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Question

A bead of mass m = 0.2 kg lies at a distance r from the axis of rotation on a rigid wire as shown in figure. Coefficient of friction between wire and bead is μ, and wire starts rotating from rest at an angular acceleration of α at t = 0. The whole arrangement is in gravity free region. r = 1m, μ=0.4 & α=0.1 r/s2


A
Relative motion between bead and wire starts at t = 2 sec
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B
Power of friction force at t = 2 is 0.0016 watt
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C
Power of normal contact at t = 2 is 0.004 watt
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D
Power of net force on bead at t = 2 is 0.004 watt
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Solution

The correct options are
A Relative motion between bead and wire starts at t = 2 sec
C Power of normal contact at t = 2 is 0.004 watt
D Power of net force on bead at t = 2 is 0.004 watt

N=mat=mrα
fr=μN=μmrα

The bead starts sliding when friction (fr) is less than centrifugal force.
fr=mω2r
μmrα =mω2r
μm/r/α=m/ω2r/
ω2=μα and we know, ω=αt
α2t2=μαt=μα=0.40.1=2 sec
v=rω=rαt at t = 2, v=1×0.1×2=0.2m/s
fl=μN=μmrα=0.4×0.2×1×0.1=0.008
N=mrα=0.2×1×0.1=0.02
Power of friction = 0
Power of normal = N×v=0.004 Watt
 

Physics

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