Centripetal is the name of a direction, not the name of a kind of force.
If I ask you to list for me, all of the forces acting on a tether ball as it rotates on the string around the pole, "centripetal force" IS NOT one of them. It is the tension and the gravity that act on the ball, and possibly air drag if you care. Gravity pulls the ball downward, and tension pulls the ball upward and inward. The net force is inward, in a direction perpendicular to velocity, and hence the ball changes direction due to it rather than speed.
For a force to count as one of the forces acting on an object, it must be a real force. As in, you can identify an object which is the agent of that force, from which it is exerted. And you can identify why that force is exerted, such as in the case of tension, it is a force that restores the rope back to its initial length, after it gets stretched.
No such "centripetal force" is "acting as" the net force. It is a net force that is in the centripetal direction. As opposed to the tangential direction, which is what you call the direction along an object's velocity (since it is tangent to any curvature that there may be).
The net force either acts in the centripetal direction, or the tangential direction, or some combination of both. A centripetally directed net force causes a change in direction. A tangentially directed net force causes a change in speed.
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In your example problem, there are two forces acting on the car that we care about. We neglect friction and air drag, for simplicity. The forces we care about, are GRAVITY and the NORMAL FORCE.
Gravity will act according to the object's proximity to a nearby dominant mass, namely Earth. We simplify this, by talking about Earth's gravitational field g, of 9.8 N/kg.
The normal force, is the force of the track pushing upward on the roller coaster car. It will be as large as necessary, to prevent the roller coaster car from crumbling through the structure. That is, provided that it doesn't exceed the strength of the steel track rails. If your track and wheels were built in a bi-directional manner, then the track could also pull downward on the car. It is seldom that this is the case, and usually you don't want to let this happen. As it can make the ride unsafe for passengers to go airborne in their seats.
At point A, the bottom of the track, the car must accelerate upward to avoid crumbling through the structure below. It is in a locally circular path, and thus the acceleration upward, is a centripetal acceleration, of value v^2/r. Normal force acts up, gravity acts down. N - m*g = m*v^2/r. Solve for N, to get the force from the track on the car.
N = m*(g + v^2/r)
At point B, the top of the track, we hope that the car remains in contact with the track. If it looses contact with the track, the ride is unsafe. We could make for bidirectional constraint wheels, but that still will make for an awkward ride, and an unsafe ride. So, the maximum speed will be such that it just barely unloads the force from the track. In otherwords, N=0, and not a single Newton less.
Normal force acts up
Gravity acts down
The acceleration is centripetal acceleration, thus directed inward to the curve center, which is thus downward.
m*g - N = m*a
m*g - N = m*v^2/R
Set N=0:
m*g = m*v^2/R
Solve for v:
v = sqrt(g*R)