What exactly is the reactive centrifugal force (split)

[A.T.]

L = constant

Here is your error. Only the L of the entire system must be constant. The L of the astronaut doesn't have to be constant.

 

[Dale]

? The reaction force completely disappears if the centripetal force disappears, which is what happens when the floor completely dissolves. So how can the reaction force cause the outward motion?

We are talking about the force ON the floor (the reactive centrifugal force), so of course it disappears if the floor completely dissolves. The deformation of the floor is in the outward direction which requires a force on the floor in the outward direction*. This force is the reactive centrifugal force.

*EDIT: actually, on further thought I realized that I am making an incorrect generalization from statics. What is required is a shear stress in the radial direction and/or a tension in the circumferential direction. In this specific case the stress is caused, in part, by the reactive centrifugal force, but it is not a general requirement.

I know it seems counter-intuitive, but with no external torque the centripetal force decreases as r increases.

You are taking one specific case and making an incorrect generalization. In this case, we are dealing with a massive space station, not a hula hoop. The ω of the space station is nearly independent of the astronaut's r, even in the absence of an external torque. Thus, the formula I posted (F=mω²r) is the correct one for this situation. The centripetal force increases with increasing r.

We see this from the fact that gravity, which supplies the centripetal force for orbiting bodies, varies as 1/r^2.

Again, you are taking one specific case and making an incorrect generalization. In this case, we are not dealing with gravity and a 1/r² force, we are dealing with material deformation which essentially follows Hooke's law (a force proportional to Δr) up to the point where you start getting plastic deformation (a force independent of r).

 

[Andrew Mason]

Here is your error. Only the L of the entire system must be constant. The L of the astronaut doesn't have to be constant.

The space station would have to apply a torque to the astronaut to increase his angular momentum effectively maintaining the astronaut's angular speed. So the space station would have to have a large moment of inertia compared to the astronaut: Iss >> R^2Mast. In that case you are right.

Let's suppose there is one astronaut who weighs 100 kg including his suit, and he is lying on the floor of a circular rotating space station of radius R and mass 1100 kg. It is made of aluminum except for the section directly opposite him which is made of lead that has a mass of exactly 100 kg more than the aluminum floor under/outside the astronaut.

The centre of mass of the space station is not the geometric centre. Let's say its centre of mass is Δr from the geometric centre. But the centre of rotation, with the astronaut, is the geometric centre.

The question is: what provides the centripetal force that causes the centre of mass of the space station to rotate about the geometric centre?

AM

 

[Andrew Mason]

We are talking about the force ON the floor (the reactive centrifugal force), so of course it disappears if the floor completely dissolves. The deformation of the floor is in the outward direction which requires a force on the floor in the outward direction*. This force is the reactive centrifugal force.

*EDIT: actually, on further thought I realized that I am making an incorrect generalization from statics. What is required is a shear stress in the radial direction and/or a tension in the circumferential direction. In this specific case the stress is caused, in part, by the reactive centrifugal force, but it is not a general requirement.

So you seem to be saying that the reason the astronaut makes the dent is, at least in part, because of his inertia. This would be shown by the dent itself. If the dent is not a perfectly radial dent then it has to be caused by inertia. I say the dent will be on an angle, indicating that with the momentary loss of force of the floor the astronaut simply continued moving in a straight tangential line until the solid part of the floor moved in and deflected him.

You are taking one specific case and making an incorrect generalization. In this case, we are dealing with a massive space station, not a hula hoop. The ω of the space station is nearly independent of the astronaut's r, even in the absence of an external torque. Thus, the formula I posted (F=mω²r) is the correct one for this situation. The centripetal force increases with increasing r.

Again, you are taking one specific case and making an incorrect generalization. In this case, we are not dealing with gravity and a 1/r² force, we are dealing with material deformation which essentially follows Hooke's law (a force proportional to Δr) up to the point where you start getting plastic deformation (a force independent of r).

Actually, I should have said that Fc is proportional to 1/r^3, if angular momentum is conserved: Fc = L^2/mr^3. But your point is well taken: with a massive space station where Iss >> r^2Mast the angular speed would not decrease significantly with the astronaut's increase in r. So, AFTER the Astronaut's feet again reached a solid floor the centripetal force would be greater by a factor of Δr/r because ω does not change.

But surely the essential physics doesn't depend on how massive the space station is compared to the astronaut. Let's say there are 2 floors, the inner one being 10 cm thick and made of material that will turn to jello 1 cm at a time when someone presses a switch. Each time the astronaut completes a circle, someone presses the switch. What you will end up with is a series of dents in the jello moving at an angle opposite to the direction of rotation. Each time the floor turns to jello, the Astronaut keeps moving on a tangent until the next non-jello layer moves inward 1 cm to stop him. If it was the centrifugal reaction force that caused the dents, the dents would be completely radial would they not?

AM

 

[Dale]

So you seem to be saying that the reason the astronaut makes the dent is, at least in part, because of his inertia.

No, that is not what I am saying at all. A dent forms due to stresses in the material. In dynamics there can be stresses due to the floor's own inertia, but any stresses attributable to the astronaut will be through the reaction force.

This would be shown by the dent itself. If the dent is not a perfectly radial dent then it has to be caused by inertia. I say the dent will be on an angle, indicating that with the momentary loss of force of the floor the astronaut simply continued moving in a straight tangential line until the solid part of the floor moved in and deflected him.

How do you propose that the astronaut's inertia will cause stress in the material? Please provide a reference to support the claim.

Actually, I should have said that Fc is proportional to 1/r^3, if angular momentum is conserved: Fc = L^2/mr^3. But your point is well taken: with a massive space station where Iss >> r^2Mast the angular speed would not decrease significantly with the astronaut's increase in r. So, AFTER the Astronaut's feet again reached a solid floor the centripetal force would be greater by a factor of Δr/r because ω does not change.

Thank you!

Let's say there are 2 floors, the inner one being 10 cm thick and made of material that will turn to jello 1 cm at a time when someone presses a switch. Each time the astronaut completes a circle, someone presses the switch. What you will end up with is a series of dents in the jello moving at an angle opposite to the direction of rotation. Each time the floor turns to jello, the Astronaut keeps moving on a tangent until the next non-jello layer moves inward 1 cm to stop him. If it was the centrifugal reaction force that caused the dents, the dents would be completely radial would they not?

No. When the astronaut is co-rotating the reaction force is purely centrifugal, but if the astronaut is not co-rotating then the reaction force may have some tangential component as well (think about the Coriolis force in the rotating frame). Any deformation of the material due to the Astronaut will always be via the reaction force, whether that is purely centrifugal or otherwise.

 

[A.T.]

The question is: what provides the centripetal force that causes the centre of mass of the space station to rotate about the geometric centre?

Acceleration of the centre of mass is a function of the net force, not of an individual force provided by something. An individual force has a point of attack, which together with its direction determines if it is centrifugal or centripetal. The net force doesn't have a physical point of attack, but can be thought of as acting on the center of mass. Therefore you can have one individual centrifugal force acting, and yet a centripetal acceleration of the center of mass.

 

[A.T.]

So you seem to be saying that the reason the astronaut makes the dent is, at least in part, because of his inertia.

Read post #69.

It doesn't matter how many indirect causes you can list for the deformation. The direct cause of the deformation is the local centrifugal force exerted by the astronaut on the wall. So that centrifugal force has a direct physical effect.

 

[Dale]

Andrew Mason,

In A.T.'s drawing below, do you disagree about the existence of any of the forces, or is your disagreement entirely about the labeling?

 

[A.T.]

In A.T.'s drawing below, do you disagree about the existence of any of the forces, or is your disagreement entirely about the labeling?

It seems to be about labeling and the problem that: You can have an individual force acting outwards and inward acceleration of the CoM of some arbitrarily chosen part of the system. But at the same time that outwards force can have real, physical outwards effects, like outwards acceleration/deformation of some other arbitrarily chosen part of the system.

The best way to avoid this conflict, is to ignore all the effects the force might have on arbitrarily chosen parts, and simply consider what is independent of how you split up the system: The force's point of attack and direction. This leads to the label "centrifugal".

 

[Andrew Mason]

No, that is not what I am saying at all. A dent forms due to stresses in the material. In dynamics there can be stresses due to the floor's own inertia, but any stresses attributable to the astronaut will be through the reaction force.

How do you propose that the astronaut's inertia will cause stress in the material? Please provide a reference to support the claim.

A dent signifies that there has been movement. If the dent is due to a force overcoming stress, the motion results from that force exceeding the stress momentarily. That can never happen here. My point in using the jello was to eliminate the force and the stress and show that there would be a dent in the jello that essentially traces the inertial path of the astronaut.

Our disagreement is not that there is a force between the astronaut and the space station. We don't disagree on its direction either. We just disagree on what that reaction force does. You say it is just a force that results in centrifugal tension in the space station. I say that it actually accelerates the space station in the direction opposite to the direction that the astronaut is accelerating. In other words, it accelerates the space station toward the centre of rotation.

My position is that by calling it a centrifugal reaction force giving rise only to a tension is incorrect and it also makes it extremely difficult to distinguish from the fictitious centrifugal force. The centrifugal force or pull from the outside is postulated as the source of the tension in the space station.

AM

 

[Andrew Mason]

Acceleration of the centre of mass is a function of the net force, not of an individual force provided by something. An individual force has a point of attack, which together with its direction determines if it is centrifugal or centripetal. The net force doesn't have a physical point of attack, but can be thought of as acting on the center of mass. Therefore you can have one individual centrifugal force acting, and yet a centripetal acceleration of the center of mass.

Ok. Then you are saying that the reaction force causes centripetal acceleration but you still want to call it a centrifugal force. The centripetal acceleration multiplied by the mass of the spacestation is equal and opposite to the centripetal acceleration x the mass of the astronaut. So the astronaut's centripetal force and its reaction force are both "net" forces.

The direction is what it is. It is opposite to the centripetal acceleration of the astronaut, which is the direction of the acceleration of the space station. It cannot ever cause motion to occur outward from the centre of rotation so I am not sure why anyone would want to call it centrifugal. While it acts, it accelerates mass toward the centre of rotation.

When Newton describes forces moving things he is implicitly if not explicitly referring to the accelerations of their centres of mass or centres of gravity, not the direction of tensions within the bodies themselves. Those are trivial details and they don't matter - until one gets into the world of rotating masses.

AM

 

[A.T.]

We just disagree on what that reaction force does...it accelerates the space station ...

Read post #76. Learn to distinguish between an individual force and the net force.

 

[A.T.]

Acceleration of the centre of mass is a function of the net force, not of an individual force provided by something.

Ok. Then you are saying that the reaction force causes centripetal acceleration ...

No. Read it again.

 

[Dale]

A dent signifies that there has been movement. If the dent is due to a force overcoming stress, the motion results from that force exceeding the stress momentarily.

Sure, Newton's 2nd law.

That can never happen here.

Why not?

My point in using the jello was to eliminate the force and the stress and show that there would be a dent in the jello that essentially traces the inertial path of the astronaut.

Even in jello there is a reaction force. Jello has a very low yield strength, so the force will be small, but it is still that reaction force which causes the deformation of the Jello. By using jello you have made the reaction force small (not zero), but you have also made it particularly sensitive to the force.

Our disagreement is not that there is a force between the astronaut and the space station. We don't disagree on its direction either.

OK. That force that we both agree exists is called the "centrifugal reaction force". It is how that term is defined. You may not like the name, and you may have excellent reasons for disliking the terminology (e.g. for extended bodies a centrifugal reaction force can cause centripetal acceleration), but nevertheless that is the standard terminology.

We just disagree on what that reaction force does. You say it is just a force that results in centrifugal tension in the space station. I say that it actually accelerates the space station in the direction opposite to the direction that the astronaut is accelerating. In other words, it accelerates the space station toward the centre of rotation.

I don't disagree about the acceleration which falls under Newton's 2nd law (although in A.T.'s example the space station's COM is not accelerating).

I disagree only with your interpretation of Newton's 3rd law where you try to claim that the 3rd law reaction to the centripetal force on one astronaut is the centripetal force on the other astronaut. All of the remaining discussion has been about your attempts to justify that interpretation, either by re-defining Newton's third law with reference to his use of the word "action" or by asserting that the centrifugal reaction force does not have any physical effects besides centripetal acceleration.

I think that it is clear that the reactive centrifugal force exists in some cases, in those cases it is the 3rd law pair of a centripetal force, it is always a real force, it exists in all frames and can do all of the things that you would expect of a real force including material deformations and other such things.

I also agree that the terminology can be confusing. It is clearly a topic that many students struggle with. Personally, I don't even like the "action/reaction" terminology, but it is out there and people should know what it means.

 

[A.T.]

(although in A.T.'s example the space station's COM is not accelerating).

Unless you include the other astronaut in the space station. So it depends on an arbitrary definition of objects, what the "effect" on an object's COM might be. That's why it is not a good idea to base a general naming on some object’s COM acceleration. The logic behind the centrifugal-name does not depend on how you cut the system into pieces.

And if more forces are acting, it is even more difficult to attribute a particular effect, to a certain force. That's why it is not a good idea to base the naming on effects in general. The logic behind the centrifugal-name does not depend on other forces, and what effects they might cause together.

 

[Andrew Mason]

No. Read it again.

You have quoted only part of what I said. The "net" force IS ALWAYS the mass x its acceleration. The astronaut's entire reaction force is equal and opposite to its mass x its acceleration and that is exactly equal to the mass x acceleration of the space station. It is by its very nature a "net" force.

AM

 

[A.T.]

The astronaut's entire reaction force is equal and opposite to its mass x its acceleration and that is exactly equal to the mass x acceleration of the space station. It is by its very nature a "net" force.

No, it is not the same. They have a different point of attack. See post #76.

 

[Andrew Mason]

No, it is not the same. They have a different point of attack. See post #76.

I don't see the significance of the point of attack.

Maybe I am missing something here. Stop me if you think I am saying anything that is incorrect.

1. For a rigid body that is not rotating and whose centre of mass is not accelerating, the sum all forces acting on it is 0. Since no part of the body is accelerating, the sum of all forces acting on each part of such a body is 0.

2. For a rigid body that is rotating and whose centre of mass is not accelerating, the sum of all forces acting on it is 0. Since each part of the body is accelerating, the sum of all forces acting on each part of the body is equal to the mass of such part multiplied by its (centripetal) acceleration. (Since the sum of all such mass x accelerations must be 0, a rotating free body always rotates about an axis through its centre of mass).

3. The space station with the single astronaut lying on the floor (as I described in my post #73) is a rotating rigid body whose centre of mass is not accelerating. Therefore:

• the sum of all forces acting on each part is equal to the centripetal acceleration of that part multiplied by the mass of such part.
• For any arbitrary division of the space station (including contents) into two parts, the sum of all forces acting on each part is equal to the mass of such part multiplied by its (centripetal) acceleration and
• The sum of the mass x (centripetal) acceleration of each of the two parts = 0.
• Thus, the mass x (centripetal) acceleration of any part is equal and opposite to the mass x centripetal acceleration of the other part.

4. Therefore the mass x acceleration of the astronaut = -mass x acceleration of the rest of the space station

5. Since the force applied by the space station to the astronaut = the mass x the (centripetal) acceleration of the astronaut, the equal and opposite force applied by the astronaut to the space station = the mass x (centripetal) acceleration of the rest of the space station = sum of all the forces acting on all the parts of the rest of the space station.

AM

 

[A.T.]

I don't see the significance of the point of attack.

It's a naming convention, not a matter of great significance. See post #85 for my reasons to prefer the common convention over yours.

 

[Dale]

The sum of the mass x (centripetal) acceleration of each of the two parts = 0.

This is not generally true. If you take two parts which are near each other or two parts which are opposite but not equally massive then their change in momentum may not be equal and opposite.

5. Since the force applied by the space station to the astronaut = the mass x the (centripetal) acceleration of the astronaut, the equal and opposite force applied by the astronaut to the space station = the mass x (centripetal) acceleration of the rest of the space station = sum of all the forces acting on all the parts of the rest of the space station.

This is true if you consider the opposite astronaut to be part of the space station (which is valid). In that case the centrifugal reaction force applied by the right astronaut to the space station is indeed equal to the mass x centripetal acceleration of the "space station & left astronaut".

The statement is not true if you consider the opposite astronaut not to be part of the space station (which is also valid). In that case the centrifugal reaction force applied by the right astronaut to the spact station is not equal to the mass x centripetal acceleration of the space station.

 

[A.T.]

for extended bodies a centrifugal reaction force can cause centripetal acceleration

Since the extended body aspect seems to be a reason for confusion, I propose to use an even simpler example:

A spaceship is moving on a circular path, by firing its engines continuously to provide the centripetal acceleration. The astronaut inside the ship exerts a reactive centrifugal force on the ship's wall.

 

[Dale]

Since the extended body aspect seems to be a reason for confusion, I propose to use an even simpler example:

A spaceship is moving on a circular path, by firing its engines continuously to provide the centripetal acceleration. The astronaut inside the ship exerts a reactive centrifugal force on the ship's wall.

I think that we should go ahead with the current example in your drawing. Andrew Mason has a valid point that in certain perfectly legitimate analyses of perfectly reasonable circumstances the centrifugal reaction force can cause centripetal acceleration.

The point remains that whenever a centripetal force has a 3rd law pair which is directed away from the center of rotation that force is called the "centrifugal reaction force". That is the definition of the term and it is a common enough term that people should know what it means.

In some cases the centrifugal reaction force causes only centrifugal effects (material stresses, accelerations, etc.), but sometimes it causes centripetal effects. The naming convention refers to its direction, and not to its effects. Andrew Mason has every justification to dislike the naming convention (I dislike it for a different reason), but nevertheless it is well defined and well established.

 

[Andrew Mason]

Unless you include the other astronaut in the space station. So it depends on an arbitrary definition of objects, what the "effect" on an object's COM might be. That's why it is not a good idea to base a general naming on some object’s COM acceleration. The logic behind the centrifugal-name does not depend on how you cut the system into pieces.

The ONLY effect that the centripetal force applied by a part of a rotating free body can have on the COM of the rest of the body is a centripetal acceleration of that COM. There is no other possible effect. It does not matter how you cut it or how the force is applied. There can NEVER be a centrifugal acceleration of the other part or ANY part of that other part.

And if more forces are acting, it is even more difficult to attribute a particular effect, to a certain force. That's why it is not a good idea to base the naming on effects in general. The logic behind the centrifugal-name does not depend on other forces, and what effects they might cause together.

In any rigid body there are, by definition, static tension forces everywhere. We don't concern ourselves with those forces. We cannot possibly know what those forces are doing at any moment because they operate at atomic levels and the atoms are constantly undergoing thermal motion. We can only look at the forces that cause acceleration.

It's a naming convention, not a matter of great significance. See post #85 for my reasons to prefer the common convention over yours.

Whose "naming convention"? Give me a cite. No physics text that I have ever read refers to "centrifugal reaction force". The only book that I have found that references it is Mook and Vargish "Inside Relativity" (1987):

"As a result of the force acting on the ball, the ball deviates from uniform motion and follows a circular path, just as the planet docs when acted upon by the sun's gravity, But by Newton's third law, if you exert a force on the ball, the ball must exert an equal and opposite force on your hand, and it does. You feel this force as the " tug" of the ball on the string you hold. It is not necessary to posit a centrifugal force. What is sometimes called a "centrifugal force" is a reflection of the force you are exerting on the ball to keep it in a circular path. Similarly, the sun will feel such a reactive, centrifugal force from each of the planets that it holds in an orbit by its force of gravity."​

There is absolutely no difference between this "reactive centrifugal force" ("rcf") and the "fictitous centrifugal force" ("fcf"). The rcf/fcf is the outward pull on the other end of the system (person) which is created to explain the fact that the guy is pulling the ball but the ball does not appear to be accelerating it toward him. The reaction to that fictitious force - the pull of the guy on the ball - is real. In actual fact, both bodies are accelerating about a common centre of rotation and there is no outward force.

AM

 

[A.T.]

I think that we should go ahead with the current example in your drawing.

I think it is beaten to death. Everyone agrees what happens there. The disagreement about the naming convention (by force direction or by force effect) cannot be resolved. It is a matter of personal preference.

The naming convention refers to its direction, and not to its effects.

Yes. It doesn't depend on how you define the objects. It doesn't depend on other forces that are eventually contributing to the acceleration. It is local and doesn't force you to analyze the acceleration of a larger part. It is more general and practical.

 

[A.T.]

It does not matter how you cut it

It does. The COM of the space station doesn't accelerate, if you treat it as 3 objects. And your convention depends on this acceleration.

In any rigid body there are, by definition, static tension forces everywhere. We don't concern ourselves with those forces.

What is "in the body" and what is an external force depends on how you cut it.

We can only look at the forces that cause acceleration.

Acceleration of what? Where the COMs are, and how they accelerate, depends on how you cut it.

Whose "naming convention"?

Don't look at me. I call my forces: F₁, F₂... But if I had to choose, I would prefer this simple convention, over your effect reasoning.

There is absolutely no difference between this "reactive centrifugal force" ("rcf") and the "fictitous centrifugal force" ("fcf").

You are back to page one of the previous thread. Here the differences between FCF and RCF again:
http://en.wikipedia.org/wiki/Reactive_centrifugal_force#Relation_to_inertial_centrifugal_force

 

[Andrew Mason]

This is not generally true. If you take two parts which are near each other or two parts which are opposite but not equally massive then their change in momentum may not be equal and opposite.

I should have made it clear that I was referring to any arbitrary division of the space station and contents into two parts.

This is true if you consider the opposite astronaut to be part of the space station (which is valid). In that case the centrifugal reaction force applied by the right astronaut to the space station is indeed equal to the mass x centripetal acceleration of the "space station & left astronaut".

The statement is not true if you consider the opposite astronaut not to be part of the space station (which is also valid). In that case the centrifugal reaction force applied by the right astronaut to the spact station is not equal to the mass x centripetal acceleration of the space station.

I was referring to the space station with one astronaut - eg. the single astronaut lying on the "floor" as described in my post #73. In this case there is just the space station itself whose centre of mass is not the centre of rotation (the centre of rotation being the geometric centre).

In the symmetrical station with 2 identical astronauts opposite each other, the reaction force applied by the right astronaut would be equal to the mass x centripetal acceleration of the other astronaut [+ 0 (which is the total centripetal acceleration of just the space station)].

AM

 

[Andrew Mason]

The point remains that whenever a centripetal force has a 3rd law pair which is directed away from the center of rotation that force is called the "centrifugal reaction force". That is the definition of the term and it is a common enough term that people should know what it means.

Perhaps you could provide a cite for where this term is used - please do not cite Mook and Vargish, Inside Relativity.

In some cases the centrifugal reaction force causes only centrifugal effects (material stresses, accelerations, etc.), but sometimes it causes centripetal effects. The naming convention refers to its direction, and not to its effects. Andrew Mason has every justification to dislike the naming convention (I dislike it for a different reason), but nevertheless it is well defined and well established.

A cite would be helpful.

Can you give us an example of how an the centrifugal reaction force cause anything to undergo centrifugal acceleration? I don't see that. It would be like the normal force of the Earth causing a person to jump. Or the force on the seat back of a person sitting in a car that is accelerating forward causing the car or the car seat to accelerate backward. As soon as he car or car seat stops accelerating forward the reaction force ends.

AM

 

[A.T.]

I should have made it clear that I was referring to any arbitrary division of the space station and contents into two parts.

So you demand the we:
- always consider the whole isolated system
- always cut it in exactly two parts
and you call this "arbitrary"? Sorry, but this is not what everybody wants/needs to do in an analysis. So I don't think many will want to use a naming convention based on that.

Can you give us an example of how an the centrifugal reaction force cause anything to undergo centrifugal acceleration?

The nice thing about the naming convention you oppose, is that it doesn't rely on effects like acceleration. Newtonts 3rd (the current mainstream interpretation) doesn't care about accelerations and their causes. But if you want an example, here it is:

A spaceship is moving on a circular path, by firing its engine continuously to provide the centripetal acceleration. The burned fuel is exerting a centripetal force on the ship, which causes a centripetal acceleration of the ship. The ship is exerting a centrifugal force on the burned fuel, which causes a centrifugal acceleration of the burned fuel.

 

[Dale]

I should have made it clear that I was referring to any arbitrary division of the space station and contents into two parts.

OK, under that restriction your statement is true, but no longer general. In fact, even requiring the system to be isolated makes it not general.

In the symmetrical station with 2 identical astronauts opposite each other, the reaction force applied by the right astronaut would be equal to the mass x centripetal acceleration of the other astronaut [+ 0 (which is the total centripetal acceleration of just the space station)].

Yes. Note however that the reaction force is applied to the space station and not the other astronaut. The 3rd law pair is the force on the astronaut from the floor and the force on the space station floor from the astronaut, not the forces on the two astronauts.

 

[Dale]

In any rigid body there are, by definition, static tension forces everywhere. We don't concern ourselves with those forces. We cannot possibly know what those forces are doing at any moment because they operate at atomic levels and the atoms are constantly undergoing thermal motion. We can only look at the forces that cause acceleration.

Nonsense. The whole subject of statics is concerned with these forces. We know a great deal about them, and we can easily use strain gauges to know what they are doing at any moment even when there is no acceleration (i.e. constant strain).

Whose "naming convention"? Give me a cite. No physics text that I have ever read refers to "centrifugal reaction force". The only book that I have found that references it is Mook and Vargish "Inside Relativity" (1987):

"As a result of the force acting on the ball, the ball deviates from uniform motion and follows a circular path, just as the planet docs when acted upon by the sun's gravity, But by Newton's third law, if you exert a force on the ball, the ball must exert an equal and opposite force on your hand, and it does. You feel this force as the " tug" of the ball on the string you hold. It is not necessary to posit a centrifugal force. What is sometimes called a "centrifugal force" is a reflection of the force you are exerting on the ball to keep it in a circular path. Similarly, the sun will feel such a reactive, centrifugal force from each of the planets that it holds in an orbit by its force of gravity."​

There is absolutely no difference between this "reactive centrifugal force" ("rcf") and the "fictitous centrifugal force" ("fcf"). The rcf/fcf is the outward pull on the other end of the system (person) which is created to explain the fact that the guy is pulling the ball but the ball does not appear to be accelerating it toward him. The reaction to that fictitious force - the pull of the guy on the ball - is real. In actual fact, both bodies are accelerating about a common centre of rotation and there is no outward force.

You are completely misreading their statement. It is clearly not the same as the fictitious centrifugal force.

 

[Dale]

Perhaps you could provide a cite for where this term is used - please do not cite Mook and Vargish, Inside Relativity.

Sure:

http://en.wikipedia.org/wiki/Reactive_centrifugal_force
http://en.wikipedia.org/wiki/Centrifugal_force#Reactive_centrifugal_force
http://physnet.org/modules/pdf_modules/m17.pdf
http://books.google.com/books?id=Qn...Ag#v=onepage&q="reactive centrifugal"&f=false
http://books.google.com/books?id=4G...AA#v=onepage&q="centrifugal reaction"&f=false
http://books.google.com/books?id=eF...nepage&q="reactive centrifugal force"&f=false
http://books.google.com/books?id=tv...BQ#v=onepage&q="centrifugal reaction"&f=false
http://books.google.com/books?id=xv...Bw#v=onepage&q="centrifugal reaction"&f=false

Can you give us an example of how an the centrifugal reaction force cause anything to undergo centrifugal acceleration? I don't see that.

A.T. recently proposed an example with a rocket, the centrifugal reaction force of the engine on the exhaust causes centrifugal acceleration of the exhaust. I also previously proposed an example with rapidly cutting bolts on a section of the floor. I am sure other similar examples could be constructed. I don't know if you want to count your dent formation examples since in those cases the acceleration is only centrifugal in the rotating frame.

Also, in all cases the centrifugal force causes the object on which it is exerted to have less centripetal acceleration than if the centripetal force were unopposed. That in itself is a centrifugal effect although not centrifugal acceleration.

Look, Andrew, this is pointless. The terminology is well-defined. You do have good reasons for not liking it, but it is common enough that people should know about it. You don't need to use it if you don't like it, but others surely will, so you should know what it is. All of your complaining about situations where it causes centripetal acceleration, while correct, doesn't make the term go away. The term is well-defined, and sufficiently common to be taken as standard terminology.

 

[Andrew Mason]

Nonsense. The whole subject of statics is concerned with these forces. We know a great deal about them, and we can easily use strain gauges to know what they are doing at any moment even when there is no acceleration (i.e. constant strain).

Strain gauges do not measure the actual static forces between the molecules of the body. They measure the response of the material to an applied external force. For a steel beam floating in space the strain gauge would measure 0 strain. But we know that there are enormous forces holding the steel molecules together.

My point was that when analysing the physics of the rotating rigid body, we assume it is rigid and able to remain perfectly rigid as it rotates. We don't have to be concerned with the static forces within the body to analyse the physics of rotation.

You are completely misreading their statement. It is clearly not the same as the fictitious centrifugal force.

Not only is it not "clearly" not the same as the fictitious centrifugal force, there is no clarity at all to the entire explanation of what "centrifugal force" means.

The only way that a planet's force on the sun could be called centrifugal is if you ignore the fact that the sun and other planets/asteroids etc are actually rotating around the centre of mass of the solar system. The pull of the Earth on the sun is NOT centrifugal in an inertial frame of reference. So if it is "centrifugal" it is because it is seen as force in the direction away from the perceived centre of rotation (the centre of the sun) in the non-inertial frame of reference of the sun. That is the fictitious centrifugal force.

As far as the swinging ball is concerned, the authors do no talk about the force of the ball on the rope. They talk about the force of the ball on the person. For a person to be swinging a ball the person has to rotate about the common centre of mass. For a small ball it may be hard to see this. But look at how an Olympic hammer thrower has to lean backward as he swings the hammer ball. The centre of rotation is between the thrower and the ball and both rotate around it.

AM

 

[A.T.]

The pull of the Earth on the sun is NOT centrifugal in an inertial frame of reference.

That is true in this case. You have been given other examples, where the force is centrifugal in an inertial frame of reference.

That is the fictitious centrifugal force.

No, that is nonsense in any case. A fictitious force is never part of a 3rd law pair.

 

[Dale]

Strain gauges do not measure the actual static forces between the molecules of the body. They measure the response of the material to an applied external force. For a steel beam floating in space the strain gauge would measure 0 strain. But we know that there are enormous forces holding the steel molecules together.

Are you talking about quantum mechanics here? I don't think that is relevant to the discussion which is essentially classical.

My point was that when analysing the physics of the rotating rigid body, we assume it is rigid and able to remain perfectly rigid as it rotates. We don't have to be concerned with the static forces within the body to analyse the physics of rotation.

And my point is that this is not generally true. Often you cannot assume perfect rigidity and often you will want to analyze the internal stresses of a rotating body, for instance, in turbine blade design. These internal stresses are well-characterized and understood despite your unsupported assertions to the contrary. I can provide a list of references for this if you wish, essentially any statics textbook, but it is simply not true that we can only know about forces which cause accelerations.

Not only is it not "clearly" not the same as the fictitious centrifugal force, there is no clarity at all to the entire explanation of what "centrifugal force" means.

Point well taken. "Clearly" is a matter of opinion. I should have said that it was clear to me, at least the part about the ball. I agree that the gravity example is wrong, but the ball example is correct and clear.

As far as the swinging ball is concerned, the authors do no talk about the force of the ball on the rope. They talk about the force of the ball on the person. For a person to be swinging a ball the person has to rotate about the common centre of mass. For a small ball it may be hard to see this. But look at how an Olympic hammer thrower has to lean backward as he swings the hammer ball. The centre of rotation is between the thrower and the ball and both rotate around it.

A human is an extended body, even if the center of rotation is between the COM of the human and the COM of the ball the reaction force on the hand will still be centrifugal. At least, that is how it has been any time I have swung a ball (or any other object) using my hand.

 

[Andrew Mason]

Sure:

http://en.wikipedia.org/wiki/Reactive_centrifugal_force
http://en.wikipedia.org/wiki/Centrifugal_force#Reactive_centrifugal_force
http://physnet.org/modules/pdf_modules/m17.pdf
http://books.google.com/books?id=Qn...Ag#v=onepage&q="reactive centrifugal"&f=false
http://books.google.com/books?id=4G...AA#v=onepage&q="centrifugal reaction"&f=false
http://books.google.com/books?id=eF...nepage&q="reactive centrifugal force"&f=false
http://books.google.com/books?id=tv...BQ#v=onepage&q="centrifugal reaction"&f=false
http://books.google.com/books?id=xv...Bw#v=onepage&q="centrifugal reaction"&f=false

No mainstream texts, I see. The Mook and Vargish example of the planet and sun is just wrong.

I would suggest that the reason mainstream texts do not use the term "centrifugal reaction force" to describe the reaction force to a centripetal force is because it obscures the physics. If the direction of the force is determined by the direction of the acceleration of the centre of mass of the body to which the force is applied, which I would suggest is the standard convention, all forces are all centripetal.

A.T. recently proposed an example with a rocket, the centrifugal reaction force of the engine on the exhaust causes centrifugal acceleration of the exhaust. I also previously proposed an example with rapidly cutting bolts on a section of the floor. I am sure other similar examples could be constructed. I don't know if you want to count your dent formation examples since in those cases the acceleration is only centrifugal in the rotating frame.

These are both good examples of why these do NOT cause centrifugal acceleration. We have discussed the bolt cutting example and all the bolt cutting does is STOP centripetal acceleration. It does not cause any acceleration away from the centre. The centres of mass of the astronaut and the spaceship both move at constant velocity in relation to an inertial point (although both would still rotate about their respective centres of mass).

The rocket is quite a bit more complicated. I will need some time to think about it some more. But the situation described seems incomplete - there must be other forces involved in order for the centre of mass of the rocket to rotate about the centre of the space station with the centre of mass of the space station remaining inertial.

Also, in all cases the centrifugal force causes the object on which it is exerted to have less centripetal acceleration than if the centripetal force were unopposed. That in itself is a centrifugal effect although not centrifugal acceleration.

I don't see how that is a centrifugal effect at all. If it is not opposed there would be no centripetal force at all. There would be no rotation.

When I pull on a box on a frictionless surface with a force F, the box pulls back on me with force F. Both forces are directed toward the same inertial point - the centre of mass. I don't see any forces directed away from the centre of mass. Total forces add to 0. I give myself and the box, respectively, equal and opposite changes in momentum. Both changes in momentum are toward the centre of mass.

Now, if I make the box more massive by ΔM, and apply the same force to the box as before (F), the box accelerates toward the centre of mass (which is now a different point) but with less acceleration (a' = F/(M+ΔM). I accelerate toward the centre of mass at the same rate as before (a=F/m). But the changes in momentum are the same for the box and me. So what seems like an effect directed away from the centre of mass if you only look only at the reduced acceleration of the large box, is not a reduction in the change of momentum of the box. There is no "effect" that is in a direction away from the centre of mass.

If you apply that to centripetal forces due to rotation, in the first case the less massive box and I rotate on the frictionless surface with acceleration toward the centre of mass: $F_{c-box} = m_{box}r_{box}ω^2$ and $F_{c-me} = m_{me}r_{me}ω^2$

With the more massive box but with the same pulling force between me and the box, the box and I will rotate about a centre of mass of the system that is closer to the centre of mass of the box by Δr. The box will have less centripetal acceleration and I will have greater centripetal acceleration toward the centre of mass of the system (longer radius).$F_{c-box} = m_{box}(r_{box}-\Delta r)ω^2$ and $F_{c-me} = m_{me}(r_{me}+\Delta r)ω^2$. If the centripetal forces do not change (i.e. my pull force on the box does not change), ω will have to be less so that my acceleration remains the same. Again, there is no centrifugal effect at all.

AM