A spherical ball rolls on a table without slipping. Then the fraction of its total energy associated with rotation is

Total Energy
$\mathrm{E}=\frac{1}{2}{\mathrm{I\omega }}^2+\frac{1}{2}{\mathrm{mv}}^2$
$\begin{array}{l}=\frac{1}{2}\left(\frac{2}{5}{\mathrm{mr}}^2\right){\mathrm{\omega }}^2+\frac{1}{2}{\mathrm{mr}}^2{\mathrm{\omega }}^2\\ =\frac{1}{5}{\mathrm{mr}}^2{\mathrm{\omega }}^2+\frac{1}{2}{\mathrm{mr}}^2{\mathrm{\omega }}^2=\frac{7}{10}{\mathrm{mr}}^2{\mathrm{\omega }}^2\end{array}$
Rotational energy $=\frac{1}{5}{\mathrm{mr}}^2{\mathrm{\omega }}^2$
$\therefore \frac{\text{ Rotational energy }}{\text{ Total energy }}=\frac{\frac{1}{5}{\mathrm{mr}}^2{\mathrm{\omega }}^2}{\frac{7}{10}{\mathrm{mr}}^2{\mathrm{\omega }}^2}=\frac{2}{7}$