6.6 Moments and centers of mass

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• Find the center of mass of objects distributed along a line.
• Locate the center of mass of a thin plate.
• Use symmetry to help locate the centroid of a thin plate.
• Apply the theorem of Pappus for volume.

In this section, we consider centers of mass (also called centroids , under certain conditions) and moments. The basic idea of the center of mass is the notion of a balancing point. Many of us have seen performers who spin plates on the ends of sticks. The performers try to keep several of them spinning without allowing any of them to drop. If we look at a single plate (without spinning it), there is a sweet spot on the plate where it balances perfectly on the stick. If we put the stick anywhere other than that sweet spot, the plate does not balance and it falls to the ground. (That is why performers spin the plates; the spin helps keep the plates from falling even if the stick is not exactly in the right place.) Mathematically, that sweet spot is called the center of mass of the plate .

In this section, we first examine these concepts in a one-dimensional context, then expand our development to consider centers of mass of two-dimensional regions and symmetry. Last, we use centroids to find the volume of certain solids by applying the theorem of Pappus.

Center of mass and moments

Let’s begin by looking at the center of mass in a one-dimensional context. Consider a long, thin wire or rod of negligible mass resting on a fulcrum, as shown in [link] (a). Now suppose we place objects having masses ${m}_{1}$ and ${m}_{2}$ at distances ${d}_{1}$ and ${d}_{2}$ from the fulcrum, respectively, as shown in [link] (b).

The most common real-life example of a system like this is a playground seesaw, or teeter-totter, with children of different weights sitting at different distances from the center. On a seesaw, if one child sits at each end, the heavier child sinks down and the lighter child is lifted into the air. If the heavier child slides in toward the center, though, the seesaw balances. Applying this concept to the masses on the rod, we note that the masses balance each other if and only if ${m}_{1}{d}_{1}={m}_{2}{d}_{2}.$

In the seesaw example, we balanced the system by moving the masses (children) with respect to the fulcrum. However, we are really interested in systems in which the masses are not allowed to move, and instead we balance the system by moving the fulcrum. Suppose we have two point masses, ${m}_{1}$ and ${m}_{2},$ located on a number line at points ${x}_{1}$ and ${x}_{2},$ respectively ( [link] ). The center of mass, $\stackrel{–}{x},$ is the point where the fulcrum should be placed to make the system balance.

Thus, we have

$\begin{array}{ccc}\hfill {m}_{1}|{x}_{1}-\stackrel{–}{x}|& =\hfill & {m}_{2}|{x}_{2}-\stackrel{–}{x}|\hfill \\ \hfill {m}_{1}\left(\stackrel{–}{x}-{x}_{1}\right)& =\hfill & {m}_{2}\left({x}_{2}-\stackrel{–}{x}\right)\hfill \\ \hfill {m}_{1}\stackrel{–}{x}-{m}_{1}{x}_{1}& =\hfill & {m}_{2}{x}_{2}-{m}_{2}\stackrel{–}{x}\hfill \\ \hfill \stackrel{–}{x}\left({m}_{1}+{m}_{2}\right)& =\hfill & {m}_{1}{x}_{1}+{m}_{2}{x}_{2}\hfill \\ \hfill \stackrel{–}{x}& =\hfill & \frac{{m}_{1}{x}_{1}+{m}_{2}{x}_{2}}{{m}_{1}+{m}_{2}}.\hfill \end{array}$

The expression in the numerator, ${m}_{1}{x}_{1}+{m}_{2}{x}_{2},$ is called the first moment of the system with respect to the origin. If the context is clear, we often drop the word first and just refer to this expression as the moment    of the system. The expression in the denominator, ${m}_{1}+{m}_{2},$ is the total mass of the system. Thus, the center of mass    of the system is the point at which the total mass of the system could be concentrated without changing the moment.

what is the power rule
how do i deal with infinity in limits?
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?
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