# Why does a coin falls faster when it's flipping as well?

From my experiments with measuring how fast a coin falls, I have consistently measured a faster falling rate for a coin that flips as it falls.

As an example, a coin dropping on its edge from height of $45 \:\rm{cm}$ hits the ground $20 \:\rm{ms}$ later than a flipping coin falling from the same height.

Now here's the catch: I use a microphone to mark the events. I drop the coin off the edge of a table letting it slightly brush off it. The bang noise of this event combined with the noise the coin makes as it hits the hard ground let's me measure the fall duration accurately (I hope). I also take into account the time it takes the sound of coin hitting the ground to come back up to the mic.

Using $\approx340\:\rm{m/s}$ for speed of sound and $9.806\:\rm{m/s^2}$ for acceleration due to gravity, my measurement of height is dead accurate, BUT only for a coin dropped on its edge. A flipping coin constantly gives me a measurement less than correct value.

First I suspected the air resistance, but if that was the case, shouldn't the coin falling on its edge fall quicker?

Any ideas?

• 1) How are you flipping the coin?? Note that most processes that flip a coin will impart some extra translational energy to it as well. Also, if the coin has not yet begun flipping at the time it brushes the mic, then there will be a time discrepancy. – Manishearth Apr 5 '12 at 7:18
• Thanks for replacing my numbers with proper ones (I'm not fluent in TeX). Well I drop the coin from about half a cm above the edge of the table while holding it horizontally. When it hits the edge, it start flipping as well as making the noise for mic to pick up. – Mansour Apr 5 '12 at 7:26
• Can you comment a bit on the accuracy/reproducability of your set-up? Is this 20ms based on the average of 100 realizations with a standard deviation of 100 ms for example, or is $\sigma$ 0.1 ms. – Bernhard Apr 5 '12 at 7:39
• I must warn you that I'm not a physics student, so my method may not be proper. I consistently (100% of time) get a faster falling rate for the flipping coin. The difference however changes but for the 45cm fall, I recorded it between 15ms to 22ms after 20 tries. – Mansour Apr 5 '12 at 8:01
• To extend on my remark about the difference in the fall distance of the center of mass: A difference of only 1.1 millimeters in the fall distance would change the fall time by 15 milliseconds. That is on the order of the thickness of a typical small coin. – Snowhare Apr 6 '12 at 14:38

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).