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

 Page 3 / 8

## 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

where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
can you provide the details of the parametric equations for the lines that defince doubly-ruled surfeces (huperbolids of one sheet and hyperbolic paraboloid). Can you explain each of the variables in the equations? By Rebecca Butterfield By OpenStax By Stephen Voron By Richley Crapo By By OpenStax By Janet Forrester By Charles Jumper By OpenStax By Jazzycazz Jackson