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Conservation of Energy Problem - Very Tricky

  1. Nov 27, 2012 #1
    1. The problem statement, all variables and given/known data

    http://imgur.com/ZGvC1

    At what angle does the cart fly off of the track?

    No other information is provided


    2. Relevant equations

    I know that energy is always conserved (total energy before = total energy afterwards).


    3. The attempt at a solution

    Equation for energy before = energy after:

    mgR + Ek = mgh + 1/2mv^2
    mgR = mgh + 1/2mv^2 (h is the height above the ground just before the cart leaves the track)

    gR = gh + 1/2v^2

    v^2 = 2g(R-h) --> This is the velocity just before the cart leaves the track



    The dotted line that we see is equal to R.

    Thus, Fc (centripetal force) = mv^2/R

    If the cart were to leave the track, Fn (normal force) would equal 0

    Where do I go from here? Am i on the right track?
     
  2. jcsd
  3. Nov 27, 2012 #2

    haruspex

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    So far so good. At the point where the cart leaves the track, what are the forces acting on it? What can you say about its acceleration at that point?
     
  4. Nov 27, 2012 #3
    After it leaves the track the only force acting on it is gravity.


    Would I do: mg = mv^2/R

    g = v^2/R

    R = v^2/g

    Plug in v^2 = 2g(R-h)

    R = 2g(R-h)/g

    R = 2R - 2h

    R = 2h

    How do I find the angle then?
     
  5. Nov 27, 2012 #4

    Doc Al

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    Staff: Mentor

    Good.
    Almost. What direction is the centripetal acceleration at the point that it leaves the track? Consider forces along that axis.
     
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