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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?
982309_45ad1adec62247328a4a49c59a4ffb59.png

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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=μkN2mg=0...(ii)
solving these equations N1=mg1+μ2k,N2=μkmg1+μ2k
The torque on the cylinder about the axis of rotation
τCM=mg××0+N1×0+N2×0+μkN2R=μk(1+μk)(1+μk)mgR
The moment of inertia about axis of rotation ICM=12mR2
The torque equation, τ=Iα
μk(1+μk)(1+μk)mgR=(12mR2)αα=2μk(1+μk)g(1+μ2k)R
Using equation ω2=ω202αθ
calculate the angular displacement θ
02=ω202[2μk(1+μk)g(1+μ2k)R]θ
θ=(1+μ2k)ω20R4μk(1+μk)g
Revolution accomplished
n=θ2π=(1+μ2k)ω20R2π×4μk(1+μk)g=(1+μ2k)ω20R8πμk(1+μk)g

1028934_982309_ans_ec85c2926d6d48e0b7c1f020afb6c27e.png

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