# 6.6 Moments and centers of mass  (Page 2/14)

 Page 2 / 14

This idea is not limited just to two point masses. In general, if n masses, ${m}_{1},{m}_{2}\text{,…},{m}_{n},$ are placed on a number line at points ${x}_{1},{x}_{2}\text{,…},{x}_{n},$ respectively, then the center of mass of the system is given by

$\stackrel{–}{x}=\frac{\sum _{i=1}^{n}{m}_{i}{x}_{i}}{\sum _{i=1}^{n}{m}_{i}}.$

## Center of mass of objects on a line

Let ${m}_{1},{m}_{2}\text{,…},{m}_{n}$ be point masses placed on a number line at points ${x}_{1},{x}_{2}\text{,…},{x}_{n},$ respectively, and let $m=\sum _{i=1}^{n}{m}_{i}$ denote the total mass of the system. Then, the moment of the system with respect to the origin is given by

$M=\sum _{i=1}^{n}{m}_{i}{x}_{i}$

and the center of mass of the system is given by

$\stackrel{–}{x}=\frac{M}{m}.$

We apply this theorem in the following example.

## Finding the center of mass of objects along a line

Suppose four point masses are placed on a number line as follows:

$\begin{array}{cccc}{m}_{1}=30\phantom{\rule{0.2em}{0ex}}\text{kg,}\phantom{\rule{0.2em}{0ex}}\text{placed at}\phantom{\rule{0.2em}{0ex}}{x}_{1}=-2\phantom{\rule{0.2em}{0ex}}\text{m}\hfill & & & {m}_{2}=5\phantom{\rule{0.2em}{0ex}}\text{kg,}\phantom{\rule{0.2em}{0ex}}\text{placed at}\phantom{\rule{0.2em}{0ex}}{x}_{2}=3\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ {m}_{3}=10\phantom{\rule{0.2em}{0ex}}\text{kg,}\phantom{\rule{0.2em}{0ex}}\text{placed at}\phantom{\rule{0.2em}{0ex}}{x}_{3}=6\phantom{\rule{0.2em}{0ex}}\text{m}\hfill & & & {m}_{4}=15\phantom{\rule{0.2em}{0ex}}\text{kg,}\phantom{\rule{0.2em}{0ex}}\text{placed at}\phantom{\rule{0.2em}{0ex}}{x}_{4}=-3\phantom{\rule{0.2em}{0ex}}\text{m}.\hfill \end{array}$

Find the moment of the system with respect to the origin and find the center of mass of the system.

First, we need to calculate the moment of the system:

$\begin{array}{cc}\hfill M& =\sum _{i=1}^{4}{m}_{i}{x}_{i}\hfill \\ & =-60+15+60-45=-30.\hfill \end{array}$

Now, to find the center of mass, we need the total mass of the system:

$\begin{array}{cc}\hfill m& =\sum _{i=1}^{4}{m}_{i}\hfill \\ & =30+5+10+15=60\phantom{\rule{0.2em}{0ex}}\text{kg}\text{.}\hfill \end{array}$

Then we have

$\stackrel{–}{x}=\frac{M}{m}=\frac{-30}{60}=-\frac{1}{2}.$

The center of mass is located 1/2 m to the left of the origin.

Suppose four point masses are placed on a number line as follows:

$\begin{array}{cccc}{m}_{1}=12\phantom{\rule{0.2em}{0ex}}\text{kg,}\phantom{\rule{0.2em}{0ex}}\text{placed at}\phantom{\rule{0.2em}{0ex}}{x}_{1}=-4\phantom{\rule{0.2em}{0ex}}\text{m}\hfill & & & {m}_{2}=12\phantom{\rule{0.2em}{0ex}}\text{kg,}\phantom{\rule{0.2em}{0ex}}\text{placed at}\phantom{\rule{0.2em}{0ex}}{x}_{2}=4\phantom{\rule{0.2em}{0ex}}\text{m}\hfill \\ {m}_{3}=30\phantom{\rule{0.2em}{0ex}}\text{kg,}\phantom{\rule{0.2em}{0ex}}\text{placed at}\phantom{\rule{0.2em}{0ex}}{x}_{3}=2\phantom{\rule{0.2em}{0ex}}\text{m}\hfill & & & {m}_{4}=6\phantom{\rule{0.2em}{0ex}}\text{kg,}\phantom{\rule{0.2em}{0ex}}\text{placed at}\phantom{\rule{0.2em}{0ex}}{x}_{4}=-6\phantom{\rule{0.2em}{0ex}}\text{m}.\hfill \end{array}$

Find the moment of the system with respect to the origin and find the center of mass of the system.

$M=24,\stackrel{–}{x}=\frac{2}{5}\phantom{\rule{0.2em}{0ex}}\text{m}$

We can generalize this concept to find the center of mass of a system of point masses in a plane. Let ${m}_{1}$ be a point mass located at point $\left({x}_{1},{y}_{1}\right)$ in the plane. Then the moment ${M}_{x}$ of the mass with respect to the x -axis is given by ${M}_{x}={m}_{1}{y}_{1}.$ Similarly, the moment ${M}_{y}$ with respect to the y -axis is given by ${M}_{y}={m}_{1}{x}_{1}.$ Notice that the x -coordinate of the point is used to calculate the moment with respect to the y -axis, and vice versa. The reason is that the x -coordinate gives the distance from the point mass to the y -axis, and the y -coordinate gives the distance to the x -axis (see the following figure).

If we have several point masses in the xy -plane, we can use the moments with respect to the x - and y -axes to calculate the x - and y -coordinates of the center of mass of the system.

## Center of mass of objects in a plane

Let ${m}_{1},{m}_{2}\text{,…},{m}_{n}$ be point masses located in the xy -plane at points $\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)\text{,…},\left({x}_{n},{y}_{n}\right),$ respectively, and let $m=\sum _{i=1}^{n}{m}_{i}$ denote the total mass of the system. Then the moments ${M}_{x}$ and ${M}_{y}$ of the system with respect to the x - and y -axes, respectively, are given by

${M}_{x}=\sum _{i=1}^{n}{m}_{i}{y}_{i}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{M}_{y}=\sum _{i=1}^{n}{m}_{i}{x}_{i}.$

Also, the coordinates of the center of mass $\left(\stackrel{–}{x},\stackrel{–}{y}\right)$ of the system are

$\stackrel{–}{x}=\frac{{M}_{y}}{m}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\stackrel{–}{y}=\frac{{M}_{x}}{m}.$

The next example demonstrates how to apply this theorem.

## Finding the center of mass of objects in a plane

Suppose three point masses are placed in the xy -plane as follows (assume coordinates are given in meters):

$\begin{array}{c}{m}_{1}=2\phantom{\rule{0.2em}{0ex}}\text{kg, placed at}\phantom{\rule{0.2em}{0ex}}\left(-1,3\right),\hfill \\ {m}_{2}=6\phantom{\rule{0.2em}{0ex}}\text{kg, placed at}\phantom{\rule{0.2em}{0ex}}\left(1,1\right),\hfill \\ {m}_{3}=4\phantom{\rule{0.2em}{0ex}}\text{kg, placed at}\phantom{\rule{0.2em}{0ex}}\left(2,-2\right).\hfill \end{array}$

Find the center of mass of the system.

First we calculate the total mass of the system:

$m=\sum _{i=1}^{3}{m}_{i}=2+6+4=12\phantom{\rule{0.2em}{0ex}}\text{kg}\text{.}$

Next we find the moments with respect to the x - and y -axes:

$\begin{array}{}\\ \\ {M}_{y}=\sum _{i=1}^{3}{m}_{i}{x}_{i}=-2+6+8=12,\hfill \\ {M}_{x}=\sum _{i=1}^{3}{m}_{i}{y}_{i}=6+6-8=4.\hfill \end{array}$

Then we have

$\stackrel{–}{x}=\frac{{M}_{y}}{m}=\frac{12}{12}=1\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\stackrel{–}{y}=\frac{{M}_{x}}{m}=\frac{4}{12}=\frac{1}{3}.$

The center of mass of the system is $\left(1,1\text{/}3\right),$ in meters.

#### Questions & Answers

what is the power rule
how do i deal with infinity in limits?
Add the functions f(x)=7x-x g(x)=5-x
f(x)=7x-x g(x)=5-x
Awon
5x-5
Verna
what is domain
difference btwn domain co- domain and range
Cabdalla
x
Verna
The set of inputs of a function. x goes in the function, y comes out.
Verna
where u from verna
Arfan
If you differentiate then answer is not x
Raymond
domain is the set of values of independent variable and the range is the corresponding set of values of dependent variable
Champro
what is functions
give different types of functions.
Paul
how would u find slope of tangent line to its inverse function, if the equation is x^5+3x^3-4x-8 at the point(-8,1)
pls solve it i Want to see the answer
Sodiq
ok
Friendz
differentiate each term
Friendz
why do we need to study functions?
to understand how to model one variable as a direct relationship to another variable
Andrew
integrate the root of 1+x²
use the substitution t=1+x. dt=dx √(1+x)dx = √tdt = t^1/2 dt integral is then = t^(1/2 + 1) / (1/2 + 1) + C = (2/3) t^(3/2) + C substitute back t=1+x = (2/3) (1+x)^(3/2) + C
navin
find the nth differential coefficient of cosx.cos2x.cos3x
determine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse
f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
can you please post the question again here so I can see what your talking about
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana
Multiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x
Melin
thank you ladies and gentlemen I appreciate the help!
Robert
keep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.
Andrew
please how do I learn integration
they are simply "anti-derivatives". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.
Andrew
best way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.
Andrew
example 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2
Andrew
great noticeable direction
Isaac
limit x tend to infinite xcos(π/2x)*sin(π/4x)
can you give me a problem for function. a trigonometric one