How to find centroid of a hemisphere using Pappus's centroid theorem?

  • #1
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Homework Statement:
To find the centroid of a hollow and solid hemisphere.
Relevant Equations:
Pappus's centroid theorem
I recently learned how to calculate the centroid of a semi-circular ring of radius ##r## using Pappus's centroid theorem as
##\begin{align}
&4 \pi r^2=(2 \pi d)(\pi r)\nonumber\\
&d=\frac {2r}{\pi}\nonumber
\end{align}##
Where ##d## is the distance of center of mass of the ring from its base.

Similarly its second theorem can also be used to calculate C.O.M of semi-circular disc of radius ##r##,
##\begin{align}
&\frac{4}{3} \pi r^3=(2 \pi d)(\dfrac {\pi r^2}{2})\nonumber\\
&d=\frac {4r}{3 \pi}\nonumber
\end{align}##

Now this is great because now I don't have to do the lengthy calculations to find the C.O.M using ##d=\dfrac{\int ydm}{M}##

So I wanted to extend this to find the centroid of hemispheres and cones but then how do I find the area or volume traced by rotating a hemisphere?
How do I extend this to 3D object or can I extend this to 3D objects?
 

Answers and Replies

  • #2
haruspex
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One way to understand Pappus' centroid theorems is in terms of wedge elements.
If we take a semicircular disc and rotate it only through a small angle ##d\theta## we produce a wedge element. Now copy this, not rotated around the axis but into parallel planes. Like laying out the segments of an orange, core piece uppermost on a flat table.
We can think of the thickness of a wedge at any point as a weighting. This is proportional to the distance from the core. E.g. compress each segment into a semicircular lamina of some uniform thickness but maintaining the mass, so the further from the core the greater the density. The flattening is arranged so that the thickness is preserved at the centroid.
This weighting pattern is how the centroid of the original semicircle is defined. It is the "torque arm" of each bit of the semicircle about the axis.

The purpose of writing all that is to find a way to think about it in higher dimensions.
For the centroid of a hemisphere, the 'axis' is the flat surface. The weighting is the distance from that surface, not from a diameter of it.
Starting with a hemisphere, we can create a row of hemispheric elements. In each element, the density distribution is weighted to represent the distance from the flat surface.
Each hemisphere has now a thickness in thefourth dimension, and these total ##2\pi d##.

On this basis, the distance from the flat surface of a hemisphere to its centroid should be (vol of 4D ball)/((2##\pi##)(vol of hemisphere))##=\frac{\frac 12\pi^2r^4}{2\pi.\frac 23\pi r^3}=\frac 38r##, which I believe is correct.

Note that a 4D ball is termed a 3-sphere, not a 4-sphere.
 
Last edited:
  • #3
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One way to understand Pappus' centroid theorems is in terms of wedge elements.
If we take a semicircular disc and rotate it only through a small angle ##d\theta## we produce a wedge element. Now copy this, not rotated around the axis but into parallel planes. Like laying out the segments of an orange, core piece uppermost on a flat table.
We can think of the thickness of a wedge at any point as a weighting. This is proportional to the distance from the core. E.g. compress each segment into a semicircular lamina of some uniform thickness but maintaining the mass, so the further from the core the greater the density. The flattening is arranged so that the thickness is preserved at the centroid.
This weighting pattern is how the centroid of the original semicircle is defined. It is the "torque arm" of each bit of the semicircle about the axis.

The purpose of writing all that is to find a way to think about it in higher dimensions.
For the centroid of a hemisphere, the 'axis' is the flat surface. The weighting is the distance from that surface, not from a diameter of it.
Starting with a hemisphere, we can create a row of hemispheric elements. In each element, the density distribution is weighted to represent the distance from the flat surface.
Each hemisphere has now a thickness in thefourth dimension, and these total ##2\pi d##.

On this basis, the distance from the flat surface of a hemisphere to its centroid should be (vol of 4D ball)/((2##\pi##)(vol of hemisphere))##=\frac{\frac 12\pi^2r^4}{2\pi.\frac 23\pi r^3}=\frac 38r##, which I believe is correct.

Note that a 4D ball is termed a 3-sphere, not a 4-sphere.
I don't understand what you said there about understanding Pappus theorem, but I just wanted to find the area/volume enclosed by a hollow hemisphere, solid hemisphere, hollow cone and a solid cone on rotating these about their base, i.e., how did you get
Volume of 4D ball = ##\frac 12\pi^2r^4## ?

How do you manage to imagine all these in your head without a figure or anything? I cannot understand even the simplest things without a diagram!
 
  • #4
haruspex
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I don't understand what you said there about understanding Pappus theorem, but I just wanted to find the area/volume enclosed by a hollow hemisphere, solid hemisphere, hollow cone and a solid cone on rotating these about their base, i.e., how did you get
Volume of 4D ball = ##\frac 12\pi^2r^4## ?

How do you manage to imagine all these in your head without a figure or anything? I cannot understand even the simplest things without a diagram!
For the volumes and surface volumes of n-spheres, see https://en.m.wikipedia.org/wiki/N-sphere.
It would be difficult to illustrate what I wrote with diagrams. Visual imagination takes practice. Here's one you could try:
Suppose I were to paint a blue line around the inside of a hollow rubber torus. I mean the 'long way round', so it surrounds the hole in the middle if the torus. Now I paint a red line on the outside, the short way around. These two coloured circles are linked.
Now I make a small hole in the wall of the torus and pull it inside out through the hole. So the blue line is now outside and the red line inside. Are they still linked?
 
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  • #5
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For the volumes and surface volumes of n-spheres, see https://en.m.wikipedia.org/wiki/N-sphere.
It would be difficult to illustrate what I wrote with diagrams. Visual imagination takes practice. Here's one you could try:
Suppose I were to paint a blue line around the inside of a hollow rubber torus. I mean the 'long way round', so it surrounds the hole in the middle if the torus. Now I paint a red line on the outside, the short way around. These two coloured circles are linked.
Now I make a small hole in the wall of the torus and pull it inside out through the hole. So the blue line is now outside and the red line inside. Are they still linked?
By those two circles being linked do you mean that they are concentric??

If so, then to answer your problem I think maybe that first we extend that small hole we made so that we can slit the tube (torus) along its length (along its circumference) such that now instead of a tube we are left with a strip of rubber and then after we turn the slitted tube (strip) inside out we stitch it back together along its length such that now we are only left with the small hole which we started with and so our entire arrangement is the same just the only difference is that the blue line is now outside and the red one is inside and they are still concentric or linked.

I assumed that changing the size of the hole will not change our problem so we can increase the hole such that its long enough to cover the entire length of the tube along the circumference such that we can slit the whole tube with that small hole.
 
  • #6
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I assumed that changing the size of the hole will not change our problem
I don't have a proof but I have heard that changing the size of a hole in any shape doesn't change it homeomorphically (idk if that's the word)
that is why a ring is homomorphic to a cylinder or a disc is homomorphic to a cone or a cup
 
  • #7
haruspex
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do you mean that they are concentric??
No. Lnked like links in a chain. Each passes through the other.
 
  • #8
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No. Linked like links in a chain. Each passes through the other.
What?? How are they linked like a chain? I thought that those two circles are like rings one surrounded by the other and there is a layer of rubber in between them separating each other
 
  • #9
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Suppose I were to paint a blue line around the inside of a hollow rubber torus. I mean the 'long way round', so it surrounds the hole in the middle if the torus
That means that you painted a circle along the outer circumference (because its the long way round) of the tube but on the inside surface of the tube
Now I paint a red line on the outside, the short way around.
And this means that you painted the circle on the inside circumference of the tube but this time on the outside surface
 
  • #10
haruspex
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And this means that you painted the circle on the inside circumference of the tube but this time on the outside surface
No, think of a bicycle tube. The short way around is the way you would pick it up with one hand, fingers wrapped around it.
 
  • #11
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No, think of a bicycle tube. The short way around is the way you would pick it up with one hand, fingers wrapped around it.
Oh I get it now! (hopefully)
 
  • #12
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Okay so here is my second attempt

I use the same process as earlier that is I increase the size of the hole such that the tube is now a strip, and the hole is increased such that blue circle remains intact but the red circle is cut into a line, now if we flip this strip and stitch it back together we can see that red circle which earlier was wrapped around the blue circle is now wrapped outside the blue circle.

And the blue circle which was earlier on the inside surface after flipping comes on the outside surface and the red circle which was out is now on the inside surface thus they are untangled! i.e., they are no longer linked
 
  • #13
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Okay so here is my second attempt

I use the same process as earlier that is I increase the size of the hole such that the tube is now a strip, and the hole is increased such that blue circle remains intact but the red circle is cut into a line, now if we flip this strip and stitch it back together we can see that red circle which earlier was wrapped around the blue circle is now wrapped outside the blue circle.

And the blue circle which was earlier on the inside surface after flipping comes on the outside surface and the red circle which was out is now on the inside surface thus they are untangled! i.e., they are no longer linked
So it should look somewhat like this,
001.png
 
  • #14
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they are no longer linked
Which of course is impossible. How did they pass through each other?
I think I got this from a Martin Gardner book many years ago. Took me some mental wrestling to evert the torus in my head and finally see what happens.
 
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  • #15
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Which of course is impossible. How did they pass through each other?
I think I got this from a Martin Gardner book many years ago. Took me some mental wrestling to evert the torus in my head and finally see what happens.
So that means that we cannot increase the length of the hole such that the tube is converted to a strip?

Because by this method it does completely makes sense why the red circle passed through the blue one (because we literally cut and stitched the red circle in the process)

But yes if that is not allowed then it is way too hard for me to imagine how a tube is inverted inside out through a small hole, infact I cannot believe how can people do this in their head! 🤯
 
  • #16
haruspex
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So that means that we cannot increase the length of the hole such that the tube is converted to a strip?
Yes and no.
Let's take the bicycle tube model and say the two radii are r<R.
If you cut it right around the direction of rotation of the wheel you get a short fat cylinder, length ## 2 \pi r## and radius R. If you cut it instead the other way you get a long narrow tube, length ## 2 \pi R## radius r.
If you do both you get a rectangle ## 2 \pi R## by ## 2 \pi r##
Sticking it back together, you have a choice of which to do first. Does the order matter?
 
  • #17
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Yes and no.
Let's take the bicycle tube model and say the two radii are r<R.
If you cut it right around the direction of rotation of the wheel you get a short fat cylinder, length ## 2 \pi r## and radius R. If you cut it instead the other way you get a long narrow tube, length ## 2 \pi R## radius r.
If you do both you get a rectangle ## 2 \pi R## by ## 2 \pi r##
Sticking it back together, you have a choice of which to do first. Does the order matter?
Wait, can we cut it such that we get a rectangle? doesn't that break any rules? is a toroid with single hole isomorphic to rectangle?

I agree that with can cut it in both ways that you mentioned, i.e., cutting it such that we get a short fat cylinder and cutting it such that we get a long narrow tube (as it makes sense that a short fat cylinder when stretched along its length gives a long narrow pipe) but how can we do both of these together to get a rectangle?

Consider a short fat cylinder, which can be thought of as a rectangular strip of paper whose shorter sides are joined together, now if we want to get a rectangle from this strip, we will have to cut this strip and cutting is not equivalent to increasing a size of hole. As I understand increasing or decreasing the size of hole is allowed but cutting or stitching two surfaces aren't allowed.

I only increased the size of the hole in my attempt, and then flipped the strip (short fat cylinder) inside out and decreased the size of hole in reverse such that we got back the original shape with which we started with.
 
  • #18
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Which of course is impossible. How did they pass through each other?
I think I got this from a Martin Gardner book many years ago. Took me some mental wrestling to evert the torus in my head and finally see what happens.
I was thinking about this a lot and even though I don't know how it will happen but I think that it should happen, i.e., the red and blue circles shouldn't be linked after inverting the tube inside out.

This is because, the red circle which was initially outside the tube, after flipping has to be inside that tube, and similarly blue circle which was initially inside, has to be on the outside after flipping. Now it doesn't matter where the blue circle is on the outside, as long as the red circle is inside the tube, both of these circles cannot be linked!

And this makes me think that those two circles weren't linked in the first place itself! And that must have something to do with the fact that our tube had a hole in it. If there was no hole then it makes sense that both circles are linked and cannot pass through each other.

Now what I think is happening is that, as we are pulling the inside surface of the tube out it passes through the inside of the red circle, now again I am not clear about how is that happening, I was finding toroids in my home and the only similar thing I found is my bicycle tube but to access that I'll have to rip off my tyres! Why don't we produce more things that are toroids? The other alternative is that I have to cut my shirt's sleeves and then stitch it end to end but I don't think even this is a good idea.
 
  • #19
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Given that we are allowing cutting the strip and laying it out as a rectangle we can dispense with the hole; except that when we reassemble it as a torus it must be equivalent to having pulled it through a hole.
As I mentioned, there are two ways we can stitch the rectangle back to be a torus. Obviously one way will get exactly back to where we started. What happens if we do it the other way?
 
  • #20
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it doesn't matter where the blue circle is on the outside, as long as the red circle is inside the tube, both of these circles cannot be linked!
and yet, they must be.
 
  • #21
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Given that we are allowing cutting the strip and laying it out as a rectangle we can dispense with the hole; except that when we reassemble it as a torus it must be equivalent to having pulled it through a hole.
As I mentioned, there are two ways we can stitch the rectangle back to be a torus. Obviously one way will get exactly back to where we started. What happens if we do it the other way?
But why is cutting the strip allowed? wouldn't that change the shape we are dealing with?
 
  • #22
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Obviously one way will get exactly back to where we started. What happens if we do it the other way?
As I said before that now it makes sense why the circles won't be linked after we stitch it back the other way because we are literally cutting both the circle such that they changed from a circle to a line, but the cutting isn't allowed
 
  • #23
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001.png

After stitching it back I cannot believe how the red circle will still be linked with the blue one?
 
  • #24
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View attachment 287813
After stitching it back I cannot believe how the red circle will still be linked with the blue one?
Try it with a sheet of paper. Draw a line across the width on one side and along the length on the other.
You can turn it into a cylinder by bringing together two opposite edges and gluing them together, then into a torus by doing the same with the opposite ends of the cylinder.
You can do it either so that the first fold leaves the line inside as a circle and the outside line straight, or so that the outside line is a circle around the cylinder and the inside line is straight.
Check linkage of the lines in the finished torus in each case.
 
  • #25
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Try it with a sheet of paper. Draw a line across the width on one side and along the length on the other.
You can turn it into a cylinder by bringing together two opposite edges and gluing them together, then into a torus by doing the same with the opposite ends of the cylinder.
You can do it either so that the first fold leaves the line inside as a circle and the outside line straight, or so that the outside line is a circle around the cylinder and the inside line is straight.
Check linkage of the lines in the finished torus in each case.
I think that the linkage should be different in both cases but I should check it myself
 

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