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Conservation of mechanical energy for skier on sphere

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data
    A skier of mass m starts from rest at the top of a solid sphere of radius r and slides down its frictionless surface. At what angle θ will the skier leave the sphere?

    2. Relevant equations
    KE= 0.5mv^2
    PE = mgh
    Fc = (mv^2)/r


    3. The attempt at a solution
    I am really quite confused and don't even know how to begin. I know that the skier will fall off when the normal force is 0 but I'm not sure how to even get to that point.
    I thought that PE=KE so 0.5mv^2 = mgh which simplifies to 0.5v^2 = gh
     
  2. jcsd
  3. Mar 30, 2013 #2

    Doc Al

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    Staff: Mentor

    That's the key point. To make use of it, set up a force equation with Newton's 2nd law.
    Good. You'll need that too. (Express h in terms of θ.)
     
  4. Mar 30, 2013 #3
    So rewriting the energy equations gives me
    v^2 / 2 = grsin(θ).

    When I write the force equations, I get
    N-mg = ma
    N = mg + mg
    N = m(a+g)
    m = N/(a+g)

    If I plug that in to the centripetal force equation, I get
    F = (mv^2)/r
    F = (Nv^2)/((r)(a+g))

    Is this right? If so, where do I go from here? Sorry to ask such dumb questions, I'm just very confused on this problem.
     
  5. Mar 30, 2013 #4

    Doc Al

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    Staff: Mentor

    What you want is Δh, the drop from the original position at the top. Δh ≠ r sinθ.

    Careful! Forces are vectors.

    Hint: Consider force components perpendicular to the surface at any point.
     
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