Question

A uniform rod of mass $m$ and length $L$ is hinged about one end and can freely rotate in a vertical plane. The angular velocity of the rod, when it falls from position $P$ to $P$ through an angle $\alpha$ starting from rest, is:

• A. $\sqrt{\frac{6g}{5L}}\sin \alpha$
• B. $\sqrt{\frac{6g}{L}}\sin \frac{\alpha }{2}$
• C. $\sqrt{\frac{6g}{L}}\cos \frac{\alpha }{2}$
• D. $\sqrt{\frac{6g}{L}}\sin \alpha$

Answer 1

The correct option is B $\sqrt{\frac{6g}{L}}\sin \frac{\alpha }{2}$

$\begin{array}{c}mg\frac{l}{2}\left(1-\cos \alpha \right)=\frac{1}{2}\left(\frac{m{l}^2}{3}\right)w^2\\ \frac{3g}{l}\left(1-\cos \alpha \right)=w^2\\ \frac{3g}{l}\left(22{\sin }^2\frac{\alpha }{2}\right)=w^2\\ w=\sqrt{\frac{6g}{l}\sin \frac{\alpha }{2}}\end{array}$

$\therefore$ Option $B$ is correct.