Question

A rod of length L is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod, when it is in vertical position is

• A. $\sqrt{\frac{2g}{L}}$
• B. $\sqrt{\frac{3g}{L}}$
• C. $\sqrt{\frac{g}{2L}}$
• D. $\sqrt{\frac{g}{L}}$

Answer 1

The correct option is B $\sqrt{\frac{3g}{L}}$

The loss in gravitational potential energy of the rod's center of mass is equal to the gain in rotational kinetic energy of the rod.

Hence $mg\frac{L}{2}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{m{L}^2}{3}\right){\omega }^2$

$⟹\omega =\sqrt{\frac{3g}{L}}$