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

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

Answer 1

The correct option is D ${\left(\frac{3g}{2L}\right)}^{\frac{1}{2}}$

Let COM= c and $\theta ={60}^o$
Change in height of COM due to turning of rod , $\left(h\right)=\frac{l}{2}-\frac{l}{2}cos{60}^{o}h=\frac{l}{4}$
then,
Total potential energy $\Rightarrow mgh=\frac{mgl}{4}$
Rotational kinetic energy $\Rightarrow =\frac{1}{2}I{\omega }^2$
By energy conservatiion,
$\frac{mgl}{4}=\frac{1}{2}I{\omega }^2$
$\frac{6g}{4l}={\omega }^2\omega =\sqrt{\frac{3g}{2l}}$
Hence, option D is correct.