Question

A uniform rod of length $l=3\text{m}$ free to rotate in a vertical plane about a fixed horizontal axis passing through $O$. The rod is allowed to rotate by releasing from it's vertical position. Find the angular velocity of the rod when rod makes an angle $\theta ={60}^{\circ }$ with vertical. (take $g=10{\text{m/s}}^2$)

• A. $5\text{rad/sec}$
• B. $\sqrt{5}\text{rad/sec}$
• C. $\frac{5}{\sqrt{2}}\text{rad/sec}$
• D. $5\sqrt{2}\text{rad/sec}$

Answer 1

The correct option is B $\sqrt{5}\text{rad/sec}$

Let the rod having mass $m\text{kg}$ and moment of inertia $I$ about the one end $O$.

When rod rotates through an angle $\theta$, it's centre of mass $\left(c\right)$ falls through the distance $\left(h\right)$.
So, Loss in potential energy $\Delta PE$:
$\Delta PE=mgh$
$h=\frac{l}{2}(1-\cos \theta )$
$\Delta PE=mg\frac{l}{2}(1-\cos \theta )\dots \dots \left(1\right)$
$\because$ Rod is hinged, hence constrained to rotate only.

Gain in rotational kinetic energy of rod
${KE}_{Rot}=\left(\frac{1}{2}I{\omega }^2\right)$
$\Rightarrow$By applying mechanical energy conservation:
$\text{Loss in PE}=\text{Gain in}{KE}_{Rot}$
$(KE{)}_{\text{rotational}}=\frac{1}{2}\times \left(\frac{1}{3}m{l}^2\right){\omega }^2\dots \dots (2\left)\right[\because I=\frac{m{l}^2}{3}]$
$mg\frac{l}{2}(1-\cos \theta )=\frac{1}{6}m{l}^2{\omega }^2$

$\Rightarrow \frac{mgl}{2}\times 2{\sin }^2\frac{\theta }{2}=\frac{1}{6}m{l}^2{\omega }^2$
$\Rightarrow \omega =\sqrt{\frac{6g}{l}\times {\sin }^2\frac{\theta }{2}}$
Substitute the values of $l=3\text{m}$ and $\theta ={60}^{\circ }$
$g=10{\text{m/s}}^2$
$\omega =\sqrt{\frac{6\times 10}{3}}\sin {30}^{\circ }$
$\therefore \omega =\sqrt{20}\times \frac{1}{2}=\sqrt{5}\text{rad/sec}$