Question

Two particles $A$ and $B$ move on a circle. Initially particles $A$ and $B$ are diagonally opposite to each other. Particle $A$ moves with angular velocity $\pi rad{s}^{-1}$ and angular acceleration $\pi /2rad{s}^{-2}$ and particle B move with constant angular velocity $2\pi rad{s}^{-2}$. Find the time after which both the particles $A$ and $B$ will collide.

Answer 1

Suppose the angle between $OA$ and $OB$ = $\theta$. Then the rate of change of $\theta$,
$\theta ={\omega }_B-{\omega }_A=2\pi -\pi =\pi rad{s}^{-2}$

$\theta ={\alpha }_B-{\alpha }_A=-\frac{\pi }{2}rad{s}^{-2}$

If the angular displacement is $\Delta \theta$, then
$\Delta \theta =\dot{\theta }t+\frac{1}{2}\ddot{\theta }{t}^2$
For $A$ and $B$ to collide angular displacement, $\Delta \theta =\pi$.
$\Rightarrow \pi =\pi t+\frac{1}{2}\left(\frac{-\pi }{2}\right)t^2\Rightarrow t^2-4t+4=0\Rightarrow t=2s$