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Simulated circular motion on a roller coaster

  1. Feb 23, 2016 #1
    1. The problem statement, all variables and given/known data
    A roller coaster at the Six Flags Great America amusement park in Gurnee, Illinois, incorporates some clever design technology and some basic physics. Each vertical loop, instead of being circular, is shaped like a teardrop. The cars ride on the inside of the loop at the top, and the speeds are fast enough to ensure that the cars remain on the track. The biggest loop is 40.0 mhigh. Suppose the speed at the top is 10.0 m/s and the corresponding centripetal acceleration is 2g.

    (b) If the total mass of a car plus the riders is M, what force does the rail exert on the car at the top?

    choices are:
    a) Mg (down)
    b) 2Mg (up)
    c) M(v2/r + 2g) (up)
    d) Mg (up)
    e) M(v2/r + 2g) (down)
    f) 2Mg (down)

    (d) Comment on the normal force at the top in the situation described in part (c) and on the advantages of having teardrop-shaped loops

    2. Relevant equations
    Fnet = ma

    3. The attempt at a solution
    I'm trying to create a free body diagram of the roller coaster, but I can't seem to understand how the normal force can even exist if Mg and Ma(c) (the gravity force and centripetal force) would be pointing straight down. This seems to be my weak point in this chapter. I can't comprehend how simulated circular motion works, especially with vertical circles.
     
  2. jcsd
  3. Feb 24, 2016 #2

    haruspex

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  4. Feb 25, 2016 #3
    Well, I kind of get that it's a resultant force, but I don't get how, especially in this problem,

    Basically, if you applied Newton's 2nd law, Fnet = ma, you would get N - mg = mac. Right?

    I just don't understand how you get the normal force, if it's not equal to mg in this situation.
     
  5. Feb 25, 2016 #4

    haruspex

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    That depends how you are defining the constant g. I expect you are defining it such that its value is positive, so the equation is wrong.
    From the correct version of that equation. You are told the value of ac.
     
  6. Feb 26, 2016 #5
    Does it make a difference, though? I've been doing it this way, and it seems to make more sense for me. Are you saying it should be N + mg = mac because gravity is negative? It's just more intuitive for me to do it the other way.
     
  7. Feb 26, 2016 #6

    haruspex

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    Which way it should be depends on the convention you are adopting. Which way is positive for N, for g, for ac?
     
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