Two particles A and B are moving on two conce | vedclass

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(C) At $t = 0$,particle $A$ is at $(R_1, 0)$ moving in the $+y$ direction,so $\overrightarrow{V_A} = \omega R_1 \hat{j}$. Particle $B$ is at $(R_2, 0)

Vedclass · Accepted Answer

$\omega(R_2 - R_1) \hat{i}$

Vedclass · Answer

(C) At $t = 0$,particle $A$ is at $(R_1, 0)$ moving in the $+y$ direction,so $\overrightarrow{V_A} = \omega R_1 \hat{j}$. Particle $B$ is at $(R_2, 0)$ moving in the $-y$ direction,so $\overrightarrow{V_B} = -\omega R_2 \hat{j}$.After time $t = \frac{\pi}{2\omega}$,the angle rotated is $	heta = \omega t = \omega \left( \frac{\pi}{2\omega} ight) = \frac{\pi}{2}$ (i.e.,$90^\circ$).Particle $A$ moves from $(R_1, 0)$ to $(0, R_1)$,and its velocity vector rotates by $90^\circ$ counter-clockwise. Thus,$\overrightarrow{V_A} = -\omega R_1 \hat{i}$.Particle $B$ moves from $(R_2, 0)$ to $(0, R_2)$,and its velocity vector rotates by $90^\circ$ clockwise. Thus,$\overrightarrow{V_B} = -\omega R_2 \hat{i}$.The relative velocity is $\overrightarrow{V_{rel}} = \overrightarrow{V_A} - \overrightarrow{V_B} = (-\omega R_1 \hat{i}) - (-\omega R_2 \hat{i}) = \omega(R_2 - R_1) \hat{i}$.

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