Q:

Rigid body dynamics of tossing of a coin

While tossing a coin, it is commonly experienced that you get a head, if you toss it up with the head side up, and a tails if you toss with the tails side up. Is there a mathematical proof of this using classical mechanics? I would like to see a simple model of the coin as a symmetric top, and consider the precision of the body axis of symmetry about the angular momentum. 

A:

This paper (http://www-stat.stanford.edu/~cgates/PERSI/papers/dyn_coin_07.pdf) shows that the probability distribution of getting a head, if I toss with the head side up is given by:
$p(ψ, φ) =\frac{1}{2}+\frac{1}{\pi}
\sin^{-1}
(\cot(φ) \cot(ψ))$ if $(\cot φ)(\cot ψ) ≤ 1$,
=1 if $\cot(φ) \cot(ψ) ≥ 1$
where $\phi, \psi$ are the Euler angles. 

A:

I will give it a shot.  Spoiler: I did this in the body frame so that the moment of inertia is time independent, before you get excited...
Starting with Euler's equations:
$$
I_i\dot{\Omega}_i+(I_j - I_k)\Omega_j \Omega_k = 0
$$
and taking cyclic permutations of $i,j,k$ to get the three of them; and in the absence of torques (I ignore air friction).  It's a symmetric top so $I=I_1=I_2 \neq I_3$ so write
$$
\dot{\Omega}_1 = -\frac{(I_3-I)}{I}\Omega_2 \Omega_3
$$
$$
\dot{\Omega}_2 = -\frac{(I - I_3)}{I}\Omega_1\Omega_3 
$$
$$
\dot{\Omega}_3=0 \implies \Omega_3=k_1
$$
Now for this problem the coin is spinning about one of the first two symmetric axies.  I chose 1.  Then consider small variations on the other two angular velocities from zero: $\Omega_2 = \delta\Omega_2$, $\Omega_3 = \delta\Omega_3$, and $\Omega_1 \rightarrow \Omega_1$.  So we make small changes in how the coin is rotating about a line through its center perpendicular to the coin, and about the other symmetric axis.  In other words, it was spinning ideally like a coin would, then we changed the ideal to a little weird spinning.  Making the changes, and ignoring second order in perturbations:
$$
\dot{\Omega}_1=0 \implies \Omega_1 = k_1
$$
$$
\frac{d}{dt}(\delta\Omega_2)=-\frac{(I-I_3)}{I}\Omega_1 (\delta\Omega_3)
$$
$$
\frac{d}{dt}(\delta\Omega_3)=0 \implies \delta\Omega_3 = k_2
$$
Then we can write
$$
\frac{d}{dt}(\delta\Omega_2)=-\frac{(I-I_3)}{I}k_1 k_2
$$
Everything on the r.h.s is a number so 
$$
\delta\Omega_2 = -\left( \frac{(I-I_3)}{I}k_1 k_2 \right) t
$$
so depending on how big $I$ is compared to $I_3$ will determine how $\delta\Omega_2$ changes during the flip.  If one uses a radius of $r=0.014$ m and $h=0.0015$ m for the hight of the coin, one gets a moment of inertia tensor like the following:
$$
I=M(0.0000491875) \quad I_3 = M(0.000098)
$$
which tells me that the variations are unstable... which I don't really believe since I have seen a coin in real life.  So look this over.  But I can't find anything wrong so I'm going with it, and thinking that I can't really see a coin in real life up close while it's spinning...  Hope this helps.

