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 $60^{\circ}$ with the vertical?

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

Answer 1

The Correct Option is D

The fall of centre of gravity h
$\frac{\left(\frac{1}{2}-h\right)}{\frac{1}{2}}=\cos 60^{\circ}$
$h=\frac{1}{2}\left(1-\cos 60^{\circ}\right)$
Decrease in potential energy
$M g h=M g \frac{1}{2}\left(1-\cos 60^{\circ}\right)$
Kinetic energy of rotation
$=\frac{1}{2} I \omega^{2}$
$=\frac{1}{2} \times \frac{M I^{2}}{3} \omega^{2}$
$M g=\frac{1}{2}\left(1-\cos 60^{\circ}\right)=\frac{M I^{2}}{6} \omega^{2}$
$\Rightarrow \omega=\frac{\sqrt{3 g}}{2 L}$