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A Penny Falling of a Staionary/Fixed Position Sphere

  1. Mar 20, 2013 #1
    1. The problem statement, all variables and given/known data

    A penny is released from the top of a very smooth sphere of radius 1.3 meters. The sphere is fixed to a platform and doesn't move. The penny slides down from rest and leaves the sphere at a certain point. How far will the penny fall away from the point of contact of the sphere and the platform?

    The penny just kind of slides off on it's own, and what needs to be found is where exactly it leaves the surface of the sphere.

    2. Relevant equations

    mgh = 1/2*m*(V^2)
    h can be found using the angle.

    3. The attempt at a solution

    From the drawing I drew, I had the penny start from the top of the sphere and picked a spot where it falls off. I drew a line through the center of the sphere and the point where the penny falls off, but I'm at a bit of loss of where exactly to begin.
     
  2. jcsd
  3. Mar 20, 2013 #2

    Doc Al

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    Staff: Mentor

    Analyze the forces acting on the penny and apply Newton's 2nd law. (Don't forget that the sphere is a curved surface.)
     
  4. Mar 20, 2013 #3

    rude man

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    Think centripetal force.
     
  5. Mar 21, 2013 #4
    I know the forces working on the penny in the y-axis are the normal F = Weight. What I've been thinking about doing is using

    mgh = (1/2)m(V^2)

    My problem is the h. I'm thinking about saying that the center of the sphere is 0, and that the initial height of the penny is the radius. After that, I feel like I should use

    h' = h - y

    with h' being the height the penny falls off. I just don't know how to get the angle I need.
     
  6. Mar 22, 2013 #5

    Doc Al

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    Staff: Mentor

    At any point, the normal force is perpendicular to the surface. Hint: Analyze force components perpendicular to the surface. What's the acceleration in that direction?
     
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