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Small sphere rolling off the top of a large sphere

  1. Aug 10, 2014 #1
    [>A heavy sphere of radius r = 1.00 meter is fixed with respect to the ground. A small uniform solid sphere is placed at the top of the larger sphere. After a slight disturbance, the smaller sphere begins to roll downward without slipping. How high h is the small sphere above the ground at the instant it loses contact with the large sphere

    My atempt of solution:

    I think, that at the moment when it loses contact, speed of the small sphere is higher than the one in $\Sigma F=mg\cos\theta=mv^2/r$. $\theta r$ is an arc traveled on the sphere. Also since it falls without slipping, $v=\omega r$, where $\omega$ is the angular velocity of the small sphere. The speed at a height h can be found from $$mg(2r)=1/2mv^2+mgh.$$

    From foce equation I would get $$v=\sqrt{rg\cos\theta}$$
    And from energy conservation:
    substituting the expretion for velocity

    I am not sure if this is correct or no. And I can't express $\theta$ in some other way. Could anyone help?
  2. jcsd
  3. Aug 10, 2014 #2


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    Have you forgotten about rotational kinetic energy? The small sphere rotates; this will consume energy. Your conservation of energy equation will be slightly different.
  4. Aug 10, 2014 #3


    also $$h=1+cos\theta$

    and now I would substitute v. Is that right?
    Last edited: Aug 10, 2014
  5. Aug 10, 2014 #4


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    Let ##m## and ##r## be the mass and radius of the small sphere. You need to express ##I## and ##\omega## in terms of ##m##, ##r## and ##v##.

    Also, working with ##h## is more complicated; find ##\theta##, the angle at the center of the large sphere such that the spheres just begin to separate, then convert that to a height.
  6. Aug 10, 2014 #5
    Ok, ty. I got it.
  7. Aug 21, 2015 #6
    I'm working in the same problem and tnx i got it
    Last edited: Aug 21, 2015
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