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

• Correct answer
  A

  $\sqrt{\left(\frac{6 g}{1}\right)} \sin \frac{\theta}{2}$
• B

  $\sqrt{\left(\frac{6 g}{1}\right)} \cos \frac{\theta}{2}$
• C

  $\sqrt{\left(\frac{6 g}{1}\right)} \sin \theta$
• D

  $\sqrt{\left(\frac{6 g}{1}\right)} \cos \theta$

Solution

When the rod rotates through an angle $\theta$, the centre of gravity falls through a distance h. From $\triangle B C G^{\prime}$ $\begin{aligned} & \cos \theta=\frac{(l / 2)-h}{\mu_2} \\ & or, h=\frac{1}{2}(1-\cos \theta)\end{aligned}$ Decrease in PE $=m g \frac{1}{2}(1-\cos \theta)...(i)$ The decrease in P.E. is equal to the kinetic energy of rotation $\left(\frac{1}{2} I \omega^2\right)$ $(\mathrm{KE})_{\text {rotational }}=\frac{1}{2}\left(\frac{m l^2}{3}\right) \omega^2...(ii)$ From Eqs. (i) and (ii), we get $\begin{aligned} \frac{1}{2}\left(\frac{m l^2}{3}\right) \omega^2 & =m g \frac{1}{2}(1-\cos \theta) \\ \omega & =\sqrt{\frac{6 g}{l}} \sin \frac{\theta}{2}\end{aligned}$

Hence, the correct answer is option: 'A'.