Center of mass is described as “the average position of all the parts of the system, weighted according to mass” (Khan Academy). But what does this mean in non-technical terms? The best way to think about it is the point on an object where it should perfectly balance no matter its distribution. So, if we were to try to balance an object on a toothpick, it would balance perfectly at its center of mass.
It is simple to think about on a uniform or symmetric object because it would balance in the middle. However, what about when an object isn’t uniform or symmetric or has holes removed from the center, then the center of mass needs to be adjusted.
When I think about center of mass, I always picture the classic challenge of trying to balance a broom on two fingers. If you hold the broom in the middle, it will immediately fall because the mass is not distributed evenly. Therefore, you must move your fingers closer to side of the bristles to even out the mass. After trial and error, you will eventually find that sweet spot where the broom will balance perfectly on your fingers. That spot is the center of mass.
Trial and error can work and is a great challenge, but we can also calculate the center of mass. The equation for center of mass is the sum of location times the mass divided by the total mass.
(mass)}{total&space;mass} "\sum\frac{(location)(mass)}{total mass}")
We can then break this down to the moment of x and y. The moment is calculated using the location (which will be the f(x) and g(y) equations) multiplied by the mass (which is x or y).
&space;dx "\mu_{y} = \int \rho x f(x) dx")
&space;dy "\mu_{x} = \int \rho y g(y) dy")
And the total mass is a similar equation without the additional x and y in the integral.
&space;dx "mass = \int \rho f(x) dx")
&space;dy "mass = \int \rho g(y) dy")
Example:
When deciding what shape to use for my example, I wanted to focus on real world examples for when we use center of mass. I immediately pictured trying to hang things on a wall which usually needs to be done at its center in one direction. Another thought was, I love making DIY objects for around my apartment, so I want to make my own magnet. However, it needs to be a strong magnet that remains in place on my white board without falling or swinging.
So, for this example, we will be working with an arrow with the goal to turn it into a magnet. To do so, we will find the center of mass, and this will be the location to glue the magnet, so it does not fall. The object today is an arrow represented by these equations. The arrow will have a total height of 2 inches and a width of 4 inches.
Next to make the object less symmetric and display how the center of mass is affected by different details, we placed a circle on the far-right side with the center at (7/2, 1) and radius 3/4 inches.
First, we will calculate the center of mass for the entire object (without the hole) and the center of mass for the hole. Then we will subtract the results of the hole from the entire object.
Whole Object Calculations
To start calculating the center of mass we need to calculate the moment of x and y.
-(1-x))dx&space;+&space;\rho&space;\int_{1}^{4}x(\frac{3}{2}-\frac{1}{2})dx "\mu _{y} = \rho \int_{0}^{1}x((x+1)-(1-x))dx + \rho \int_{1}^{4}x(\frac{3}{2}-\frac{1}{2})dx")
dx&space;+&space;\rho&space;\int_{1}^{4}xdx "= \rho \int_{0}^{1}x(2x)dx + \rho \int_{1}^{4}xdx")
dy&space;+&space;\rho&space;\int_{1}^{2}y(y-1)dy "\mu _{x} = \rho \int_{0}^{1}y(1-y)dy + \rho \int_{1}^{2}y(y-1)dy")
dy+\rho&space;\int_{1}^{2}(y^{2}-y)dy "= \rho \int_{0}^{1}(y - y^{2})dy+\rho \int_{1}^{2}(y^{2}-y)dy")
Since the moment of x is 1, we know the center of mass will be on the line y = 1. Therefore, the y-coordinate will be 1.
 "COM = (??, 1)")
Next, calculate the total mass. This can be done in terms of x or y.
dx "mass = \rho \int_{0}^{1}2xdx + \int_{1}^{4}(\frac{3}{2}-\frac{1}{2})dx")
Hole of Object Calculations
Continuing the same thing, the moment and mass of the hole is
 "\mu = r(\pi r^{2}\rho )")
 "\mu = \frac{3}{4}(\pi \frac{3}{4}^{2}\rho )")
^{2}\rho "mass = \pi (\frac{3}{4})^{2}\rho")
Entire Object including Hole
Finally, we take the moment of the entire object minus the moment of the hole all divided by the total mass of object minus the mass of the hole. We get the following coordinate for the center of mass.
 "COM =(\frac{\frac{49}{6}\rho }{4\rho } - \frac{\frac{27}{64}\pi \rho }{\frac{9}{16}\pi \rho }, 1)")
 "COM =(1.29, 1)")
This coordinate makes sense because when removing the hole, there becomes less mass on the right side compared to the already dominate mass left side. Therefore, we can conclude the center of mass will favor the left side of the object.
It is simple to think about on a uniform or symmetric object because it would balance in the middle. However, what about when an object isn’t uniform or symmetric or has holes removed from the center, then the center of mass needs to be adjusted.
When I think about center of mass, I always picture the classic challenge of trying to balance a broom on two fingers. If you hold the broom in the middle, it will immediately fall because the mass is not distributed evenly. Therefore, you must move your fingers closer to side of the bristles to even out the mass. After trial and error, you will eventually find that sweet spot where the broom will balance perfectly on your fingers. That spot is the center of mass.
Trial and error can work and is a great challenge, but we can also calculate the center of mass. The equation for center of mass is the sum of location times the mass divided by the total mass.
We can then break this down to the moment of x and y. The moment is calculated using the location (which will be the f(x) and g(y) equations) multiplied by the mass (which is x or y).
And the total mass is a similar equation without the additional x and y in the integral.
Example:
When deciding what shape to use for my example, I wanted to focus on real world examples for when we use center of mass. I immediately pictured trying to hang things on a wall which usually needs to be done at its center in one direction. Another thought was, I love making DIY objects for around my apartment, so I want to make my own magnet. However, it needs to be a strong magnet that remains in place on my white board without falling or swinging.
So, for this example, we will be working with an arrow with the goal to turn it into a magnet. To do so, we will find the center of mass, and this will be the location to glue the magnet, so it does not fall. The object today is an arrow represented by these equations. The arrow will have a total height of 2 inches and a width of 4 inches.
y = x + 1 y = 1 - x y = 3/2 y = 1/2
Whole Object Calculations
To start calculating the center of mass we need to calculate the moment of x and y.
Since the moment of x is 1, we know the center of mass will be on the line y = 1. Therefore, the y-coordinate will be 1.
Next, calculate the total mass. This can be done in terms of x or y.
Hole of Object Calculations
Continuing the same thing, the moment and mass of the hole is
Entire Object including Hole
Finally, we take the moment of the entire object minus the moment of the hole all divided by the total mass of object minus the mass of the hole. We get the following coordinate for the center of mass.
This coordinate makes sense because when removing the hole, there becomes less mass on the right side compared to the already dominate mass left side. Therefore, we can conclude the center of mass will favor the left side of the object.
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