What is the terminal velocity of coin Very little work has been done at my office for the past few weeks, as we've been bogged down in a debate on the terminal velocity of a coin (the coin in question is a local one, of 9.2g, 28.5mm diameter, 2mm thickness, with a hole in the middle).
We have two experimental observations, where we've dropped the coin from 1.8 meters and then filmed with the slow-motion setting on a phone (240 fps, however the metadata in the resulting file said 208 fps). On both the recordings, 17 frames elapse between the 120cm and 160cm (i.e. 60cm above the ground to 20cm).
Our logic has been the following:
In the 40 cm between the two marks, $\frac{17frames}{208fps} = 0.0817 s$ elapsed.
The average velocity in that period must then be $\frac{40cm}{0.0817s} = 4.896 m/s$
We then used the WolframAlpha terminal velocity calculator (here). We assumed that the coin would fall straight on its edge (although according to the video it didn't), and thus calculated the projected surface area to be the thickness of the coin times its diameter.
The unknown variable is the drag coefficient - we tried adjusting it up until the calculated final speed matched the observed one, which gave us a terminal velocity of 9.9m/s, but then the drag coefficient was 25, which doesn't feel right.
Where did we go wrong?
 A: There are a couple issues

*

*Your coin might not have reached terminal velocity yet. Assuming your value of the drag coefficient, how long would it take to reach terminal velocity? Experimentally it would be nice to calculate the velocity over time by calculating the velocity for a number of frames. You can then plot it. Does it level out to a straight line?

*The coin might rotate during its fall. Try to eliminate this as much as possible, for example by trying to make the coin fall with its edge down. If the coin rotates too much the drag coefficient will be very ambiguous.

EDIT: to get a feeling for 1. we can first calculate the Reynolds Number using this site

Since the flow is most likely turbulent the friction is quadratic, i.e.
$$ma=-C v^2+mg.$$
Using $v(0)=0$ and a little help of Mathematica we can solve this to get
$$v(t)=v_t\tanh\left(\frac{t}\tau\right)$$
where I define $\tau=\sqrt{\frac{m}{gC}}$ as the characteristic time and $v_t=g\tau=\sqrt{\frac{mg}C}$ as the terminal velocity. Let's now plot this function

From this plot we can conclude that if the time is greater than about 2-3 characteristic times then the object has reached terminal velocity.
