When we spin a coin on a table, we observe 2 things:

  1. It slows down and stops after sometime.

  2. It does not stay at just one point on the table but its point of contact with table changes with time.

I was trying to explain quantitatively this but I am stuck at how to take frictional torques into account. Any help will be appreciated.

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    $\begingroup$ Precession is one of the main phenomena at play: en.wikipedia.org/wiki/Precession $\endgroup$ Commented Dec 4, 2013 at 14:22
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    $\begingroup$ I'm expectantly waiting for a brilliant answer from alemi or floris (or both! :D), but maybe they're too busy to give an insightful and well-developed response. We'll see :) $\endgroup$ Commented Sep 11, 2014 at 2:23
  • $\begingroup$ related: physics.stackexchange.com/q/68676 and mathandcode.com/disk I wanted to solve this problem in full generality. So my solution is undoubtedly more complicated than it needs to be. But you should be especially interested in "Partial constraint 2" $\endgroup$
    – user12029
    Commented Nov 20, 2014 at 3:52

4 Answers 4


I think that if you spin "perfectly" (i.e., such that the rotational axis is normal to the surface and goes through he centre of the coin), is only a rotation movement with friction. This motion is unstable though, so, the axis tilt a little bit and this cause a rotation in the axis itself, the precession. The point of contact will be moving with the precession, maybe you can calculate its position by geometrical arguments, although it should be a circular/spiral/cycloid movement (if you see in the coin a movement towards a given direction, this is solely because of the way you made if spin or the coin or because the table has a tilt or imperfections).

I don't know your level of knowledge, but for a complete description you need knowledge of Hamiltonian dynamics, rigid body and Euler angles, so basically a course of classical (a.k.a. analytical) mechanics. A very common, related, problem is the problem of the spinning top, the difference here is that the contact point is material, so there you have to see if you have to see if the contact point slips or not (if not, it creates a rotation in the axis normal to the coin).

Personally I think that it is a complicated but somehow treatable problem (with a lot of patience).


There is no easy way to model a spinning coin and calculate these observations. It slows down mostly because of air resistance and friction(here you must take velocity dependent friction-angular velocity in your case-) and it moves due to the combination of torque of gravity(a.k.a. precession) and friction. Velocity dependent frictions generally gives you non-linear differential equations which are often very hard to handle. When you write hamiltonian and canonical equations probably you will get some coupled non-linear partial differential equations which are worst combinations to solve.

Moreover, after it slows enough, contact point(on the coin) will start to move and after that time you should consider rolling friction.

  • $\begingroup$ Where do partial differential equations come in? Unless you're at the level of the Hamilton-Jacobi equation and looking for a full solution of the problem (which is evidently impossible), everything else is ODEs. $\endgroup$ Commented Dec 4, 2013 at 15:38
  • $\begingroup$ Canonical equations, themshelves are pde . you will always encounter partial differentials. $\endgroup$
    – maynak
    Commented Dec 4, 2013 at 15:55
  • $\begingroup$ If by "canonical equations" you mean Hamilton's equations, then those are not PDEs. The hamiltonian itself is known. The only derivatives of unknown functions are time derivatives of coordinate and momentum, so they're (coupled, possibly nonlinear) ODEs. $\endgroup$ Commented Dec 4, 2013 at 16:14
  • $\begingroup$ yeah you may be right.. anyways coupled, nonlinear ODEs are enough to be a nightmare... $\endgroup$
    – maynak
    Commented Dec 5, 2013 at 2:37
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    $\begingroup$ To a point. Coupled nonlinear ODEs are the stuff of chaos. Coupled nonlinear PDEs are the stuff of we-just-can't-solve-it, millenium-prize-style problems. $\endgroup$ Commented Dec 5, 2013 at 11:59

if u consider an idealised case, with no friction and air resitance and the coin is a perfect circle that only a minute elementary part of it is touching the ground. then u can sinply treat it as a disc rotating about the diameter.

But ofcourse thats not what you asked for :P u want the general case where all the forces matter since thats what happens in real life.

well one way u can take is to consider any elementary pary at distance r from diameter and add an additional force -bv (since air resistance proportional to velocity of that point) and explain why it slows down.

But as to why it doesnt stay at the same place is because it is not a perfect circle nor is it a 2-D disc.. its a cylinder and its point of contact is not just a single point. and the friction coeff on the floor is not same everywhere causing a net translation.


Analytically when the coin is static the gravity makes it fall to one side. But when it is spinning and gravity pulls each point of contact towards Down and Inside the circle of rotation.

Here the inside pull component is cancelled out at high speed as each point of contact has it. This is verified as the axes of the coin is always tilted and rotates forming a cone.

When the speed reduces the inside force makes it bend slightly more than before effectively increasing the circumference of points of contact. This increases the time to cancel out inside force and slowly the coin falls down.


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