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Roller Coaster Loop Pbm - Pls Help.

  1. Jan 9, 2006 #1
    A roller coaster car of mass 1216 kg slides on
    a frictionless track starting at a distance 32 m
    above the bottom of a loop 25 m in diameter.
    The acceleration of gravity is 9.81 m/s^2

    If friction is negligible, what is the magni-
    tude of the force of the track on the car when
    the car is at the top of the loop? Answer in
    units of N.

    Fn=(mv^2) /R - Fg

    mgh = .5mv^2

    v^2= 2gh = 2*9.81*32 = 627.84

    Fn = (1216*627.84)/12.5m - (1216*9.81)

    = 49147.3152N ???

    apparently this is wrong. What did i do wrong?
     
  2. jcsd
  3. Jan 9, 2006 #2

    Doc Al

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    Staff: Mentor

    Your error was in calculating the KE at the top of the loop:
    [tex]mg \Delta h = 1/2 m v^2[/tex]

    [tex]\Delta h[/tex] is not 32 m!
     
  4. Jan 9, 2006 #3

    mukundpa

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    Homework Helper

    is this the velocity at the top.
     
  5. Jan 9, 2006 #4

    mukundpa

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    Homework Helper

    A bit slow in typing then Doc Al.:smile:
     
  6. Jan 9, 2006 #5
    I presume delta h is then 32m - 25m ?

    Thanks for the help.
     
  7. Jan 10, 2006 #6

    Doc Al

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    Staff: Mentor

    That's correct.
     
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