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Finding an object's center of mass using integration

Thinking about center of mass

As a kid, I would sit in class and balance my pencil on my finger. What I didn't know then was that I was finding the pencil's center of mass. An object's center of mass is a point where the average weight of the object lies. In other words, the weighted position vectors of the object at this point sum to zero (source: https://www.dictionary.com/browse/center-of-mass).
The exact location of this point can be found using calculus.

First, it is important to understand the moment. The moment for a two-dimensional object is given with respect to both the x- and y-axes. It is a physical characteristic of the object a certain distance from these axes. It's easiest to think about the axes as reference points for where the characteristic acts; in this case, where the mass is distributed.

When computing these moments, the hardest concept to grasp is that the moment with respect to y, for example, is given by the x-coordinates of the object. These coordinates are how far away it is from the y-axis. As for the moment with respect to x, we use the y-coordinates for how far away it is. This concept is illustrated below:
For more information on moments, see: https://en.wikipedia.org/wiki/Moment_(physics)#:~:text=The%20concept%20of%20moment%20in%20physics%20is%20derived,to%20an%20arm%2C%20a%20beam%20of%20some%20sort.

When there are more than just point masses, things get trickier to compute! For our purposes, we will use laminas of constant density \(\rho\). The moment with respect to y is given as follows: \[My=\int_{a}^{b} \rho*x*f(x) dx\] Then, the moment with respect to x is: \[Mx=\int_{c}^{d} \rho*y*g(y) dy\] The last thing we will need to compute the center of mass is the mass of the object given by: \[M=\int_{a}^{b} \rho*f(x) dx\] The center of mass is reported as a coordinate (x, y), where x and y are \((\frac{My}{mass}, \frac{Mx}{mass})\). It's important to note that the constant \(\rho\) in each integral will cancel when the moments are divided by the mass, so its value is irrelevant as long as it is constant.

Computing the center of mass

For an example, consider the following:
As you can see, the moment with respect to y is zero. This is to be expected because the object is symmetric across the y-axis. The most noteworthy thing is that the center of mass indicated by a red star is outside of the object! This makes the example interesting because despite symmetry across the y-axis, the object's center of mass is not in the middle of it.

Why I chose this solid

I chose this example because of its interesting center of mass. Additionally, when I was first thinking of an object, I thought of the coins we made for our first assignment. I thought a disk would be a good choice, and especially a disk with a piece missing. However, I did not want to print another coin. I settled for a half of a disk with the center removed. It has a width of 1 cm from r = 3 cm to r = 4 cm. It is 0.1 cm thick.

A similar problem was on an old exam from MA 213, but it was asking for the area. I wanted to further apply it to center of mass because I understood the problem then and wanted to take my work from three years ago and build off of what I learned. When I learned Calc III, I remember polar coordinates being my favorite topic. They made integrating disks very straightforward, so hopefully it's easiest to understand this example in terms of polar coordinates, too.

For a review of polar coordinates, I used APEX Calculus, a free online calculus textbook, section 9.4.
Hartman, Gregory and Department of Mathematics, University of North Dakota, "APEX Calculus: UND Edition" (2017). Open Educational Resources. 2. https://commons.und.edu/oers/2

Word count: 588

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