This week, we are diving into the Center of Mass of an object! The method of finding the Center of Mass of an object goes with our theme of integration. The Center of Mass of an object is very literal in its meaning, it is the position in which the relative distance of masses from that point sum to zero. For example, a square with uniform mass has a center of mass directly in the middle of the square.
The Center of Mass of an object can be calculated by finding the Moments about the X and Y axis of the object and dividing by the total mass of the system. This is true for something on a plan, if you are dealing with just a line then you only have moments at the points on the line. The Moment about the X-Axis, denoted as $M_x$, is the sum of each mass at each point in terms of the y values. Following from this, the Moment about the Y-Axis, denoted as $M_y$, is the sum of each mass at each point in terms of the x values. To find the Center of Mass, or COM,, you use the following formula, where M is the total mass of the system:
$COM = (\frac{M_y}{M}, \frac{M_x}{M})$
If we take a region with a uniform density, lets call it $\rho$, then we calculate the Mass, $M_x$ and $M_y$ by the given formulas.
Mass = $\int_{a}^{b} \rho * (f(x) - g(x)) dx$ (You can also use the y's instead!)
$M_x = \int_{c}^{d} \rho * y * (f(y) - g(y)) dy $
$M_x = \int_{a}^{b} \rho * x * (f(x) - g(x)) dx $
Here, we get the Center of Mass, COM, is a point in the X-Y Plane. From now on, we will refer to the x-coordinate and the y-coordinate of the COM by $\bar{x},
\bar{y}$ respectively. Now that we have most of the ground work done, we will look at the equation I decided to dive into this week. I did not want to do anything super crazy, since coming up with interesting shapes can be hard when doing COM. So instead, I wanted to look at the difference of shaving a little bit off of a symmetrical shape. Today, we will look at the following equations:
$2|x| -2x = 16$ and $x=4$ enclosed by the lines $y=-4,
y=4$
Below is the graph of these! The area with the dashed dark lines showcase the area that we are interested in!
These equations make a square that has had a smaller square and two triangles cut out. In order to find the Center of Mass of this object, we will have to split up the integrals based on the shapes of that region. For $M_y$, we can see that we have one rectangle, and two different triangles. Thus, we will take the integrals of those regions and sum them up.
$\it{Math Time!}$
$M_y = \int_{-4}^{\frac{-4}{3}} (\rho * 8x )dx + \int_{\frac{-4}{3}}^{0} (\rho * x(-3x))dx + \int_{0}^{4} (\rho * x(-x) )dx$
Here, we will discuss each of the three integrals individually. The 8 in the first integral is the area of the rectangle enclosed by the bounds given. The second integral is based on the line containing the points: ($\frac{-4}{3}, 4$) & (0,0) which gives us the line $y= -3x$. Same idea is the reason for the third integral, where the line contains points (0,0) & (4, -4), giving us the line $y= -x$.
By solving through the three integrals and adding them up, we find that:
$M_y = \frac{-2048\rho}{27}$.
Now lets switch to $M_x$, which is found by the following four integrals:
$M_x = \int_{-4}^{4} (\rho *y * \frac{8}{3})dy + \int_{-4}^{0} (\rho * y * \frac{4}{3})dy + \int_{0}^{4} (\rho * y * \frac{y}{-3})dy + \int_{-4}^{0} (.5 * \rho * y * 4)dy$
We follow this process from before of finding the areas of the shapes now with respect to the x-axis with the constraints of the bounds. Solving through these four integrals, we arrive at following:
$M_x = \frac{-304\rho}{9}$
Our last step is to solve for the Mass! We will do the four integrals we just worked through for this, besides having that outside y. So the Mass has the following sum of integration's:
Mass $= \int_{-4}^{4} (\rho * \frac{8}{3})dy + \int_{-4}^{0} (\rho * \frac{4}{3})dy + \int_{0}^{4} (\rho *\frac{y}{-3})dy + \int_{-4}^{0} (.5 * \rho * 4)dy$
Solving through, we arrive at:
Mass $= 32\rho$
Now we have everything to have an actual point on the object for its center of mass! By substituting in the formula from before, the COM is:
COM $ = (\frac{\frac{-2048\rho}{27}}{32\rho}, \frac{\frac{-304\rho}{9}}{32\rho})$
COM $ = (\frac{-64}{27}, \frac{-19}{18})$
Below is the OnShape File of the construction of this, along with a small hole where the Center of Mass is!
So, we see that the center of mass is slightly below $y = -1$ and a little past $x = -2$, which makes sense because we have that larger mass on the bottom right of the object. If we did not have the smaller triangle in the top left of the object and instead the triangle was the same size as the larger one, the COM would have been (-1.33, -1.33). By making the triangle smaller on the top right right, we see that the COM gets shift up and to the left since we lost mass at the top left.
The Center of Mass of an object can be calculated by finding the Moments about the X and Y axis of the object and dividing by the total mass of the system. This is true for something on a plan, if you are dealing with just a line then you only have moments at the points on the line. The Moment about the X-Axis, denoted as $M_x$, is the sum of each mass at each point in terms of the y values. Following from this, the Moment about the Y-Axis, denoted as $M_y$, is the sum of each mass at each point in terms of the x values. To find the Center of Mass, or COM,, you use the following formula, where M is the total mass of the system:
$COM = (\frac{M_y}{M}, \frac{M_x}{M})$
If we take a region with a uniform density, lets call it $\rho$, then we calculate the Mass, $M_x$ and $M_y$ by the given formulas.
Mass = $\int_{a}^{b} \rho * (f(x) - g(x)) dx$ (You can also use the y's instead!)
$M_x = \int_{c}^{d} \rho * y * (f(y) - g(y)) dy $
$M_x = \int_{a}^{b} \rho * x * (f(x) - g(x)) dx $
Here, we get the Center of Mass, COM, is a point in the X-Y Plane. From now on, we will refer to the x-coordinate and the y-coordinate of the COM by $\bar{x},
\bar{y}$ respectively. Now that we have most of the ground work done, we will look at the equation I decided to dive into this week. I did not want to do anything super crazy, since coming up with interesting shapes can be hard when doing COM. So instead, I wanted to look at the difference of shaving a little bit off of a symmetrical shape. Today, we will look at the following equations:
$2|x| -2x = 16$ and $x=4$ enclosed by the lines $y=-4,
y=4$
Below is the graph of these! The area with the dashed dark lines showcase the area that we are interested in!
These equations make a square that has had a smaller square and two triangles cut out. In order to find the Center of Mass of this object, we will have to split up the integrals based on the shapes of that region. For $M_y$, we can see that we have one rectangle, and two different triangles. Thus, we will take the integrals of those regions and sum them up.
$\it{Math Time!}$
$M_y = \int_{-4}^{\frac{-4}{3}} (\rho * 8x )dx + \int_{\frac{-4}{3}}^{0} (\rho * x(-3x))dx + \int_{0}^{4} (\rho * x(-x) )dx$
Here, we will discuss each of the three integrals individually. The 8 in the first integral is the area of the rectangle enclosed by the bounds given. The second integral is based on the line containing the points: ($\frac{-4}{3}, 4$) & (0,0) which gives us the line $y= -3x$. Same idea is the reason for the third integral, where the line contains points (0,0) & (4, -4), giving us the line $y= -x$.
By solving through the three integrals and adding them up, we find that:
$M_y = \frac{-2048\rho}{27}$.
Now lets switch to $M_x$, which is found by the following four integrals:
$M_x = \int_{-4}^{4} (\rho *y * \frac{8}{3})dy + \int_{-4}^{0} (\rho * y * \frac{4}{3})dy + \int_{0}^{4} (\rho * y * \frac{y}{-3})dy + \int_{-4}^{0} (.5 * \rho * y * 4)dy$
We follow this process from before of finding the areas of the shapes now with respect to the x-axis with the constraints of the bounds. Solving through these four integrals, we arrive at following:
$M_x = \frac{-304\rho}{9}$
Our last step is to solve for the Mass! We will do the four integrals we just worked through for this, besides having that outside y. So the Mass has the following sum of integration's:
Mass $= \int_{-4}^{4} (\rho * \frac{8}{3})dy + \int_{-4}^{0} (\rho * \frac{4}{3})dy + \int_{0}^{4} (\rho *\frac{y}{-3})dy + \int_{-4}^{0} (.5 * \rho * 4)dy$
Solving through, we arrive at:
Mass $= 32\rho$
Now we have everything to have an actual point on the object for its center of mass! By substituting in the formula from before, the COM is:
COM $ = (\frac{\frac{-2048\rho}{27}}{32\rho}, \frac{\frac{-304\rho}{9}}{32\rho})$
COM $ = (\frac{-64}{27}, \frac{-19}{18})$
Below is the OnShape File of the construction of this, along with a small hole where the Center of Mass is!
So, we see that the center of mass is slightly below $y = -1$ and a little past $x = -2$, which makes sense because we have that larger mass on the bottom right of the object. If we did not have the smaller triangle in the top left of the object and instead the triangle was the same size as the larger one, the COM would have been (-1.33, -1.33). By making the triangle smaller on the top right right, we see that the COM gets shift up and to the left since we lost mass at the top left.
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