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Circular motion with forces

  1. Feb 20, 2008 #1
    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
     
  2. jcsd
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