Question

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

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

Answer 1

The correct option is B $\sqrt{\frac{6g}{l}}\sin \frac{\theta }{2}$


Initial total energy of the rod:

$E_i=P.E+K.E=mg\frac{l}{2}+0$

Final total energy of the rod:

$E=P.E+K.E=mg\frac{l}{2}\cos \theta +\frac{1}{2}\frac{m{l}^2}{3}{\omega }^2$

From law at conservation of energy,

$E_i=E_f$

$\Rightarrow mg\frac{l}{2}=mg\frac{l}{2}\cos \theta +\frac{1}{2}\frac{m{l}^2}{3}{\omega }^2$

$\Rightarrow mg\frac{l}{2}(1-\cos \theta )=\frac{m{l}^2}{6}{\omega }^2$

$\Rightarrow \frac{g}{2}\times 2{\sin }^2\frac{\theta }{2}=\frac{l{\omega }^2}{6}$

$\therefore \omega =\sqrt{\frac{6g}{l}}\sin \frac{\theta }{2}$

Hence, $\left(\text{B}\right)$ is the correct answer.