# 5.6 Calculating centers of mass and moments of inertia  (Page 3/8)

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## Finding a centroid

Find the centroid of the region under the curve $y={e}^{x}$ over the interval $1\le x\le 3$ (see the following figure). Finding a centroid of a region below the curve y = e x .

To compute the centroid, we assume that the density function is constant and hence it cancels out:

$\begin{array}{}\\ \\ \\ \\ {x}_{c}=\frac{{M}_{y}}{m}=\frac{\underset{R}{\iint }x\phantom{\rule{0.2em}{0ex}}dA}{\underset{R}{\iint }dA}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{y}_{c}=\frac{{M}_{x}}{m}=\frac{\underset{R}{\iint }y\phantom{\rule{0.2em}{0ex}}dA}{\underset{R}{\iint }dA},\hfill \\ {x}_{c}=\frac{{M}_{y}}{m}=\frac{\underset{R}{\iint }x\phantom{\rule{0.2em}{0ex}}dA}{\underset{R}{\iint }dA}=\frac{\underset{x=1}{\overset{x=3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{y=0}{\overset{y={e}^{x}}{\int }}x\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx}{\underset{x=1}{\overset{x=3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{y=0}{\overset{y={e}^{x}}{\int }}dy\phantom{\rule{0.2em}{0ex}}dx}=\frac{\underset{x=1}{\overset{x=3}{\int }}x{e}^{x}dx}{\underset{x=1}{\overset{x=3}{\int }}{e}^{x}dx}=\frac{2{e}^{3}}{{e}^{3}-e}=\frac{2{e}^{2}}{{e}^{2}-1},\hfill \\ {y}_{c}=\frac{{M}_{x}}{m}=\frac{\underset{R}{\iint }y\phantom{\rule{0.2em}{0ex}}dA}{\underset{R}{\iint }dA}=\frac{\underset{x=1}{\overset{x=3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{y=0}{\overset{y={e}^{x}}{\int }}y\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx}{\underset{x=1}{\overset{x=3}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{y=0}{\overset{y={e}^{x}}{\int }}dy\phantom{\rule{0.2em}{0ex}}dx}=\frac{\underset{x=1}{\overset{x=3}{\int }}\frac{{e}^{2x}}{2}dx}{\underset{x=1}{\overset{x=3}{\int }}{e}^{x}dx}=\frac{\frac{1}{4}{e}^{2}\left({e}^{4}-1\right)}{e\left({e}^{2}-1\right)}=\frac{1}{4}e\left({e}^{2}+1\right).\hfill \end{array}$

Thus the centroid of the region is

$\left({x}_{c},{y}_{c}\right)=\left(\frac{2{e}^{2}}{{e}^{2}-1},\frac{1}{4}e\left({e}^{2}+1\right)\right).$

Calculate the centroid of the region between the curves $y=x$ and $y=\sqrt{x}$ with uniform density in the interval $0\le x\le 1.$

${x}_{c}=\frac{{M}_{y}}{m}=\frac{1\text{/}15}{1\text{/}6}=\frac{2}{5}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{y}_{c}=\frac{{M}_{x}}{m}=\frac{1\text{/}12}{1\text{/}6}=\frac{1}{2}$

## Moments of inertia

For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition in Section $6.6.$ The moment of inertia of a particle of mass $m$ about an axis is $m{r}^{2},$ where $r$ is the distance of the particle from the axis. We can see from [link] that the moment of inertia of the subrectangle ${R}_{ij}$ about the $x\text{-axis}$ is ${\left({y}_{ij}^{*}\right)}^{2}\rho \left({x}_{ij}^{*},{y}_{ij}^{*}\right)\text{Δ}A.$ Similarly, the moment of inertia of the subrectangle ${R}_{ij}$ about the $y\text{-axis}$ is ${\left({x}_{ij}^{*}\right)}^{2}\rho \left({x}_{ij}^{*},{y}_{ij}^{*}\right)\text{Δ}A.$ The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis.

The moment of inertia ${I}_{x}$ about the $x\text{-axis}$ for the region $R$ is the limit of the sum of moments of inertia of the regions ${R}_{ij}$ about the $x\text{-axis}\text{.}$ Hence

${I}_{x}=\underset{k,l\to \infty }{\text{lim}}{\sum _{i=1}^{k}\sum _{j=1}^{l}\left({y}_{ij}^{*}\right)}^{2}{m}_{ij}=\underset{k,l\to \infty }{\text{lim}}{\sum _{i=1}^{k}\sum _{j=1}^{l}\left({y}_{ij}^{*}\right)}^{2}\rho \left({x}_{ij}^{*},{y}_{ij}^{*}\right)\text{Δ}A=\underset{R}{\iint }{y}^{2}\rho \left(x,y\right)dA.$

Similarly, the moment of inertia ${I}_{y}$ about the $y\text{-axis}$ for $R$ is the limit of the sum of moments of inertia of the regions ${R}_{ij}$ about the $y\text{-axis}\text{.}$ Hence

${I}_{y}=\underset{k,l\to \infty }{\text{lim}}{\sum _{i=1}^{k}\sum _{j=1}^{l}\left({x}_{ij}^{*}\right)}^{2}{m}_{ij}=\underset{k,l\to \infty }{\text{lim}}{\sum _{i=1}^{k}\sum _{j=1}^{l}\left({x}_{ij}^{*}\right)}^{2}\rho \left({x}_{ij}^{*},{y}_{ij}^{*}\right)\text{Δ}A=\underset{R}{\iint }{x}^{2}\rho \left(x,y\right)dA.$

Sometimes, we need to find the moment of inertia of an object about the origin, which is known as the polar moment of inertia. We denote this by ${I}_{0}$ and obtain it by adding the moments of inertia ${I}_{x}$ and ${I}_{y}.$ Hence

${I}_{0}={I}_{x}+{I}_{y}=\underset{R}{\iint }\left({x}^{2}+{y}^{2}\right)\rho \left(x,y\right)dA.$

All these expressions can be written in polar coordinates by substituting $x=r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,$ $y=r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta ,$ and $dA=r\phantom{\rule{0.2em}{0ex}}dr\phantom{\rule{0.2em}{0ex}}d\theta .$ For example, ${I}_{0}=\underset{R}{\iint }{r}^{2}\rho \left(r\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\theta ,r\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta \right)dA.$

## Finding moments of inertia for a triangular lamina

Use the triangular region $R$ with vertices $\left(0,0\right),\left(2,2\right),$ and $\left(2,0\right)$ and with density $\rho \left(x,y\right)=xy$ as in previous examples. Find the moments of inertia.

Using the expressions established above for the moments of inertia, we have

$\begin{array}{ccc}\hfill {I}_{x}& =\hfill & \underset{R}{\iint }{y}^{2}\rho \left(x,y\right)dA=\underset{x=0}{\overset{x=2}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{y=0}{\overset{y=x}{\int }}x{y}^{3}dy\phantom{\rule{0.2em}{0ex}}dx=\frac{8}{3},\hfill \\ \hfill {I}_{y}& =\hfill & \underset{R}{\iint }{x}^{2}\rho \left(x,y\right)dA=\underset{x=0}{\overset{x=2}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{y=0}{\overset{y=x}{\int }}{x}^{3}y\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx=\frac{16}{3},\hfill \\ \hfill {I}_{0}& =\hfill & \underset{R}{\iint }\left({x}^{2}+{y}^{2}\right)\rho \left(x,y\right)dA=\underset{0}{\overset{2}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{0}{\overset{x}{\int }}\left({x}^{2}+{y}^{2}\right)xy\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx\hfill \\ & =\hfill & {I}_{x}+{I}_{y}=8.\hfill \end{array}$

Again use the same region $R$ as above and the density function $\rho \left(x,y\right)=\sqrt{xy}.$ Find the moments of inertia.

${I}_{x}=\underset{x=0}{\overset{x=2}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{y=0}{\overset{y=x}{\int }}{y}^{2}\sqrt{xy}\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx=\frac{64}{35}$ and ${I}_{y}=\underset{x=0}{\overset{x=2}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{y=0}{\overset{y=x}{\int }}{x}^{2}\sqrt{xy}\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx=\frac{64}{35}.$ Also, ${I}_{0}=\underset{x=0}{\overset{x=2}{\int }}\phantom{\rule{0.2em}{0ex}}\underset{y=0}{\overset{y=x}{\int }}\left({x}^{2}+{y}^{2}\right)\sqrt{xy}\phantom{\rule{0.2em}{0ex}}dy\phantom{\rule{0.2em}{0ex}}dx=\frac{128}{21}.$

As mentioned earlier, the moment of inertia of a particle of mass $m$ about an axis is $m{r}^{2}$ where $r$ is the distance of the particle from the axis, also known as the radius of gyration    .

Hence the radii of gyration with respect to the $x\text{-axis,}$ the $y\text{-axis,}$ and the origin are

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