Question

Two particles $A$ and $B$ are moving on two concentric circles of radii $R_1$ and $R_2$ with equal angular speed $\omega$. At $t=0$, their positions and direction of motion are shown in the figure. The relative velocity $\overrightarrow{v_A}-\overrightarrow{v_B}$ at $t=\frac{\pi }{2\omega }$ is given by:

• 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 $\omega (R_2-R_1)\hat{i}$

${\theta }_A={\omega }_{A}t=\omega t$ and
${\theta }_B={\omega }_{B}t=\omega t$

At $t=\frac{\pi }{2\omega }$, angular displacements of A and B are
${\theta }_A=\frac{\pi }{2\omega }\times \omega =\frac{\pi }{2}\text{rad}$ anticlockwise
and ${\theta }_B=\frac{\pi }{2}\text{rad}$ clockwise

Final positions of $A,B$ are shown in the figure.

${\overrightarrow{v}}_A=-\omega R_1\hat{i}$ and ${\overrightarrow{v}}_B=-\omega R_2\hat{i}$
$\Rightarrow \overrightarrow{v_A}-\overrightarrow{v_B}=\omega (R_2-R_1)\hat{i}$