Two particles A and B are moving on two concentric circles of radii R1 and R2 with equal angular speed - At t-0- their positions and direction of motion are shown in the figure -The relative velocity vA-vB at t-2- is given by -

Using, θ=ω t we get, θ=ωπ2ω=π2 So, both have completed quarter circle Relative velocity, v_A v_B=ω R_1( î) ω R_2( i) ∴ nbsp;v_A v_B nbsp;=ω(R_2 R_1)i ...

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Byju's Answer

Standard XII

Physics

Net Acceleration for Non Uniform Circular Motion

Two particles...

Question

Two particles $A$ and $B$ are moving on two concentric circles of radii $R_{1} and R_{2}$ with equal angular speed $ω$ . At $t = 0$ , their positions and direction of motion are shown in the figure :

The relative velocity $_{A} -_{B} at t = \frac{π}{2 ω}$ is given by :

ω (R_{1} + R_{2})^i

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- ω (R_{1} + R_{2})^i

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ω (R_{2} - R_{1})^i

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ω (R_{1} - R_{2})^i

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Solution

The correct option is C $ω (R_{2} - R_{1})^i$
Using, $θ = ω t$ we get,

$θ = ω \frac{π}{2 ω} = \frac{π}{2}$

So, both have completed quarter circle

Relative velocity,

${\to v}_{A} - {\to v}_{B} = ω R_{1} (-^i) - ω R_{2} (- i)$

$∴ {\to v}_{A} - {\to v}_{B} = ω (R_{2} - R_{1}) i$

Hence, $(C)$ is the correct answer.

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Two particles A and B are moving on two concentric circles of radii R1 and R2 with equal angular speed ω. At t=0, their positions and direction of motion are shown in the figure : The relative velocity −→vA−−→vB at t=π2ω is given by :

The correct option is C ω(R2−R1)^iUsing, θ=ωt we get, θ=ωπ2ω=π2 So, both have completed quarter circle Relative velocity, →vA−→vB=ωR1(−^i)−ωR2(−i) ∴→vA−→vB=ω(R2−R1)i Hence, (C) is the correct answer.

Two particles $A$ and $B$ are moving on two concentric circles of radii $R_{1} and R_{2}$ with equal angular speed $ω$ . At $t = 0$ , their positions and direction of motion are shown in the figure :

The relative velocity $_{A} -_{B} at t = \frac{π}{2 ω}$ is given by :