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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 (figure shown above).The coefficient of friction between the corner walls and the cylinder is equal to k. The number of turns the cylinder accomplish before it stops is given by n=ω20(1+k2)Rxπk(1+k)g. Find the value of x.
141510_4facbff0b0cd4b4db577cf92aad23bec.png

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Solution

In the problem, the rigid body is in translation equilibrium but there is an angular retardation. We first sketch the free body diagram of the cylinder. Obviously the friction forces, acting on the cylinder, are kinetic. From the condition of translational equilibrium for the cylinder,
mg=N1+kN2; N2=kN1
Hence, N1=mg1+k2; N2=kmg1+k2
For pure rotation of the cylinder about its rotation axis, Nz=Iβz
or, kN1RkN2R=mR22βz
or, kmgR(1+k)1+k2=mR22βz
or, βz=2k(1+k)g(1+k2)R
Now, from the kinematical equation,
ω2=ω20+2βzΔφ we have,
Δφ=ω20(1+k2)R4k(1+k)g, because ω=0
Hence, the sought number of turns,
n=Δφ2π=ω20(1+k2)R8πk(1+k)g
229167_141510_ans_f6d9d15ed8224f949605489b7ab709df.png

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