A uniform cylinder of radius `R` is spinned about it axis to the angular velocity `omega_(0)` and then placed into a corner,. The coeficient of friction between the corner walls and the cylinder is `mu_(k)` How many turns will the cylinder accomplish before it stops?

Explanation

In the problem, the rigid body is in translation equlibrium 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=N_1+kN_2`, `N_2=kN_1`
Hence, `N_1=(mg)/(1+k^2)`, `N_2=k(mg)/(1+k^2)`
For pure rotation of the cylinder about its rotation axis, `N_z=Ibeta_z`
or, `-kN_1R-kN_2R=(mR^2)/(2)beta_z`
or, `-(kmgR(1+k))/(1+k^2)=(mR^2)/(2)beta_z`
or, `beta_z=-(2k(1+k)g)/((1+k^2)R)`
Now, from the kinematical equation,
`omega^2=omega_0^2+2beta_xDeltavarphi` we have,
`Deltavarphi=(omega_0^2(1+k^2)R)/(4k(1+k)g)`, because `omega=0`
Hence, the sought number of turns,
`n=(Deltavarphi)/(2pi)=(omega_0^2(1+k^2)R)/(8pik(1+k)g)`