Question

Two particles $A$ and $B$ are moving with equal angular speed $\omega$ as shown in figure. At $t=0$, their positions are shown. Then, relative velocity $\overrightarrow{v_A}-\overrightarrow{v_B}$ at $t=\frac{\pi }{2\omega }$ is

• A. $\omega (R_1+R_2)\hat{i}$
• B. $-\omega (R_1+R_2)\hat{i}$
• C. $\omega (R_1-R_2)\hat{i}$
• D. $\omega (R_2-R_1)\hat{i}$

Answer 1

The correct option is D $\omega (R_2-R_1)\hat{i}$

Angle traced in $t=\frac{\pi }{2\omega }$ is
$\theta =\omega t=\omega \times \frac{\pi }{2\omega }=\frac{\pi }{2}$
Since both particles will execute a quarter revolution in time $t=\frac{\pi }{2\omega }$, final positions will be as shown in figure.


$\overrightarrow{v_A}-\overrightarrow{v_B}=-\omega R_1\hat{i}-(-\omega R_2\hat{i})$
[Since tangential velocity for a circular motion is,
$v=\omega R$]
$\Rightarrow \overrightarrow{v_A}-\overrightarrow{v_B}=\omega (R_2-R_1)\hat{i}$
Hence the relative velocity $(\overrightarrow{v_A}-\overrightarrow{v_B})$ at $t=\frac{\pi }{2\omega }$ is $\omega (R_2-R_1)\hat{i}$