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 by.

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

Answer 1

The correct option is A $\sqrt{\left(\frac{6g}{l}\right)}\sin \frac{\theta }{2}$

When the rod rotates through an angle $\theta$, the centre of gravity falls through a distance h. From $\Delta BC{G}^{\prime }$
$\cos \theta =\frac{\left(1/2\right)-h}{{\mu }_2}$
$h=\frac{l}{2}(1-\cos \theta )$
Decrease in PE
$=mg\frac{l}{2}(1-\cos \theta )$ ....(i)
The decrease in P.E. is equal to the kinetic energy of rotation $\left(\frac{1}{2}I{\omega }^2\right)$
$(KE{)}_{rotational}=\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2$ .....(ii)
From Eqs. (i) and (ii), we get
$\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2=mg\frac{l}{2}(1-\cos \theta )$
$\omega =\sqrt{\frac{6g}{l}}\sin \frac{\theta }{2}$.