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 average 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 height ${}^{\prime }{h}^{\prime }$ through which centre of gravity falls is given by
$\frac{L/2-h}{L/2}=\cos \theta$
or $h=\frac{L}{2}(1-\cos \theta )$
$\therefore$ Decrease in potential energy
$=Mgh=Mg\frac{L}{2}(1-\cos \theta )$
Now, $KE$ of rotation $=\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}\times \frac{M{L}^2}{3}{\omega }^2$
[$I=M{L}^2/3$ (because rod is rotating about an axis passing through its one end)]


According to law of conservation of energy,
$Mg\frac{L}{2}(1-\cos \theta )=\frac{M{L}^2}{6}{\omega }^2$
$\Rightarrow Mg\frac{L}{2}\left(2{\sin }^2\frac{\theta }{2}\right)=\frac{M{L}^2}{6}{\omega }^2$
$\therefore \omega =\sqrt{\frac{6g}{L}}\sin \left(\frac{\theta }{2}\right)$