Rotational motion with friction I am quoting this question from IE Irodov Dynamics of solid body q 248.

A uniform cylinder of radius R is spinned about its axis to the angular velocity Wo and then placed into a corner. The coefficient of friction between the corner walls and the cylinder is equal to K. How many turns will the cylinder accomplish before it stops.

I have conceptual problem in this question.
I want to know why the centre of mass of the cylinder cannot move away from the wall and upward direction?
why should the normal force from vertical wall be equal to the frictional force from the ground?
Any help will be appreciated.
 A: Here, I assume that by right, you mean away from the wall.
The cylinder will not do so if it is placed in such a way that it tends to move towards (or rather into the wall, if there was a cavity) i.e. if it rotates in an anticlockwise manner. Placing it in a clockwise manner makes it a simple problem (relatively easy). So, we're just going to omit that possibility.
Now, as to why the force applied by the wall should equal the frictional force on the ground, the answer is simple. If it were more, the cylinder will move to the right, which makes no sense, if you think about it. If it were less, the ball will go into the wall, which too should not take place. (A Free Body Diagram would help you)
As to why it does not go up, there is only a single upward force acting i.e the frictional force of the wall. The frictional force on the wall is the friction coefficient (<1) times the force exerted by the wall, which is less than the weight of the cylinder. So, it cannot possibly raise. By the way, it can at most produce angular retardation.
I hope you can understand 
