Question

A solid cylinder of mass $m$ and radius $r$ starts rolling down an inclined plane of inclination $\theta$. Friction is sufficient to prevent slipping. Find the speed of its centre of mass when its centre of mass has fallen a height $h$.

Answer 1

Consider the two shown position of the cylinder. As it does not slip, its total mechanical energy will be conserved.
Energy at position $1$ is $E_1=mgh$
Energy at position $2$ is
$E_2=\frac{1}{2}m{v}_{c.m.}^2+\frac{1}{2}{I}_{c.m.}{\omega }^2$
$\because \frac{v_{c.m.}}{r}=\omega$ and $I_{c.m.}=\frac{m{r}^2}{2}$
$\Rightarrow E_2=\frac{3}{4}{mv}_{c.m}^2$
From law of conservation of energy, $E_1=E_2$
$\Rightarrow v_{c.m.}=\sqrt{\frac{4}{3}gh}$