If I read the description correctly, your experiment is set up to impart an initial velocity to the coin before your timing starts - as you drop it from slightly above the edge of the table, but don't start timing until you hit the edge (hear the impact).

At the moment of impact, the center of mass of the coin will slow down a little bit - but it will still have some residual velocity. An example calculation:

If you drop the coin from distance $h_0$ above the table edge, it will have a certain velocity $v_0$ by the time it hits the edge. If you make a "perfect" hit between the edge of the coin and the edge of the table, the coin will start spinning while continuing to fall; for a coin with mass $m$ and radius $r$, the moment of inertia about the axis is $I = \frac14 m r^2$. The change in linear momentum $\Delta p$ will give rise to a change in angular momentum $\Delta L = r\Delta p$; conservation of energy tells us that the sum of rotational and translational kinetic energy after the impact are the same as before. Finally, the contact force will go to zero when the velocity of the edge of the coin is zero, so $r\omega_1=v_1$. We can now write conservation of energy:

$$\frac12 m v_0^2 = \frac12 m v_1^2 + \frac12 I \omega_1^2$$
which we solve by substituting for $\omega_1$ and $I$ per the above. We are left with the relationship between $v_0$ and $v_1$:

$$v_1 = \sqrt{\frac{4}{5}}v_0$$

so the coin will lose a fraction of its initial velocity when it hits the table. And then it continues to fall distance $h$.

Now a coin that starts with an initial velocity will reach the floor more quickly than a coin starting from rest. The time taken is obtained by solving

$$h = v_1 t + \frac12 g t^2\\
t = \frac{-v_1 +\sqrt{v_1^2 + 2 g h}}{g}$$

Sanity check: when $v_1=0$ this reduces to the usual result $t = \sqrt{\frac{2h}{g}}$, and when $v_1$ is quite small, we can do an expansion of the expression to obtain

$$ t = \sqrt{\frac{2h}{g}} - \frac{v_1}{g}\left(1+\frac{v_1}{4gh}\right)$$

Substituting things in to find the effect on calculated height quickly gets very messy, but it's straightforward to plot the relationship between height above the table from which the coin was dropped, and calculated height difference (where the true value = 45 cm):

Note that the difference in calculated height is a fraction of the height from which you dropped the coin:
\frac{-v+\sqrt{1+\frac{v_1^2}{2gh}}}{g}$$

Also note that the velocity of the coin when hitting the floor (approximately 3 m/s after a 45 cm drop) means that a 20 ms timing difference corresponds to a 6 cm difference in apparent height: that's a massive number.

I conclude that there is another problem with your setup that is not properly described - "something" you are not telling us, or some way in which you are not interpreting your data correctly... In particular I wonder whether your interpretation of the sounds as they relate to the start of the drop are correct. I would like to repeat this experiment using high speed video (most modern cameras and even phone can film 240 fps which should be plenty to see what is causing a 20 ms difference).

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).

## Best Answer

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.

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).

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.