Question

A uniform rod of length $2\, L$ is placed with one end in contact with the horizontal and is then inclined at an angle $\alpha$ to the horizontal and allowed to fall without stopping at contact point when it becomes horizontal. Its angular velocity will be

• A. $\omega=\sqrt{\frac{3 g \sin \alpha}{2 L}}$
• B. $\omega=\sqrt{\frac{2 L}{3 g \sin \alpha}}$
• C. $\omega=\sqrt{\frac{6 g \sin \alpha}{L}}$
• D. $\omega=\sqrt{\frac{L}{g \sin \alpha}}$

Answer 1

Answer: (A)

By the conservation of energy,
Loss in $PE$ of rod= gain of rotational $KE$
$m g \frac{1}{2} \sin \alpha=\frac{1}{2} l \omega^{2}$
$m g \frac{1}{2} \sin \alpha=\frac{1}{2} \frac{m l^{2}}{3} \omega^{2}$
$\Rightarrow \omega=\sqrt{\frac{3 g \sin \alpha}{l}}$
But in the problem length of the rod $2\, L$ is given
$\therefore \omega=\sqrt{\frac{3 g \sin \alpha}{2 L}}$