Question

Two particles $A$ and $B$ are moving on two concentric circles of radii $R_1$ and $R_2$ respectively with equal angular speed $\omega \text{rad/s}$. At $t=0$, their positions and linear velocities are shown in the figure. Then relative velocity $(\text{m/s}$) of $A$ with respect to $B$ (i.e $\overrightarrow{v_{AB}})$ at $t=\frac{\pi }{2\omega }\text{s}$ 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 $\omega (R_2-R_1)\hat{i}$

Both $A$ and $B$ are moving on the circular path with same angular speed $\left(\omega \right)$
$\therefore$ Angular displacement of both particles in interval of $\Delta t=\frac{\pi }{2\omega }\text{s}$ is:

$\theta =\omega \times \Delta t=\omega \times \frac{\pi }{2\omega }=\frac{\pi }{2}\text{rad}$

Both particles have their angular displacement in opposite directions.
$\therefore$ Position of particles at $t=\frac{\pi }{2\omega }\text{s}$ is as shown in figure:


Velocities of particles at $t=\frac{\pi }{2\omega }\text{s}$ are:
$\overrightarrow{v_A}=-\omega R_1\hat{i}$
$\overrightarrow{v_B}=-\omega R_2\hat{i}$

The relative velocity is given by:
${\overrightarrow{v}}_{AB}=$ $({\overrightarrow{v}}_A-{\overrightarrow{v}}_B$)
$=-\omega R_1\hat{i}-(-\omega R_2\hat{i})$

$\therefore {\overrightarrow{v}}_{AB}=\omega (R_2-R_1)\hat{i}$