A Really Hard Piece of Swiss Cheese

When thinking about center of mass, the first thing that comes to mind is a seesaw. You can usually find a seesaw on a children's playground and if perfectly balanced, the center of mass is in the center of the plank of wood. Once a child sits on the seesaw, the center of mass immediately changes. Another way the center of mass can change on a seesaw is if the wood gets damaged or dented over the years from weather and rough children. Because of this, I'm not sure I have ever seen a seesaw perfectly balanced.

When continuing to think about objects with odd centers of mass, I started to think about objects with pieces missing. I immediately thought of a piece of swiss cheese, one because I really like cheese, and two because it's usually fairly holey. Now, I understand that a thin slice of swiss cheese is usually fairly flimsy, but for our sake we are going to pretend it's a really hard piece of swiss cheese.

In order to construct my slice of swiss cheese, I used a large triangle and then I removed a series of smaller circles and half circles from the triangle. The equations used to construct my slice of cheese are as follows: \begin{aligned}y=-x+10\\ y=0\\ x=0\\ \left( x-5\right) ^{2}+\left( y-2\right) ^{2}=2\\ \left( x-7.5\right) ^{2}+\left( y-1\right) ^{2}=0.4\\ \left( x-2\right) ^{2}+\left( y-1.2\right) ^{2}=1\\ x^{2}+\left( y-3\right) ^{2}=1;x\geq 0\\ \left( x-3\right) ^{2}+\left( y-5.5\right) ^{2}=0.4\end{aligned} The following solid was created.

When thinking about the math for how to calculate the center of mass, if it was a symmetric object, such as a perfect seesaw, the center of mass would be directly in the center. If it's not perfectly symmetric, like swiss cheese, it takes a little more work. Center of mass can be calculated by taking the moment of a particle and dividing it by the mass. Since our slice of cheese has several holes in it, the easiest way to calculate the center of mass would be to calculate the moment for the triangle and subtract the moments of all the circles and then divide by all the masses together. One thing we want to be sure to avoid is just combining all the centers of mass together because even though the individual properties can be added, the overall centers cannot. The following calculations show how to compute the moments and mass of the triangle: \begin{aligned}M_{y}=\int _{0}^{10}\rho*x\left( -x+10\right) dx\\ =\rho \int _{0}^{10}-x^{2}+10xdx\\ =\rho \left( -\dfrac{1}{3}x^{3}+5x^{2}\right) | _{0}^{10}\\ =\rho \left( -\dfrac{1000}{3}+500\right) -0\\ =\rho \left( \dfrac{500}{3}\right) \end{aligned} \begin{aligned}M_{x}=\int _{0}^{10}\rho*y\left( -y+10\right) dy\\ =\rho \int _{0}^{10}-y^{2}+10ydy\\ =\rho \left( -\dfrac{1}{3}y^{3}+5y^{2}\right) | _{0}^{10}\\ =\rho \left( -\dfrac{1000}{3}+500\right) -0\\ =\rho \left( \dfrac{500}{3}\right) \end{aligned} \begin{aligned}Mass=\int _{0}^{10}\rho \left( -x+10\right) dx\\ =\rho \left( -\dfrac{1}{2}x^{2}+10x\right) | _{0}^{10}\\ =\rho \left( -50+100\right) \\ =50\rho \end{aligned} Now, although we can assume the center of mass for each circle is the center of the circle, we must calculate the moments and masses separately in order to subtract them from our triangle. Below are the calculations for the moments and masses of each circle I cut out of the triangle: \begin{aligned}M_{y}=\int ^{5+\sqrt{2}}_{5-\sqrt{2}}\rho x\left( \left( 2+\sqrt{2-\left( x-5\right) ^{2}}\right) -\left( 2-\sqrt{2-\left( x+5\right) ^{2}}\right) \right) dx\\ =\rho \int _{5-\sqrt{2}}^{5+\sqrt{2}}2x\sqrt{2-\left( x+5\right) ^{2}}dx\\ u=x+5\\ du=dx\\ x=u-5\\ M_{y}=2\rho \int ^{10+\sqrt{2}}_{10-\sqrt{2}}\left( u-5\right) \sqrt{2-u^{2}}du\\ =289.41\rho \end{aligned} After attempting to compute a moment for my first circle equation, I can clearly see that I am doing something wrong because my center is not where it's supposed to be. Since I know where my center is supposed to be, I am going to attempt to work backwards and see if I can calculate the moment that way. The calculations below show moment and mass calculations for all of the full circles I cut out of the triangle: \begin{aligned}center=\left( \dfrac{My}{mass},\dfrac{Mx}{mass}\right) \\ mass=\rho \left( \sqrt{2}\right) ^{2}\pi \\ =\rho 2\pi \\ \left( 5,2\right) =\left( \dfrac{My}{\rho 2\pi },\dfrac{Mx}{\rho 2\pi }\right) \\ M_{y}=10\pi \rho \\ M_{x}=4\pi p\end{aligned} \begin{aligned}center=\left( \dfrac{My}{mass},\dfrac{Mx}{mass}\right) \\ mass=\rho \left( \sqrt{.4}\right) ^{2}\pi \\ =\rho (.4)\pi \\ \left( 7.5,1\right) =\left( \dfrac{My}{\rho (.4)\pi },\dfrac{Mx}{\rho (.4)\pi }\right) \\ M_{y}=3\pi \rho \\ M_{x}=.4\pi p\end{aligned} \begin{aligned}center=\left( \dfrac{My}{mass},\dfrac{Mx}{mass}\right) \\ mass=\rho \left( \sqrt{1}\right) ^{2}\pi \\ =\rho \pi \\ \left( 2,1.2\right) =\left( \dfrac{My}{\rho 2\pi },\dfrac{Mx}{\rho 2\pi }\right) \\ M_{y}=2\pi \rho \\ M_{x}=1.2\pi p\end{aligned} \begin{aligned}center=\left( \dfrac{My}{mass},\dfrac{Mx}{mass}\right) \\ mass=\rho \left( \sqrt{.4}\right) ^{2}\pi \\ =\rho (.4)\pi \\ \left( 3,5.5\right) =\left( \dfrac{My}{\rho (.4)\pi },\dfrac{Mx}{\rho (.4)\pi }\right) \\ M_{y}=1.2\pi \rho \\ M_{x}=2.2\pi p\end{aligned} Since the fifth circle is a semi-circle, I am going to try our integral method again and see if I have better luck. Below are my calculations: \begin{aligned}M_{y}=\int ^{1}_{0}\rho x\left( 3+\sqrt{1-x^{2}}\right) dx\\ =\rho \int _{0}^{1}x\left( 3+\sqrt{1-x^{2}}\right) dx\\ u=1-x^{2}\\ du=-2xdx\\ -\dfrac{1}{2}du=xdx\\ M_{y}=-\dfrac{1}{2}\rho \int \left( 3+\sqrt{u}\right) du\\ =-\dfrac{1}{2}\rho \left( 3u+\dfrac{2}{3}u^{3/2}\right) \\ =-\dfrac{1}{2}\rho \left( 3\left( 1-x^{2}\right) +\dfrac{2}{3}\left( 1-x^{2}\right) ^{3/2}\right) _{0}^{1}\\ =0-\left( -\dfrac{1}{2}\rho \left( 3+\dfrac{2}{3}\right) \right) =\dfrac{11}{6}\rho \end{aligned} \begin{aligned}mass=\int ^{1}_{0}p\left( 3+\sqrt{1-x^{2}}\right) dx\\ =3.785\end{aligned} \begin{aligned}center=\left( \dfrac{My}{mass},\dfrac{Mx}{mass}\right) \\ =\left( \dfrac{\dfrac{11}{0}}{3.785},\dfrac{Mx}{3.785}\right) \\ =\left( 0.484,3\right) \\ M_{x}=11.355\rho \end{aligned} Now that we have all of the moments and masses, we can proceed to subtract the moments and masses of the circles from our big triangle. Below shows our math: \begin{aligned}totalM_{y}=\rho \left( \dfrac{500}{3}\right) -10\pi \rho -3\pi \rho -2\pi \rho -1.2\pi \rho -\dfrac{11}{6}\rho \\ =113.94\rho \\ totalM_{x}=\rho \left( \dfrac{500}{3}\right) -4\pi \rho -0.4\pi p-1.2\pi \rho -2.2\pi \rho -11.355\rho \\ =130.81\rho \end{aligned} \begin{aligned}totalmass=50\rho -2\pi \rho -0.4\pi \rho -\pi \rho -0.4\pi \rho -3.785\rho \\ =34.28\rho \end{aligned} \begin{aligned}center=\left( \dfrac{M_{y}}{mass},\dfrac{M_{x}}{mass}\right) \\ =\left( \dfrac{113.94\rho }{34.28\rho },\dfrac{130.81\rho }{34.28\rho }\right) \\ =\left( 3.324,3.816\right) \end{aligned} The center of mass of my slice of cheese is located at \((3.324,3.816)\) which is honestly not as off centered as I hoped it would be. Next time, I would probably centralize my holes more to one side to affect the center of mass more dramatically. I have marked my center of mass on my object in the stl file, so once I print my object, I'll be able to test my calculated center of mass. My estimated print time is only 20 minutes so I may scale the size up so I can have a larger slice of cheese.