A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy $\left(K_t\right)$ as well as rotational kinetic energy $\left(K_r\right)$ simultaneously. The ratio $K_t:\left(K_t+K_r\right)$ for the sphere is

A rolling body has two KE;

translational $\mathrm{KE}={\mathrm{K}}_{\mathrm{T}}=\frac{1}{2}\mathrm{m}{v}^2$

and rotational $\mathrm{KE}={\mathrm{K}}_{\mathrm{R}}=\frac{1}{2}\mathrm{I}{\omega }^2$

where I is the moment of inertia and $\omega$ is the angular velocity.

Total energy  ${\mathrm{K}}_{\mathrm{T}}+{\mathrm{K}}_{\mathrm{R}}=\frac{1}{2}\mathrm{m}{v}^2+\frac{1}{2}\mathrm{I}{\omega }^2$

$\mathbf{I}={\mathrm{MK}}^2$ where K = radius of gyration

$\omega =v/R$ where R is the radius of the sphere

$\text{so,}{K}_T+K_R=\frac{1}{2}m{v}^2+\frac{1}{2}m{k}^2\frac{v^2}{R^2}$

$=\frac{1}{2}{\mathrm{mv}}^2\left(1+\frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}\right)=\frac{1}{2}{\mathrm{mv}}^2\left(\frac{7}{5}\right)$  for a sphere

$\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}=\frac{2}{5}$

Hence, $\frac{{\mathrm{K}}_{\mathrm{T}}}{{\mathrm{K}}_{\mathrm{R}}+{\mathrm{K}}_{\mathrm{T}}}=\frac{\frac{1}{2}\mathrm{m}{v}^2}{\frac{1}{2}\mathrm{m}{v}^2\left(\frac{7}{5}\right)}=\frac{5}{7}.$