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Spinning Sphere

  1. Apr 17, 2009 #1
    1. The problem statement, all variables and given/known data
    A solid sphere of mass M and radius R rotates freely in space with an angular velocity w about a fixed diameter. A particle of mass m, initially at one pole, movies with a constant velocity v along a great circle of the sphere. Show that when the particle has reached the other pole, the rotation of the sphere will have been retarded by an angle
    [tex]\alpha = \omega T ( 1 - \sqrt{\frac{2M}{2M+5m}}) [/tex]


    2. Relevant equations



    3. The attempt at a solution

    So I have a picture of a rotating sphere with a mass on it. I know that [tex]I_{total} = 2/5 M R^2 + mr^2[/tex] where r is the vector from the initial pole to the position of the particle on the sphere. I believe I also need to take into account the polar angle phi, and the azimuthal angle theta.

    I know that angular momentum is conserved. So dL/dt = 0
    I also know that L = Iw

    I was thinking of doing the following: d/dt(Iw) = 0 = d/dt(I) w + I d/dt(w)
    so far that hasn't worked, but I think it may be because I haven't been able to come up with a decent guess for the angular velocity w. My best guess so far is w = v/(r*sin(phi)), which is clearly incorrect.

    Anyone have any hints?
     
    Last edited: Apr 18, 2009
  2. jcsd
  3. Apr 17, 2009 #2
    So I think I have narrowed it down to:

    [tex]L = (2/5 M R^2 + m r^2) (d \phi / dt)[/tex]
    [tex]\phi = \int_0^T \frac{L dt}{2/5MR^2 + mr^2}[/tex]

    Let
    [tex]\omega = 5/2 L / (MR^2)[/tex]
    Unfortunately, I can't seem to find a relationship between r(t) and t? In other words, how do I integrate that? I can't seem to recall anything that would work? I think I am getting close, I just can't get this darn integral to work out.

    I know v is constant, but I can't seem to make anything out of that work.
    if v is constant then dv/dt = 0.
    I was thinking that [tex]v = r \theta[/tex] where theta is the azimuthal angle. But this implies that [tex]d^2( \theta)dt^2 = 0[/tex] which means that [tex]d/dt (\theta ) = constant[/tex] but that means that my expression for v doesn't work, so I am going to guess that it is incorrect.
     
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