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Question

Position vector of a particle moving in xy plane at time t is r=a(1cosωt)^i+asinωt^j and the angular velocity of the particle is ωo^k. The magnitude of the linear velocity of particle at t=π2ω is 


A
ωoa2
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B
ωoa
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C
ωoa
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D
2ωoa
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Solution

The correct option is D 2ωoa
We know that, v=w×r 
Given, ω=ωo^k  and r=a(1cosωt)^i+asinωt^j
v=ω×r
=∣ ∣ ∣^i^j^k00ωoa(1cosωt)asinωt0∣ ∣ ∣
=^i(ωoa sinωt)^j(ωoa(1cosωt))+0^k
=ωoasinωt^i+ωoa(1cosωt)^j

Hence, the magnitude is given by
|v|=(ωoasinωt)2+(ωoa(1cosωt))2
=(ωoa)2(sin2ωt+1+cos2ωt2cosωt)
=(ωoa)2×2(1cosωt)
=2ωoasin(ωt2)
At t=π2ω;  |v|=2ωoasin(π4)=2ωoa

Physics

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