Question

Two particles A and B move anticlockwise with the same speed $v$ in a circle of radius $R$ and are diametrically opposite to each other. At $t=0$, A is given a constant tangential acceleration $a_t=\frac{72v^2}{25\mathrm{\pi R}}$. Calculate the time in which A collides with B.

• A. $\frac{25\pi R}{3v}$
• B. $\frac{5\pi R}{6v}$
• C. $\frac{3\pi R}{2v}$
• D. $\frac{7\pi R}{6v}$

Answer 1

The correct option is B $\frac{5\pi R}{6v}$

Since both the particles have equal velocity
$v_A=v_B$$\Rightarrow v_{AB}=0$
Relative angular velocity, ${\omega }_{rel}=\frac{v_{AB}}{R}=0$
Angular displacement,
${\theta }_{rel}={\omega }_{rel}t+\frac{1}{2}\alpha t^2$
Since the bodies are diametrically opposite to each other initially, the relative angular displacement ${\theta }_{rel}$ is equal to $\pi$
Angular acceleration, $\alpha =\frac{a_t}{R}=\frac{72v^2}{25\pi R^2}$
Hence,

${\theta }_{rel}=\pi =0+\frac{1}{2}\times \frac{72v^2}{25\pi R^2}\times t^2$
$\Rightarrow t^2=\frac{50{\pi }^2{R}^2}{72v^2}=\frac{25{\pi }^2{R}^2}{36v^2}$
Hence, time taken by A to collide with B is $t=\frac{5\pi R}{6v}$