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Homework Help: Dynamics/kinematics of rotating sphere

  1. Dec 1, 2015 #1
    I'm wondering about how one would describe the dynamics of a rotating sphere. Consider this: a solid sphere of mass "m" and radius "r" is set to rotate about a tangent to its surface. If it is released from the horizontal position such that it swings like a pendulum, what would be the force acting on the sphere at the lowest point (force from pivot, i.e. centripetal force)

    First of all, one could find the moment of inertia using the parallel axis theorem correct? So

    I = Icm + mr²
    I = (7/5)mr²

    Now we could say that its being released from a height "h" of

    h = 2r

    So the potential energy is converted to rotational kinetic energy:

    mgh = ½Iω²
    mgh = ½((7/5)mr²)ω²
    gh = (7/10)r²ω²

    Now I'm puzzled, the force acts on the body's center of mass or the surface? Could we replace ωr = v to find the velocity about the center of mass and use that velocity to find the centripetal force? I.e.;

    v² = (10/7)gh


    F - mg = mv²/r
    F = mv²/r + mg
    F = m(v²/r + g)
    F = m((10/7)gh/r + g)
    F = mg((10/7)h/r + 1)

    This makes sense right? Since the velocity at the center of mass of the sphere is simply v = rω where r is the radius of the circular path of rotation, which happens to be the radius of the sphere? I need some help with this, thanks!
  2. jcsd
  3. Dec 1, 2015 #2


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    Hello cat of Faraday, :welcome:

    You really want to benefit from the homework template ! This post fits that very nicely and the problem statement is crystal clear to me (I think...:smile:)

    You are doing well, but I think you can benefit from a few comments:
    My picture is this (in a side view):


    The red dot is the end view of the rotation axis A that is glued tangent to the ball. The axis is fixed.
    In "1" the ball is released from a 'horizontal' postion and you have to calculate the force in "2".

    So the center of mass describes an arc with a radius of "not 2r". And at point 2 the ball doesn't just have kinetic energy from that center-of-mass motion....
  4. Dec 1, 2015 #3
    Thank you for your response, I will definitely post these kinds of questions in the homework section in the future (if that is what you're implying).

    So what you are saying is the radius of the arc is not 2r, ah yes I must have made a mistake there, my bad... But you are also saying that the center of mass doesn't only have rotational kinetic energy (I think)? So I should add that it has translational as well? That is, the gravitational potential energy is converted to:

    mgr = ½Iω² + ½mv²

    Is that it? I'm sorry if I'm completely misunderstanding what you are saying, could you please elaborate? :smile:
  5. Dec 1, 2015 #4


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    What do you think ?
    My five cents worth is that you have already accounted for that by using the parallel axis theorem.

    ( in other words: I my have confused you -- :frown: ).
  6. Dec 1, 2015 #5


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    Yes, in this case it is correct to find the centripetal force by using the motion of the center of mass like that. The key reason is that it is purely rotating about that axis. This basically means every 'piece of mass' goes in some circle centered on that axis. The sphere is taken to be rigid, therefore each 'piece of mass' must also have the same angular speed ω.

    So let's see why it's okay to use the center of mass' motion. If we don't use the center of mass motion, we would have to find the net centripetal force needed by summing up the centripetal force on each 'piece of mass:'
    but since we reasoned that ω is the same for each part, we can factor it out of the sum,
    which is by definition the same as ω2MRCoM
    Last edited: Dec 1, 2015
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