An insect of mass m=3 kg is inside a vertical drum of radius 2 m that is rotating with an angular velocity of 5 rad s-1.The insect doesn’t fall off.Then the minimum coefficient of friction required is
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Asked by nipunverma59
11th November 2018,
10:49 PM
Answered by Expert
Answer:
As shown in figure, the forces acting on insect when the drum rotates with the speed w rad/s is
(1) centrifugal force = mω2r
(2) Reaction force R
(3) friction force μR
(4) weight of insect mg
where m is mass of insect, r is radius of drum and μ is friction coefficient.
Reaction force is same as centrifugal force. hence friction coefficient is obtained from,
μR = μmω2r = mg or μ = g/(ω2r) = 9.8/(5×5×2) ≈ 0.2
Answered by Expert
12th November 2018,
1:00 AM
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