Where Does Ball Lose Contact on Ramp?

[sfsy1]

A ball is rolled down a smooth ramp, that curls upwards.
Where will the ball lose contact with the surface?

According to the conservation of energy law, the ball is able to reach its original height, which means the ball should stop at A. However, after the ball passes point B, won't the normal contact force be acting downwards on the ball? I'm unsure whether this will affect the point where the ball loses contact.

Also, I've tried using Normal force \leq 0 to determine the position where the ball loses contact, but i don't seem to be getting anywhere.

[PLAIN]http://img15.imageshack.us/img15/4441/17201057.gif

Am I approaching it the correct method by using the normal force, if not what will be more appropriate for this question? Many thanks in advance!

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[uart]

Assuming no friction then the ball will lose contact somewhere between B and A. It will then follow a parabolic trajectory that will at some point re-collide with the track.

To find where the ball first loses contact you should try to find where the normal (to the curve) component of gravity first exceeds the required centripetal force.

 

[Petr Mugver]

Let's suppose that the curved line is a circle of radius R. Use an angle \alpha to describe the position of the ball on the circle, so that the point B is \alpha=0 and point A is \alpha=\pi/2. Conservation of energy says

\frac{v^2}{2}+gR\sin\alpha=gR

(we suppose the ball starts at rest). We derive the centripetal force

\frac{mv^2}{R}=2mg(1-\sin\alpha)

The component of the gratitational force normal to the trajectory is

mg\sin\alpha

Now we write F=ma:

mg\sin\alpha+T=2mg(1-\sin\alpha)

where T is the reaction of the circle on the ball (inwards going). Since T can only be positive, we have

3\sin\alpha<2

That is \alpha < 42 degrees (almost half way between B and A).

 

[sfsy1]

thanks, very detailed workings.