Rotational Dynamics — Physics STD 12 Science — Question

Total Rolling kinetic energy = Translational K. E. + Rotational K. E.
\(=\frac{1}{2} M V^2+\frac{1}{2} I \omega^2\)
But
\(\omega=\frac{V}{R}\)
Total Rolling K.E \(=\frac{1}{2} M V^2+\frac{1}{2} I\left(\frac{V^2}{R^2}\right)\)
for solid sphere \(I=\frac{2}{5} M R^2\)
Total Rolling K.E. \(=\frac{1}{2} M V^2+\frac{1}{2} \frac{2 M R^2}{5}\left(\frac{V^2}{R^2}\right)\)
\(=\frac{1}{2} M V^2+\frac{1}{5} M V^2\)
\(=\frac{7}{10} M V^2\)