Center of Mass

Introduction
Imagine you are given a random object and also asked to balance it on your finger. How would you go about finding out where to place your finger to balance it? You may try just making a judgement based on what the object looks like and choosing the best spot. However, this will not always be exact and you might not be able to balance the object. The proper way to find the correct spot to balance the object is using the idea of center of mass. In short, the center of mass is the region of a solid that the region will balance perfectly horizontally if suspended from that point.
Center of Mass?
The process of finding the center of mass is a relatively straight forward process. It involves finding two moments of the region and the entire mass of the region. Since the center of mass is found on the 2-Dimensional plane, we only have to find the moments for x and y, and the mass. These moments are \(M_x\) and \(M_y\) and the mass is \(M\). These are all used in finding the exact x and y coordinate for the center of mass. I will go into the calculations for the moments and the mass later, but first lets take a look at the region. For my example, I have chosen the functions: \(f(x)=-3x^3+4\) and \(g(x)=-2x^3-x^\frac{1}{3}+4\)

First lets analyze the graph of \(f(x)\) and \(g(x)\). Since we have to find both the x and the y coordinate of this region, we have to find the bounds for our integration. Since the functions are nice, we will have the same bounds which are \(0\leq x \leq1\). Now that we have our functions, and we know the bounds for integration, we can now go into how to find the moments and the mass of this region.
Computing the Center of Mass
Theoretically, you can find the center of mass of any solid as long as you are given the respective functions of the solid. Further, by using the moments of the region which are usually denoted by \(M_x\) and \(M_y\) and the mass of the region \(M\) we will be able to find the center of mass of the region. The mass is the easiest, as it is found from subtracting \(f(x)\) by \(g(x)\) and it is found from the integration \[M=\rho\int_a^bf(x)-g(x)dx\]The moment \(M_x\) is found through the integral \[M_y=\rho\int_a^bx(f(x)-g(x))x\] Further, the center of mass for the x-cordinate \(\displaystyle\bar{x}\) is found by dividing the moment by the mass or \(\bar{x}=\displaystyle\frac{M_y}{M}\). Likewise, to find the moment of y, the process is almost the same. \(M_x\) is found through the integral \[M_x=\rho\int_a^b\frac{1}{2}(f(x)^2-g(x)^2)dx\] Further, the center of mass for the y-coordinate \(\displaystyle\bar{y}\) is found by dividing the moment by the mass or \(\bar{y}=\displaystyle\frac{M_x}{M}\). This will give us the x-coordinate and the y-coordinate of where the center of mass will be in our region of the solid. \[(\bar{x},\bar{y})=(\frac{M_y}{M},\frac{M_x}{M})\]

Now that we know the basics of center of mass, we can go into actually computing the center of mass. First, lets start with computing the mass of the region. We are going to use the integral that we found earlier. Take the integral \[M=\rho\int_a^bf(x)-g(x)dx\] Next, substitute the functions into the integral and integrate. From this, we get the integral \[M=\rho\int_0^1(-3x^3+4)-(-2x^3-x^\frac{1}{3}+4)dx \rightarrow \rho\int_0^1(-3x^3+4+2x^3+x^\frac{1}{3}-4)dx \rightarrow \rho\int_0^1(-x^3+x^\frac{1}{3}dx) \rightarrow\] \[\rho(\frac{-(1)^4}{4}+\frac{3}{4}(1)^\frac{4}{3}) \rightarrow \rho(\frac{-1}{4}+\frac{3}{4}) = \frac{\rho}{2}\] Now that we have found the mass of our region, we can move on to finding the individuals moments for both \(M_x\) and \(M_y\). We will use the integrals that we found above and substitute our functions into them respectively. First, lets do \(M_x\). \[M_x = \rho\int_a^b\frac{1}{2}(f(x)^2-g(x)^2)dx\] Subsituting our functions in, the integral then becomes \[M_x = \frac{1}{2}\rho\int_0^1(-3x^3+4)^2-(-2x^3-x^\frac{1}{3}+4)^2)dx \rightarrow \rho\frac{1}{2}\int_0^1(9x^6-24x^3+16)-(4x^6+4x^\frac{10}{3}-16x^3+x^\frac{2}{3}-8x^\frac{1}{3}+16)dx\] \[\rho\frac{1}{2}\int_0^1(5x^6-8x^3-4x^\frac{10}{3}+8x^\frac{1}{3}-x^\frac{2}{3})dx \rightarrow \rho\frac{1}{2}(\frac{5x^7}{7}-2x^4-\frac{12}{13}x^\frac{13}{3}+6x^\frac{4}{3}-\frac{3}{5}x^\frac{5}{3})\] \[\frac{1}{2}\rho(\frac{5(1)^7}{7}-2(1)^4-\frac{12}{13}(1)^\frac{13}{3}+6(1)^\frac{4}{3}-\frac{3}{5}(1)^\frac{5}{3})=\frac{726\rho}{455}\] From this calculation, we now have our \(M_x\) and using the mass that we previously found, we can find our \(\bar{y}\) coordinate. However, let's first computate \(M_y\) so we can find both coordinates. Next, we will repeat the same process as \(M_x\), but we will use the other integral that we found for \(M_y\). Recall that the intregral for \(M_y\) was \[M_y=\rho\int_a^bx(f(x)-g(x))dx \rightarrow \rho\int_0^1x((-3x^3+4)-(-2x^3-x^\frac{1}{3}+4))dx \rightarrow \rho\int_0^1x(-x^3+x^\frac{1}{3})dx \rightarrow \rho\int_0^1(-x^4+x^\frac{4}{3})dx\] \[\rho(-\frac{x^5}{5}+\frac{3}{7}x^\frac{7}{3}) \rightarrow \rho(-\frac{(1)^5}{5}+\frac{3}{7}(1)^\frac{7}{3}) = \frac{8\rho}{35}\] Now that we have both moments for \(M_x\) and \(M_y\) we can find both coordinates \(\bar{x}, \bar{y}\). To do this we are going to divide each moment by the mass of the region. The formula is \[(\bar{x},\bar{y})=(\frac{M_y}{M},\frac{M_x}{M})\] We can now subsitute the values that we found to find the exact center of mass of the region! Notice that there is \(\rho\) in both the numerator and the denominator. This means that density was insignficant since it cancels when finding the center of mass. The exact coordinates of the center of mass are: \[(\bar{x},\bar{y})=\displaystyle(\frac{\frac{8\rho}{35}}{\frac{\rho}{2}},\frac{\frac{726\rho}{455}}{\frac{\rho}{2}}) \rightarrow (\frac{16}{35},\frac{1452}{455})\] Finally, the center of mass! Below is an image both on the graph and on the solid itself of where the center of mass would be exactly located!

Why These Functions?
To be completely honest, there is not a direct reason to why I chose these functions. I wanted my shape to be something weird with curves and not just points, thus challenging myself to brainstorm. I came to these functions after just typing in random numbers till I had an interesting shape that I couldn't just eyeball where the center of mass would be. However, the integration isn't too difficult, so this would be great practice for someone who is just learning the topic. Overall, these functions can be helpful for someone who is just learning center of mass and would like a question that isn't easy to guess where the center of mass would be!