A solid cylinder of mass ' \(M\) ' and radius ' \(R\) ' rolls down an inclined plane of height ' \(h\) '. When it reaches the foot of the plane, its rotational kinetic energy is ( \(\mathrm{g}=\) acceleration due to gravity)

From the law of conservation of energy, we have Potential energy \(=\) Translational kinetic energy + Rotational kinetic energy
or \(\quad \mathrm{mgh}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I} \omega^2\)
or \(\quad \mathrm{mgh}=\frac{1}{2} m v^2 \omega^2+\frac{1}{2}\left(\frac{1}{2} m r^2\right) \omega^2=\frac{3}{4}{m r^2 \omega^2}^2\)
or \(\quad \omega^2=\frac{4 \mathrm{gh}}{3 \mathrm{r}^2}\)
Now the rotational kinetic energy \(=\frac{1}{2} \mathrm{I} \omega^2\)
\(\therefore \quad\) Substituting for \(\omega^2\) and I , we have,
\(\begin{aligned}
\text { Rotational kinetic energy } & =\frac{1}{2}\left(\frac{1}{2} \mathrm{mr}^2\right) \frac{4 \mathrm{gh}}{3 \mathrm{r}^2} \\
& =\frac{\mathrm{Mgh}}{3}
\end{aligned}\)
... \((\because \mathrm{M}=\mathrm{m})\)