Question

A thin uniform rod of mass ${}^{\prime }{m}^{\prime }$ and length ${}^{\prime }{l}^{\prime }$ is standing on a smooth horizontal surface. A slight disturbance causes the lower end to slip on the smooth surface. Find the velocity of centre of mass of the rod at the instant when it makes an angle ${60}^{\circ }$ with the vertical.

Answer 1

$\sin 60=\frac{h}{\frac{L}{2}}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{h}{L}2$

$h=\frac{\sqrt{3}}{4}L$

Change in PE of COM $=$ change in KE of COM

$\Rightarrow mg[\frac{L}{2}-\frac{\sqrt{3L}}{\frac{4}{2}}]=\frac{1}{2}m{v}^2$

$\Rightarrow V=\sqrt{gL(1-\frac{\sqrt{3}}{L})}$

Hence, solved.