# Two solid hemispheres, one resting on top of the other

1. Dec 1, 2016

### Dustgil

1. The problem statement, all variables and given/known data
A solid homogeneous hemisphere of radius a rests on top of a rough hemispherical cap of radius b, the curved faces being in contact. Show that the equilibrium is stable if a is less than 3b/5.

2. Relevant equations
V = mgh

3. The attempt at a solution

So the center of mass of a solid hemisphere is 3a/8. This hemisphere is upside down and resting on top of another hemisphere of radius b. Therefore if we define the zero point of the potential energy to be level with the bottom of the lower hemisphere, the height at equilibrium of the center of mass is h = 5a/8 + b.

As hemisphere a rotates, the center of mass is displaced an angle theta to the initial point of contact. theta is our generalized coordinate. This is where things get shaky. the line from the center of mass to the original point of contact and the line normal to the surface form a right triangle with angle theta. the normal line is part of the line contributing to the height of the center of mass. It seems like this could be equal to

$$\frac {5a} {8} * cos\theta$$

but i'm not so sure. maybe its the way i've drawn the picture, but it seems like a chunk is still missing out of the line if I go with that.

For the other contribution, since the point of contact is at a different location on hemisphere b, this part cannot just be equal to b. I make the supposition (that I feel unsure of) that the distance from point of contact the center of curvature makes the same angle theta with the normal. I reason this because they are both spherical in shape and their center of curvatures are located at their centers of mass. But I feel like this only holds true when the spheres have the same radius..so I'm not sure on that. At any rate, I get

$$bcos\theta$$

for the height contribution at this point. Taken together, the potential energy function is

$$V = mgcos(\theta)(\frac {5a} {8} + b)$$

which agrees with my original thoughts on when theta = 0 but doesn't give me the correct answer when I determine the minimum of the function. So, I messed up in some spots (probably at least the two I mentioned). Any help?

As an addendum: If anyone has any helpful resources for solving advanced trigonometry problems I'm all eyes. That'd be much appreciated.

2. Dec 1, 2016

### PeroK

You need two variables here. The angle of rotation of the upper hemispehere and the point of contact of the two spheres, which can be represented by another angle..

3. Dec 3, 2016

### Dustgil

yeah, i think the other angle is the angle between the distance from the point of contact to the center of curvature of the bottom hemisphere and the distance between the initial contact and the center of curvature. is that correct?

sorry, had a project to take care of for a couple days there. im back.

4. Dec 3, 2016

### PeroK

You need to relate the angle of rotation of the upper hemisphere to the angle of the point of contact on the lower hemisphere.

Hint: how are these related when the radii are the same or the lower one is, say, four times larger? It might help to think of a sphere rolling all the way round an inner sphere to work out the relationship of the angles.

5. Dec 3, 2016

### Dustgil

well, when considering two spheres rolling around each other of the same size, they both have the same circumference so they rotate through the same angle. if the lower one is four times larger then it takes the outer sphere four full rotations to get all the way around, so inner angle = 4 * outer angle. so its just inner angle = n * outer angle, where n is the relative size between the two, which is the ratio of the two radii. is that close?

6. Dec 3, 2016

### PeroK

You need to think about it more carefully. For spheres of the same size, the outer one will rotate twice on its way round.

7. Dec 3, 2016

### Dustgil

Theres rotation occurring too, I know that now. You can tell this by keeping the point of contact fixed and just sliding along the edge of the coin. If they are of the same size, it will rotate once just by doing this, and this holds regardless of the relative size of the two coins. one full rotation always = one full rotation if they are just sliding. So for two coins of the same size, it's two rotations. for the inner coin being 4 times larger, the outer undergoes five rotations and so on. If phi is the inner angle and theta is the outer,

$$\phi = \frac {a} {b} \theta + \theta$$

I'm not sure if that's dead-on, but is it closer? I ran out of time to think about it so I'm going with my best guess for the time being. I'll think about it more at work.

8. Dec 3, 2016

### PeroK

I'd say that $\theta = \frac{a+b}{a}\phi$.

it's late here, so perhaps someone else can help, as I'm off to bed.

9. Dec 3, 2016

### Dustgil

Thats essentially what I have, I just have everything backwards. You defined theta to be the outer angle and a to be the outer radius, correct?

10. Dec 3, 2016

### haruspex

Not sure what you mean by inner and outer there.
The question specifies a as the radius of the upper hemisphere.
If we define θ as the angle that the arc described by the moving point of contact on the lower hemisphere subtends at the centre of curvature of that hemisphere, and ψ as the angle the corresponding arc on the upper hemisphere subtends at its centre of curvature, we have
aψ=bθ

11. Dec 4, 2016

### PeroK

I was taking $\phi$ as the angle on the lowest hemisphere as described by @haruspex - what he calls $\theta$.

And I was taking $\theta$ to be the angle through which the upper hemisphere has rotated.

12. Dec 4, 2016

### Dustgil

k, that makes sense. I'm not sure why I had that backwards..it definitely makes sense to have it the other way around because theta as defined should always rotate through more than phi if b is larger.

$$a\theta=b\phi$$
$$\theta = \frac {b} {a} \phi$$

but thats just the rotation, we add in another phi term to account for the translation around the inner hemisphere.

$$\theta = \frac {b} {a} \phi + \phi$$
$$\theta = \frac {a+b} {a} \phi$$

13. Dec 4, 2016

### PeroK

The next step is to work out the locus of the centre of mass of the upper hemisphere.