Question

A uniform rod of length $L$ is free to rotate in a vertical plane about a fixed horizontal axis through $B$. The rod begins rotating from rest from its unstable equilibrium position. When it has turned through an angle $\theta$, its angular velocity $\omega$ is given as:

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

Answer 1

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


When the rod rotates through angle $\theta$, the fall $h$ of centre of gravity is given by,
$h=\frac{L}{2}(1-\cos \theta )$
$\therefore$ Decrease in potential energy
$=Mgh=Mg\frac{L}{2}(1-\cos \theta )$
Now, K.E. of rotation $=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times \frac{M{L}^2}{3}{\omega }^2$

According to law of conservation of energy,

Decrease in potential energy = Kinetic energy of rotation
$Mg\frac{L}{2}(1-\cos \theta )=\frac{1}{2}\times \frac{M{L}^2}{3}{\omega }^2$
$\omega =\sqrt{\frac{6g}{L}}\sin \left(\frac{\theta }{2}\right)$