# 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.

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I have to say that your claim "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" does not match my experience. have you tried this for yourself? If you toss a coin from heads 100 times how many heads and tails do you get? – John Rennie Nov 23 '12 at 11:28

## 2 Answers

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.

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If the product $k_1 k_2<0$ is negative then the spin will be stable. Maybe your axis conventions need more careful consideration. – ja72 Dec 15 '12 at 14:43
Maybe the torque due to aerodynamic drag stabilizes things. – ja72 Dec 15 '12 at 14:51

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.

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