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Centripetal Forces and Roller Coasters

  1. Oct 10, 2015 #1
    1.
    A roller coaster is designed with a clothoid loop that has a radius of 12 m at the top. For comfort, the apparent weight of a rider at the top of the loop must be 0.400 normal weight. What is the speed of the car at the top of the loop?


    2.
    Fn = 0.400(Fg)
    Fn = 0.400(mg)
    mv2 / r = Fn + Fg (Since the Fn and the Fg face the same direction.)

    3. The attempt at a solution

    So my attempt at this was to draw a FBD, and I noticed that the Fn and the Fg face the same direction. From that I could come up with the equation ΞFy = Fn + Fg and then using things I could sub in I came up with;

    mv2 / r = 0.400(mg) + mg
    mv2 / 12.0m = 1.400mg
    v = √((1.400mx9.81m/s2)(12.0m) / m)
    v = 12.8m/s

    My question along with if this is correct. Is why is there no force in the positive direction? I understand that the cart is resting on the tracks and is held in there. But shouldn't there be a force going upward? Wondering if someone could explain that and tell me if/where I went wrong.
     
  2. jcsd
  3. Oct 10, 2015 #2
    You could work your way around the track to see how the normal force changes direction from level ground to travelling within the loop.
    mg would always point downwards. Fn will change direction.
    I don't see why you would be incorrect.

    A FBD displays forces acting ONLY on the cart. - not any forces that the cart has on the track.

    By the way, The question doesn't state that cart travels on the inside or outside of the loop.
    Does anything change for the FBD in either case?
    Would you then call force of the track on the cart the normal force or the centripetal force if on the outside?

    Do one for the cart going over a hill with radius 12 m at the top.
    How does the FBD look now?
    Does the velocity differ?
     
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