Question

A uniform thin rod of length $l$ and mass $m$ is hinged at a distance $\frac{l}{4}$ from one of the end and released from horizontal position as shown in the figure. The angular velocity of the rod as it passes the vertical position is : (acceleration due to gravity $=g$)

• A. $2\sqrt{\frac{5g}{7l}}$
• B. $2\sqrt{\frac{6g}{7l}}$
• C. $\sqrt{\frac{3g}{7l}}$
• D. $2\sqrt{\frac{g}{l}}$

Answer 1

The correct option is B $2\sqrt{\frac{6g}{7l}}$

Using conservation of energy.
The cm of the rod moves by a distance $\frac{l}{4}$ when the rod is vertical.
Decrease in potential energy is converted into rotational kinetic energy.
$\Delta \left(PE\right)=+K_{rotation}=\frac{1}{2}I{\omega }^2=\frac{1}{2}(\frac{m{l}^2}{12}+m(\frac{l}{4}{)}^2)\times {\omega }^2=\frac{1}{2}m{l}^2\times \frac{7}{48}{\omega }^2$
$\therefore mg\frac{l}{4}=\frac{1}{2}m{l}^2\times \frac{7}{48}{\omega }^2$

$\Rightarrow {\omega }^2=\frac{24g}{7l}$

$\Rightarrow \omega =2\sqrt{\frac{6g}{7l}}$