# Thread: Cones - Center of Gravity, Center of Mass, etc.

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## Cones - Center of Gravity, Center of Mass, etc.

In wikipedia and elsewhere it is stated that: The center of mass of a conic solid of uniform density lies one-quarter of the way from the center of the base to the vertex, on the straight line joining the two.

But when I set up the problem as shown in this image, with "x" being the location of the center of mass then I get:

x = h*(1 - 1/(2)1/3) = 0.2063*h

I essentially set the volume (it is uniform density) of the small cone, V2 = V3, the mass of the frustum, then solve for "x"

Known: V3 = V1 - V2

I used centroids and areas to create a solid/volume of revolution: http://en.wikipedia.org/wiki/Pappus's_centroid_theorem

What am I missing or is wikipedia wrong?

2. I'm way out of practice with this stuff, but for fun tried a different approach. I figured, apply a scale factor (f) to the height and radius of a cone, to make the volume half what it was.

V1 = (0.333...).Pi.r2.h

V2 = (0.333...).Pi.(f.r)2.(f.h) = 0.5 V1

f3.(0.333...).Pi.r2.h = 0.5 (0.333...).Pi.r2.h

f3 = 0.5

f = 0.793701

Which would place x, 0.206 from the base of the cone. That's "close" to the "one quarter" on that Wiki page, but not very close. (But does match your number).

(Googling, it seems the answer involves integration...)
Last edited by pzkpfw; 2012-Mar-15 at 12:07 AM. Reason: Add last brackets, then more

3. It's been a while since I solved for center of gravity in for something other than a bar or a sphere, but I think that the center of gravity is not just a question of finding the dividing line about which there are two equal masses, but that the length of the lever arm is also a factor. Hence the volume v0 at the tip has a greater influence than the same volume in the thin wafer at the bottom which is closer to the CG.

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Isn't the center of gravity of an isoceles triangle at the same point
as the center of mass of a cone with the same cross-section?
If so, you can cut a triangle out of cardboard and balance it on
a rounded point to find the center by trial and error. That should
be more than accurate enough to distinguish between 0.206 h
and 0.25 h.

-- Jeff, in Minneapolis

5. Originally Posted by Jeff Root
Isn't the center of gravity of an isoceles triangle at the same point
as the center of mass of a cone with the same cross-section?
No the Third dimension makes this comparison wrong.

6. Here's a site with it all worked out: http://math.feld.cvut.cz/mt/txtd/5/txe3dc5i.htm

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At least my first guess for the figure that I thought would apply to
the cone appears to apply to the triangle: I get 1/3 h with a paper
cutout. That would apply to prisms, too. Funny it doesn't work
for cones.

-- Jeff, in Min Minneapolis

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Aha! The triangle has 1/2 the area of the enclosing rectangle,
but the cone has only 1/3 the volume of the enclosing cylinder!

-- Jeff, in Minneapolis

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## Center of Gravity vs Center of Mass

Originally Posted by antoniseb
It's been a while since I solved for center of gravity in for something other than a bar or a sphere, but I think that the center of gravity is not just a question of finding the dividing line about which there are two equal masses, but that the length of the lever arm is also a factor. Hence the volume v0 at the tip has a greater influence than the same volume in the thin wafer at the bottom which is closer to the CG.
For a cone I figured that the center of mass was different than the center of gravity due to the moment arm being greater at the point than at the base so I did my calculation to exactly divide the mass into two parts - an equal number of atoms on each side of that point on the x-axis (I already surmised that the CoM was on the x-axis).

When I use the centroid of a triangle and revolve it then it seems that the center of gravity is at h/3

I can derive the formula for the volume of the cone using the second theorem of Pappus' centroid theorem and I can determine that the cone has only 1/3 the volume of the enclosing cylinder ... I don't see why this method does not give the center of gravity for the cone.

10. Originally Posted by xylophobe
For a cone I figured that the center of mass was different than the center of gravity
In a constant gravity field (approximately true for small objects near the surface of the earth), the center of gravity and the center of mass are the same, by definition. The center of mass is the point where, to first degree, you can treat the entire mass as a point mass.

Your approach is problematic anyway. Consider three points, A, B, and C, having the same mass. The center of mass, as you've taken it, would be somewhere on the line including the one in the middle, B. However, if the points form a triangle, you can move your point-of-view so that the two points A and C align so your center of mass must be along the line between them--and you can do this for any of the pairs of points.

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I have a test tomorrow in calculus regarding iterated integration. Once i've written the test, ill come back and show that method for solving this problem. I suspect you have the right answer with 0.206 from the base, height=1. Note that a multiple integral has density as a function of xyz, so we can solve for non uniform density cones as well.

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Originally Posted by swampyankee
Here's a site with it all worked out: http://math.feld.cvut.cz/mt/txtd/5/txe3dc5i.htm
Correct .
The formula in this link simply says : X cgravity = integral ( x * dm ) / Total mass , where dm is an infetisimal mass increase over the section ( area) with a thickness dx .
dm here can be written as rho * Area * dx , where Area on its turn again is a function of x .
This formula applies for every centroid .
As different centroids have other functions of "area" as function of x the height of the center of mass is different for each centroid . In case of a cone we get indeed 0.75 of the height , if we measure from the point .

13. Reason
markup problems

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## Engineering Analysis

Okay, I set up the problem as shown in the image, where x=% of h (% in decimal format) and I get the following polynomial:

2x3-7x2+9x-2=0

that results in x=0.2773

I also did an engineering analysis (using real values instead of variables), that gives the same value for x=0.2773 , by finding the composite centroid of the frustum area and the centroid of the Small Cone area. Further, as a check, I used these centroids to create volumes that sum to the same Frustum volume as achieved using the regular cone volume formula:

Frustum Volume = Large Cone Volume - Small Cone Volume

Cone Volume Formula = pi()r2h/3

To get the above polynomial I multiplied the volume of the Small Cone times the moment arm of the Small Cone the result of which should equal the volume of the Frustum times the moment arm of the Frustum. I then solved for "x".

M2=M3

where
:
M2=V2*C2
M3=V3*C3
V2=pi()r2h(1-x)3
V3=pi()r2h(1-(1-x)3)
C2=h(1-x)/3 --- (location of the centroid of the Small Cone)
C3=xh*(3-x)/3(2-x) --- (location of the centroid of the Frustum)

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I thnk the flaw in the approach above lies in the fact of assuming that the center of gravity of a centroid lies at 1/3 of the hieght , just as it does in a flat figure :
The assumption : CG = h(1-x)/3 for instance applies for a flat triangle , not essentially for a centroid .

16. You're trying to find the centroid of the large cone, and assume it is at xh. That's fine, but then the centroid of the small cone should be at a distance from there of h(1-x)x, whereas you label it h(1-x)/3.

How you're going to determine the formula to use for the frustrum is beyond me.

However, the link provided by swampyankee has the steps all in one line, and it looks like it might be correct.

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Originally Posted by frankuitaalst
I thnk the flaw in the approach above lies in the fact of assuming that the center of gravity of a centroid lies at 1/3 of the hieght , just as it does in a flat figure :
The assumption : CG = h(1-x)/3 for instance applies for a flat triangle , not essentially for a centroid .
I like to be able to solve things in more than one way so that I can use the second method to check the result of the first method which is why I am trying to solve the above differently than what is given in this here post.

As far as your statement it seems to be at odds with this statement: If a physical object has uniform density, then its center of mass is the same as the centroid of its shape.

So I'm thinking the CoG should be on the axis (x-axis) of the cone where the perpendicular line from the area-centroid meets the x-axis.

As far as formula given given in this here website why is the upper integral multiplied by "x" (I understand that the lower (denominator) integral works out to be the volume)?

18. Originally Posted by xylophobe
I like to be able to solve things in more than one way so that I can use the second method to check the result of the first method which is why I am trying to solve the above differently than what is given in this here post.
He meant, that the centroid of a cone is different than the centroid of a triangle.
So I'm thinking the CoG should be on the axis (x-axis) of the cone where the perpendicular line from the area-centroid meets the x-axis.

As far as formula given given in this here website why is the upper integral multiplied by "x" (I understand that the lower (denominator) integral works out to be the volume)?
That's the moment arm.

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Originally Posted by xylophobe
I like to be able to solve things in more than one way so that I can use the second method to check the result of the first method which is why I am trying to solve the above differently than what is given in this here post.

As far as your statement it seems to be at odds with this statement: If a physical object has uniform density, then its center of mass is the same as the centroid of its shape.

So I'm thinking the CoG should be on the axis (x-axis) of the cone where the perpendicular line from the area-centroid meets the x-axis.

As far as formula given given in this here website why is the upper integral multiplied by "x" (I understand that the lower (denominator) integral works out to be the volume)?
The "x" is there to account for the moment arm for the same reason as you do in your formulae .
Further I agree with Grapes when he states that when you assume the CG to be at x for the whole then this also applies for the upper part of the cone , being h(1-x) x instead of h(1-x)/3

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## Circular Reasoning?

Originally Posted by grapes
..., but then the centroid of the small cone should be at a distance from there of h(1-x)x, whereas you label it h(1-x)/3. ...
I know what you are getting at because it seems to be a circular logic - I guess I could substitue h(1-x)/4 to see if it all works out to be x=4.0

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## A Simple Graphical Representation

Ok, my math skills are limited so I shall resort to a graphical reproof of the cone CoG=1/4h

In the picture below I show a progression from a cylinder, to half a cylinder, to the cone in question.

We can all agree that the CoG for the cylinder is in the middle.

If I cut the cylinder in half then the CoG for the half moves to 1/4h.

So if I cut off the outer portion shaded red and add the inner portion shaded green then, by area, I have added the same amount but since these areas are rotated around the centerline then I am subtracting much more volume with the red than is added by the green so therefore the CoG for the cone can not remain at the 1/4h position as shown when the cylinder is cut in half.

The CoG for a cone is at the 1/3h position.

22. Originally Posted by xylophobe
The CoG for a cone is at the 1/3h position.
But, it's actually at the 1/4h position??

23. I just performed the integration, using the formula given here,
http://en.wikipedia.org/wiki/Center_of_mass#Definition
and got exactly 1/4 of the height.

I am completely confident that Wiki has it right, in accordance with the definition of center of mass. Can anyone who got a different result explain, in appropriate mathematical detail, why they think otherwise?

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## Differences in Volume(Mass)

According to this picture and using Pappus's 2nd centroid theorem the volume of the red area is 5x the volume of the green area.

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xylophobe,

Your graphical analysis starts out by saying that you are looking
at a cylinder cut in half, and ends by saying that you locate the
centroid of a cone, but the intervening steps appear to be looking
only at plane figures, not solids. As was pointed out to me early
in the thread, the centroid of a cone is different from the centroid
of the triangle which is the cross-section of the cone. That became
apparent to me when I saw that the triangle has 1/2 the area of the
enclosing rectangle, but the cone has only 1/3 the volume of the
enclosing cylinder. The difference shifts the location of the cone's
centroid proportionally toward the base.

I just don't see that your diagrams show what you describe them
as showing.

Correction:

For some reason I said "and ends by saying that you locate the
centroid of a cone" But actually xylophobe's post #20 ends by
saying that he locates the "CoG" (center of gravity) of the cone.
Not that there's any difference between the centroid and the CoG --
there isn't -- but I said what I was thinking instead of what I read.

-- Jeff, in Minneapolis
Last edited by Jeff Root; 2012-Mar-17 at 07:38 PM. Reason: see above

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Originally Posted by Jeff Root
xylophobe,

..., and ends by saying that you locate the
centroid of a cone, ...

-- Jeff, in Minneapolis
Hi Jeff,

I did not say that I located the center of gravity of the cone ... but I do know where the centroid of the area is which is needed in order to use Pappus's 2nd theorem to find the volumes.

That is why I show the r/6 and the 5r/6 dimensions to the centroids because these y-axis locations are the "r" in the Pappus calculation.

Because 5/6 is 5 times larger than 1/6 then I know that the mass of the red area is 5 times more than the mass of the green area (after revolving around the x-axis).

27. I stand behind my conclusion in post 22, and nothing you have said since then shakes my confidence.

Originally Posted by xylophobe
Ok, my math skills are limited so I shall resort to a graphical reproof of the cone CoG=1/4h

In the picture below I show a progression from a cylinder, to half a cylinder, to the cone in question.

We can all agree that the CoG for the cylinder is in the middle.

If I cut the cylinder in half then the CoG for the half moves to 1/4h.

So if I cut off the outer portion shaded red and add the inner portion shaded green then, by area, I have added the same amount but since these areas are rotated around the centerline then I am subtracting much more volume with the red than is added by the green so therefore the CoG for the cone can not remain at the 1/4h position as shown when the cylinder is cut in half.

The CoG for a cone is at the 1/3h position.

My bold. From that remark, along with everything else you have posted, I infer that you do not know how to do the calculus that yields an exact analytic solution. You appear to be attempting rough-and-dirty numerical integrations on much too coarse a scale. For problems that do not have analytic solutions, and there are such things, numerical integration is a good plan B, but the increments have to be very small in proportion to the whole body to get reasonably good approximations.

There is no shame in having trouble with calculus. I had some rough moments before progressing from a C after one quarter in my senior year in high school to an A in college.

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Cone CG.pdf

One way to calculate the CM "graphically" or "numerically" is to divide the cone in several slices .
A division in two sections is shown in annex . We get a ratio of 0.7H .
As we divide the cone section in more slices the ratio increases ....
If we would do the same for 100 ,1000,... sections we get close to the 0.75 H .
In fact this is what numerical integration or calculus does...

29. what a funny thread! the integral of x^3 divided by the integral of x^2 as given earlier in the posts is clearly the way to calculate the moment of each disc of the cone divided by the volume of each disc; so the 3/4 result is obviously right.

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O/T but there was an interesting shape called a superegg that looked interesting...

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Is a supergg based on the superellipse? The superellipse
was given its name while I was in junior high school.

I'm surprised and rather shocked that finding the center of
a cone apparently requres calculus. Finding the center of
a cylinder is as easy as falling off a cylindrical log. Why
is finding the center of a cone so much more involved?

-- Jeff, in Minneapolis

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