Question

Two particles $A$ and $B$ are moving in a horizontal place anticlockwise on two different concentric circles with different constant angular velocities $2\omega$ and $\omega$ respectively. Find angular velocity (in $\text{rad/s}$) of $B$ w.r.t. $A$ after time $t=\frac{\pi }{\omega }$. They both start at the position as shown in figure (Take $\omega =3\text{rad /sec},r=2\text{m}$)

• A. $4$ rad/s
• B. $5$ rad/s
• C. $16$ rad/s
• D. $25$ rad/s

Answer 1

The correct option is A $4$ rad/s

Position of particle A after time $t=\frac{\pi }{\omega }$,
${\theta }_A={\omega }_{A}t$
$=2\omega \times \frac{\pi }{\omega }=2\pi$
Position of particle B after time $t=\frac{\pi }{\omega }$,
${\theta }_B={\omega }_{B}t$
$=\omega \times \frac{\pi }{\omega }=\pi$
Final position after time $t=\frac{\pi }{\omega }s$ are shown in figure.
Velocity of particle A, ${\overrightarrow{v}}_A=2\omega r\left(\hat{j}\right)$
Velocity of particle B, ${\overrightarrow{v}}_B=\omega 2r(-\hat{j})$
${\overrightarrow{v}}_B-{\overrightarrow{v}}_A=-4\omega r\hat{j}=-4\times 3\times 2\hat{j}$
$=-24\hat{j}\text{m/s}$
Angular velocity of B w.r.t A, ${\omega }_{AB}=\left|\frac{v_{AB}}{AB}\right|=\left|\frac{-24}{3\left(2\right)}\right|=|-4|=4\text{rad/s}$