Say I have a hemisphere (A solid sphere, cut exactly in half), and I have an experement in which, when flipped, it can land either as:
- Flat Face Up
- Flat Face Down
How would one go about calculating the probability of each outcome?
*Assume that when flipped, it is flipped fairly
This might be more of a physics question, than a mathematics one.
To make this more of a mathematics question, let's minimize the physics aspects first:
Very low drop height
As the object is asymmetric, air will push the object asymmetrically, and tend to turn it. This not only induces rotation (angular momentum), but it also makes the different orientations of the object at the time it touches the surface (the final time) no longer equally likely.
Dropped straight, with no rotation
If there is any angular momentum when the object (last) touches the surface with its round part, the added momentum may cause it to flip over to the flat side. I wouldn't want to have to calculate the probabilities of this one. (I'm sure it can be taken into account, it's just complex.)
Dropped to flat, horizontal surface
If the surface is not flat, or it is not horizontal, we again get back to the above point: extra angular momentum. We don't want that.
No bounce
After our object touches the surface, it will just come to rest, instead of bouncing back up in the air. If it were to bounce, the fact that the object is asymmetric, would mean that the object would not bounce like a ball, but induces rotation. (You can see this with dice with sharp corners: when they hit the surface, they sometimes fly off in one direction, rotating furiously. Typically off the table. Dice with rounded corners are less violent.)
Object has continuous, uniform density distribution (constant density)
If we agree with the conditions above, then the orientation in which the object hits the table determines the final, resting orientation. There is no angular momentum, and all hitting orientations are equally likely.
(Note that thus far, the shape of the object does not matter. If it was a coin, the symmetries would mean that the hard part would be to find out the fraction of edge-on cases.)
For a half-sphere:
If the center of mass is on the same side as the round hemisphere, from the point of contact at the moment of contact, it will end up resting on the round side towards the surface. Otherwise, the object will tip on the flat side. (If there is no added angular momentum, due to the symmetries the object cannot roll far enough to tip over on the other side: there is not enough kinetic energy to overcome the potential energy barrier of the tipping point.)
Because the center of mass is inside the object, not on the surface, this means that the probability of the object ending up with its round side to the surface is more likely than it sitting flat.
Let's assume a coordinate system for the object, where the flat part is on the $xy$ plane, and the hemisphere being the positive $z$ halfspace of the unit sphere (radius $1$). Rotation around the $z$ axis does not matter, due to symmetries (and because we assume zero angular momentum at the point of contact with the surface). Due to symmetries, the coordinates for the center of mass must be $(0, 0, z)$, with $0 \le z \le 1$.
To find out the probability of the object ending up sitting flat (flat face towards the surface), we need to calculate the fraction in which the center of mass is not supported by the contact point. Consider the following diagram:
Here, $\Omega$ is the solid angle extended by the flat side, when observed from the center of mass.
The probability of the object ending up sitting flat (flat side down) is $$p(\text{flat}) = \frac{\Omega}{4\pi}$$ and sitting on the round side (flat side up) $$p(\text{round}) = \frac{4\pi - \Omega}{4\pi} = 1 - p(\text{flat})$$
This should now be a rather easy mathematical problem to solve.