Question

A rod of length $L$ is hinged from one end. It is brought to the horizontal position and released. The angular velocity of the rod when it is in the vertical position is $\sqrt{Kg/L}$. The find the value of $K$.

Answer 1

Change in Potential energy$=-mg\frac{L}{2}$

Change in Kinetic energy$=\frac{1}{2}I{\omega }^2$

Conservation of energy $\Rightarrow \frac{1}{2}I{\omega }^2-mg\frac{L}{2}=0$

$\Rightarrow \frac{m{L}^2}{3}\cdot {\omega }^2=mgL$

$\Rightarrow {\omega }^2=\frac{3g}{L}$ or $\omega =\sqrt{\frac{3g}{L}}$

$\therefore K=3$