Question

A uniform rod of length $l$ is free to rotate in a vertical plane about a fixed horizontal axis through $O$. The rod is allowed to rotate from rest from its unstable vertical position. Then, the angular velocity of the rod when it has turned through an angle $\theta$ is

• A. $\sqrt{\frac{3 g}{l}} \sin (\theta / 2)$
• B. $\sqrt{\frac{6 g}{l}} \sin (\theta / 2)$
• C. $\sqrt{\frac{3 g}{l}} \cos (\theta / 2)$
• D. $\sqrt{\frac{6 g}{l}} \cos (\theta / 2)$

Answer 1

Answer: (B)

Height descended by centre of mass
$h=\frac{l}{2}-\frac{l}{2} \cos \theta$
Loss in $PE =$ gain in $KE$
$m g h=\frac{1}{2}\left(\frac{m l^{2}}{3}\right) \omega^{2}$
put the value of $h$ and solve to get the answe