is

Professor PC Thomas Classes · Answer

For rolling condition, the velocity of center of mass is $v=\omega_0 R$ Since the inclined surface is smooth therefore, the rotational kinetic energy of the sphere is not converted into potential energy. If $l$ is the distance moved along the inclined plane, then using energy conservation $ \begin{aligned} & -\Delta K_{	ext {trans }}=\Delta U \ & \frac{1}{2} m v^2=m g l \sin 	heta \end{aligned} $ \or $\quad l=\frac{\omega_0^2 R^2}{2 g \sin 	heta}$