A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane is: 

$\mathrm{I}=\frac{1}{2}{\mathrm{MR}}^2$

When the cylinder rails down the inclined plane from a vertical height h its potential energy Mgh is partly convened into translations kinetic energy $\frac{1}{2}{\mathrm{Mv}}^2$ and partly into rotational kinetic energy $\frac{1}{2}\mathrm{I}{\omega }^2$. Therefore,

$\mathrm{Mgh}=\frac{1}{2}{\mathrm{Mv}}^2+\frac{1}{2}{\mathrm{I\omega }}^2$

$=\frac{1}{2}{\mathrm{Mv}}^2+\frac{1}{2}\left(\frac{1}{2}{\mathrm{MR}}^2\right){\omega }^2$

$=\frac{1}{2}\mathrm{M}(\mathrm{R\omega }{)}^2+\frac{1}{2}{\mathrm{MR}}^2{\mathrm{\omega }}^2$

$(\because R\omega )$

$\mathrm{Mgh}=\frac{3}{4}{\mathrm{MR}}^2{\mathrm{\omega }}^2$

$\mathbf{\omega }=\sqrt{\frac{4\mathrm{gh}}{3{\mathrm{R}}^2}}=\frac{2}{\mathrm{R}}\sqrt{\frac{\mathrm{gh}}{3}}$