How does friction cancel out in this problem?

[Vitani11]

Homework Statement

Consider a uniform solid disk of mass m and radius R, rolling without slipping down an inclined plane with an angle γ to the horizontal. The instantaneous point of contact between the disk and the incline is called P.

Homework Equations

ma = mgsinγ - μmgcosγ
Γ = Rmgsinγ-Rμmg
Moment of inertia about P = (3/2)MR² from parallel axis theorem
Γ is torque

The Attempt at a Solution

Are these the correct equations to start out from? Probably not. My professor said you are to assume that there is friction and it will end up cancelling. I can do the problem without friction but not including it.
Γ = Rmgsinγ-Rμmg = (3/2)MR²α
gsinγ-μg = (3/2)a
a = (2/3)(gsinγ-μg)

 

[BvU]

Can you be a bit more explicit ? What does $\Gamma = R mg\sin\gamma - R\mu mg$ stand for ?

 

[TomHart]

It looks like you are defining the friction force to be equal to μN, but that only applies when the object is ready to slip. The problem only says that it is rolling without slipping. It doesn't give you any sense of how close it is to slipping. Also, I think it would be helpful if you could post your free body diagram showing how you defined your axes.

 

[rcgldr]

The problem statement implies that the instantaneous torque is to be calculated about the instantaneous point P, in which case friction keeps the disk from sliding so that it does instantaneously rotate about point P, but friction is not related to any torque about point P, since that is the pivot point.

I don't understand why your professor stated that friction will end up cancelling. Although the torque about point P related to friction is zero, the point of application of linear forces doesn't matter, and the friction force opposes the component of gravity in the direction of the inclined plane, reducing the linear rate of acceleration of the disk (versus a frictionless plane).

 

[Vitani11]

Got it - thanks