A coin spun on a table will eventually end up flat on the table, ignoring the cases where it remains balanced and comes to rest on its edge (or falls off etc).

Before the coin comes to rest on its side, there will be a stage where the coin has fallen and is now 'rolling' in a tight circle, with decreasing period (increasing frequency of the "ringing" sound associated with this).

Consider a point on the coin's edge $P$ - every 'cycle' of the coin rolling around on its edge, the coin as a whole rotates (with respect from above). $P$ will move around the axis of rotation in the same direction as the point of contact between the coin edge and the table, relative to the center of the coin in the table's rotational frame. I assume this is because the circle drawn out by the point of contact is smaller in circumference than the coin itself, approaching the coin's circumference as the coin comes to rest.

I am ignoring air resistance for convenience.

  1. If the coin and table were made of materials such that the coin bouncing off the table is a perfectly elastic collision, would it be possible for the coin to ever slow down, or would it be able to 'roll on its side' forever like that? Intuition says no, but I can't explain why that is.
  2. The rotational speed of the coin when viewed from above (average angular speed of P) - does this change or remain constant? If it remains constant, why the discontinuity when the coin comes to rest?
  3. The point $P$, every round-trip of the contact point between the coin and the table, will bounce off the table. This vertical speed appears to increase without bound as the period of the round-trip decreases, which in turn means the acceleration due to each bounce is much higher (from $-v$ to $v$ with increasing frequency and increasing $|v|$). This would imply the forces in the coin are growing as it's slowing down, but that can't be right... I can only assume the speed doesn't increase, merely the distance in the bounce decreases? Wouldn't that also mean that the bouncing speed actually decreases as the coin slows down?
  • $\begingroup$ What makes you think that the coin/$P$ bounces? Because $P$ briefly makes contact with the surface of the table and then leaves it again does not imply a bounce, this is similar to a rolling wheel. $\endgroup$
    – fibonatic
    Commented May 6, 2015 at 13:58
  • $\begingroup$ This is true, it is very much like a rolling wheel, and I guess that's part of the confusion - that it rolls under gravity means there is a normal force on the table pushing against the coin; each point instantaneously must experience this force per cycle / roll / revolution (whatever is most appropriate), but I'm at odds to call it a bounce too, so unsure... $\endgroup$ Commented May 6, 2015 at 23:14
  • $\begingroup$ If a point on the edge of the coin was bouncing, it would move in a series of parabolic arcs with abrupt changes in direction at the bounce. Since the coin is rotating with the center of mass at a slowly descending height, the motion of a point at the edge is (approximately) sinusoidal. $\endgroup$
    – mmesser314
    Commented Jun 4, 2015 at 13:20

2 Answers 2

  1. The spinning coin is not a collision. Elastic Collisions involve the linear motion of two or more objects and the transfer or transformation of their kinetic energy. The situation described involves rotational motion, the contact forces between the table and coin, and angular velocity. I don't see a connection between the condition imposed of materials that allow a perfectly elastic collision and the coefficients of friction between the two materials. The coins falls flat because the direction of its axis of rotation (spin) and the coin itself are not perfectly normal to the table surface. Because of this, gravity acting at the center of mass and the normal force acting at the point of contact create a torque in a plane normal to gravity that leads to precession of the spin axis. This external torque also breaks conservation of angular momentum.

  2. The angular velocity of the point P is not constant with respect to the table since the axis of rotation changes but the angular speed can be, neglecting friction. When it finally comes to rest it isn't a discontinuity but a very rapid change (at the right time-scale you could watch it slow, fall flat and spin flat on the table before stopping).

  3. The point P is not bouncing off of the table. If it appears that way it is just that the coin's angular velocity and angular momentum vectors are not perpendicular to the table. Friction between the table and coin slows its spin. As the period of the spin around the spin axis increases, the period of the precession increases. The friction also complicates the precession picture but in short it also increases the angle between the spin axis and the precession axis until the coin falls flat and spinning stops.


This is a very, very good question, and the answer is not nearly as obvious as one might think. In fact, this system was the subject of a Nature article published in 2000, and of several other journal articles since then. To summarise, relating to the original question:

  1. Dissipation of the energy comes from rolling friction and from slippage between the coin and the surface. Viscous effects in the layer of air between the coin and the table are of less importance (in contrast to Moffatt's original finding).
  2. There is a relationship between the angular velocity of a fixed point on the coin (i.e. the 'spin' about the symmetry axis of the coin) and the precession velocity (rolling motion). The two are proportional in steady-state, with the constant of proportionality being given by the ratio of the horizontal distance between the axis of rotation and the centre of mass of the coin to the radius of the coin. This distance tends to zero as the coin settles, cancelling the divergent frequency of the precession.
  3. As previously discussed, this is not a bouncing but rather a rolling motion. Nonetheless, it helps to draw parallels with bouncing to understand why the period decreases with time. Picture a bouncing ball; with each bounce, it loses energy just as the coin continually loses energy as it rolls. Because the amplitude of the bounces gets smaller, the period of the bouncing motion gets smaller too. This is complemented in the case of the coin by the increasing vertical acceleration as the precession frequency increases.

A nice, accessible outline of the theory is provided here

  • 1
    $\begingroup$ You can also find a listing of some more articles on the phenomenon here. $\endgroup$ Commented Jun 4, 2015 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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