How To Find Center Of Mass Calculus

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Department 2-3 : Center Of Mass

In this section nosotros are going to find the center of mass or centroid of a sparse plate with uniform density $\rho $. The center of mass or centroid of a region is the bespeak in which the region volition be perfectly balanced horizontally if suspended from that indicate.

Then, let's suppose that the plate is the region bounded by the two curves $f\left( ten \right)$ and $g\left( x \correct)$ on the interval $\left[ {a,b} \right]$. So, nosotros want to notice the center of mass of the region below.

$This is the graph of two unknown functions on the domain a<10<b that are mostly in the 1st quadrant. The graph of f(x) is always over the graph of g(x) and the graph of g(x) dips briefly into the 4th quadrant just to make the point that the quadrant we are in does not matter. The area between the two functions has been shaded in.$

Nosotros'll first need the mass of this plate. The mass is,

\[\brainstorm{align*}Yard & = \rho \left( {{\mbox{Area of plate}}} \right)\\ & = \rho \int_{{\,a}}^{{\,b}}{{f\left( 10 \right) - g\left( x \correct)\,dx}}\end{marshal*}\]

Next, nosotros'll need the moments of the region. There are 2 moments, denoted by ${M_x}$ and ${M_y}$. The moments measure out the trend of the region to rotate nearly the $x$ and $y$-axis respectively. The moments are given past,

Equations of Moments

\[\begin{marshal*}{M_x} & = \rho \int_{{\,a}}^{{\,b}}{{\frac{one}{2}\left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {k\left( x \correct)} \right]}^ii}} \right)\,dx}}\\ {M_y} & = \rho \int_{{\,a}}^{{\,b}}{{x\left( {f\left( x \correct) - g\left( x \right)} \right)\,dx}}\end{align*}\]

The coordinates of the center of mass, $\left( {\overline{x},\overline{y}} \correct)$, are so,

Center of Mass Coordinates

\[\begin{align*}\overline{x} & = \frac{{{M_y}}}{M} = \frac{{\int_{{\,a}}^{{\,b}}{{x\left( {f\left( 10 \correct) - thou\left( x \correct)} \right)\,dx}}}}{{\int_{{\,a}}^{{\,b}}{{f\left( x \right) - g\left( x \right)\,dx}}}} = \frac{1}{A}\int_{{\,a}}^{{\,b}}{{x\left( {f\left( x \correct) - g\left( 10 \right)} \right)\,dx}}\\ \overline{y} & = \frac{{{M_x}}}{M} = \frac{{\int_{{\,a}}^{{\,b}}{{\frac{1}{two}\left( {{{\left[ {f\left( x \right)} \right]}^two} - {{\left[ {g\left( x \right)} \right]}^ii}} \right)\,dx}}}}{{\int_{{\,a}}^{{\,b}}{{f\left( ten \right) - g\left( x \right)\,dx}}}} = \frac{one}{A}\int_{{\,a}}^{{\,b}}{{\frac{1}{2}\left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( ten \right)} \correct]}^2}} \correct)\,dx}}\terminate{marshal*}\]

where,

\[A = \int_{{\,a}}^{{\,b}}{{f\left( ten \correct) - g\left( x \right)\,dx}}\]

Note that the density, $\rho $, of the plate cancels out and so isn't really needed.

Let'due south work a couple of examples.

Instance one Make up one's mind the eye of mass for the region bounded by $y = two\sin \left( {2x} \correct)$, $y = 0$ on the interval $\left[ {0,\displaystyle \frac{\pi }{two}} \right]$.

Show Solution

Hither is a sketch of the region with the center of mass denoted with a dot.

$This is the graph of $y=ii\sin \left( 2x \correct)$ on the domain 0<10<$\frac{\pi}{2}$ and the area between the curve and the x-axis is shaded. There is also a dot at the center of mass whose coordinates will be determined below.$

Let's first get the area of the region.

\[\begin{align*}A & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{2\sin \left( {2x} \right)\,dx}}\\ & = \left. { - \cos \left( {2x} \correct)} \right|_0^{\frac{\pi }{2}}\\ & = 2\finish{align*}\]

Now, the moments (without density since information technology will just drop out) are,

\[\brainstorm{array}{*{20}{c}}\brainstorm{aligned}{M_x} & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{ii{{\sin }^2}\left( {2x} \right)\,dx}}\\ & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{ane - \cos \left( {4x} \right)\,dx}}\\ & = \left. {\left( {x - \frac{1}{4}\sin \left( {4x} \right)} \right)} \right|_0^{\frac{\pi }{two}}\\ & = \frac{\pi }{2}\stop{aligned}& \hspace{0.5in} &\brainstorm{aligned}{M_y} & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{2x\sin \left( {2x} \right)\,dx}}\hspace{0.25in}{\mbox{integrating by parts}}...\\ & = - \left. {x\cos \left( {2x} \right)} \correct|_0^{\frac{\pi }{2}} + \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{\cos \left( {2x} \correct)\,dx}}\\ & = - \left. {ten\cos \left( {2x} \right)} \right|_0^{\frac{\pi }{2}} + \left. {\frac{1}{2}\sin \left( {2x} \right)} \right|_0^{\frac{\pi }{ii}}\\ & = \frac{\pi }{2}\end{aligned}\stop{array}\]

The coordinates of the heart of mass are and then,

\[\brainstorm{marshal*}\overline{x} & = \frac{{{}^{\pi }/{}_{two}}}{ii} = \frac{\pi }{iv}\\ \overline{y} & = \frac{{{}^{\pi }/{}_{two}}}{ii} = \frac{\pi }{4}\stop{align*}\]

Again, note that we didn't put in the density since it volition cancel out.

So, the center of mass for this region is $\left( {\frac{\pi }{four},\frac{\pi }{4}} \right)$.

Example ii Decide the heart of mass for the region bounded past $y = {x^iii}$ and $y = \sqrt ten $.

Evidence Solution

The 2 curves intersect at $x = 0$ and $x = 1$ and here is a sketch of the region with the eye of mass marked with a box.

$This is the graph of $y={{x}^{iii}}$ and $y=\sqrt{x}$ on the domain 0<x<1. The graph of $y=\sqrt{x}$ is always larger than the graph of $y={{x}^{3}}$ and the area between the curves is shaded. There is also a dot at the center of mass whose coordinates will be determined below.$

We'll first get the area of the region.

\[\begin{align*}A & = \int_{{\,0}}^{{\,1}}{{\sqrt x - {x^3}\,dx}}\\ & = \left. {\left( {\frac{2}{3}{ten^{\frac{3}{2}}} - \frac{1}{iv}{10^4}} \right)} \right|_0^1\\ & = \frac{5}{{12}}\end{align*}\]

Now the moments, again without density, are

\[\begin{array}{*{20}{c}}\begin{aligned}{M_x} & = \int_{{\,0}}^{{\,1}}{{\frac{1}{2}\left( {x - {x^6}} \right)\,dx}}\\ & = \left. {\frac{1}{2}\left( {\frac{one}{2}{x^2} - \frac{ane}{vii}{x^7}} \correct)} \right|_0^one\\ & = \frac{5}{{28}} \\ & \finish{aligned}& \hspace{0.5in} &\brainstorm{aligned}{M_y} & = \int_{{\,0}}^{{\,1}}{{x\left( {\sqrt 10 - {10^3}} \right)\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{{10^{\frac{3}{ii}}} - {10^four}\,dx}}\\ & = \left. {\left( {\frac{2}{5}{x^{\frac{5}{2}}} - \frac{1}{five}{ten^v}} \right)} \right|_0^ane\\ & = \frac{1}{v}\terminate{aligned}\terminate{assortment}\]

The coordinates of the center of mass is then,

\[\begin{align*}\overline{x} & = \frac{{{one}/{5}\;}}{{{5}/{{12}}\;}} = \frac{{12}}{{25}}\\ \overline{y} & = \frac{{{5}/{{28}}\;}}{{{5}/{{12}}\;}} = \frac{3}{7}\finish{marshal*}\]

The coordinates of the eye of mass are then,$\left( {\frac{{12}}{{25}},\frac{three}{vii}} \correct)$.

Source: https://tutorial.math.lamar.edu/classes/calcii/centerofmass.aspx

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