The center of mass marks the position defined relative to an object or system of objects. It is the average position of all parts of the system showing us the balance or pivot point for objects. For objects of uniform density such as the object we are dealing with, the center of mass is located at the centroid. You can think of the center of mass of an object as weights on an axis or a number line. Where would a fulcram be placed on the axis to balance the line? It's calculated in this case by adding the masses multiplied by their distance from the origin all divided by the total mass, resulting in a coordinate on the axis where the line would balance if a fulcram were placed there.
The total of all the masses multiplied by their distance from the origin is called the moment of the system about that origin. The moment of the system measures the tendency of the system to rotate about the origin.For a planar region the object can rotate about either the x or y axis so there are two moments, one about the x axis and one about the y axis. The x coordinate of the center of mass location is the moment of the y axis divided by the area or total mass of the object or total mass, and the y coordinate of the center of mass location is the moment of the x axis divided by the area of the object or total mass. Using integrals and moments, we can find moments about the y and x axis and mass of the total system to find the coordinates of the center of mass location in the object.
The following graph shows the shape created by the lines
\[y=x^5, y=x^\frac{1}{5}+\frac{1}{3}\]
The green dot is the center of mass. When we laser cut the shape using an stg file (shown below on tinkercad) we will see how accurate this calcultion is. I have a model with the center of mass hole and a model without the center of mass hole. My plan is to laser cut both, and use the model with the hole to mark the center of mass on the laser cut without the hole in it because I couldn't fill the hole with a cylinder.
I chose these equations and the shape created from them somewhat arbitrairly. I knew I wanted some sort of clover or petal shape that had two different sized shapes on either side of the diagnol. I picked \(y=x^5\) first and then its inverse, \(y=x^\frac{1}{5}\) and added numbers and fractions to its inverse until I got two different sized shapes on either side of the diagnol. The shape that looked best to me was created from adding \(\frac{1}{3}\) resulting in the shape created from
\(y=x^5\) and \(y=x^\frac{1}{5}+\frac{1}{3}\).
Now, how did I compute the center of mass to be that point?
I used the formula for the center of mass of a thin plate with uniform density p: \[M=p(Area\ of\ plate)\] \[=p\int_{a}^{b}f(x)-g(x) dx\]
The moments of this region also need to be calculated. The moment of x and the moment of y are notated by \(M_x\) and \(M_y\) respectively. They are calculated like so: \[M_x=p\int_{a}^{b}\frac{1}{2} ([f(x)]^2-[g(x)]^2) dx\] \[M_y=p\int_{a}^{b} x(f(x)-g(x)) dx\]
The center of mass coordinates are \((\frac{M_y}{M}\),\(\frac{M_x}{M}\)).
\[M=Area=upper\ region\ area + lower\ region\ area\] \[p\int_{0}^{1.061}(x^\frac{1}{5}+\frac{1}{3})-(x^5) dx\] \[+p\int_{-0.917}^{0}(x^5)-(x^\frac{1}{5}+\frac{1}{3}) dx\] \[=1.35687\]
\[M_x=p\int_{a}^{b}\frac{1}{2} ([f(x)]^2-[g(x)]^2) dx\] \[M_x=p\int_{0}^{1.061}\frac{1}{2}[(x^\frac{1}{5}+\frac{1}{3})^2-(x^5)^2]dx\] \[+ p\int_{-0.917}^{0}\frac{1}{2}[(x^5)^2-(x^\frac{1}{5}+\frac{1}{3})^2]dx\] \[=0.65799+(-0.09942)\] \[=0.55857\]
\[M_y=p\int_{a}^{b} x(f(x)-g(x)) dx\] \[M_y=p\int_{0}^{1.061}x((x^\frac{1}{5}+\frac{1}{3})-(x^5)) dx\] \[+p\int_{-0.917}^{0}x((x^5)-(x^\frac{1}{5}+\frac{1}{3}))dx\] \[=0.48918+(-0.43791)\] \[=0.05127\] \[(\frac{M_y}{M}),(\frac{M_x}{M})\]
\[(\frac{0.05127}{1.35687}),(\frac{0.55857}{1.35687})\] \[(0.037785492),(0.41166066)\] \[(0.0378),(0.4117)\]
The Equation for the moment about the x axis is found in many center of mass calculations. It is usually easier than converting everything into being in terms of y. [Shown below] It comes from the fact that the coordinate of the center of mass of x is equal to x, and the coordinate of the center of mass of y is equal to the function of f plus the function of g all dvided by 2. Then the integral with respect to mass is equal to pdA or the density with respect to the change in area which is equl to the density of x multiplied by the function of f minus the function of g with respect to x. \[\bar{x}=x\] \[\bar{y}=(\frac{f(x)+g(x)}{2})\] \[\bar{d}m=pdA\] \[=p(x)[f(x)-g(x)]dx\]
So in order to get everything in terms of x, we do \[M_x=p\int_{a}^{b}\frac{1}{2} [(f(x)+g(x)]p(x)[f(x)-g(x)]dx\] \[=p\int_{a}^{b}\frac{1}{2} ([f(x)]^2-[g(x)]^2) dx\]
Now, how did I compute the center of mass to be that point?
I used the formula for the center of mass of a thin plate with uniform density p: \[M=p(Area\ of\ plate)\] \[=p\int_{a}^{b}f(x)-g(x) dx\]
The moments of this region also need to be calculated. The moment of x and the moment of y are notated by \(M_x\) and \(M_y\) respectively. They are calculated like so: \[M_x=p\int_{a}^{b}\frac{1}{2} ([f(x)]^2-[g(x)]^2) dx\] \[M_y=p\int_{a}^{b} x(f(x)-g(x)) dx\]
The center of mass coordinates are \((\frac{M_y}{M}\),\(\frac{M_x}{M}\)).
\[M=Area=upper\ region\ area + lower\ region\ area\] \[p\int_{0}^{1.061}(x^\frac{1}{5}+\frac{1}{3})-(x^5) dx\] \[+p\int_{-0.917}^{0}(x^5)-(x^\frac{1}{5}+\frac{1}{3}) dx\] \[=1.35687\]
\[M_x=p\int_{a}^{b}\frac{1}{2} ([f(x)]^2-[g(x)]^2) dx\] \[M_x=p\int_{0}^{1.061}\frac{1}{2}[(x^\frac{1}{5}+\frac{1}{3})^2-(x^5)^2]dx\] \[+ p\int_{-0.917}^{0}\frac{1}{2}[(x^5)^2-(x^\frac{1}{5}+\frac{1}{3})^2]dx\] \[=0.65799+(-0.09942)\] \[=0.55857\]
\[M_y=p\int_{a}^{b} x(f(x)-g(x)) dx\] \[M_y=p\int_{0}^{1.061}x((x^\frac{1}{5}+\frac{1}{3})-(x^5)) dx\] \[+p\int_{-0.917}^{0}x((x^5)-(x^\frac{1}{5}+\frac{1}{3}))dx\] \[=0.48918+(-0.43791)\] \[=0.05127\] \[(\frac{M_y}{M}),(\frac{M_x}{M})\]
\[(\frac{0.05127}{1.35687}),(\frac{0.55857}{1.35687})\] \[(0.037785492),(0.41166066)\] \[(0.0378),(0.4117)\]
The Equation for the moment about the x axis is found in many center of mass calculations. It is usually easier than converting everything into being in terms of y. [Shown below] It comes from the fact that the coordinate of the center of mass of x is equal to x, and the coordinate of the center of mass of y is equal to the function of f plus the function of g all dvided by 2. Then the integral with respect to mass is equal to pdA or the density with respect to the change in area which is equl to the density of x multiplied by the function of f minus the function of g with respect to x. \[\bar{x}=x\] \[\bar{y}=(\frac{f(x)+g(x)}{2})\] \[\bar{d}m=pdA\] \[=p(x)[f(x)-g(x)]dx\]
So in order to get everything in terms of x, we do \[M_x=p\int_{a}^{b}\frac{1}{2} [(f(x)+g(x)]p(x)[f(x)-g(x)]dx\] \[=p\int_{a}^{b}\frac{1}{2} ([f(x)]^2-[g(x)]^2) dx\]
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