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

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## Experiments

Has there ever been experiments conducted to see if the CoG of an open cone is gravitationally special?

For instance using a set-up similar to the Cavendish experiment but using hollow cones as attractors instead of spheres?

It would seem that the test spheres would be pulled into the cone(s) until the CoG is reached and then the test spheres would be repelled outwardly from the cone(s) after passing the CoG.

2. Originally Posted by xylophobe
Has there ever been experiments conducted to see if the CoG of an open cone is gravitationally special?

For instance using a set-up similar to the Cavendish experiment but using hollow cones as attractors instead of spheres?

It would seem that the test spheres would be pulled into the cone(s) until the CoG is reached and then the test spheres would be repelled outwardly from the cone(s) after passing the CoG.
Science and math have shown over and over again that "it would seem" is not always reliable. It would take some serious calculus to determine the orientation and magnitude of the gravity vector at various points inside a hollow cone. The forces are just too weak for any practical test here. It is hard enough with simple solid lead balls.

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## Mass Per Unit Area

If two gravitating objects are to exert the same force upon a test mass then the mass per unit area (or mass per surface element) should be equal, correct?

Considering a gravitating mass to be a point mass then a spherical shell around that point with a radius equal to the distance between the point mass and the test mass will have a definite mass per unit area.

If there are two gravitating masses

m1=1.000

and

m2=4.000

and

r1=1.0

while

r2=2.0

then the

spherical area1 = 4*pi()*12

while

spherical area2 = 4*pi()*22

and the

mass per unit area1 = 1.000 / (4*pi()*12) = mass per unit area2 = 4.000 / (4*pi()*22)

4. Originally Posted by xylophobe
If two gravitating objects are to exert the same force upon a test mass then the mass per unit area (or mass per surface element) should be equal, correct?
If I'm understanding what you're saying (and it's certainly possible that I'm misunderstanding you), then I believe that you're correct. It's pretty much a direct consequence of the fact that the gravitational force on a test mass is proportional to the mass of the source and inversely proportional to the square of the distance.

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## CoG for Tilted Cone

Will the cone balance on the CoG if it is tilted in the uniform gravity field (shown by blue arrows) as shown by the green cone on the right in this image?

We already determined that the cone will balance on the CoG if it is positioned as shown by the red cone on the left.

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Yes, of course.

When I found that the COG is 1/4 of the way up, the solution is actually a point, and not a disk in the cone. Imagine the cone is standing on its tip. For it to be balanced, it must be perfectly vertical. Then we can see from symmetry that the same distribution is on the left and right. Likewise, front to back. As a matter of fact, the cone could be spinning like a top and nothing changes. This rotational symmetry means that the COG has to be on the line that passes right down the center.

With the integration of the cone on its side complete, we have that the COG is somewhere on a line that is parallel to the base, at 1/4 of the height from the base, and passes through the center of any circular cross section.

The two lines intersect at a single point, the COG. Imagine that you could take a very thin circular slice out of the cone, at 1/4 of the height from the base. The center of gravity is the point at the origin of that circle.

In physics, you learn that as long as the force is acting through the cog, then only its linear momentum is affected.

In your diagram, the third blue arrow from the left(for the green 'cone') hits the point marked COG. If you extended this line straight down through the other side of the cone, that is the place where a needle must be held vertically to balance the cone.

edit: with a ruler to the screen, I see that you carefully put the COG at 1/4(h). Very nice!

7. Yes, it will balance at the COM

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Below is a snapshot of my MS Excel spreadsheet that explores cone volumes, masses, CoGs, and the gravitational effect of cone segments.

It is interesting to note that the mass of a cone increases by the cube of the factor of the change in cone height (h); whereas, the CoG location ("r" squared) increases by the square of the factor of the change in the cube height.

For instance the difference between V1 and V2 is an increase in height by a factor of 2 while the mass increases by a factor of 8 but the r-squared of the CoG changes by a factor of 4.

Then for the frustums, which are labeled: "segments"; the masses do not increase by the same rate as the r-squared location of the CoG.

This indicates to me that the "segments" do not possess the same gravitational influence upon the vertex point "P" so if we double the height of the cone then the gravitational influence of the cone does not double.

I also looked at the mass per unit area for the vertex-side of the frustums.

9. Originally Posted by xylophobe
It is interesting to note that the mass of a cone increases by the cube of the factor of the change in cone height (h); whereas, the CoG location ("r" squared) increases by the square of the factor of the change in the cube height.
Not sure where you're going with this.

Volume increases by the cube of one dimension, if all other increase is proportional--because there are three dimensions. That's true for any shape, not just cones.

The CoG location doesn't increase by the square, it is directly proportional. But it looks like you may have confused it with the effect of gravity, which does decrease as the square of distance. Which means, if you double the size of the object, its mass will increase 2x2x2, 8 times, but since the distance doubles also, the gravitational effect will only be 2x2x2/2x2, or twice.
This indicates to me that the "segments" do not possess the same gravitational influence upon the vertex point "P" so if we double the height of the cone then the gravitational influence of the cone does not double.
Maybe we're talking about two different things, but the same holds true for a cone--the gravitational strength at a particular point will double.

ETA: I retract my last statement, after looking over your diagram. The force of gravity is not necessarily towards the center of mass, of course.

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Originally Posted by grapes
...The CoG location doesn't increase by the square, it is directly proportional. But it looks like you may have confused it with the effect of gravity, which does decrease as the square of distance. ...
Yeah, it was kinda confusing what I was trying to point out - the effect of gravity decreases as the square of the distance, with the distance being from the vertex of the cone (or where the vertex of the frustrum would be) and the CoG of either the cone or the frustum.

I guess the point I am making is that the segments (frustums) do not have the same gravitational effect upon the point "P" which is at the vertex (unlabeled in my previous picture).

Below is a picture that makes it more clear, I think. From my calculations it would be segment 4 (frustum 4) that would exert the greatest gravitational force upon point "P" and in fact none of the segments on either side has the same gravitational effect as any of the others.

Part of this is due to the CoGs for the frustums not following a strict increment; whereas, for the cones it progresses in a manner directly proportional to the increase in the cone size.

11. The gravitational effect at the top point of the cone cannot be calculated by G times the mass of the cone, divided by the square of the distance to the COM/COG. It should be

if I've set it up right. Then you can divide by the mass of the cone times G

and take the inverse square root of that to get the effective distance
Last edited by grapes; 2012-Apr-10 at 08:33 AM. Reason: Added G to formulas

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I'm studying iterated integrals right now as I have a final on wednesday.

That looks like two things at once. It looks like a surface area for a single integral since I see 2*pi*y in it. But that can't be right. It's a double integral in dydx, so I can't see how it's the gravitational potential for a 3d object.

I need formula for the region of integration to be able to visualize it. It's actually a straightforward evaluation of the integral. The x and y upstairs should disappear. Looks like the original function is 8*pi*sqrt(x^2 + y^2) derived twice.

13. Originally Posted by ShinAce
I'm studying iterated integrals right now as I have a final on wednesday.

That looks like two things at once. It looks like a surface area for a single integral since I see 2*pi*y in it. But that can't be right. It's a double integral in dydx, so I can't see how it's the gravitational potential for a 3d object.
It's not potential, it's acceleration (I should have included G in there). The 2*pi*y is circumference of a ring whose cross section is dxdy--opposite sides of the ring cancel, except for the component of the acceleration parallel to the cone axis.
I need formula for the region of integration to be able to visualize it. It's actually a straightforward evaluation of the integral. The x and y upstairs should disappear. Looks like the original function is 8*pi*sqrt(x^2 + y^2) derived twice.
Wolframalpha says it ( Integrate[ G delta * 2 * Pi * x * y /(x^2+y^2)^(3/2), {x,0,h},{y,0,r*x/h}] ) evaluates to

Dividing that by

leaves

So the effective distance depends upon the ratio between h and r, and not just h. If r = h/2, which would be a cone as wide as it is tall, and taking the inverse square root, we get .628h.

So, if I've done nothing wrong (what are the odds of that?), such a cone acts on its point as if its entire mass were concentrated at .628h, and not at the COM at .75h. I must've done something wrong.

ETA: The integral is easy, by hand. One interesting result is that when r = 1.0096h, the mass of the cone acts as if it were concentrated at .75h.

EETA: of course I made a mistake--ignore that last result! It would seem there is no ratio that would allow that.
Last edited by grapes; 2012-Apr-10 at 02:50 PM. Reason: Editted To Add

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Originally Posted by grapes
The gravitational effect at the top point of the cone cannot be calculated by G times the mass of the cone, divided by the square of the distance to the COM/COG. It should be

if I've set it up right. Then you can divide by the mass of the cone times G

and take the inverse square root of that to get the effective distance
Shouldn't the nominator of the first integral not read as ...y²dydx instead of ...yxdydx ?
EDIT : Ignore my remark . The integral is correct ...
Last edited by frankuitaalst; 2012-Apr-10 at 06:39 PM. Reason: correction

15. Thanks for checking!

If r=xh, then the effective distance seems to be sqrt( (1+x^2+sqrt(1+x^2))/6), which agrees with my answer for x=1/2, but also allows an effective distance of .75 when x=.985422

I dunno what I did wrong, yet. But the effective distance grows without bound as the ratio increases, essentially as the cone becomes more like a large disk next to the point. I guess that makes sense.

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I got the same result Grapes , after calculating the integral . So nothing is wrong with your approach or calculation :-).
The only thing I can think of is the assumption that the force exerted on a mass this close does not behave as 1/r² .
Remember : only at "big distances" a body may be considered to have it's mass centered in the COG .
I think this was proven above :-)

17. Originally Posted by grapes
But the effective distance grows without bound as the ratio increases, essentially as the cone becomes more like a large disk next to the point. I guess that makes sense.
Actually, I think that does make sense. The gravitational force of an infinite flat plane is constant, with no dependence on distance. As the cone flattens out, it approaches an infinite plane as a limit. (I think you have to let the mass go infinite, too, or else it approaches an infinite plane of zero density, which has no gravitational effect at all).

18. Originally Posted by frankuitaalst
I got the same result Grapes
I appreciate the cross check!
Remember : only at "big distances" a body may be considered to have it's mass centered in the COG .
I think this was proven above :-)
Proven, and I've mentioned several times this thread.
Originally Posted by Grey
Actually, I think that does make sense. The gravitational force of an infinite flat plane is constant, with no dependence on distance. As the cone flattens out, it approaches an infinite plane as a limit. (I think you have to let the mass go infinite, too, or else it approaches an infinite plane of zero density, which has no gravitational effect at all).
Yeah, that's what I was thinking. We let stay constant, instead of the total mass.

A nice thing about doing the integration "by hand" is that I noticed that the contribution of each disk is the same--just another example of the "area principle": if two similar (shape, density) objects subtend the same area in the sky, their gravitational effect is the same. The x integral is of just a constant--that's where the factor of h comes from.

19. Originally Posted by grapes
A nice thing about doing the integration "by hand" is that I noticed that the contribution of each disk is the same--just another example of the "area principle": if two similar (shape, density) objects subtend the same area in the sky, their gravitational effect is the same.
That's very cool. That only works for flat objects, though, right? So that the total mass is proportional to the area subtended. For nonflat objects, the mass increases faster than the area, so it's no longer exactly balanced by the distance squared factor. For example, the Sun and the Moon subtend almost precisely the same angle in the sky, and the Moon actually has a higher mean density, but the gravitational force from the Sun is much higher.

20. surely the gravitational effect inside a mass is only inversely proportional to the distance rather than the square. That's what I was taught for calculating forces inside the Earth.

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## Clarify?

Originally Posted by grapes
...

A nice thing about doing the integration "by hand" is that I noticed that the contribution of each disk is the same--just another example of the "area principle": if two similar (shape, density) objects subtend the same area in the sky, their gravitational effect is the same. ...
If I am reading this correctly then because, from point "P", the cone segments subtend the same area of the sky then they should have the same gravitational effect upon point "P" even though the masses of the cone segments are differing by factors not directly related to the distance-squared? Or does this area principle only apply to spherical objects?

22. As Grey points out, they'd have to have the same depth.

23. Originally Posted by profloater
surely the gravitational effect inside a mass is only inversely proportional to the distance rather than the square. That's what I was taught for calculating forces inside the Earth.
That would be inside a sphere of constant density, which the earth is not--the downward force of gravity stays pretty constant, almost to the core-mantle boundary, about halfway to the center of the earth.

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Originally Posted by xylophobe
If I am reading this correctly then because, from point "P", the cone segments subtend the same area of the sky then they should have the same gravitational effect upon point "P" even though the masses of the cone segments are differing by factors not directly related to the distance-squared? Or does this area principle only apply to spherical objects?
Imagine staring down at a small pizza. Under it is a medium and under that, a large. If they're lined up eith matching edges, like an eclipse, then the force of gravity from any one pizza on you is a constant. Same thickness, different radius.

Radius goes linearly with distance, but area would be proportional to distance squared. Area is mass in this case so mass goes up by d^2 while gravity goes down by d^2. Unity.

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Originally Posted by ShinAce
Imagine staring down at a small pizza. Under it is a medium and under that, a large. If they're lined up eith matching edges, like an eclipse, then the force of gravity from any one pizza on you is a constant. Same thickness, different radius.

Radius goes linearly with distance, but area would be proportional to distance squared. Area is mass in this case so mass goes up by d^2 while gravity goes down by d^2. Unity.
If I understand your description correctly you are describing cylinders with equal thicknesses (or "depths" as grapes put it) but the radius varies directly proportional with distance which then means that the volume/mass varies in a manner directly proportional to the diameter of the pizza-cylinders.

There is a difference, I think, between the frustums and the cylinders because the frustums have a triangular cross-section on the periphery of the cylinder which changes the amount of mass per cone segment (or frustum). And as the diameters increase the mass-contribution of the triangular section increases more due to the larger radius of rotation. I tried to show that by showing the mass per unit area --- the area increases proportionally to the square of the distance but the mass increases even more than that.

As for the sun and moon analogy descibed by Grey the gravitational flux of both bodies goes through essentially the same-size subtended area but the mass per unit area is much greater for the sun than the moon (even though the sun is much farther away) so the sun is gravitationally stronger than the moon, from our Earth-based perspective.

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Originally Posted by grapes
The equipotential surfaces of the gravity field of a cone are not even a cone at the very surface of the cone! It's close to spherical even before the surfaces stop intersecting the cone.

As an example, on the earth's surface, ocean water tends to conform to an equipotential surface (the geoid), which has little islands sticking out of it (and large continents).
Indeed , it is . Equipotential lines generally do not follow the surface . Exception is the well known spherical surface .

Maybe somewhat out of topic relative to the OP :
In annex a visualisation of the equipotential lines of a simple rod freely floating in space .
The red line represents the rod .
The green lines represent the gravitational potential of the rod . The animation starts near the rod . Each frame increases the potential
In the rod the gravitational potential is -infinity .
The surface of the equipotential field may represent the shape of a water volume on the rod ( assuming we neglect the self gravitation of the water ) .
It's nice to see how a small volume of water stays centered around the center of the rod .

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## Picture

Originally Posted by xylophobe
...

There is a difference, I think, between the frustums and the cylinders because the frustums have a triangular cross-section on the periphery of the cylinder which changes the amount of mass per cone segment (or frustum). And as the diameters increase the mass-contribution of the triangular section increases more due to the larger radius of rotation. I tried to show that by showing the mass per unit area --- the area increases proportionally to the square of the distance but the mass increases even more than that.

...
I hate to quote myself but ...... to illustrate the above here is a picture that shows the radius of revolution of the triangular portions, per Pappus's centroid theorem, which shows that the centroids of the triangles do not line up (red lines) with respect to point "P" which means the mass of the cone segments is not directly proportional to the distance from point "P".

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Originally Posted by xylophobe
If I understand your description correctly you are describing cylinders with equal thicknesses (or "depths" as grapes put it) but the radius varies directly proportional with distance which then means that the volume/mass varies in a manner directly proportional to the diameter of the pizza-cylinders..
No. Mass varies as distance^2.

Edit: see posts 108 and 114

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## Big Cone minus Small Cone

Originally Posted by ShinAce
No. Mass varies as distance^2.

Edit: see posts 108 and 114
I did the math and the mass of the frustums does not vary as distance2

Refer back to the very first post in this thread and it outlines a simple way to calculate the mass of a frustum (V3) = large cone (V1) minus small cone (V2)

30. Originally Posted by xylophobe
I did the math and the mass of the frustums does not vary as distance2
You must be talking about two different things
Originally Posted by xylophobe
If I understand your description correctly you are describing cylinders with equal thicknesses (or "depths" as grapes put it) but the radius varies directly proportional with distance which then means that the volume/mass varies in a manner directly proportional to the diameter of the pizza-cylinders.
If, as you say, the radius varies directly proportional with distance, then since mass is proportional to the square of the radius, mass must be proportional to the square of distance. No? Why would mass be directly proportional to diameter?

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