Solving for Min Speed on Loop-the-Loop Rides

[Anne Armstrong]

Homework Statement

In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 260 kg and moves with speed v = 15 m/s. The loop-the-loop has a radius of R = 10 m. What is the minimum speed of the car so that it stays in contact with the track at the top of the loop?

Homework Equations

a_c=v²/r
F=ma

The Attempt at a Solution

At the top of the loop, the forces acting on the car are F_gravity, F_Normal, and F_centrifugal (I think). So I think the minimum speed would be one that made all the forces cancel to zero (aka, F_c is just strong enough to counteract F_gravity and F_Normal). If that's true, then F_c=F_N+F_g. Since F_c=m*a_c=v²/r , so far I have: m*a_c=v²/r = m*g+m*g.
..but I don't think that makes sense... Is F_N in this case equal and opposite to F_c?

 

[haruspex]

If the car is only just staying in contact, what will the normal force equal? Remember, this the force the track and car exert on each other.

 

[Anne Armstrong]

The normal force will equal the force of gravity, m*g?

 

[haruspex]

The normal force will equal the force of gravity, m*g?

Why would it have to be that?
A normal force is the reaction that results when attempting to push an object through something that resists. When you place an object on solid ground, the weight of the object acts to push the object through the floor. The normal force is the reaction necessary from the floor to prevent it. When you place on object on an incline, only part of the weight is trying to push the object into the incline, so the normal force is less. In this case, what is trying to push the object through the top of the loop?