Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
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 2

up vote 1 down vote accepted

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.

share|improve this answer
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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.