Question

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 stopping 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. $\omega =\sqrt{\frac{6g\sin \alpha }{L}}$
• 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 $KE$
$mg\frac{1}{2}\sin \alpha =\frac{1}{2}l{\omega }^2$
$mg\frac{1}{2}\sin \alpha =\frac{1}{2}\frac{m{l}^2}{3}{\omega }^2$
$\Rightarrow \omega =\sqrt{\frac{3g\sin \alpha }{l}}$
But in the problem length of the rod $2L$ is given
$\therefore \omega =\sqrt{\frac{3g\sin \alpha }{2L}}$