Introduction
Any object can be balanced using its center of mass. Theoretically, we could balance a car if we wanted to. We would look at the weight of the car and try to balance it in the middle. We know that if the weight is not balanced correctly the car would tip on one side. This process can be applied to any object we want. Fortunately, there is a formula that can help on finding the center of mass of any object making it a little easier than trial and error. Trying to find the center of mass of a car can be a bit hard since the mass can be different throughout the car. Looking at something entirely symmetric we can predict the center of mass would be in the middle. So, to start off we will look at something simple where the mass is constant throughout the entire object but not symmetric. We will be looking at the object in the image below and find the center of mass.
We can see that the object looks like a house that is mostly symmetric with a door that is slightly to the side.
To have a better understanding we will imagine the object on the coordinate plane. This allows us to see that the center of mass is a coordinate point with a x- and y- value. These points can be found by finding the moments of x and y and the mass of the object. The formula to find the center of mass is \[ Center\ of\ Mass = \left(\frac{M_y}{Mass},\frac{M_x}{Mass}\right)\] To find the x-coordinate we need to find the moment of y and the total mass of the entire object. The formula to find the moment of y is \[M_y = \int_{a}^{b}x*\rho*f(x)*dx\] \(\rho\ \) is the density, in our example the density is constant so it will eventually cancel out with the \(\rho\ \) in the mass when we calculate the x-coordinate. The points a and b are the starting and ending points corresponding to the axis. The f(x) is the total length of the object from the points a and b. To calculate the moment of x uses the same formula as above with some slight changes corresponding the y-axis.
Compute
With the image on the right, we can now see all the points that we need. We can see that this object isn't symmetric so to make this easier we can divide this object into more workable shapes. We can divide this object up into a triangle, square and a rectangle.
The first thing we need to do is find the two lines for the triangle. With the points we can use the slope formula to find the slope then use the slope-intercept form to find the b-value and the final equation for the line. The two lines that we get are \begin{align*} y &= x+1 \\ y&= -x+2 \end{align*} \(M_y\) can now be calculated. We can add and subtract the moments of y that we find. Which allows us to add \(M_y\) values of the triangle and square and subtract the rectangle. \begin{align*} M_y&= \int_{0}^{.5}(\rho*x*(x+1))dx\ + \int_{.5}^{1}(\rho*x*(-x+2)) dx\ +\ \int_{0}^{1}(\rho*x*(1-0))\ dx\ -\ \int_{.15}^{.4} (\rho*x*(.5-0))\ dx \\ &= \rho\int_{0}^{.5}(x^2+x)\ dx\ +\ \rho\int_{.5}^{1} (-x+2x)\ dx\ +\ \rho\int_{0}^{1} x\ dx\ -\ \rho\int_{.15}^{.4} .5x\ dx \\ &= \rho\left.\frac{x^3}{3}+\frac{x^2}{2}\right\vert_{0}^{.5}\ +\ \rho\left.\frac{-x^3}{3}+x^2\right\vert_{.5}^{1}\ +\ \rho\left.\frac{x^2} {2}\right\vert_{0}^{1}\ -\ \rho\left.\frac{.5x^2}{2}\right\vert_{.15}^{.4} \\ &= \rho\frac{1}{6}\ +\ \rho\left[\left(-\frac{1}{3}+1\right)-\left(\frac{5}{24}\right)\right]\ +\ \rho\frac{1}{2}\ -\ \rho\left(\frac{1}{25}- \frac{9}{1600}\right)\ \\ M_y&= \frac{349}{320}\rho\ =\ 1.090625\rho \end{align*} \(M_x\) can be found using the same process but now the bounds would be in terms of y and we will integrate with respect to x. The lines that we found earlier would now be \begin{align*} x &= y-1 \\ x&= 2-y \end{align*} \begin{align*} M_x&= \int_{0}^{1}(\rho*y*(1-0))dy\ + \int_{1}^{1.5}(\rho*y*[(2-y)-(y-1)])\ dy\ -\ \int_{0}^{.25}(\rho*y*(.4-.15)\ dy\ \\ &= \rho\int_{0}^{1}y\ dy\ +\ \rho\int_{1}^{1.5} (-2y^2+3y)\ dy\ -\ \rho\int_{0}^{.25} .25y\ dx\ \\ &= \rho\left.\frac{y^2}{2}\right\vert_{0}^{1}\ +\ \rho\left(\left.\frac{-2y^3}{3}+\frac{3y^2}{2}\right)\right\vert_{1}^{1.5}\ -\ \rho\left.\frac{.25y^2}{2}\right\vert_{0}^{.25} \\ &= \rho\frac{1}{2}\ +\ \rho\left[\left(-\frac{9}{4}+\frac{27}{8}\right)-\left(-\frac{2}{3}+\frac{3}{2}\right)\right]\ -\ \rho\frac{1}{128}\ \\ M_x&= \frac{301}{384}\rho\ =\ 0.7838541\rho \end{align*} The last thing we need to find is the mass to calculate the center of mass. \begin{align*} Mass&= \int_{0}^{.5}(\rho*(x+1))dx\ + \int_{.5}^{1}(\rho*(-x+2)) dx\ +\ \int_{0}^{1}(\rho*(1-0))\ dx\ -\ \int_{.15}^{.4} (\rho*(.5-0))\ dx \\ &= \rho\left(\left.\frac{x^2}{2}+ x \right)\right\vert_{0}^{.5}\ +\ \rho\left(\left.\frac{-x^2}{2}+2x\right)\right\vert_{.5}^{1}\ +\ \rho\left.x\right\vert_{0}^{1}\ -\ \rho\left. .5x\right\vert_{.15}^{.4} \\ &= \rho\frac{5}{8}\ +\ \rho\frac{5}{8}\ +\ \rho1\ -\ \rho\frac{1}{8} \\ Mass&= 2.125\rho \end{align*} We know have the \(M_y\), \(M_x\) and the mass so, we can now calculate the center of mass for our object. \[ Center\ of\ Mass = \left(\frac{M_y}{Mass},\frac{M_x}{Mass}\right) = \left(\frac{1.090625\rho}{2.125\rho},\frac{0.7838541\rho}{2.125\rho}\right) = (.5132,.3688)\]
Why this Object
The image on the left has marked the center of mass. I chose this shape because it was mostly symmetrical making the computations easier allowing us to focus on the process. I added the rectangle hole which further allowed us to see how that can change the center of mass from the expected one. I was also surprised that the center of mass was lower than what I expected it to be. The dimensions of this object are 1.5x1 inches with the width of 0.2 inches.
Any object can be balanced using its center of mass. Theoretically, we could balance a car if we wanted to. We would look at the weight of the car and try to balance it in the middle. We know that if the weight is not balanced correctly the car would tip on one side. This process can be applied to any object we want. Fortunately, there is a formula that can help on finding the center of mass of any object making it a little easier than trial and error. Trying to find the center of mass of a car can be a bit hard since the mass can be different throughout the car. Looking at something entirely symmetric we can predict the center of mass would be in the middle. So, to start off we will look at something simple where the mass is constant throughout the entire object but not symmetric. We will be looking at the object in the image below and find the center of mass.
We can see that the object looks like a house that is mostly symmetric with a door that is slightly to the side.
To have a better understanding we will imagine the object on the coordinate plane. This allows us to see that the center of mass is a coordinate point with a x- and y- value. These points can be found by finding the moments of x and y and the mass of the object. The formula to find the center of mass is \[ Center\ of\ Mass = \left(\frac{M_y}{Mass},\frac{M_x}{Mass}\right)\] To find the x-coordinate we need to find the moment of y and the total mass of the entire object. The formula to find the moment of y is \[M_y = \int_{a}^{b}x*\rho*f(x)*dx\] \(\rho\ \) is the density, in our example the density is constant so it will eventually cancel out with the \(\rho\ \) in the mass when we calculate the x-coordinate. The points a and b are the starting and ending points corresponding to the axis. The f(x) is the total length of the object from the points a and b. To calculate the moment of x uses the same formula as above with some slight changes corresponding the y-axis.
Compute
With the image on the right, we can now see all the points that we need. We can see that this object isn't symmetric so to make this easier we can divide this object into more workable shapes. We can divide this object up into a triangle, square and a rectangle.
The first thing we need to do is find the two lines for the triangle. With the points we can use the slope formula to find the slope then use the slope-intercept form to find the b-value and the final equation for the line. The two lines that we get are \begin{align*} y &= x+1 \\ y&= -x+2 \end{align*} \(M_y\) can now be calculated. We can add and subtract the moments of y that we find. Which allows us to add \(M_y\) values of the triangle and square and subtract the rectangle. \begin{align*} M_y&= \int_{0}^{.5}(\rho*x*(x+1))dx\ + \int_{.5}^{1}(\rho*x*(-x+2)) dx\ +\ \int_{0}^{1}(\rho*x*(1-0))\ dx\ -\ \int_{.15}^{.4} (\rho*x*(.5-0))\ dx \\ &= \rho\int_{0}^{.5}(x^2+x)\ dx\ +\ \rho\int_{.5}^{1} (-x+2x)\ dx\ +\ \rho\int_{0}^{1} x\ dx\ -\ \rho\int_{.15}^{.4} .5x\ dx \\ &= \rho\left.\frac{x^3}{3}+\frac{x^2}{2}\right\vert_{0}^{.5}\ +\ \rho\left.\frac{-x^3}{3}+x^2\right\vert_{.5}^{1}\ +\ \rho\left.\frac{x^2} {2}\right\vert_{0}^{1}\ -\ \rho\left.\frac{.5x^2}{2}\right\vert_{.15}^{.4} \\ &= \rho\frac{1}{6}\ +\ \rho\left[\left(-\frac{1}{3}+1\right)-\left(\frac{5}{24}\right)\right]\ +\ \rho\frac{1}{2}\ -\ \rho\left(\frac{1}{25}- \frac{9}{1600}\right)\ \\ M_y&= \frac{349}{320}\rho\ =\ 1.090625\rho \end{align*} \(M_x\) can be found using the same process but now the bounds would be in terms of y and we will integrate with respect to x. The lines that we found earlier would now be \begin{align*} x &= y-1 \\ x&= 2-y \end{align*} \begin{align*} M_x&= \int_{0}^{1}(\rho*y*(1-0))dy\ + \int_{1}^{1.5}(\rho*y*[(2-y)-(y-1)])\ dy\ -\ \int_{0}^{.25}(\rho*y*(.4-.15)\ dy\ \\ &= \rho\int_{0}^{1}y\ dy\ +\ \rho\int_{1}^{1.5} (-2y^2+3y)\ dy\ -\ \rho\int_{0}^{.25} .25y\ dx\ \\ &= \rho\left.\frac{y^2}{2}\right\vert_{0}^{1}\ +\ \rho\left(\left.\frac{-2y^3}{3}+\frac{3y^2}{2}\right)\right\vert_{1}^{1.5}\ -\ \rho\left.\frac{.25y^2}{2}\right\vert_{0}^{.25} \\ &= \rho\frac{1}{2}\ +\ \rho\left[\left(-\frac{9}{4}+\frac{27}{8}\right)-\left(-\frac{2}{3}+\frac{3}{2}\right)\right]\ -\ \rho\frac{1}{128}\ \\ M_x&= \frac{301}{384}\rho\ =\ 0.7838541\rho \end{align*} The last thing we need to find is the mass to calculate the center of mass. \begin{align*} Mass&= \int_{0}^{.5}(\rho*(x+1))dx\ + \int_{.5}^{1}(\rho*(-x+2)) dx\ +\ \int_{0}^{1}(\rho*(1-0))\ dx\ -\ \int_{.15}^{.4} (\rho*(.5-0))\ dx \\ &= \rho\left(\left.\frac{x^2}{2}+ x \right)\right\vert_{0}^{.5}\ +\ \rho\left(\left.\frac{-x^2}{2}+2x\right)\right\vert_{.5}^{1}\ +\ \rho\left.x\right\vert_{0}^{1}\ -\ \rho\left. .5x\right\vert_{.15}^{.4} \\ &= \rho\frac{5}{8}\ +\ \rho\frac{5}{8}\ +\ \rho1\ -\ \rho\frac{1}{8} \\ Mass&= 2.125\rho \end{align*} We know have the \(M_y\), \(M_x\) and the mass so, we can now calculate the center of mass for our object. \[ Center\ of\ Mass = \left(\frac{M_y}{Mass},\frac{M_x}{Mass}\right) = \left(\frac{1.090625\rho}{2.125\rho},\frac{0.7838541\rho}{2.125\rho}\right) = (.5132,.3688)\]
Why this Object
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