Question

Two particles $A$ and $B$ are moving in same direction on a circular path. Initially particle $A$ and $B$ are diametrically opposite to each other. Particle $A$ moves with angular velocity $\pi rad/s$ and angular acceleration $\frac{\pi }{2}rad/s^2$ and particle $B$ moves with constant angular velocity $2\pi rad/s$. Find the time after which both particles $A$ and $B$ collides.

• A. $1\text{s}$
• B. $2\text{s}$
• C. $3\text{s}$
• D. $4\text{s}$

Answer 1

The correct option is B $2\text{s}$

Suppose the angle between OA and $OB=\theta$ then the rate of change of $\theta$ is $w_{\theta }$

$w_{\theta }=w_B-w_A=2\pi -\pi =\pi \text{rad/sec}$
${\alpha }_{\theta }={\alpha }_B-{\alpha }_A=-\frac{\pi }{2}{\text{rad/sec}}^2$
It the angular displacement is $\Delta \theta$ then
$\Delta \theta =w_{\theta }t+\frac{1}{2}{\alpha }_{\theta }{t}^2$
A and B collides after angular displacement $\Delta \theta =\pi ,$
$\Rightarrow \pi =\pi t-\frac{1}{2}\times \frac{\pi }{2}\times t^2$
$\Rightarrow t^2-4t+4=0$
$\Rightarrow t=2\text{s}$