Question 194 of 308

Two particles $A,B$ 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 $\vec{{V}_{A}}-\vec{{V}_{B}}$ at $t=\dfrac{\pi }{2\omega }$ is given by:

A$-\omega \left({R}_{1}+{R}_{2}\right)\hat{i}$B$\omega \left({R}_{1}+{R}_{2}\right)\hat{i}$C$\omega \left({R}_{1}-{R}_{2}\right)\hat{i}$D$\omega \left({R}_{2}-{R}_{1}\right)\hat{i}$

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Answer: (D) $\omega \left({R}_{2}-{R}_{1}\right)\hat{i}$

$t=\dfrac{\pi }{2\omega }$


${\vec{V}}_{A}=\omega {R}_{1}\left(-\hat{i}\right)$

${\vec{V}}_{B}=\omega {R}_{2}\left(-\hat{i}\right)$

$\vec{{V}_{A}}-\vec{{V}_{B}}=\omega \left({R}_{2}-{R}_{1}\right)\hat{i}$