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Force of the track on a rollercoaster

  1. Mar 14, 2007 #1
    a roller coster car of mass 1500 kg starts at a distance of H=23m above the bottom of the loop 15 m in diameter. if friction is negligible, what is the downward force of the rails on the can when it is upside down at the top of the loop?

    so what i was thinking is this:

    f=ma and in this case a=v^2/r so i have .....

    f=mv^2/r and i know what m and r are so i need to find v

    ei=ef
    mgh=mgh+.5mv^2
    1500(9.8)23=1500(9.8)15+.5(1500)v^2

    v^2=156.8

    then i put this into f=mv^2/r and f=3.14 x10^4


    is that right?
     
  2. jcsd
  3. Mar 14, 2007 #2

    Doc Al

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    Staff: Mentor

    Note that f=mv^2/r gives the net force, but that you are asked to find the force of the rails on the car. (What other force acts on the car?)
     
  4. Mar 14, 2007 #3
    the forces that act on the car are: normal force, weight, acceleration
     
  5. Mar 14, 2007 #4

    Doc Al

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    Staff: Mentor

    Normal force and weight are the two forces acting on the car. Acceleration is not a force!
     
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