On the Densities of Some of the Philippine Coins

During this particular laboratory session, (which is along side measuring the lengths of grains of rice) we measured the diameters, heights, and masses of the different Philippine Coins. Specifically on 5¢, 10¢, 25¢, ₱1.00, and ₱5.00.

Figure 1: The Coins Measured in the Experiment
(source: http://www.bohol.ph/article34.html)

Again, we used the Vernier Caliper in measuring the said dimensions of these coins and tabulated them. After tabulating our data, we got the following:

Table 1: The Recorded Dimensions of the Philippine Coins

(Pray please don't mind the wrong placement of the title-head of the table). 

Our task for this experiment is to be able to give an approximate on the densities of these coins. Recall that the density \(\rho\) of a material with mass \(m\) and volume \(V\) is:

$$\rho=\dfrac{m}{V}$$
However we only have their diameter \(d\), height \(h\), and mass. So what we have to do is to get the volume in terms of the coin's diameter, height and mass. We assume that the coins are uniformly cylindrical. Given that, its volume would be:
$$V=\pi r^2h$$
We know that \(d=2r\), so this volume equation becomes:
$$V=\dfrac{1}{4}\pi d^2h$$
And using this equation to get the density of the coins, we essentially get:
$$\rho=\dfrac{4m}{\pi d^2h}$$
However, we have to consider that the 5¢ coin has a whole in the center. With this, we let \(D\) be the larger diameter, and \(d\) as its diameter. Given this, the volume equation we got would differ. This time it would be:
$$\rho_{5\text{c}}=\dfrac{4m}{\pi h(D^2-d^2)}$$
Ok, so now that we know the equations to use, we have to know another thing. We have to know that besides knowing the densities of the coins, we have to consider how the error propagates as we make the calculations. Therefore, as we calculate the effective densities of these coins, we also have to calculate for the error introduced as we make such calculations. As we continue to finding the approximate densities of the coins, we will see that calculating for the error would prove itself to be easy for all of the coins except the 5¢ coin. This is simply because of the small hole it has in the center. Anyway, to calculating the densities!

We start by solving for the density of the 5¢ coin \(\rho_{5\text{c}}\). The expected value we have would be:
\begin{eqnarray}
\rho_{5\text{c}}&=&\dfrac{4m}{\pi h(D^2-d^2)} \\
\rho_{5\text{c}}&=&\dfrac{4(1.9\ \text{g})}{\pi(1.44\ \text{mm})((15.46\ \text{mm})^2-(3.68\ \text{mm})^2)}=7.45\times10^{-3}\ \text{g}/\text{mm}^3 \\
\rho_{5\text{c}}&\approx&7.45\ \text{mg}/\text{mm}^3
\end{eqnarray}
Now, our only problem is the calculation of how the error propagated in that simple play of numbers.
The easy way out of it is to first calculate the error propagation in the difference, and from there, find the error propagation in the whole equation.
So in the \(D^2-d^2\) term, we have multiplication and subtraction operations, and error propagation calculation is different for the two operations. So for this, we factor the term into \((D-d)(D+d)\) and calculate their separate error propagation, before getting the error propagation for \(D^2-d^2\) and using that to get the error propagation of the whole equation. Our format for this (to cut the length of the whole blog) would be such: \(\langle Q\rangle\pm\Delta Q\) where \(\langle Q\rangle\) is the expected value, and \(\Delta Q\) is the error propagation.
So starting with \(D-d\)

\begin{eqnarray}
\langle D-d\rangle\pm\Delta(D-d) \\
(15.46-3.68)\pm(0.02+0.02)\ \text{mm} \\
11.786\pm0.04\ \text{mm}
\end{eqnarray}
Then with \(D+d\)
\begin{eqnarray}
\langle D+d\rangle\pm\Delta(D+d) \\
(15.46+3.68)\pm(0.02+0.02)\ \text{mm} \\
19.146\pm0.04\ \text{mm}
\end{eqnarray}
Therefore:
\begin{eqnarray}
\dfrac{\Delta(D^2-d^2)}{\langle D^2-d^2\rangle}&=&\dfrac{\Delta(D-d)}{\langle D-d\rangle}+\dfrac{\Delta(D+d)}{\langle D+d\rangle} \\
\Delta(D^2-d^2)&=&\left(\dfrac{\Delta(D-d)}{\langle D-d\rangle}+\dfrac{\Delta(D+d)}{\langle D+d\rangle}\right)\langle D^2-d^2\rangle \\
\Delta(D^2-d^2)&=&\left(\dfrac{0.04}{11.786}+\dfrac{0.04}{19.146}\right)(225.4692) \\ \Delta(D^2-d^2)&=&1.236262584\approx1.24
\end{eqnarray}
Therefore, \(\langle D^2-d^2\rangle\pm\Delta(D^2-d^2)=225.4692\pm1.24\) mm.
And now we use this to get the total error propagated in the whole equation.
\begin{eqnarray}
\dfrac{\Delta\rho_{5\text{c}}}{\langle\rho_{5_\text{c}}\rangle}&=&\dfrac{\Delta m}{\langle m\rangle}+\dfrac{\Delta h}{\langle h\rangle}+\dfrac{\Delta(D^2-d^2)}{\langle D^2-d^2\rangle} \\ \Delta\rho_{5\text{c}}&=&\left(\dfrac{\Delta m}{\langle m\rangle}+\dfrac{\Delta h}{\langle h\rangle}+\dfrac{\Delta(D^2-d^2)}{\langle D^2-d^2\rangle}\right)\langle\rho_{5\text{c}}\rangle \\ \Delta\rho_{5\text{c}}&=&\left(\dfrac{0.05}{1.9}+\dfrac{0.02}{1.44}+\dfrac{1.24}{225.4692}\right)(7.45\times10^{-3}) \\ \Delta\rho_{5\text{c}}&=&0.340418947\times10^{-3}\approx0.340\times10^{-3}
\end{eqnarray}
Therefore, the density of the 5¢ coin is \(7.45\pm0.340\ \text{mg}/\text{mm}^3\).

We now move to solving for the density of the 10¢ coin \(\rho_{10\text{c}}\). Since it doesn't have the small hole as the 5¢ does, our calculation simplifies. Getting its expected density:
\begin{eqnarray} \rho_{10\text{c}}&=&\dfrac{4m}{\pi d^2h} \\ \rho_{10\text{c}}&=&\dfrac{4(2.5\ \text{g})}{\pi(16.88\ \text{mm})^2(1.54\ \text{mm})}=7.25\times10^{-3} \text{g}/\text{mm}^3 \\ \rho_{10\text{c}}&\approx&7.25\ \text{mg}/\text{mm}^3 \end{eqnarray}
Solving for the error propagation:
\begin{eqnarray}
\Delta\rho_{10\text{c}}&=&\left(\dfrac{\Delta m}{\langle m\rangle}+\dfrac{\Delta h}{\langle h\rangle}+2\dfrac{\Delta d}{\langle d\rangle}\right)\langle\rho_{10\text{c}}\rangle \\ \Delta\rho_{10\text{c}}&=&\left(\dfrac{0.05}{2.5}+\dfrac{0.02}{1.54}+2\dfrac{0.02}{284.9344}\right)(7.25\times10^{-3}) \\ \Delta\rho_{10\text{c}}&=&0.240309988\times10^{-3}\approx0.240\times10^{-3}
\end{eqnarray}
Therefore, the density of the 10¢ coin is \(7.25\pm0.240\ \text{mg}/\text{mm}^3\).

And we use the same form of calculation as the 10¢ coin to that of the 25¢, ₱1.00, and ₱5.00 coins.

Skipping all of the calculations, we get that:

the density of the 25¢ is \(6.72\pm0.172\ \text{mg}/\text{mm}^3\),

the density of the ₱1.00 is \(6.38\pm0.129\ \text{mg}/\text{mm}^3\), and

the density of the ₱5.00 is \(7.27\pm0.125\ \text{mg}/\text{mm}^3\).

Now, upon searching the net, you will never see any website which displays the standard density of the Philippine coins. However, I was able to find two websites which display the coin's ideal mass and percent composition. From here I was able to calculate the effective ideal density of the coins.

For example, according to that site (please refer to the websites linked at the end of the blog), an ideal 5¢ coin has a mass of 1.9 g, is 6% Cu, and 94% Steel. From here we can calculate the ideal density of the coin through multiplying the sums of the products of the percent composition of a substance in that material and the density of that substance. Explicitly:

\begin{equation} \rho=\sum_{i=1}^n\,\text{Percent Composition}_i\cdot\rho_i \end{equation}

So if we want to get the ideal density of a 5¢ coin, we have to get the respective densities of copper and steel (Again, please refer to the websites linked at the end of the blog). Getting everything you need to calculate the density of this coin:

\begin{eqnarray}
\rho_{5\text{c,ideal}}&=&\text{Percent Composition Cu}\cdot\rho_{\text{Cu}}+\text{Percent Composition Steel}\cdot\rho_{\text{Steel}} \\
\rho_{5\text{c,ideal}}&=&0.06\cdot8.96\ \text{mg}/\text{mm}^3+0.94\cdot7.75\ \text{mg}/\text{mm}^3=7.8226\ \text{mg}/\text{mm}^3 \\
\rho_{5\text{c,ideal}}&\approx&7.82\ \text{mg}/\text{mm}^3
\end{eqnarray}
And we use this same style to get the ideal densities of the other coins. After doing so:
the density of the 10¢ is \(7.82\ \text{mg}/\text{mm}^3\),
the density of the 25¢ is \(8.32\ \text{mg}/\text{mm}^3\),
the density of the ₱1.00 is \(8.95\ \text{mg}/\text{mm}^3\), and
the density of the ₱5.00 is \(8.51\ \text{mg}/\text{mm}^3\).

From here, we can immediately see the big discrepancies between the ideal densities and the one we got in the experiment. This may be due to the degradation of the coins through time, the tarnishing, deformation, dirt that gets stuck on the coin. Even the actual measuring of the coins can be a source of error. Basing on the same site for the dimensions of the coins, the measured values we have is near the actual dimensions, however is still not equal to it. So most likely that did affect the calculated densities.

References:
- Current Philippine Coins
http://philmoney.blogspot.com/search/label/New%20BSP%20Series
- Philippine Coins
http://www.bohol.ph/article34.html
- Densities of Some Metals
http://www.coolmagnetman.com/magconda.htm
- Densities of Steel
http://www.engineeringtoolbox.com/metal-alloys-densities-d_50.html