6.6 Moments and centers of mass  (Page 6/14)

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Let R be the region bounded above by the graph of the function $f\left(x\right)=6-{x}^{2}$ and below by the graph of the function $g\left(x\right)=3-2x.$ Find the centroid of the region.

The centroid of the region is $\left(1,13\text{/}5\right).$

The symmetry principle

We stated the symmetry principle earlier, when we were looking at the centroid of a rectangle. The symmetry principle can be a great help when finding centroids of regions that are symmetric. Consider the following example.

Finding the centroid of a symmetric region

Let R be the region bounded above by the graph of the function $f\left(x\right)=4-{x}^{2}$ and below by the x -axis. Find the centroid of the region.

The region is depicted in the following figure.

The region is symmetric with respect to the y -axis. Therefore, the x -coordinate of the centroid is zero. We need only calculate $\stackrel{–}{y}.$ Once again, for the sake of convenience, assume $\rho =1.$

First, we calculate the total mass:

$\begin{array}{cc}\hfill m& =\rho {\int }_{a}^{b}f\left(x\right)dx\hfill \\ & ={\int }_{-2}^{2}\left(4-{x}^{2}\right)dx\hfill \\ & ={\left[4x-\frac{{x}^{3}}{3}\right]\phantom{\rule{0.2em}{0ex}}|}_{-2}^{2}=\frac{32}{3}.\hfill \end{array}$

Next, we calculate the moments. We only need ${M}_{x}\text{:}$

$\begin{array}{cc}\hfill {M}_{x}& =\rho {\int }_{a}^{b}\frac{{\left[f\left(x\right)\right]}^{2}}{2}dx\hfill \\ & =\frac{1}{2}{\int }_{-2}^{2}{\left[4-{x}^{2}\right]}^{2}dx=\frac{1}{2}{\int }_{-2}^{2}\left(16-8{x}^{2}+{x}^{4}\right)dx\hfill \\ & =\frac{1}{2}{\left[\frac{{x}^{5}}{5}-\frac{8{x}^{3}}{3}+16x\right]\phantom{\rule{0.2em}{0ex}}|}_{-2}^{2}=\frac{256}{15}.\hfill \end{array}$

Then we have

$\stackrel{–}{y}=\frac{{M}_{x}}{y}=\frac{256}{15}·\frac{3}{32}=\frac{8}{5}.$

The centroid of the region is $\left(0,8\text{/}5\right).$

Let R be the region bounded above by the graph of the function $f\left(x\right)=1-{x}^{2}$ and below by x -axis. Find the centroid of the region.

The centroid of the region is $\left(0,2\text{/}5\right).$

The grand canyon skywalk

The Grand Canyon Skywalk opened to the public on March 28, 2007. This engineering marvel is a horseshoe-shaped observation platform suspended 4000 ft above the Colorado River on the West Rim of the Grand Canyon. Its crystal-clear glass floor allows stunning views of the canyon below (see the following figure).

The Skywalk is a cantilever design, meaning that the observation platform extends over the rim of the canyon, with no visible means of support below it. Despite the lack of visible support posts or struts, cantilever structures are engineered to be very stable and the Skywalk is no exception. The observation platform is attached firmly to support posts that extend 46 ft down into bedrock. The structure was built to withstand 100-mph winds and an 8.0-magnitude earthquake within 50 mi, and is capable of supporting more than 70,000,000 lb.

One factor affecting the stability of the Skywalk is the center of gravity of the structure. We are going to calculate the center of gravity of the Skywalk, and examine how the center of gravity changes when tourists walk out onto the observation platform.

The observation platform is U-shaped. The legs of the U are 10 ft wide and begin on land, under the visitors’ center, 48 ft from the edge of the canyon. The platform extends 70 ft over the edge of the canyon.

To calculate the center of mass of the structure, we treat it as a lamina and use a two-dimensional region in the xy -plane to represent the platform. We begin by dividing the region into three subregions so we can consider each subregion separately. The first region, denoted ${R}_{1},$ consists of the curved part of the U. We model ${R}_{1}$ as a semicircular annulus, with inner radius 25 ft and outer radius 35 ft, centered at the origin (see the following figure).

The legs of the platform, extending 35 ft between ${R}_{1}$ and the canyon wall, comprise the second sub-region, ${R}_{2}.$ Last, the ends of the legs, which extend 48 ft under the visitor center, comprise the third sub-region, ${R}_{3}.$ Assume the density of the lamina is constant and assume the total weight of the platform is 1,200,000 lb (not including the weight of the visitor center; we will consider that later). Use $g=32\phantom{\rule{0.2em}{0ex}}{\text{ft/sec}}^{2}.$

1. Compute the area of each of the three sub-regions. Note that the areas of regions ${R}_{2}$ and ${R}_{3}$ should include the areas of the legs only, not the open space between them. Round answers to the nearest square foot.
2. Determine the mass associated with each of the three sub-regions.
3. Calculate the center of mass of each of the three sub-regions.
4. Now, treat each of the three sub-regions as a point mass located at the center of mass of the corresponding sub-region. Using this representation, calculate the center of mass of the entire platform.
5. Assume the visitor center weighs 2,200,000 lb, with a center of mass corresponding to the center of mass of ${R}_{3}.$ Treating the visitor center as a point mass, recalculate the center of mass of the system. How does the center of mass change?
6. Although the Skywalk was built to limit the number of people on the observation platform to 120, the platform is capable of supporting up to 800 people weighing 200 lb each. If all 800 people were allowed on the platform, and all of them went to the farthest end of the platform, how would the center of gravity of the system be affected? (Include the visitor center in the calculations and represent the people by a point mass located at the farthest edge of the platform, 70 ft from the canyon wall.)

why n does not equal -1
ask a complete question if you want a complete answer.
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The
What is limit
it's the value a function will take while approaching a particular value
Dan
don ger it
Jeremy
what is a limit?
Dlamini
it is the value the function approaches as the input approaches that value.
Andrew
Thanx
Dlamini
Its' complex a limit It's a metrical and topological natural question... approaching means nothing in math
Antonio
is x=2 a function?
The
3y^2*y' + 2xy^3 + 3y^2y'x^2 = 0 sub in x = 2, and y = 1, isolate y'
what is implicit of y³+x²y³=5 at (2,1)
tel mi about a function. what is it?
Jeremy
A function it's a law, that for each value in the domaon associate a single one in the codomain
Antonio
function is a something which another thing depends upon to take place. Example A son depends on his father. meaning here is the father is function of the son. let the father be y and the son be x. the we say F(X)=Y.
Bg
yes the son on his father
pascal
a function is equivalent to a machine. this machine makes x to create y. thus, y is dependent upon x to be produced. note x is an independent variable
moe
x or y those not matter is just to represent.
Bg