Q 11-06-001JEE MainTop questionMedium
A solid sphere rolls without slipping on a horizontal surface. What fraction of its total kinetic energy is rotational?
Answer: (B) $\tfrac27$
For a solid sphere $I = \tfrac25 mR^2$ and rolling gives $v = \omega R$.
$$K_{rot} = \tfrac12 \cdot \tfrac25 mR^2 \omega^2 = \tfrac15 mv^2, \qquad K_{total} = \tfrac12 mv^2 + \tfrac15 mv^2 = \tfrac{7}{10} mv^2$$
$$\frac{K_{rot}}{K_{total}} = \frac{1/5}{7/10} = \frac27$$
Solution by Sreeraj P, M.Sc Physics