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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

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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: $$h=\frac{4g-v^2}{2g}$$ substituting the expretion for velocity $$h=\frac{4-r\cos\theta}{2}$$

I am not sure if this is correct or no. And I can't express $\theta$ in some other way. Could anyone help?

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  • $\begingroup$ Well one thing is that it seems you have lost a factor of r in your final equation for h. $\endgroup$ – wgwz Aug 9 '14 at 22:28
  • $\begingroup$ hmm. I don't think so. How? @skywalker $\endgroup$ – Mykolas Aug 9 '14 at 22:32
  • $\begingroup$ How can we do 2 minus something with dimensions of length? (You final equation for h) $\endgroup$ – wgwz Aug 9 '14 at 22:36
  • $\begingroup$ I am sorry, but I am not getting what are you trying to say. Could you explaine more, plz $\endgroup$ – Mykolas Aug 9 '14 at 22:39
  • $\begingroup$ Try resolving your energy conservation equation for h. Then maybe you'll see what I mean. $\endgroup$ – wgwz Aug 9 '14 at 22:43
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We are not told whether the static coefficient of friction is infinite, of between an infinity and zero. It is a poorly written question if this information is not provided. However, it reminds me of a paper that I read last year (which addresses this occurrence very thoroughly.) The link to the paper can be found here.

Instead of regurgitating this paper (or taking credit for their findings), I think you would benefit by reading it directly from the authors.

However, I will include in this response that it would help you out a lot to consider a force diagram. The normal force is acting through a theoretical line that joins the centers of these spheres. The smaller sphere is moving through the center of mass, vertically in the negative direction. The problem states that the sphere does not slide, so the coefficient of kinetic friction, $\mu_k$ will be acting in the negative direction, through the surface of contact (opposite to the motion of the smaller sphere).

Assess the problem in planar polar coordinates, and note that the value of the angle ($\theta$) of the normal force (relative to the starting point of the sphere) that results in zero value, will be where small sphere will fall off.

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