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Question

Two particles A and B are moving on two concentric circles of radii R1 and R2 respectively with equal angular speed ω rad/s. At t=0, their positions and linear velocities are shown in the figure. Then relative velocity (m/s) of A with respect to B (i.e vAB) at t=π2ω s is given by:


A
ω(R1+R2) ^i
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B
ω(R1+R2) ^i
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C
ω(R1R2) ^i
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D
ω(R2R1) ^i
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Solution

The correct option is D ω(R2R1) ^i
Both A and B are moving on the circular path with same angular speed (ω)
Angular displacement of both particles in interval of Δt=π2ω s is:

θ=ω×Δt=ω×π2ω=π2 rad

Both particles have their angular displacement in opposite directions.
Position of particles at t=π2ω s is as shown in figure:


Velocities of particles at t=π2ω s are:
vA=ωR1 ^i
vB=ωR2 ^i

The relative velocity is given by:
vAB= (vAvB)
=ωR1 ^i(ωR2 ^i)

vAB=ω(R2R1) ^i

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