Q. A uniform rod of length $2L$ 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 slipping at contact point. When it becomes horizontal, its angular velocity will be
A) $\omega = \sqrt{\frac{3g\sin\alpha}{2L}}$
B) $\omega = \sqrt{\frac{2L}{3g\sin\alpha}}$
C) $\omega = \sqrt{\frac{6g\sin\alpha}{2L}}$
D) $\omega = \sqrt{\frac{2L}{g\sin\alpha}}$
Explanation
(a) By the conservation of energy
$P.E.$ of rod = Rotational $K.E.$
$mg\frac{l}{2}\sin{\alpha = \frac{1}{2}I\omega^{2} \Rightarrow mg\frac{l}{2}\sin{\alpha = \frac{1}{2}\frac{ml^{2}}{3}\omega^{2}}}$
$\Rightarrow \omega = \sqrt{\frac{3g\sin\alpha}{l}}$
But in the problem length of the rod $2L$ is given
$\Rightarrow \omega = \sqrt{\frac{3g\sin\alpha}{2L}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.