[Physics] Spinning, decelerating coin on a table


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?

Best Answer

  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.

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