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 Ctanθ3 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 f≤μN where N=Mgcosθ ⇒Mgsinθ3≤μMgcosθ
or μ≥tanθ3 ∴μmin=tanθ3