A solid sphere is rolling down a rough inclined plane and a block is slipping down a frictionless in

Question

A solid sphere is rolling down a rough inclined plane and a block is slipping down a frictionless inclined plane of the same angle and height. Then the velocity of both at the bottom is related as ($v_s=$ velocity of sphere and $v_b=$ velocity of block)

Professor PC Thomas Classes · Answer

(a) : For the slipping block, all potential energy converts to translational kinetic energy: $mgh = \frac{1}{2}mv_b^2 \implies v_b = \sqrt{2gh}$. \ For the rolling sphere, potential energy converts to both translational and rotational kinetic energy: $mgh = \frac{1}{2}mv_s^2 + \frac{1}{2}I\omega^2 = \frac{7}{10}mv_s^2 \implies v_s = \sqrt{\frac{10}{7}gh}$. \ Since $\frac{10}{7} < 2$, the translational velocity of the sphere $v_s$ will be less than $v_b$.