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

1. Order of Kilopi
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As is too often the case, I'm skipping the obviously involved
posts and going for the seemingly easy one...

Originally Posted by swampyankee
Originally Posted by Jeff Root
Could you describe an actual case in which tidal differences
in gravity result in a measurable difference in the location of
the center of gravity of something?
Gravitationally stabilized satellites.
So, can you describe it?

Is a satellite's center of gravity different from its center
of mass because it is gravitationally stabilized? Or is the
center of gravity of the Earth changed because satellites
are gravitationally stabilized? Or what?

-- Jeff, in Minneapolis

2. Originally Posted by Jeff Root
As is too often the case, I'm skipping the obviously involved
posts and going for the seemingly easy one...

So, can you describe it?

Is a satellite's center of gravity different from its center
of mass because it is gravitationally stabilized?
No, that would have the causality backward. The satellite is stabilized because the center of mass and the center of gravity are separated, and there is a position at which this creates a stable equilibrium.
Or is the
center of gravity of the Earth changed because satellites
are gravitationally stabilized? Or what?

-- Jeff, in Minneapolis
The center of gravity of the Earth is unaffected by the presence of satellites, whether they are stabilized or not.

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I did the integral and got 1/4 of height from the base. I setup the integral so that the line x=0 is at the center of gravity. Then torque from negative x balances torque from positive x values. So i integrated from -a to 0 on lhs and then 0 to b on right hand side. Since i took disk cross sections, the force from torque is rF. But r=x and F=mg where m=area*dx. I dropped the linear density as it is constant. We can drop acceleration due to gravity as its constant and appears on both sides. The integral is trivial but the solution gives a polynomial in a and b of degree 4.

edit: As is always the case, make a nice post and it will be destroyed. Since I lost everything, we all get shafted.

Short answer is integral is actually evaluated from (tip)-a to 0 and also from 0 to b(base).

The resulting polynomial is:
0= (a^2*b^2)/2 + 2(a*b^3)/3 + (b^4)/4 - (a^4)/12

which has two roots:
a=3b
a= -b

but the limits of integration forbid a = -b. I doubt I'm allowed to start cutting the cone and flipping pieces around.

The height h=a+b so a=3h/4 . Which means the CG is 3/4 of the way down from the tip.

QED. density is uniform, gravity is uniform, and center of gravity is along central axis by symmetry.
Last edited by ShinAce; 2012-Mar-25 at 09:06 PM.

4. congratualtions let's move to second moment of inertia and the energy in a spinning cone??

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Easy enough. Assume that the central axis is the axis of rotation. Take slices of the cone as thin cylinders concentric with the axis of rotation. You go from cylinders with height h and radius 0 to ones with height 0 and radius r. You will get (3/10)*m*r^2 . Where m is the cylinder's total mass and r is the radius. For any other axis of rotation, find a parallel one which passes through the center of mass, the distance between both axes is d. The new moment of inertia, I = (3/10)*m*r^2 + m*d^2. This is the parallel axis theorem.

Kinetic energy follows from Ke = (1/2)*I*angular velocity^2, where the angular velocity is in radians/second.

Once you have the center of mass, the only other part that requires integration is the moment of inertia about the central axis. But the relation between height and radius is a linear one.

Again, the cone is solid and of uniform density.

6. Ah but the while the M of I is always through the C of G it is different in the three axes for any non symmetrical body so there are three Is (or only two for a cone) but for a general shape, three. Then you can find a unique M of I For any axis, knowing these three.

7. We can coin the expression "center of mass" and define it anyway we wish that might be useful for a task at hand. A physicist or engineer likely will be more interested in finding a balance point than in calculating the division of an ice cream cone into equal masses of ice cream.

As has been pointed out, there is a unique point about which an extended body will be balanced in any orientation in a uniform gravitation field when suspended by that point. There is an explicit vector integration formula for calculating the location of that point if we know the mass distribution. The expression "center of gravity" is a reasonable choice of words for this point.

In the absence of external gravitation, we can use a thruster of some sort to push the body at that same point. The same sort of calculation that was done with the gravitation shows that the body will be balanced in any orientation and will not turn. Thus, the expression "center of mass". This is an example of the principle of equivalence at work.

Some further thoughts:

In a non-uniform gravitational field, the center of gravity is not necessarily a fixed point on the body. It can change with changes in orientation relative to the gravitational gradient. Suppose we hang a long dumbbell by its center of mass in the Earth's field. If the dumbbell is horizontal it will be in equilibrium, but it will be unstable. If it is tilted the least bit, the lower ball experiences more downward force than the upper one and creates a net torque that results in a runaway increase in the tilt. At some particular tilt angle we could balance it by moving the suspension toward the lower ball, but it will still be unstable. A further increase in the tilt increases the difference between the forces on the two balls and requires a further movement of the suspension point to get a momentary equilibrium. Clearly there is not a fixed point we can call the center of gravity regardless of the amount of tilt.

Of course, the torque drops to zero as the dumbbell reaches the vertical position, and this will be a stable equilibrium if it is still hanging from the center of mass. The center of gravity is now at the maximum distance below the center of mass. As long as it hangs from any point above this center of gravity it will be stable. If hung from the center of mass it will be stable with either ball down, even if they are of different masses.

8. Originally Posted by xylophobe
Maybe there is another term for what I describe as the center of mass - what I am describing is the point through which the 3 orthogonal planes pass whereby those planes evenly divide the subject mass into two equal masses. So if I melted down either half then I could pour it into a mold of the second and exactly fill it creating two exact replicas.
Possibly, there is no such term. The reason might be because it's not a unique point--that point changes position as you tilt the cone.

If you tilt it far enough, the ice cream falls out. Another complicating factor.

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Originally Posted by profloater
Ah but the while the M of I is always through the C of G it is different in the three axes for any non symmetrical body so there are three Is (or only two for a cone) but for a general shape, three. Then you can find a unique M of I For any axis, knowing these three.
Crap, that's right. I will take a pass on calculating the moment of inertia for the axis of rotation perpendicular to the central radial axis. How about I integrate e^x^2 in polar coordinates instead? All the dazzle, half the hassle.

10. Actually for a cone it's easy peasy using the perpendicular axis theorem Ix =Iy + Iz. By inspection the two orthogonal axes are the same so the I about them is half the I about the long axis. and as you said before you can then shift the centre of rotation using the parallel axis theorem.

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## Gravity Field Around a Cone

So if I were to drawn lines around the cone (depicted as a triangle) to represent equal gravitational potential/acceleration could I just scale up the cone outline using the CoG as the scaling origin?

For a sphere (depicted as a circle) these lines would be concentric circles so to double the size of the circle I can scale it to 2.0 using the center of the circle.

I would like to visualize the gravitational field around the cone (and other irregularly shaped objects).

This image is sort of what I am talking about.

12. Originally Posted by xylophobe
So if I were to drawn lines around the cone (depicted as a triangle) to represent equal gravitational potential/acceleration could I just scale up the cone outline using the CoG as the scaling origin?
Except for the fact that center of mass (and in most cases center of gravity) is quarter of the way from the base and not where you have depicted, yes (CoM not CoG) as long as scaling is uniform. This is because scaling cone about a point 1/4 of the way from the base along the centerline will result in a cone which has the same point as the 1/4 of the way from its base along the centerline.
As for CoG it will gradually move away more and more from the previous CoG as you scale it up. As an extreme visualize a super low density cone, thousands of times bigger than Earth sitting on it. The CoG will be much closer than the CoM in this case because the far points will be much less attracted than the close points.

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The gravitational field around a cone looks more like my
modification of your diagram below, but it still isn't very
accurate. I'm pretty sure the contours would not parallel
the sides of the cone. (They *would* parallel the base.)

(It looks really ugly in Firefox when the window is zoomed
to anything other than a multiple of 100%. Yechhh.)

-- Jeff, in Minneapolis

.

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This is probably closer to it.

-- Jeff, in Minneapolis

15. at a small distance from the cone like one length, the lines would start to look like a sphere centred on the c of g I would think; it seems very unlikely they would follow the cone especially at the point where there is very little mass.

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## Gravity Contour

Originally Posted by profloater
at a small distance from the cone like one length, the lines would start to look like a sphere centred on the c of g I would think; it seems very unlikely they would follow the cone especially at the point where there is very little mass.
I was thinking along those lines - that it converges to a spherical field at increasing distances from the CoG, sort of like Jeff Root drew - but then there is also more mass because of the points so I think the lines would be parallel to the original form. Also, considering the 1/r2 decrease of gravity then the surface area has to increase proportionally to the distance. Jeff Root's depiction does not meet the squaring requirement for area.

For a sphere the area increases by the square of the size which is also true for cones.

In the image below the red is the actual solid mass while the green is where the area is doubled which means the gravitational effect is 1/4.

btw: I fixed the location of the CoG per a1call's observation.

17. 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).

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## Water on the Cone Planet

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).
So you're saying that if I poured water onto our cone-planet then it would accumulate as shown here:

and leave "mountains" at points "A", "B", and "C"?

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## Evolution in action

This is probably as close as I can get. Somebody needs to
use mathcad or some such to go for the kill.

-- Jeff, in Minneapolis
Last edited by Jeff Root; 2012-Mar-30 at 11:17 PM.

20. Originally Posted by xylophobe
So you're saying that if I poured water onto our cone-planet then it would accumulate as shown here:

and leave "mountains" at points "A", "B", and "C"?
That's the general idea, but the equipotential surfaces will never be exactly spherical, no matter how far you get from the cone.

Notice, too, in your illustration, that the figure is rotated about the axis--A and B are not mountain peaks, but are both points on the edge of a "mesa" (the bottom of the cone would not be "flat", it's center would be lower than its edge.

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The bottom of the cone *would* be flat, but nevertheless it would
be uphill in every direction from the center of that flat mesa to the
circular edge.

-- Jeff, in Minneapolis

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## True Ice Cream Cone

Originally Posted by grapes
That's the general idea, but the equipotential surfaces will never be exactly spherical, no matter how far you get from the cone.

Notice, too, in your illustration, that the figure is rotated about the axis--A and B are not mountain peaks, but are both points on the edge of a "mesa" (the bottom of the cone would not be "flat", it's center would be lower than its edge.
I'm gonna think on the "... never be exactly spherical ..." for a while.

Maybe Jeff Root's latest drawing is closer to reality.

Anyhow, here is another picture of the cone which is more representative of an ice cream cone:

I'm thinking that if I drill a hole through then the water will accumulate at the CoG.

23. If the cone is a shell with a thickness of zero, and without a base, then the centroid will be 1/3rd of the way from the base rather than 1/4 of the way from the base.

http://en.wikipedia.org/wiki/Centroi...one_or_pyramid

The CoM will converge towards this point as the sell gets thinner.

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## Thanks

Originally Posted by a1call
If the cone is a shell with a thickness of zero, and without a base, then the centroid will be 1/3rd of the way from the base rather than 1/4 of the way from the base.

http://en.wikipedia.org/wiki/Centroi...one_or_pyramid

The CoM will converge towards this point as the shell gets thinner.
I found the CoG for my hollowed-out cone by multiplying the mass of the smaller subtracted cone by the CoG location for the small cone then multiplying the mass of the larger cone by its CoG location then dividing the result by the mass of the hollowed cone.

If the cone is zero thickness then it will have zero mass and no CoG - just to clarify.

Here is a corrected picture with a hole depicting the water draining to the CoG.

25. Originally Posted by xylophobe
I found the CoG for my hollowed-out cone by multiplying the mass of the smaller subtracted cone by the CoG location for the small cone then multiplying the mass of the larger cone by its CoG location then dividing the result by the mass of the hollowed cone.
Have you checked up on the accuracy of your simple method by doing the necessary calculus? Caution: For all we know some integrations might not be doable analytically. We may need to do a numerical integration in small increments with a powerful computer.

If the cone is zero thickness then it will have zero mass and no CoG - just to clarify.
We are concerned with thin walls and the limit as the thickness of the walls approaches zero.

Here is a corrected picture with a hole depicting the water draining to the CoG.

Have you done any integration to determine the gravitational vectors in different parts of the hollow. For all we know they may not all line up with a single point.

26. I have to repeat: the equipotential surface will never be a perfect sphere.

And water will not "flow" down to the COG like that. The COG doesn't gravitationally attract the water--just consider that there is *no* gravitational force inside a hollow sphere, but its COG is its center.

The formula for that "cone-hollow inside a cone" shape is , where h is the ratio of the smaller cone to the larger cone. If h is zero, the COG is .25; if h is 1, the COG is 1/3.

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## Calculus Not Necessary

Originally Posted by Hornblower
Have you checked up on the accuracy of your simple method by doing the necessary calculus? ...
Calculus is not necessary for such a simple CoG calculation - I used the method for finding a composite centroid of an area only I found the composite centroid of a volume and because it is a uniform solid then volume = mass which means the centroid of the volume is the CoG for the solid.

I get the exact same answer for the location of the CoG using my method as I get using grapes's formula in his post.

"If an object has uniform density then its center of mass is the same as the centroid of its shape."

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## CoG Attraction

Originally Posted by grapes
...And water will not "flow" down to the COG like that. The COG doesn't gravitationally attract ...
If there are two identical bodies then a spacecraft can orbit the CoG which is halfway between the two planets. The spacecraft can also "fall" towards the CoG if the spacecraft fires a retrograde burn (see picture below).

29. Originally Posted by xylophobe
If there are two identical bodies then a spacecraft can orbit the CoG which is halfway between the two planets. The spacecraft can also "fall" towards the CoG if the spacecraft fires a retrograde burn (see picture below).
Yes, but if the spacecraft is not exactly on the plane between the two spheres, it will not generally be attracted to the center of gravity (in particular, if it's fairly close to one of the two objects, it's acceleration will be toward that object, largely ignoring the effects of the other). It's not that an object is never attracted toward the center of gravity of some object or system, it's that you cannot safely assume that it will in all cases.

30. Grey is right (surely, I've said that many times before). I didnt mean to imply that the attraction is never towards the COG--there are plenty of examples where it is. Just not in this case.

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