Nonuniform Circular Motion: The Top of the Loop

[L'Arrogant]

If a ball is going through a loop-the-loop, at the top, if mg=m(v^2/r), it will continue around the loop. If mg>Fc, it will fall. In the first case, when mg=Fc, the normal force on the ball from the metal of the loop is 0. This is easy enough to see when the equation from the free body diagram of the ball is rearranged (if N=mg-Fc, and the latter two are equal, Normal is zero); I don't understand why this is, though. Rereading it, my question seems vague. To restate: the fact that N depends on mg and Fc makes sense, but it doesn't make sense to me that the fact that N=0 means that the ball is at some minimum velocity. The first relationship follows directly from the free body diagram. The second doesn't follow so clearly. Or does it?

Similarly, a go cart riding over a hill can go a maximum speed before it leaves the ground entirely. Same thing: N=0, v=(gr)^1/2. Is the fact that N=0 a result of it leaving the ground, or the cause? My professor wrote that N=mg-Fc=0 when contact lost.

Thanks for taking the time to read my question.

 

[Doc Al]

If a ball is going through a loop-the-loop, at the top, if mg=m(v^2/r), it will continue around the loop. If mg>Fc, it will fall. In the first case, when mg=Fc, the normal force on the ball from the metal of the loop is 0. This is easy enough to see when the equation from the free body diagram of the ball is rearranged (if N=mg-Fc, and the latter two are equal, Normal is zero); I don't understand why this is, though. Rereading it, my question seems vague. To restate: the fact that N depends on mg and Fc makes sense, but it doesn't make sense to me that the fact that N=0 means that the ball is at some minimum velocity. The first relationship follows directly from the free body diagram. The second doesn't follow so clearly. Or does it?

N = 0 just specifies the condition that the ball is about to lose contact with the loop. For high enough speeds, the ball will be pressed against the loop and thus N > 0. The minimum speed for maintaining contact with the loop can be determined from setting N = 0. If the speed drops below that minimum speed, the net force on the ball (which is now just mg) becomes too much--the ball is pulled away from its circular path and becomes a free projectile following a parabolic path.

Similarly, a go cart riding over a hill can go a maximum speed before it leaves the ground entirely. Same thing: N=0, v=(gr)^1/2. Is the fact that N=0 a result of it leaving the ground, or the cause? My professor wrote that N=mg-Fc=0 when contact lost.

The same idea but in reverse for the cart going over the hill. If the cart goes too fast, the force of gravity (mg) is not enough to hold the motion in its circular path over the hill, so the cart shoots off into the air as a free projectile following a parabolic path.