A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic en

Question

A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is\

Professor PC Thomas Classes · Answer

(d) : Let $K_{1}$ be the linear kinetic energy of the centre of mass of the sphere and let $K_{2}$ be the rotational kinetic energy of the solid sphere.\ Linear kinetic energy $=\frac{1}{2} m v^{2}$, where ' $m$ ' and ' $v$ ' are mass and velocity of the solid sphere.\ Rotational kinetic energy of the solid sphere $=\frac{1}{2} I \omega^{2}$, where\ ' $P$ ' is M.I and ' $\omega$ ' is angular velocity of solid sphere.\ For solid sphere, $I=\frac{2}{5} m R^{2}$\ So, $\frac{K_{1}}{K_{2}}=\frac{\frac{1}{2} m(\omega R)^{2}}{\frac{1}{2}\left(\frac{2}{5} m R^{2}ight) \omega^{2}} \quad(\because v=\omega R)$\ $\Rightarrow \frac{K_{1}}{K_{2}}=\frac{5}{2}$\