A solid cylinder of mass ' $M$ ' and radius ' $R$ ' rolls down an inclined plane of height ' $h$ '. When it…
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)
$\frac{\mathrm{Mgh}}{3}$
$\frac{\mathrm{Mgh}}{6}$
$\frac{\mathrm{Mgh}}{4}$
$\frac{\mathrm{Mgh}}{2}$
Solution
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,
$\begin{aligned}
\text { Rotational kinetic energy } & =\frac{1}{2}\left(\frac{1}{2} \mathrm{mr}^2\right) \frac{4 \mathrm{gh}}{3 \mathrm{r}^2} \\
& =\frac{\mathrm{Mgh}}{3}
\end{aligned}$
... $(\because \mathrm{M}=\mathrm{m})$