Circular motion with forces

[StephenDoty]

As a roller coaster car crosses the top of a 50.0 -diameter loop-the-loop, its apparent weight is the same as its true weight.What is the car's speed at the top?

So the car is under the circle at the top. so the forces acting on the car would be mg and the normal force acting down

My question is, is the force f=mv^2/r going down with the rest of the forces or up against the other forces?

I know that since a = 0 and that a goes toward the center, thus Fnet = 0 and going to the center. But does this mean that F=mv^2/r goes down too?

Thank you.

Stephen

 

[Nate810]

No, the force f=mv^2/r is not going down with the rest of the forces. This force is a centripetal force, which is the force that keeps the car moving in a circular path around the top of the loop. At the top of the loop, the car’s speed must be great enough to counteract gravity (mg) and the normal force (N) to keep it moving in a circular path. Thus, the speed of the car at the top can be calculated by equating the centripetal force to the sum of the other two forces: f = mv^2/r = mg + N or v = √[gr/m] where g is the acceleration due to gravity, r is the loop’s radius, and m is the mass of the car. Plugging in the given values gives us the car’s speed at the top of the loop-the-loop as v = 8.94 m/s.