A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is

$$\begin{gathered} {\mathrm{K}}_{\text{total }}={\mathrm{K}}_{\text{rotational }}+{\mathrm{K}}_{\text{Translational }} \\ {\mathrm{K}}_{\text{total }}=\frac{1}{2}{\mathrm{I}}_{\mathrm{cm}}{\mathrm{\omega }}^2+\frac{1}{2}{\mathrm{mV}}_{\mathrm{cm}}^2 \\ {\mathrm{v}}_{\mathrm{cm}}=\mathrm{R\omega }\text{ for pure rolling } \\ {\mathrm{I}}_{\mathrm{cm}}=\frac{2}{5}{\mathrm{mR}}^2 \\ {\mathrm{K}}_{\mathrm{Rot}}=\frac{1}{2}{\mathrm{I}}_{\mathrm{cm}}{\mathrm{\omega }}^2 \\ =\frac{1}{2}\times \frac{2}{5}{\mathrm{mR}}^2\times \frac{{\mathrm{v}}_{\mathrm{cm}}^2}{{\mathrm{R}}^2}=\frac{1}{5}{\mathrm{mv}}_{\mathrm{cm}}^2 \\ {\mathrm{K}}_{\text{Total }}=\frac{1}{5}{\mathrm{mv}}_{\mathrm{cm}}^2+\frac{1}{2}{\mathrm{mv}}_{\mathrm{cm}}^2=\frac{7}{10}{\mathrm{mv}}_{\mathrm{cm}}^2 \\ \frac{{\mathrm{K}}_{\mathrm{R}\circ }}{{\mathrm{K}}_{\text{Told }}}\frac{\frac{1}{5}{\mathrm{mv}}_{\mathrm{cm}}^2}{\frac{7}{10}{\mathrm{mv}}_{\mathrm{cm}}^2}=\frac{2}{7} \end{gathered}$$