# Small sphere rolling off the top of a large sphere

1. Aug 10, 2014

### myko

[>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:
$$h=\frac{4g-v^2}{2g}$$
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?

2. Aug 10, 2014

### verty

Have you forgotten about rotational kinetic energy? The small sphere rotates; this will consume energy. Your conservation of energy equation will be slightly different.

3. Aug 10, 2014

### myko

$$mg(2r)=1/2mv^2+1/2I\omega^2+mgh$$

$$h=\frac{2rg-7/10v^2}{g}$$

also h=1+cos\theta\$

and now I would substitute v. Is that right?

Last edited: Aug 10, 2014
4. Aug 10, 2014

### verty

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.

5. Aug 10, 2014

### myko

Ok, ty. I got it.

6. Aug 21, 2015

### zed-bak

Hi
I'm working in the same problem and tnx i got it

Last edited: Aug 21, 2015