\(\text { mass }=m\)

Radius \(=r\)

Now the Kinetic energy wllbe

The total Kinetic energy (KE) of an object is given by

\(KE =\frac{1}{2} mv ^2+\frac{1}{2} I \omega^2\)

Where I is the moment of inertia of the object and \(\omega\) is the angular momentum. For a solid sphere, the moment of inertia is given by

\(I=\frac{2}{5} m r^2\)

Angular velocity would be derived from its radius and linear velocity

\(\omega=\frac{V}{r}\)

So the total equation would

\(KE =\frac{1}{2} mv ^2+\frac{1}{2} \times \frac{2}{5} mr ^2\left(\frac{ v ^2}{ r }\right)\)

\(=\frac{1}{2} mv ^2+\frac{2}{10} mv ^2\)

To get the percentage attributed to rotational energy, we would divide the rotational part of the energy by the total energy

\(\frac{\frac{2}{10} m v^2}{\frac{1}{2} m v^2+\frac{2}{10} m v^2}=\frac{\frac{2}{10} m v^2}{\frac{7}{10} m v^2}\)

\(=\frac{2}{7}\)

Hence \(\frac{5}{7}\) is rotational and \(\frac{2}{7}\) translational.