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Question

Two particles A and B are moving with equal angular speed ω as shown in figure. At t=0, their positions are shown. Then, relative velocity vAvB at t=π2ω is


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
Angle traced in t=π2ω is
θ=ωt=ω×π2ω=π2
Since both particles will execute a quarter revolution in time t=π2ω, final positions will be as shown in figure.


vAvB=ωR1^i(ωR2^i)
[Since tangential velocity for a circular motion is,
v=ωR]
vAvB=ω(R2R1)^i
Hence the relative velocity (vAvB) at t=π2ω is ω(R2R1)^i

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