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Question

A uniform cylinder of radius R is spinned about its axis to the angular velocity ω0 and then placed into a corner. The coefficient of friction between the corner walls and the cylinder is μk. How many turns will the cylinder accomplish before it stops?

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Solution

As the centre of mass of the cylinder does not accelerate hence ∑F=0∑Fx=0,N2=μkN1=0...(i) ∑Fx=0,N1=μkN2−mg=0...(ii)solving these equations N1=mg1+μ2k,N2=μkmg1+μ2kThe torque on the cylinder about the axis of rotationτCM=mg××0+N1×0+N2×0+μkN2R=μk(1+μk)(1+μk)mgRThe moment of inertia about axis of rotation ICM=12mR2The torque equation, τ=Iαμk(1+μk)(1+μk)mgR=(12mR2)α⇒α=2μk(1+μk)g(1+μ2k)RUsing equation ω2=ω20−2αθcalculate the angular displacement θ02=ω20−2[2μk(1+μk)g(1+μ2k)R]θ⇒θ=(1+μ2k)ω20R4μk(1+μk)gRevolution accomplishedn=θ2π=(1+μ2k)ω20R2π×4μk(1+μk)g=(1+μ2k)ω20R8πμk(1+μk)g

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