Question

A uniform rod of length $L$ and mass $M$ is held vertical with its bottom end pivoted to the floor. The rod falls under gravity, freely turning about the pivot. If acceleration due to gravity is $g$, what is the instantaneous angular speed of the rod when it makes an angle ${60}^o$ with the vertical

• A. ${\left(\frac{g}{L}\right)}^{1/2}$
• B. ${\left(\frac{3g}{4L}\right)}^{1/2}$
• C. $\sqrt{3gl}$
• D. ${\left(\frac{3g}{2L}\right)}^{1/2}$

Answer 1

The correct option is C $\sqrt{3gl}$

Hence, option $\left(C\right)$ is correct answer.