How To Find Centroid From Center Of Mass

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Department ii-3 : Center Of Mass

In this section we are going to detect the centre of mass or centroid of a thin plate with uniform density $\rho $. The center of mass or centroid of a region is the indicate in which the region will be perfectly balanced horizontally if suspended from that point.

So, permit's suppose that the plate is the region bounded by the two curves $f\left( 10 \right)$ and $grand\left( x \correct)$ on the interval $\left[ {a,b} \right]$. So, we desire to notice the center of mass of the region below.

$This is the graph of 2 unknown functions on the domain a<10<b that are mostly in the 1st quadrant. The graph of f(x) is always over the graph of g(x) and the graph of g(x) dips briefly into the 4th quadrant just to make the point that the quadrant we are in does not matter. The area between the two functions has been shaded in.$

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

\[\brainstorm{align*}G & = \rho \left( {{\mbox{Area of plate}}} \right)\\ & = \rho \int_{{\,a}}^{{\,b}}{{f\left( ten \right) - chiliad\left( x \right)\,dx}}\stop{align*}\]

Adjacent, we'll need the moments of the region. There are 2 moments, denoted past ${M_x}$ and ${M_y}$. The moments measure the tendency of the region to rotate nigh the $10$ and $y$-centrality respectively. The moments are given by,

Equations of Moments

\[\begin{align*}{M_x} & = \rho \int_{{\,a}}^{{\,b}}{{\frac{1}{2}\left( {{{\left[ {f\left( ten \right)} \correct]}^two} - {{\left[ {one thousand\left( 10 \correct)} \right]}^2}} \correct)\,dx}}\\ {M_y} & = \rho \int_{{\,a}}^{{\,b}}{{10\left( {f\left( ten \right) - thousand\left( x \correct)} \right)\,dx}}\end{marshal*}\]

The coordinates of the centre of mass, $\left( {\overline{10},\overline{y}} \correct)$, are so,

Center of Mass Coordinates

\[\begin{align*}\overline{x} & = \frac{{{M_y}}}{Grand} = \frac{{\int_{{\,a}}^{{\,b}}{{10\left( {f\left( x \right) - m\left( x \correct)} \correct)\,dx}}}}{{\int_{{\,a}}^{{\,b}}{{f\left( 10 \right) - g\left( x \right)\,dx}}}} = \frac{i}{A}\int_{{\,a}}^{{\,b}}{{x\left( {f\left( x \right) - m\left( x \right)} \right)\,dx}}\\ \overline{y} & = \frac{{{M_x}}}{M} = \frac{{\int_{{\,a}}^{{\,b}}{{\frac{1}{2}\left( {{{\left[ {f\left( x \right)} \correct]}^2} - {{\left[ {1000\left( 10 \right)} \correct]}^two}} \correct)\,dx}}}}{{\int_{{\,a}}^{{\,b}}{{f\left( ten \right) - g\left( x \right)\,dx}}}} = \frac{1}{A}\int_{{\,a}}^{{\,b}}{{\frac{one}{2}\left( {{{\left[ {f\left( x \right)} \correct]}^2} - {{\left[ {one thousand\left( x \right)} \correct]}^2}} \right)\,dx}}\end{align*}\]

where,

\[A = \int_{{\,a}}^{{\,b}}{{f\left( 10 \right) - thousand\left( 10 \right)\,dx}}\]

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

Allow'south piece of work a couple of examples.

Example i Determine the centre of mass for the region divisional by $y = 2\sin \left( {2x} \correct)$, $y = 0$ on the interval $\left[ {0,\displaystyle \frac{\pi }{2}} \correct]$.

Bear witness Solution

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

$This is the graph of $y=2\sin \left( 2x \correct)$ on the domain 0<ten<$\frac{\pi}{2}$ 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 kickoff get the surface area of the region.

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

Now, the moments (without density since it volition just drop out) are,

\[\begin{array}{*{20}{c}}\begin{aligned}{M_x} & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{two{{\sin }^2}\left( {2x} \right)\,dx}}\\ & = \int_{{\,0}}^{{\,\frac{\pi }{two}}}{{i - \cos \left( {4x} \correct)\,dx}}\\ & = \left. {\left( {x - \frac{i}{four}\sin \left( {4x} \right)} \right)} \right|_0^{\frac{\pi }{two}}\\ & = \frac{\pi }{2}\stop{aligned}& \hspace{0.5in} &\brainstorm{aligned}{M_y} & = \int_{{\,0}}^{{\,\frac{\pi }{two}}}{{2x\sin \left( {2x} \right)\,dx}}\hspace{0.25in}{\mbox{integrating by parts}}...\\ & = - \left. {x\cos \left( {2x} \correct)} \correct|_0^{\frac{\pi }{2}} + \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{\cos \left( {2x} \right)\,dx}}\\ & = - \left. {10\cos \left( {2x} \correct)} \right|_0^{\frac{\pi }{2}} + \left. {\frac{one}{2}\sin \left( {2x} \right)} \right|_0^{\frac{\pi }{2}}\\ & = \frac{\pi }{2}\terminate{aligned}\end{array}\]

The coordinates of the centre of mass are so,

\[\brainstorm{align*}\overline{x} & = \frac{{{}^{\pi }/{}_{2}}}{2} = \frac{\pi }{iv}\\ \overline{y} & = \frac{{{}^{\pi }/{}_{2}}}{ii} = \frac{\pi }{4}\terminate{align*}\]

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

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

Example 2 Determine the eye of mass for the region bounded by $y = {x^iii}$ and $y = \sqrt x $.

Evidence Solution

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

$This is the graph of $y={{x}^{three}}$ 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.$

Nosotros'll first get the surface area of the region.

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

Now the moments, again without density, are

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

The coordinates of the center of mass is and then,

\[\begin{align*}\overline{x} & = \frac{{{1}/{5}\;}}{{{5}/{{12}}\;}} = \frac{{12}}{{25}}\\ \overline{y} & = \frac{{{5}/{{28}}\;}}{{{five}/{{12}}\;}} = \frac{3}{7}\terminate{align*}\]

The coordinates of the middle of mass are then,$\left( {\frac{{12}}{{25}},\frac{3}{7}} \right)$.

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

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