Gravity effects on a solid rigid ring around earth

[AGADOTTI]

I was trying this mental exercise and I need help and confirmations... so please launch your comments

Imagine (as an assumption) that the surface of the Earth is completely flat with no disturbances. A perfect sphere.
Imagine that is possible to build a solid ring made of a completely rigid material all around one meridian of the earth. The ring will have a perimeter a little longer of the Earth perimeter so that it hovers 1 meter over the surface of the earth.

Now assuming that it is possible to build the ring to be a completely rigid structure and because we build it to have it's center of gravity corresponding with the center of gravity of the earth, the following assumptions should be right:

1) The total weight (measured relatively to earth) of the ring is ZERO. The mass will be enormous, but of course if the weight is relative to the gravity, then for each point of the ring where a gravity vector is applied we will always have the opposite point with a completely opposite and equal vector is applied. The total weight should be therefore ZERO

2) In a stable configuration (i mean with no any other force in action) the ring will hover 1 meter over the Earth surface

3) If we apply a force on one side of the ring and imagine we move the ring half meter closer on that side, then ideally we are moving the center of gravity of the ring of the same half meter on the side. This will imply that after some oscillations the ring will come back to the original position hovering 1 meter over the earth

4) The configuration of hovering over the Earth is the configuration with the lowest potential energy

Is this correct or I'm losing something ?

Alex

 

[mgb_phys]

The ring is fundementaly unstable (in the same way as a pencil balancing on it's tip) any small perturbance will cause it crash into the surface of the Earth.

The proof is a little involved - it's a result famous because of a sci-fi novel called Ringworld by Larry Niven about a solar system sized ring around a sun.

 

[AGADOTTI]

Do you know where i can find the involved proof ?

I've heard about Ringworld before. The question is, the two center of gravity of the Ring and the Earth in the starting configuration are exactly at the same place, aren't they forming a system with a minimum potential energy ?
if not why ?

Thanks

 

[DocZaius]

I find no reason why the oscillation should ever stop, if all the forces involved are conservative.

edit: Looks like the ring isn't stable so nevermind!

 

[AGADOTTI]

What if you do teeter the ring with cables to the Earth surface (imagining the cables are strong enough to stop the ring from moving ?

 

[Vanadium 50]

mgb_phys is right - the situation is unstable.

There is no restoring force if you move the ring slightly - that's what "zero weight" means. Since there's no restoring force, it won't oscillate. If you push one end, it will keep going until some part of the ring hits the ground.

 

[DocZaius]

mgb_phys is right - the situation is unstable.

There is no restoring force if you move the ring slightly - that's what "zero weight" means. Since there's no restoring force, it won't oscillate. If you push one end, it will keep going until some part of the ring hits the ground.

Is that really what "zero weight" means though? Consider a tunnel bored through the earth, passing through its center. An object at the center of the Earth in that tunnel could be considered to have "zero weight" and yet if pushed in either direction will have a restoring force which will make that object oscillate indefinitely, right?

 

[Dale]

Now assuming that it is possible to build the ring to be a completely rigid structure and because we build it to have it's center of gravity corresponding with the center of gravity of the earth, the following assumptions should be right:

The problem is that the center of gravity is not quite the same as the center of mass. In most cases it doesn't make a difference, but in this case it does. The ring does not actually have a single point that can be called the center of gravity in this case. That is where the instability arises.

 

[russ_watters]

1) The total weight (measured relatively to earth) of the ring is ZERO. The mass will be enormous, but of course if the weight is relative to the gravity, then for each point of the ring where a gravity vector is applied we will always have the opposite point with a completely opposite and equal vector is applied. The total weight should be therefore ZERO

No. If you and your twin are on opposite sides of the earth, it would be nonsensical to say your total weight is 0. The total weight is the total of the two weights. Similarly, the total weight of the structure is the total of the weights of individual elements.

2) What if you do teeter the ring with cables to the Earth surface (imagining the cables are strong enough to stop the ring from moving ?

Yes, you could tether such a structure to Earth and it wouldn't take much force to keep it positioned.

However, the structure would need to be enormously strong to support its enormous weight. Essentially, you're talking about a bridge that spans the circumference of the earth.

 

[AGADOTTI]

I do not agree on your example of the two identical twin. We are not connected together.
While the mass of the structure is certainly huge, the weight on Earth is defined by the combined effect of the gravity on the total body, or in other words the force you need to apply to the body in order to hold it at rest in a gravitational field.
Assuming that the ring is in equilibrium, then the force that you need to apply to hold it at rest in the gravitational field of Earth is ZERO, therefore I would conclude that the weight is ZERO.

True the reasoning on the structural integrity of the ring, that can be comparable to a bridge with 40,000 km span. But here we are not talking about the technical feasibility of it's construction.

 

[Vanadium 50]

Consider a tunnel bored through the earth, passing through its center. An object at the center of the Earth in that tunnel could be considered to have "zero weight" and yet if pushed in either direction will have a restoring force which will make that object oscillate indefinitely, right?

Yes, but once pushed in another direction, the object no longer has zero weight and now there is a restoring force. The ring as described does not have that property.

 

[A.T.]

The ring is fundementaly unstable (in the same way as a pencil balancing on it's tip) any small perturbance will cause it crash into the surface of the Earth.

Is the ring unstable like a balancing pencil (local energy maximum), or indifferent like a ball on a plane (energy plateau)?

 

[Caesar_Rahil]

I really don't get the point that if you move the ring by half meter, the center of gravity will move by half meter.
Won't it be a little complicated to calculate center of gravity as now gravity is not uniform over the circumference of the ring?

Hey, and by the way..
If i really get this thing, there is tremendous force exerted on every part of the ring (internal tension), but the ring being rigid, overcomes this as forces on opposite sides cancel each other. So now, if you move it, there will not be a balance and the resulting force will pull it on the surface.
I guess this is what is called an unstable equilibrium, force zero but potential energy maximum

 

[DocZaius]

There is no restoring force if you move the ring slightly - that's what "zero weight" means. Since there's no restoring force, it won't oscillate. If you push one end, it will keep going until some part of the ring hits the ground.

I always took the definition of weight as meaning that the net gravitational force acting on an object is non zero. I didn't know it had anything to do with whether that force would also be restoring or not, and would lead to oscillations.

So as soon as this unstable ring is pushed, and it is falling towards the planet, I think that it does have weight, just as my example tunnel object does, when pushed either way. Of course I agree that the ring would crash into the planet and would not oscillate, while my tunnel object would oscillate indefinitely. However my point is that the system's stability (due to a restoring force) isn't merely based on the presence or absence of weight as you seem to contend, but rather on the way the system is configured.

Of course I'm probably wrong, but would love to hear why!

 

[AGADOTTI]

After some investigation I agree with the fact that the system is unstable. If it was a sphere then the balance would have restored always, but being a ring it is not true.
Now the problems is still the weight in the "equilibrium" state. So imagining we tether the ring with cables to avoid it to move (and please forget the feasibility issue), what is the weight of the entire structure.
If it is in the equilibrium state I would say ZERO.

 

[DaveC426913]

After some investigation I agree with the fact that the system is unstable. If it was a sphere then the balance would have restored always, but being a ring it is not true.

No. A sphere is in the same boat. There is no restorative force.

Now the problems is still the weight in the "equilibrium" state. So imagining we tether the ring with cables to avoid it to move (and please forget the feasibility issue), what is the weight of the entire structure.
If it is in the equilibrium state I would say ZERO.

What is the weight of the centre span of a bridge? The fact that it is braced so it does not fall down does not mean it has no weight. It must support its own weight (probably from compression forces, akin to a giant keystone) to prevent it from collapsing.

The ring has the same problem except that it is simply a "centre span" and is 40,000km long.

 

[AGADOTTI]

If the ring was a sphere then If the sun was at the center of the sphere it would have equal amounts of mass distributed at the same distance from it in all directions, this all cancels to a net force of zero. While if the sun was off center then the side it is closest to would have a stronger field strength due to less distance, but at the same time it the opposing side now has more mass behind the sun and though it is further away these effects perfectly cancel each other out.

Considering that the gravitational field is r^2 inverse proportional, we have the same situation of a flux, therefore if the sphere is equally dense and thick, then the gravitational attraction is proportional to the surface of the sphere again r^2 proportional to the distance from the center point mass. Easy to demonstrate that 1/r^2 and r^2 will elide each other for any given small portion of the sphere, and therefore integrating it for the entire sphere. Moving the sphere will not cause it to crash on the planet.
This is of course not true for the ring, and hence the ring instability.

Also the analogy with the Span of the bridge does not work because the bridge is hold in place by pillars that provide the force to sustain it. In this case the only force is the structural integrity of the material the ring is made of (imagining possible a material enough strong to sustain it). The case is very much different. Obviously the sphere and the ring will be subject to the gravitational pull, that will strain the structure, but we need to convene of what is the definition of weight. If it is the sum of all the force vectors that are applied to the body, then in the case of the ring and of the sphere the sum of the vectors is exactly ZERO (in a equilibrium position).

 

[D H]

All this talk about "center of gravity" makes me think quite a few of you have a misconception of Newton's law of gravitation. Newton's law of gravitation, F=GmM/r², pertains to point masses only. For non-point masses blinding applying Newton's law will in general lead to the wrong answer. The gravitational force between two non-point masses is the superposition of the gravitational forces between all pairs of infinitesimal point masses that comprise the two objects.

Thanks to Newton's shell theorem, the gravitational force between a point mass and some non-point mass that has a spherical mass distribution reduces to Newton's law. The ring does not have a spherical mass distribution. Newton's law does not apply per se. You have to do an integral over all the elements of the ring.

This is a bit messy in general. The gravitational force is fairly easy to calculate when the center of mass of the Earth is on the plane of the ring and on the axis of the ring. The first case is particularly easy: The gravitational force between the ring and the (spherical) Earth is zero whenever the Earth's center of mass is on the plane of the ring. There is no restoring force in this case, nor is there a disturbing force. There is no force, period.

 

[Malarksb]

I am thinking about atmospheric pressure and its variance being a force that would be applied to said ring. What would building a chamber that could have a vacuum drawn on to it do to equalize that pressure and other forces. My thought is that it would isolate this to the one element of gravity.

 

[DaveC426913]

I am thinking about atmospheric pressure and its variance being a force that would be applied to said ring. What would building a chamber that could have a vacuum drawn on to it do to equalize that pressure and other forces. My thought is that it would isolate this to the one element of gravity.

What atmospheric pressure? This thing's in space, right? If not, well, we've changed the parameters.

 

[AGADOTTI]

Yes this thing originally is with no atmosphere... so only gravity matters... do not complicate the problem...

 

[sophiecentaur]

Two points:
the 'span' is in compression, not tension
and
the term 'weight' merely describes the force on an object. It says nothing about how the force varies with position. By going to an appropriate distance from any planet you can have the same weight (or gravitational field) as on the surface of the Moon. The gradient of the field will be different in each case.

 

[AGADOTTI]

Ok I agree the Span is in compression
The point here is not to have the same weight in different gravitational fields, this is obvious that any gravitational field has it's own gradient.
The key question here is :
"Has the Ring so designed a Weight equal to ZERO or not ?"
we know that :
- The ring is unstable, but who cares because we imagine to answer the question in a case where the ring is in equilibrium
- The ring is stressed by a compression force due to gravity, but also here, who cares, because we imagine that the material the ring is composed is strong enough to maintain its structural integrity and its rigidity also in this case
- Different Planets have different gravity. Ok. Who cares we are experimenting the ring around earth, even if the same question about the weight can be valid around any other planet
- There is no atmosphere and any other force applied to the ring
- We know that the mass of the ring is huge. Who cares also in this case. The ring is constructed to withstand. and it's mass is still ridiculously small compared to the Earth mass.

so the question for who wants to answer is :
IS THE WEIGHT OF THE RING ZERO ?

 

[A.T.]

IS THE WEIGHT OF THE RING ZERO ?

If "weight" is the net gravitational force on the entire ring, then yes it's zero. But locally the ring parts are not "weightless", as accelerometer attached to them would show. The ring is not different from the Earth's crust in that respect.

 

[AGADOTTI]

Any other definition of WEIGHT for the structure ?

If not then the Weight is ZERO.

Of course locally there are gravitational interactions, this is why the ring should be strong enough not to be crushed by these forces. But this is merely a term of feasibility.

 

[Caesar_Rahil]

Looks like our concepts of weight are not still built enough...
Lets see.. how do you define weight..
If I am in a lift, I stand on a scale, it shows my basic weight (mg)
If lift is moving up, don't we say weight increased..
and if lift is moving down, we say our weight decreased.
which means all this while I have really thought of weight as some quantity which i can measure.
Correct me if I am wrong, but i really believe that weight is:
The normal reaction exerted on a scale (or tension in a string) that is measured when an object is kept upon it and is non-accelerating with respect to the observer.

If you go with that concept, the weight of the ring will be zero.

but i just made that up, so i may be wrong..its up to you people now

 

[A.T.]

Any other definition of WEIGHT for the structure ?

Sure, for example the weight scalar : integral of m*|g| over the entire structure.

Usually the term "weight" is used for an object in an uniform gravitational field or an object so small that the relevant part of the gravitational field can be assumed uniform. In such a case the weight scalar is the magnitude of the weight vector, and the term 'weight' is used for both interchangeably.

 

[Dadface]

Although Newtons law applies to point objects we can apply it to extended objects if any distances are referred to the centres of gravities of the objects.With a perfectly spherical Earth of uniform density its centre of gravity is at its geometrical centre and with a perfectly circular ring of uniform density its centre of gravity is also at its geometrical centre.The system of Earth and ring which has been described will tend to minimum potential energy,in this case zero in that the resultant force on either object will tend to become zero, the two centres of gravities tending to become coincident.If the system is displaced a restoring force will act and bring the system back to equilibrium i.e. with the C of Gs coincident.Having written this I have an uncomfortable feeling that there may be a flaw in my reasoning somewhere, but here goes anyway.

 

[AGADOTTI]

Well, we have determined that the weight is zero when in equilibrium, but unfortunately the restoring force is only valid for a sphere and not a ring. As many others commented there is a famous demonstration of this for RingWorld.

 

[D H]

There is no restoring force for a sphere either. Google "Newton's shell theorem."

 

[DaveC426913]

There is no restoring force for a sphere either. Google "Newton's shell theorem."

What I was not sure about was this:

Newton's Shell Theorem is normally applied to a massive shell of material and its effects on a body of negligible internal mass.

Does it also apply inside out? i.e. a negligible shell outside under the influence of a massive inside?

 

[D H]

What I was not sure about was this:

Newton's Shell Theorem is normally applied to a massive shell of material and its effects on a body of negligible internal mass.

Does it also apply inside out? i.e. a negligible shell outside under the influence of a massive inside?

The shell theorem places absolutely no constraint on the mass of the object or of the shell. It doesn't matter if the object's mass is one one-millionth of the mass of the shell or a million times the mass of the shell. The gravitational acceleration due to the shell on any point mass located anywhere inside the shell is zero. For a non-point mass, use the superposition principle: Integrate the gravitational force over the volume of the object. Integrate a zero vector integrated over the volume of the object and you get zero.

 

[DaveC426913]

The shell theorem places absolutely no constraint on the mass of the object or of the shell. It doesn't matter if the object's mass is one one-millionth of the mass of the shell or a million times the mass of the shell. The gravitational acceleration due to the shell on any point mass located anywhere inside the shell is zero.

Ah but again, that is not what is in question. What is in question is whether the gravitational acceleration due to the object is nonzero on the shell.

However, upon reflection, I realize they must be the same. You simply cannot have unequal gravitational forces between two bodies.

i.e. F(m1)=G*m2/r^2 == F(m2)=G*m1/r^2

If this were not true, you would suddenly have yourself a free propulsion system!

So, OK. The shell theorem must work both ways.

 

[Dadface]

The shell theorem places absolutely no constraint on the mass of the object or of the shell. It doesn't matter if the object's mass is one one-millionth of the mass of the shell or a million times the mass of the shell. The gravitational acceleration due to the shell on any point mass located anywhere inside the shell is zero. For a non-point mass, use the superposition principle: Integrate the gravitational force over the volume of the object. Integrate a zero vector integrated over the volume of the object and you get zero.

I can take this point but only if the surroundings are ignored.In his original post Agadotti did not ignore the surroundings in that he described applying a force on one side of the ring.To apply a force, by whatever mechanism, must bring the surroundings into the analysis and the force needed to change the location of the Earth within the ring/shell depends,amongst other things, upon the geometry and structure of the whole system including that of the surroundings.If the force is applied to the Earth the same reasoning applies.If the surroundings are ignored no force can be applied and the system remains in stable equilibrium.Muddled thinking?Possibly, I am still thinking this through.

 

[sophiecentaur]

Isn't the answer to this really "Zero, using the standard definition of weight but so what?"?
It's zero when in equilibrium but is that not only an unstable equilibrium. Why try to make anything more out of it? The definition of weight was not made to cope with such a strange circumstance any more than it was intended to cope with 'weight' inside an orbiting spacecraft or a falling lift. You can work out 'what will happen' and the forces involved in any similar situation without using the term 'weight'.

Edit, a bit later ::I don't think the shell theorem applies to a ring - but it may apply to a spherical shell around the earth,

 

[D H]

I can take this point but only if the surroundings are ignored.In his original post Agadotti did not ignore the surroundings in that he described applying a force on one side of the ring.To apply a force, by whatever mechanism, must bring the surroundings into the analysis and the force needed to change the location of the Earth within the ring/shell depends,amongst other things, upon the geometry and structure of the whole system including that of the surroundings.If the force is applied to the Earth the same reasoning applies.If the surroundings are ignored no force can be applied and the system remains in stable equilibrium.Muddled thinking?Possibly, I am still thinking this through.

Sorry to say this, but this is indeed muddled thinking.

In the original post Agadotti misapplied Newton's theory of gravitation. He made the same mistake in post #15 with regard to a sphere as opposed to a ring. The ring is metastable if the Earth's center of mass lies along the ring axis, flat if it lines in the ring plane, plain unstable if it lies elsewhere. The sphere configuration is flat. A pencil standing upright balanced on its tip is a metastable configuration. The pencil can stay upright if it is perfectly upright. The slightest of nudges and it will fall over. A perfectly smooth air hockey table with a perfectly smooth puck is a flat configuration. The puck will not move if it is stationary, but the slightest touch sends it toward an edge of the table. Metastable and flat configurations are unstable. There is no restoring force.

 

[Dadface]

Sorry to say this, but this is indeed muddled thinking.

In the original post Agadotti misapplied Newton's theory of gravitation. He made the same mistake in post #15 with regard to a sphere as opposed to a ring. The ring is metastable if the Earth's center of mass lies along the ring axis, flat if it lines in the ring plane, plain unstable if it lies elsewhere. The sphere configuration is flat. A pencil standing upright balanced on its tip is a metastable configuration. The pencil can stay upright if it is perfectly upright. The slightest of nudges and it will fall over. A perfectly smooth air hockey table with a perfectly smooth puck is a flat configuration. The puck will not move if it is stationary, but the slightest touch sends it toward an edge of the table. Metastable and flat configurations are unstable. There is no restoring force.

Rather than describe this as "muddled thinking" I prefer now to describe it as "thinking in progress aided by the thoughts of others".
My post number 28 was an example of "forgetful thinking".I am familiar with the shell theorem and have indirectly referred to it on a few occasions in this forum.I just forgot it. I agree with the metastability equilibria you have described but all are sort of thought experiments and I am just taking them a bit further.
The pencil example is a good one but can apply to the real world because,in principle, a pencil can be balanced on its tip.The ice hockey puck is a real thought experiment but can easily be extended to the real world by taking resistive forces into account.All good but consider again the thought experiment as originally described by agadotti and later extended to take into account shells as well as rings.Everything has been idealised ,a perfectly spherical earth,uniform density and the rest of it and I can accept these things,its a nice thought experiment and I agree with the analysis you have presented and your conclusion about metastability.What I am thinking about now is the situation as originally described namely the system being subjected to a force.From where comes this force? I can think of many examples but all of them disrupt the idealised structure that has been analysed.Is the force applied externally?if so then other gravitational fields need to be brought into the analysis.Is it internal for example could a super fruit fly take off from the ring imparting an impulse to the system and disrupting its equilibrium as well as its idealised structure.In short I think it's a very nice thought experiment,I think you have come to a good conclusion but to do the original question full justice we have to take into account the whole thing as described and not ignore the force and its effects.Roughly speaking Agadotti described three main things,earth,ring and force.Are we allowed to ignore the earth?No there is nothing left to analyse.Are we allowed to ignore the ring? Ditto.Are we allowed to ignore the force? If we do we are not answering the question.

 

[DaveC426913]

OK, one last thing to analyze.

When we talk about this "restoring force" (or lack of), we are talking about a negative feedback loop. In a negative feedback loop, evolution of the system in some direction results in a force opposing the evolution (speed governors on steam engines do this) Dadface called this arrangement metastable.

A positive feedback loop is when the evolution of the system results in a force encouraging that evolution (the balanced pencil does this: the farther off-centre it is, the more it is encouraged to continue falling off-centre.)

Now, the object-in-the-sphere must be neither, to satisfy Newton's Shell Theorem. ("Flat", as per Dadface's terminology.)

But what about the ring? We've established that there is no negative feedback loop. Is there a positive feedback loop? If nudged off-centre, does the ring go farther off-centre?

 

[Dale]

A pencil standing upright balanced on its tip is a metastable configuration. The pencil can stay upright if it is perfectly upright. The slightest of nudges and it will fall over.

I thought that was the definition of "unstable". I am not familiar with the term "metastable", what makes the upright pencil metastable rather than unstable.

 

[Vanadium 50]

But what about the ring? We've established that there is no negative feedback loop. Is there a positive feedback loop? If nudged off-centre, does the ring go farther off-centre?

It continues to move farther off center, but it does not accelerate. An object in motion remains in motion and all that.

 

[DaveC426913]

It continues to move farther off center, but it does not accelerate. An object in motion remains in motion and all that.

OK. So no positive feedback loop then.

Hm. This means that the ring and the sphere are identical. In both cases, there is no net force under any circumstances.

The implication is that Newton's Shell Theorem is actually Newton's Ring Theorem (which includes the Shell Theorem as the special case where ring's breadth = 100% ).

Is this what we're all agreeing on?

 

[willem2]

Sorry to say this, but this is indeed muddled thinking.

In the original post Agadotti misapplied Newton's theory of gravitation. He made the same mistake in post #15 with regard to a sphere as opposed to a ring. The ring is metastable if the Earth's center of mass lies along the ring axis, flat if it lines in the ring plane, plain unstable if it lies elsewhere.

This isn't true. If the center of the Earth is in the plane of the ring, but not in the center, there
is a disturbing force. See here:

http://testservice-eprints.gla.ac.uk/38/1/JIBS_C_McInnes_56_308.pdf

If the disturbance is along the axis, it's easy to see that there will be a restoring force.

 

[DaveC426913]

If the disturbance is along the axis, it's easy to see that there will be a restoring force.

Y'all need to read Larry Niven's essay "Bigger Than Worlds". He goes into all sorts of ramifications. He puts the ring around the star rather than around the planet, then sets the star bobbing. Now he's got seasons.

He uses the ring to make a giant magnetic pinch, which causes the sun to lase, throwing out gouts of plasma along its axis. Now he's got a solar system-sized world with an engine. He can cruise the galaxy from the comfort of his planet-ring. (Just don't accelerate too fast or the engine will go exploring - sans planet-ring!)

And he's just getting warmed up!

Wait'll he starts talking about living on the surface of a galaxy-sized Dyson-sphere. The gravitational gradient and atmosphere is measured in light-years. You really can fly to the Moon on gossamer wings!

 

[sophiecentaur]

I don't think the term 'feedback' is appropriate here. That term refers to control loops involving amplification. All that is at work here is the gradient of the potential with displacement and the shape of the potential surface. If the surface is a dip, then the force is restorative after displacement, if it is a peak, the force will be away from the initial position. There can be other shapes, such as saddles and plateaux but the same principle applies; it's nothing to do with feedback because there is not control loop..

 

[Dadface]

I thought that was the definition of "unstable". I am not familiar with the term "metastable", what makes the upright pencil metastable rather than unstable.

For the pencil I am more familiar with the term unstable equilibrium,it can in principle balance but the slightest wobble and over it goes. According to my dictionary the balancing pencil can be described as being in a metastable state because it is capable of changing to a more stable state but I do feel ,however,that "metastable" is not such an appropriate word to describe the other situations described here because any changes are not necessarily towards greater stability.

 

[willem2]

I don't think the term 'feedback' is appropriate here. That term refers to control loops involving amplification. All that is at work here is the gradient of the potential with displacement and the shape of the potential surface. If the surface is a dip, then the force is restorative after displacement, if it is a peak, the force will be away from the initial position. There can be other shapes, such as saddles and plateaux but the same principle applies; it's nothing to do with feedback because there is not control loop..

there is certainly a control loop involving amplification there

distance from the center - > acceleration away from the center -> speed away from the center ->
even larger distance from the center.

To counteract this you need to provide an artificial negative feedback mechanism.

 

[DaveC426913]

I don't think the term 'feedback' is appropriate here. That term refers to control loops involving amplification.

It refers to many other situations as well. Your choice is a little too specious.

Feedback occurs where the output of one state is used as part of the input for the next state.

In this case, the evolving position (of pencil or ring) is a factor in the forces applied.

 

[sophiecentaur]

I agree that a control loop is needed to keep the ring in position, artificially (if you don't want to support it on pillars. But the natural forces do not constitute a feedback loop and I thought that was what the OP was discussing.

 

[sophiecentaur]

Re feedback.
Would you also say that a mass hanging on a spring constitutes a negative feedback loop?
That would mean that any simple mechanical system involves 'feedback'. Is that really what feedback is?

 

[D H]

This isn't true. If the center of the Earth is in the plane of the ring, but not in the center, there
is a disturbing force. See here:

http://testservice-eprints.gla.ac.uk/38/1/JIBS_C_McInnes_56_308.pdf

You're right. I did the integration wrong. The in-plane configuration is metastable.

If the disturbance is along the axis, it's easy to see that there will be a restoring force.[/QUOTE]
Thanks; I stand corrected.

For the pencil I am more familiar with the term unstable equilibrium,it can in principle balance but the slightest wobble and over it goes.

That is precisely what I was taught that metastable means: An equilibrium point that is unstable.