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

 A: 
As the ball starts rolling (without slipping)  down the disc, potential energy is converted to kinetic energy:
$$\frac12 mv^2+\frac12 I\omega^2=mgR(1-\sin\theta)\tag{1}$$
With:
$$I=\frac25 mr^2$$
And:
$$v=\omega r$$
$$\frac{7}{10}mv^2=mgR(1-\sin\theta)\tag{2}$$
To keep the ball on its circular trajectory, a centripetal force is needed:
$$F_c=\frac{mv^2}{R}=\frac{10}{7}mg(1-\sin\theta)$$
This centripetal force is provided by the normal component of the weight $mg$:
$$mg\sin\theta$$
So:
$$mg\sin\theta=\frac{10}{7}mg(1-\sin\theta)$$
$$\sin\theta=\frac{10}{7}-\frac{10}{7}\sin\theta$$
So to keep the ball stays on the sphere as long as:
$$\sin\theta\leq\frac{10}{17}$$
A: 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.
A: As the ball rolls down the the disc it is moving along the arc of a circle of radius equal to that of the disc and so is undergoing a centripetal acceleration.
To do this the ball must have a net force towards the centre of the disc.
The net force on the ball is $mg\cos \theta -N$ where $mg \sin \theta$ is the component of the weight of the ball towards the centre of the disc.
You can think of the normal force $N$ as being of such a magnitude as to make sure the net force towards the centre of the disc is just enough to provide the centripetal acceleration.
As the ball moves down its speed $v$ increases and angle $\theta$ increases.
This means that the centripetal acceleration of the ball increases and so a larger force towards the centre of the disc is needed to provide the larger centripetal acceleration.
However at the same time the component of the weight of the ball $mg \sin \theta$ has decreased.
This means that the normal force has to be lower.
As the ball moves down its speed increases and so the normal force gets smaller and smaller until it becomes zero.
At that moment the component of the weight of the ball provides just enough force to produce the centripetal acceleration of the ball.
The ball drops a little more and it moves faster and now the component of the ball's weight towards the centre of the disc is not large enough for the ball to move along an arc of a circle of radius which is the radius of the disc.
So what happens?
The ball loses contact with the disc as it follows a less curved trajectory than that which would be required to keep in contact with the disc.
