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Frictionless sphere, falling particle

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle initially sits on top of a large smooth sphere of radius R. The
    particle begins to slide down the sphere without friction. At what angle .
    does the particle lose contact with the sphere?

    2. Relevant equations
    g=9.8m/s2 perhaps? (Is this dependent on the acceleration?)

    3. The attempt at a solution
    It seems like the angle that the particle loses contact with the sphere is possibly dependent on the acceleration of the particle. If so, I don't know how to mathematically explain this relationship. If the angle is equal to x, sinx will give a tangent vector of a magnitude proportional to the size of the angle, but I don't know if that gets me anywhere.

    My second guess is that the particle either loses contact with the sphere at 90 degrees, or 45 degrees. I know that the particle cannot possibly hold on to the sphere without an adhesive property after 90 degrees. I do not know if a particle would be able to stay on the sphere past 45 degrees though.
  2. jcsd
  3. Sep 29, 2008 #2


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    Homework Helper

    As the particle slides down it will gain speed according to mgh being converted to kinetic energy. You may recognize that h is a function of θ.
    You may also want to recognize that when mv2/R exceeds a certain point that it will lose contact with the sphere.
  4. Sep 30, 2008 #3
    Thanks, I got it now, the angle is the arccos of 2/3. I just used conservation of energy for the particle, and get mgh+mv^2/r =mgh+mv^2/r, then solved fo V^2 later by equating centripetal force with mgcos(x) so the net force would be 0, where the particle fell off of the sphere.
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