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

`SigmaF_(x)=0, N_(1)-mu_(k)N-1=0`……….i
`SigmaF_(y)=0, N_(1)+mukN_(2)-mg=0`……..ii
Solving these equations `N_(1)=(m)/(1+mu_(k)^(2)):N_(2)=(mu_(k)mg)/(1+mu_(k)2)`
The torque on the cylinder about the axis of rotation
`tau_(CM)=mgxx0+N_(1)xx0+N_(2)xx0+mu_(k)N_(2)R`
`=(mu_(k)(1+mu_(k)))/((1+mu_(k)^(2)))mgR`
The moment of inertia about axis of rotation `I_(CM)=1/2mR^(2)`
The torque equation `tau=Ialpha`
`(mu_(k)(1+mu_(k)))/((1+mu_(k)^(2)))mgR=(1/2mR^(2))alphaimplies =(2mu_(k)(1+mu_(k))g)/((1+mu_(k)^(2))R)`
Using equation `omega^(2)=omega_(0)^(2)+2atheta`, calculate the angular displacement `theta`.
`0^(2)=omega_(0)^(2)-2[(2mu_(k)(1+mu_(k))g)/((1+mu_(k)^(2))R)]theta`
(`omega` and `alpha` are in the opposite sense)
`implies theta=((1+mu_(k)^(2))omega_(0)^(2)R)/((4mu+k(1+mu_(k))g)`
Revolution accomplished
`n=theta/(2pi)=((1+mu_(k)^(2))omega_(0)^(2)R)/(2pixx4mu_(k)(1+mu_(k))g)=((1+mu_(k)^(2))omega_(0)^(2)R)/(8pimu_(k)(1+mu_(k))g)`