[Physics] Why does the normal force go to zero when a vertically rotating object slips off the surface it is rolling along

newtonian-mechanicsrotationrotational-kinematics

In the classic problem of a ball rolling down a disk with constant gravity from rest at the top, it is known that the angle transversed before the ball loses contact with the surface (i.e. where the net force on the object is no longer $mv^2/r$) is $\cos^{-1}(2/3)$. This is trivial once you assume the normal force goes to zero at this point. My question is, how do we know the normal force must go to zero for the ball to slip off?

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Best Answer

The answer seems rather obvious to me : That an object "slips off the surface" means that there is no longer any contact between them. When there is no contact, there is no contact force. This applies to the normal force and friction, which are both contact forces.

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