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

Answer 1

Answer: (B)

When the rod rotates through angle $\theta$ , the fall $h$ of centre of gravity is given by $\theta$
$\frac{\frac{L}{2}-h}{\frac{L}{2}}=cos\left(\theta \right)$
or $h=\frac{L}{2}(1-cos\theta )$

Decrease in potential energy
$=Mgh=Mg\frac{L}{2}(1-cos\theta )$
Now, $KE$ of rotation $=\frac{1}{2}I{\omega }^2$
$=\frac{1}{2}\times \frac{M{L}^2}{3}{\omega }^2$
$=\frac{1}{2}\times \frac{M{L}^2}{3}{\omega }^2$
$[I=M{L}^2/3$ (because rod is rotating about an axis passing through its one end)
According to law of conservation of energy, decrease in potential energy will be equal to gain in kinetic energy.
$Mg\frac{L}{2}(1-cos\theta )=\frac{M{L}^2}{6}{\left(\omega \right)}^2$
$\omega =\sqrt{\frac{3g}{L}(1-Cos\theta )}=\sqrt{\frac{6g}{L}}Sin\frac{\theta }{2}$