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Roller Coaster Hills and Loops

  1. Nov 8, 2008 #1
    1. The problem statement, all variables and given/known data

    A roller coaster begins with a steep downhill descent followed by a vertical loop. How much higher does the initial hill have to be to than the height of the loop so that the riders experience a force of 1G keeping them in their seats at the top of the loop? Assume that the rolling friction involved is negligible.

    2. Relevant equations

    Fc = mV2/r
    Fc = FT - Fg (at the top of the loop)
    ac = V2/r
    Ep = mgh
    Ek = mV2/2
    ET1 = ET2

    3. The attempt at a solution

    For starters, I imagine there's going to be a lot of algebra involved here as no information is given. I've tried working under the assumption that m=1 and r=1. I'm fairly certain I can do the same thing with V, but I'm not going to risk sabotaging my work any more than I might have already :redface: Also, I figure Fc must be 9.8 N to account for the 1G, although I'm not convinced I'm right there. I've tried determining the velocity through centripetal force and I've tried applying the law of conservation of energy. Unfortunately, it's been about 4 months since my last physics course, so a lot of this stuff has left me already.

    Thanks :smile:
     
  2. jcsd
  3. Nov 8, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi JebsCripedìas! Welcome to PF! :smile:
    Come on … think!

    How much force (they mean acceleration, of course) are you experiencing keeping you in your seat at the moment? :wink:
     
  4. Nov 8, 2008 #3
    Honestly, I'm ashamed of myself. I know this is an easy question. Mass is cancelled (at least when applyign the LCE), that much is obvious. The problem for me is that my h1 always ends up being a fraction of h2. I can't wrap my head around this and it's really getting on my nerves. I've somehow ended up with different answers, too. I've got h1 = 1/20, 1/2, and 1/r. The latter seems like the most likely to me, but I just don't see it.
     
    Last edited: Nov 8, 2008
  5. Nov 8, 2008 #4

    LowlyPion

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    I think you are thrashing about a bit. Draw a simple force diagram for the moment at which they want the passenger to be experiencing a 1g force into his seat. He's upside down at the top of the loop isn't he? His upward force needs to be g into the seat and there is also gravity working against that, so the radial acceleration must be ... what?

    Now look at Kinetic energy he would be carrying at that point. From the height of the drop how much would that be?

    Don't you now have a pretty good idea of how to express the height he started above that point in terms of r?
     
  6. Nov 8, 2008 #5
    Well here's what I'm doing, and I'm always getting h1 = h2/4. It's what I had done when I said one of my results for h1 was 1/2

    Fc = FT - Fg = mV2/r
    FT = m(V2/r + g)

    Since FT - Fg has to yield 1g, or 9.8, I know FT = 19.6

    19.6 = m(V2/r + 9.8)

    I'm operating under the assumption that m = 1

    19.6 = V2/r + 9.8
    9.8 = V2/r

    Now if r = 1 (in which case h2 = 2)

    9.8 = V2 and 3.1 = V

    So now I apply the law of conservation of energy.

    Ep = Ek
    mgh1 = mV2/2
    gh1 = V2/2
    9.8h1 = 9.8/2
    h1 = 1/2 = 0.5m

    Now I have h1 = 0.5m and h2 = 2m. I have to be missing something when I do the LCE. Actually, I think I see it now.

    Ep1 = Ek + Ep2
    mgh1 = mV2/2 + mgh2
    gh1 = V2/2 + gh2
    9.8h1 = 4.9 + 19.6
    9.8h1 = 24.5
    h1 = 2.5m

    So now I have h1 = 5h2/4. It looks good to me. Tell me this is right please :smile:
     
  7. Nov 8, 2008 #6

    tiny-tim

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    HI JebsCripedìas! :smile:

    Yes, that's right …

    but it would be a lot simpler if you left g as "g", and r as "r" rather than 1:

    mgh = mv2/2, and v2/r = g, so h = … :wink:
     
  8. Nov 8, 2008 #7
    Well I want to write h1 as a function of h2, so I don't see any other way to say it. I guess I must not be trying hard enough.
     
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