A thin spherical shell rolls on a horizontal table without slipping. The precise ratio of its transl

Question

A thin spherical shell rolls on a horizontal table without slipping. The precise ratio of its translational kinetic energy to its total kinetic energy is

Professor PC Thomas Classes · Answer

(d) : The translational kinetic energy is $KE_{trans} = \frac{1}{2}Mv^2$. For a thin spherical shell, the moment of inertia is $I = \frac{2}{3}MR^2$. Its rotational kinetic energy is $KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}\left(\frac{2}{3}MR^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{3}Mv^2$. The total kinetic energy is the sum: $KE_{total} = \frac{1}{2}Mv^2 + \frac{1}{3}Mv^2 = \frac{5}{6}Mv^2$. The required ratio is $\frac{KE_{trans}}{KE_{total}} = \frac{1/2}{5/6} = \frac{3}{5}$.