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Two particles A, B are moving on two concentric circles of radii R₁ and R₂ 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 $t = \frac{\pi}{2\omega}$ is given by :
Option 1)
$\omega \left ( R_{1} - R _{2} \right ) \hat{i}$
Option 2)
$\omega \left ( R_{2} - R _{1} \right ) \hat{i}$
Option 3)
$\omega \left ( R_{1} + R _{2} \right ) \hat{i}$
Option 4)
$-\omega \left ( R_{1} + R _{2} \right ) \hat{i}$

Answers (1)

best_answer

Angular velocity -

Denoted by $\omega$ (omega)

S.I. units- Radian per second (rad s^-1 )

$\omega$ is a vector quantity

$\omega$ -Rate of change of angular displacement.

$\omega = \frac{\Theta }{t}$ or $\omega = \frac{d\Theta }{dt}$

- wherein

Fig. Shows angular velocity

So W_l = w

W₀ = 0

So

$\theta =w_{0}+w_{t}.t$

$\theta =wt = w\left ( \frac{2\pi}{2w} \right )= \frac{\pi }{2}$

so after $\theta =\frac{\pi }{2}$

Position of A and B are shown in figure

$V_{A} = wR_{1}\left ( -\widehat{i} \right )$

$V_{B} = wR_{2}\left ( -\widehat{i} \right )$

$V_{A}- V_{B} = wR_{1}\left ( -\widehat{i} \right ) - \left ( wR_{2}\left ( -\widehat{i} \right ) \right )=w\left ( R_{2}-R_{1} \right )\widehat{i}$

Option 1)

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

Option 2)
$\omega \left ( R_{2} - R _{1} \right ) \hat{i}$
Option 3)

$\omega \left ( R_{1} + R _{2} \right ) \hat{i}$

Option 4)

$-\omega \left ( R_{1} + R _{2} \right ) \hat{i}$

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