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Roller coaster at bottom of dip weight increases 50%?

  1. Dec 11, 2008 #1
    1. The problem statement, all variables and given/known data

    A roller coasters weight increases 50% through the center of a dip in the track. The radius of the dip is 30m. What is the speed at the bottom?

    2. Relevant equations

    This is a uniform circular motion problem correct? But how do I calculate if I dont know the weight?

    v = [tex]\omega[/tex]r
    Fnet = mv2/r
    3. The attempt at a solution

    So is weight twice that of the normal force? Or the normal force has to balance it out by increasing by 50% as well?

    Or should I be thinking in terms of [tex]\omega[/tex] instead of forces?
     
  2. jcsd
  3. Dec 11, 2008 #2

    mgb_phys

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    You are on the right track ;-)
    At the bottom of the curve their weight due to gravity is 'mg'
    You are simply looking for the centrifugal force that equals half this.

    Then the total force downward is 1.5mg
     
  4. Dec 11, 2008 #3
    So right now I have F = N - (1.5mg) ...

    which is the only forces acting on the car...

    but I thought the force has to point to the middle of the circle?
     
  5. Dec 11, 2008 #4
    N - W = (mv2)/r

    but in the book it says the normal force is greater than the weight, resulting in a force pointing towards the center...
     
  6. Dec 11, 2008 #5

    mgb_phys

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    The normal force acting on the car must equal the force of the car on the track.
    At the bottom of the curve there is weight acting down and centrifugal acting outward - which = down at the bottom of the curve.
     
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