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?

Answer

8

Explanation

The free body diagram is shown in the figure, The initial energy =`1/2Iomega_(0)^(2)`
Where `I` is the moment of inertia of the cylinder and is given by `I=1/2Mr^(2)`
(`M` =Mass of the cylinder and `r`=Radius)
`:.` Initial Kinetic energy of cylinder
`=1/4 Mr^(2)omega_(0)^(2).....(1)`

Here, there is no motion of the centre of gravity of the cylinder, hence,
`R+muN=Mg.....(2), N=muR....(3)`
Solving for the `R` and `N`,
`R=(Mg)/((1+mu^(2))).....(4), N=(muMg)/((1+mu^(2))).....(5)`
The total initial energy is dissipated against frictional forces,
`:. 1/4 Mr^(2)omega_(0)^(2)=(muN +muR).2pin, n=(r^(2)omega_(0)^(2))/(8 pig)`
Where `n` is the number of turns made by the cylinder before it stops. Putting the values of `N` and `R`, and solving for `n` gives the final results.