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Q.
A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$. What is the speed of its centre of mass when the cylinder reaches its bottom?
Solution:
$m g h=\frac{1}{2} M v^{2}+\frac{1}{2} \frac{M R^{2}}{2} \times \frac{v^{2}}{R^{2}}$
$\Rightarrow g h=\frac{2 v^{2}+v^{2}}{4} $
$\Rightarrow \frac{3 v^{2}}{4}=g h $
$\Rightarrow v=\sqrt{\frac{4}{3}\, g h}$