Determing the center of gravity of a shaded section

• Guillem_dlc
In summary, the conversation discusses the use of the theorem of Pappus Guldinus to determine the volume of a shaded area around the Y-axis. The steps involved include finding the area of the shaded section, calculating the center of gravity, and using the theorem to determine the volume. The conversation also mentions the assumption that the Y-axis and the side of length 2R are parallel.
Guillem_dlc
Homework Statement
Determine the volume of the shaded area around the Y-axis by using the theorem of Pappus Guldinus, where value of R = 143,3 cm.

a) Determine the area of the shaded section.

b) Determine the center of gravity of the shaded section.

c) Detrmine the volume by using the theorem of Pappus Guldinus.
Relevant Equations
theorem of Pappus Guldinus, center of gravity
Determine the volume of the shaded area around the Y-axis by using the theorem of Pappus Guldinus, where value of R = 143,3 cm.
a) Determine the area of the shaded section.
b) Determine the center of gravity of the shaded section.
c) Detrmine the volume by using the theorem of Pappus Guldinus.

My attempt at the solution:

a) ##A_T=2R\cdot 2R-\dfrac14 \cdot \pi \cdot R^2+\dfrac{R\cdot 2R}{2}##
##A_T=4R^2-\dfrac{\pi \cdot R^2}{4}+R^2=4\cdot 143,3^2-\dfrac{\pi \cdot 143,3^2}{2}+143,3^2=86546,39\, \textrm{cm}^2=\boxed{8,65\, \textrm{m}^2}##

b) I consider three figures to calculate the centres of gravity:
Figure 1 (Rectangular triangle with side R and hypotenuse 2R) --> ##\bar{x}=2R+\dfrac{R}{3}=2\cdot 143,3+\dfrac{143,3}{3}=334,36\, \textrm{cm}##
##\bar{y}=\dfrac{2R}{3}=\dfrac{2\cdot 143,3}{3}=95,53\, \textrm{cm}##
Figure 2 (Square of side 2R) --> ##\bar{x}=\dfrac{2R}{2}=R=143,3\, \textrm{cm}##
##\bar{y}=\dfrac{2R}{2}=R=143,3\, \textrm{cm}##
Figure 3 (sector of circle)--> ##\bar{x}=\dfrac{4r}{3\pi}=\dfrac{4\cdot 143,3}{3\pi}=60,82\, \textrm{cm}##
##\bar{y}=2R-\dfrac{4R}{3\pi}=2\cdot 143,3-\dfrac{4\cdot 143,3}{3\pi}=225,78\, \textrm{cm}##
Then:
##\bar{x}=\dfrac1A\sum_{i=1}^n \bar{x_i}A_i=\dfrac{334,36\cdot 143,3^2+143,3\cdot 4\cdot 143,3^2-60,82\cdot \frac{\pi \cdot 143,3^2}{2}}{86546,39}=192,67##

This last result would be correct?

Thanks

The image seems to show the Y axis as at some angle to the side of length 2R. But there does not seem to be enough information to determine that angle.
Are they parallel really, and the image is taken at some angle to the page?

haruspex said:
The image seems to show the Y axis as at some angle to the side of length 2R. But there does not seem to be enough information to determine that angle.
Are they parallel really, and the image is taken at some angle to the page?

I think it's constructed from a rectangle of dimensions 3R x 2R attached to the y-axis like a flag, then with a right triangle removed from the right and a quarter circle removed from the top left.

etotheipi said:
I think it's constructed from a rectangle of dimensions 3R x 2R attached to the y-axis like a flag, then with a right triangle removed from the right and a quarter circle removed from the top left.
Yes, if the 2R and Y axis are indeed parallel.
@Guillem_dlc , you make life hard for yourself and anyone reading your work by plugging in numbers much too soon. Write it out again all in terms of R. Don't plug in its value until the very end.

etotheipi
haruspex said:
Yes, if the 2R and Y axis are indeed parallel.

Ah, I see what you mean. I guess since no further information is given we just have to assume it's sloppiness on the question setter's part!

As an aside, I must admit I'd never heard of 'Pappus-Guldinus' but the result looks almost too simple to be true! I haven't read through the proof yet but at the moment it seems a bit like magic. Add that to the fact that when I said the name out loud my furniture started floating...

etotheipi said:
assume it's sloppiness on the question setter's part!
No, I wouid guess it is reasonably obvious in the original diagram, but the distortion in the way the photo has been taken makes it hard to believe.

etotheipi
haruspex said:
The image seems to show the Y axis as at some angle to the side of length 2R. But there does not seem to be enough information to determine that angle.
Are they parallel really, and the image is taken at some angle to the page?
Yes, Y and 2R are parallel.

haruspex said:
Yes, if the 2R and Y axis are indeed parallel.
@Guillem_dlc , you make life hard for yourself and anyone reading your work by plugging in numbers much too soon. Write it out again all in terms of R. Don't plug in its value until the very end.
Okay, I'll do it by leaving R indicated and I'll pass it around. You're absolutely right.

etotheipi said:
Ah, I see what you mean. I guess since no further information is given we just have to assume it's sloppiness on the question setter's part!

As an aside, I must admit I'd never heard of 'Pappus-Guldinus' but the result looks almost too simple to be true! I haven't read through the proof yet but at the moment it seems a bit like magic. Add that to the fact that when I said the name out loud my furniture started floating...
Yes

etotheipi said:
As an aside, I must admit I'd never heard of 'Pappus-Guldinus' but the result looks almost too simple to be true! I haven't read through the proof yet but at the moment it seems a bit like magic. Add that to the fact that when I said the name out loud my furniture started floating...

etotheipi
Writing in terms of ##R##:

a) ##A_T=2R\cdot 2R-\dfrac14 \pi R^2+\dfrac{R2R}{2}=4R^2-\dfrac{\pi R^2}{4}+R^2=\dfrac{16R^2-\pi R^2+4R^2}{4}=\dfrac{R^2\cdot (20-\pi)}{4}=\dfrac{20R^2-\pi R^2}{4}=143,3\, \textrm{cm}=1,433\, \textrm{m}##

b) Centers of gravity:
- Figure 1: ##\bar{x}=2R+\dfrac{R}{3}=\dfrac{6R+R}{3}=\dfrac{7R}{3}##
- Figure 2: ##\bar{x}=\dfrac{2R}{2}=R##
- Figure 3: ##\bar{x}=\dfrac{4R}{3\pi}##
Then:
##\bar{x}=\dfrac1A \sum_{i=1}^n \bar{x_i}\cdot A_i=\dfrac{\frac{7R}{3}+R+\frac{4R}{3\pi}}{\frac{20R^2-\pi R^2}{4}}=\dfrac{\frac{7R+3R}{3}+\frac{4R}{3\pi}}{\frac{20R^2-\pi R^2}{4}}=\dfrac{\pi (7R+3R)+4R}{3\pi}: \dfrac{20R^2-\pi R^2}{4}=\dfrac{4\pi \cdot (7R+3R)+16R}{3\pi \cdot (20R^2-\pi R^2)}=\dfrac{R\cdot (4\pi \cdot (7+3)+16)}{3\pi R^2(20-\pi)}=\dfrac{40\pi +16}{3\pi R(20-\pi)}=\dfrac{40\pi +16}{3\pi \cdot 143,3(20-\pi)}=0,622\, \textrm{m}##

So I know the result is not right but I can't find the mistake.

Guillem_dlc said:
##\bar{x}=\dfrac1A \sum_{i=1}^n \bar{x_i}\cdot A_i=\dfrac{\frac{7R}{3}+R+\frac{4R}{3\pi}}{\frac{20R^2-\pi R^2}{4}}##
What you have written in the numerator is ##\sum_{i=1}^n \bar{x_i}##, not ##\sum_{i=1}^n \bar{x_i}\cdot A_i##.

haruspex said:
What you have written in the numerator is ##\sum_{i=1}^n \bar{x_i}##, not ##\sum_{i=1}^n \bar{x_i}\cdot A_i##.
##\bar{x}=\dfrac{\frac{7R}{3}\cdot R^2+R\cdot 4R^2-\frac{4R}{3\pi}\cdot \frac{\pi R^2}{4}}{\frac{20R^2-\pi R^2}{4}}=\dfrac{\frac{7R^3}{3}+4R^3-\frac{4\pi R^3}{12\pi}}{\frac{R^2\cdot (20-\pi)}{4}}=\dfrac{\frac{7R^3}{3}+4R^3-\frac{R^3}{3}}{\frac{R^2\cdot (20-\pi)}{4}}=\dfrac{\frac{7R^3+12R^3-R^3}{3}}{\frac{R^2\cdot (20-\pi)}{4}}=\dfrac{18R^3}{3}:\dfrac{R^2\cdot (20-\pi)}{4}=\dfrac{72R^3}{3R^2\cdot (20-\pi)}=\dfrac{24R}{20-\pi}##
Then,
##\bar{x}=\dfrac{24\cdot 143,3}{20-\pi}=204,005\rightarrow \boxed{2,04\, \textrm{m}}##

Now, is it correct?

Guillem_dlc said:
##\bar{x}=\dfrac{\frac{7R}{3}\cdot R^2+R\cdot 4R^2-\frac{4R}{3\pi}\cdot \frac{\pi R^2}{4}}{\frac{20R^2-\pi R^2}{4}}=\dfrac{\frac{7R^3}{3}+4R^3-\frac{4\pi R^3}{12\pi}}{\frac{R^2\cdot (20-\pi)}{4}}=\dfrac{\frac{7R^3}{3}+4R^3-\frac{R^3}{3}}{\frac{R^2\cdot (20-\pi)}{4}}=\dfrac{\frac{7R^3+12R^3-R^3}{3}}{\frac{R^2\cdot (20-\pi)}{4}}=\dfrac{18R^3}{3}:\dfrac{R^2\cdot (20-\pi)}{4}=\dfrac{72R^3}{3R^2\cdot (20-\pi)}=\dfrac{24R}{20-\pi}##
Then,
##\bar{x}=\dfrac{24\cdot 143,3}{20-\pi}=204,005\rightarrow \boxed{2,04\, \textrm{m}}##

Now, is it correct?
Looks right.

sysprog and Guillem_dlc
Please think of Pappus' simple and brilliant idea as a clue to doing the triple integral that you have to do when you are modeling Homer Simpson eating a donut. When does the centroid exit the empty space at the donut hole and actually enter the material of the remaining portion of the donut (torus/toroid)? The theorem is like if an anaconda swallowed a DVD and had to pass it at maximum area all the way down its alimentary canal.

1. How is the center of gravity of a shaded section determined?

The center of gravity of a shaded section is determined by finding the point where the weight of the section is evenly distributed in all directions. This can be done by using the formula for calculating the center of mass, which takes into account the weight and location of each individual part of the shaded section.

2. Why is it important to determine the center of gravity of a shaded section?

Determining the center of gravity is important because it helps in understanding the stability and balance of an object or structure. It also allows for accurate predictions of how the object will behave when subjected to external forces such as gravity or wind.

3. What factors affect the location of the center of gravity?

The location of the center of gravity is affected by the shape, size, and weight distribution of the shaded section. It can also be influenced by external factors such as the orientation of the section and any additional weight added to it.

4. Can the center of gravity of a shaded section be outside of the object?

Yes, the center of gravity can be outside of the shaded section if the weight distribution is uneven or if the section is irregularly shaped. In these cases, the center of gravity may be located in empty space or even outside of the object entirely.

5. How can the center of gravity of a shaded section be used in real-world applications?

The center of gravity is used in various industries, such as engineering and construction, to ensure the stability and safety of structures. It is also important in fields such as sports and transportation, where balance and weight distribution play a crucial role in performance and efficiency.

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