Question

The lower end of a $4\text{m}$ long uniform rod $AB$ is pulled with constant speed $v=4\text{m/s}$. The speed of the centre of mass of the rod when $\theta ={60}^{\circ }$ will be

• A. $\frac{4}{\sqrt{3}}\text{m/s}$
• B. $2\sqrt{3}\text{m/s}$
• C. $4\text{m/s}$
• D. $4\sqrt{3}\text{m/s}$

Answer 1

The correct option is A $\frac{4}{\sqrt{3}}\text{m/s}$

In the figure,

$OB=x$ (say) $=AB\cos \theta$
$\Rightarrow$ $x=4\cos \theta$

$\therefore \frac{dx}{dt}=(-4\sin \theta )\frac{d\theta }{dt}.....\left(1\right)$


Given, velocity along $\text{x-direction}$$\frac{dx}{dt}=4\text{m/s}$
From equation $\left(1\right)$,
${\left(\frac{d\theta }{dt}\right)}_{\text{at}\theta ={60}^{\circ }}=-\text{cosec}{60}^{\circ }=-\frac{2}{\sqrt{3}}\text{rad/s}$

Radius of centre of mass of the rod,
${\overrightarrow{r}}_c=\left(2\cos \theta \right)\hat{i}+\left(2\sin \theta \right)\hat{j}$

Velocity of centre of mass of the given rod
${\overrightarrow{v}}_c=\frac{d\overrightarrow{r_c}}{dt}=(-2\sin \theta \frac{d\theta }{dt})\hat{i}+\left(2\cos \theta \frac{d\theta }{dt}\right)\hat{j}$

At $\theta ={60}^{\circ }$,

${\overrightarrow{v}}_c=(-2\times \frac{\sqrt{3}}{2}\times \frac{-2}{\sqrt{3}})\hat{i}+(2\times \frac{1}{2}\times \frac{-2}{\sqrt{3}})\hat{j}$

$=2\hat{i}-\frac{2\hat{j}}{\sqrt{3}}$

$\therefore v_c=\sqrt{4+\frac{4}{3}}=\frac{4}{\sqrt{3}}\text{m/s}$

Why this question? This question will help students to grasp the concept of motion of COM and the concept of rigid body