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Loop de loop normal force roller coaster

  1. Mar 18, 2009 #1
    hello, i'm trying to study for the mcat, and I have a conceptual question about normal force, mg, and centripetal force during a loop-de-loop on a roller coaster.

    Could you validate these force equations?

    1. At the very bottom of the loop:
    N - mg = ma = mv^2 / r
    N = mg + ma

    2. At the side of the loop:
    N = ma = mv^2 / r
    the normal force is providing all of the centripetal acceleration

    3. At the very top of the loop:
    N + mg = ma = mv^2 / r
    N = ma - mg

    Thus the normal force would be the greatest at the bottom of the loop, and least at the top of the loop.

    Is all of this correct?

    also, for the very top of the loop, since normal force and weight are directed downward, what force prevents the cart from just dropping off the tracts?
    in relation to the previous question, what is happening on the side of the loop?

    thank you very much !
     
  2. jcsd
  3. Mar 18, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Good.

    Yes, the normal force provides the centripetal acceleration. But it's not the only force acting on the coaster. So N = ma_c ≠ ma

    Good.

    Good.

    The fact that it's moving is what prevents the cart from falling off, not any upward force. (Just like when you toss a ball in the air. What force makes the ball rise? :wink:)
    Same basic idea, but see my note above regarding #2.
     
  4. Mar 18, 2009 #3
    Hmm, I don't think its highly important, but when the roller coaster is on the side, and the Normal force is not the only force acting.. what would the force equations look like?

    ie:

    Fy : N = ma_c
    Fx : mg = ma

    ??

    also, are centripetal acceleration and linear acceleration related?

    if so, how? what variables would I need?

    thank you very much!!
     
  5. Mar 19, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Good. (Where x is vertical and y is horizontal.)
    Not fundamentally.
     
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