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Question

A solid cylinder of mass M and radius R rolls (without slipping) down an inclined plane of inclination θ. The minimum coefficient of friction μ between the cylinder and the plane so that it rolls without slipping must be

A
tanθ
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B
tanθ2
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C
tanθ3
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D
tanθ4
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Solution

The correct option is C tanθ3
The acceleration of the cylinder down the incline is
a=gsinθ1+IMR2 ...(i)
Equation for torque about centre of mass of cylinder,
τ=Iα
fR=Iα=IaR ...(ii)
[a=αR for pure rolling]
From Eq (i) & (ii),
f=IR2×gsinθ(1+IMR2)
Putting I=MR22,
f=Mgsinθ3 ...(iii)

To avoid slipping, required value of f must be lesser than limiting value of friction i.e fμN where N=Mgcosθ
Mgsinθ3μMgcosθ
or μtanθ3
μmin=tanθ3

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