Question

Two particle A and B move on a circle. Initially Particle A and B are diagonally opposite to each other. Particle A moves with angular velocity $\pi \frac{rad}{sec}$, angular acceleration $\frac{\pi }{2}\frac{rad}{se{c}^2}$ and particle B moves with constant angular velocity $2\pi \frac{rad}{sec}$. Find the time after which both the particle A and B will collide.

• A. $t=2sec$
• B. $t=1sec$
• C. $t=4sec$
• D. $t=\frac{4\pm \sqrt{14}}{2}sec$

Answer 1

The correct option is A

$t=2sec$

$\begin{array}{ccc}& \omega & angularacceleration\\ ParticleA& \pi \frac{rad}{sec}& \frac{\pi }{2}\frac{rad}{se{c}^2}\\ ParticleB& 2\pi \frac{rad}{sec}& 0\\ ParticleBwithrespecttoparticleA& 2\pi -\pi ;\pi \frac{rad}{sec}& 0-\frac{\pi }{2};-\frac{\pi }{2}\frac{rad}{se{c}^2}\end{array}$

If I keep A constant and see the motion of B, B will look like coming towards me and in the process will have to cover an angular displacement of $\pi rad$.

$\Rightarrow \theta =\pi rad$

$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$

$\theta =\omega t+\frac{1}{2}\alpha t^2$

$\Rightarrow \pi =\pi t+\frac{1}{2}\left(\frac{-\pi }{2}\right)t^2$

$\Rightarrow t^2-4t+4=0$

$\Rightarrow t=2sec$