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 −→vA−−→vB at t=π2ω is
A
ω(R1+R2)^i
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B
−ω(R1+R2)^i
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C
ω(R1−R2)^i
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D
ω(R2−R1)^i
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Solution
The correct option is Dω(R2−R1)^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.
−→vA−−→vB=−ωR1^i−(−ωR2^i) [Since tangential velocity for a circular motion is, v=ωR] ⇒−→vA−−→vB=ω(R2−R1)^i Hence the relative velocity (−→vA−−→vB) at t=π2ω is ω(R2−R1)^i