What exactly is the reactive centrifugal force (split)


by A.T.
Tags: centrifugal, force, reactive, split
A.T.
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Feb5-13, 09:34 AM
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Quote Quote by DaleSpam View Post
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 space ship 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.
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Feb5-13, 01:56 PM
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Quote Quote by A.T. View Post
Since the extended body aspect seems to be a reason for confusion, I propose to use an even simpler example:

A space ship 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.
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Feb5-13, 02:27 PM
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Quote Quote by A.T. View Post
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 objectís 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.

Quote Quote by A.T. View Post
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.

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A.T.
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Feb5-13, 02:34 PM
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Quote Quote by DaleSpam View Post
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.
Quote Quote by DaleSpam View Post
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.
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Feb5-13, 03:27 PM
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Quote Quote by Andrew Mason View Post
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.
Quote Quote by Andrew Mason View Post
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.
Quote Quote by Andrew Mason View Post
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.
Quote Quote by Andrew Mason View Post
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.
Quote Quote by Andrew Mason View Post
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/Reactiv...trifugal_force
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Feb5-13, 06:33 PM
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Quote Quote by DaleSpam View Post
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)].

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Andrew Mason
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Feb5-13, 07:20 PM
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Quote Quote by DaleSpam View Post
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.

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A.T.
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Feb6-13, 01:41 AM
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Quote Quote by Andrew Mason View Post
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.

Quote Quote by Andrew Mason View Post
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 space ship 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.
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Feb6-13, 06:28 AM
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Quote Quote by Andrew Mason View Post
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.

Quote Quote by Andrew Mason View Post
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.
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Quote Quote by Andrew Mason View Post
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).

Quote Quote by Andrew Mason View Post
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.
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Feb6-13, 06:58 AM
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Quote Quote by Andrew Mason View Post
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/Centrif...trifugal_force
http://physnet.org/modules/pdf_modules/m17.pdf
http://books.google.com/books?id=QnJ...gal%22&f=false
http://books.google.com/books?id=4Gm...ion%22&f=false
http://books.google.com/books?id=eF-...rce%22&f=false
http://books.google.com/books?id=tvk...ion%22&f=false
http://books.google.com/books?id=xvS...ion%22&f=false

Quote Quote by Andrew Mason View Post
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.
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Feb6-13, 10:59 AM
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Quote Quote by DaleSpam View Post
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.

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A.T.
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Feb6-13, 12:23 PM
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Quote Quote by Andrew Mason View Post
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.
Quote Quote by Andrew Mason View Post
That is the fictitious centrifugal force.
No, that is nonsense in any case. A fictitious force is never part of a 3rd law pair.
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Feb6-13, 03:29 PM
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Quote Quote by Andrew Mason View Post
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.

Quote Quote by Andrew Mason View Post
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.

Quote Quote by Andrew Mason View Post
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.

Quote Quote by Andrew Mason View Post
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.
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Feb7-13, 07:48 PM
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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 space ship 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: [itex]F_{c-box} = m_{box}r_{box}ω^2[/itex] and [itex]F_{c-me} = m_{me}r_{me}ω^2[/itex]

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).[itex]F_{c-box} = m_{box}(r_{box}-\Delta r)ω^2[/itex] and [itex]F_{c-me} = m_{me}(r_{me}+\Delta r)ω^2[/itex]. 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.

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DaleSpam
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Feb7-13, 09:01 PM
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Quote Quote by Andrew Mason View Post
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.
Then please provide some references which demonstrate that this is the standard convention. Note, that being a standard convention is a much stronger claim than merely that it is one opinion or that it is an alternate terminology. I don't know how you can possibly support this, but then again, you haven't supported any of your claims.

Quote Quote by Andrew Mason View Post
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.
Yes, it does. Between the sudden cutting of the bolts and the propagation of the shear wave to the feet of the astronaut there is still a centrifugal reaction force and this centrifugal reaction force accelerates the plate away from the center. I already explained this in detail, since you did not respond further I thought that you had understood it.

Quote Quote by Andrew Mason View Post
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.
No other force is needed. Please analyze it in depth. If a rocket is spinning and burning its engines to produce a constant magnitude of acceleration then there exists some inertial reference frame where it is in uniform circular motion.

Quote Quote by Andrew Mason View Post
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.
Where did you get this idea? The direction of the force is not automatically towards the COM. If you pull horizontally then the force is only directed towards the COM if it is vertically aligned with the COM. Otherwise the force could be directed above or below the COM.

Andrew, this continued discussion is pointless. The terminology exists, is well defined, and commonly accepted. You are absolutely correct that the centrifugal reaction force can cause centripetal acceleration in some circumstances, but that doesn't change a thing.
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Feb7-13, 11:02 PM
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Quote Quote by DaleSpam View Post

Yes, it does. Between the sudden cutting of the bolts and the propagation of the shear wave to the feet of the astronaut there is still a centrifugal reaction force and this centrifugal reaction force accelerates the plate away from the center. I already explained this in detail, since you did not respond further I thought that you had understood it.
Here is what you said:
"I specified that the bolts were "suddenly" cut for a very important reason. As the astronaut is standing on the floor the floor is under stress with centripetal forces from the bolts and a centrifugal reaction force from the astronaut. The centripetal force is greater than the centrifugal reaction force so there is a net acceleration towards the center.

When the bolts are suddenly cut the stress is relieved from the outside of the section of floor, but the inner part of the floor (where the astronaut is standing) is still under stress. This sets up a shear wave where the floor material transitions from stress to stress-free. During the time between when the bolts are suddenly cut and when that shear wave reaches the feet of the astronaut the centrifugal reaction force still exists, the feet and floor are still in contact, and the floor is accelerating in a direction away from the center. It may help to think of the floor as being made of a stretchy rubber material.

The centrifugal force is every bit as "centrifugal" as the centripetal force is "centripetal". The centrifugal force points away from the center, the centripetal points towards the center. If either is unbalanced then it will result in acceleration in the corresponding direction. If there are other forces involved then the actual acceleration depends on the net force, per Newton's 2nd law.
Let's deal with the first paragraph:
"I specified that the bolts were "suddenly" cut for a very important reason. As the astronaut is standing on the floor the floor is under stress with centripetal forces from the bolts and a centrifugal reaction force from the astronaut. The centripetal force is greater than the centrifugal reaction force so there is a net acceleration towards the center."
This makes no sense to me. How can the centripetal force (presumably by the force exerted by the floor on the astronaut) be greater than the "centrifugal reaction force" (the force exerted by the astronaut on the floor, I assume) if they are equal and opposite 3rd law pairs? There is something wrong here because the centripetal force is not opposed by any force. The centripetal force, by definition, is mac = ω2r. What force is opposing it?

Now your second paragraph:
"When the bolts are suddenly cut the stress is relieved from the outside of the section of floor, but the inner part of the floor (where the astronaut is standing) is still under stress. This sets up a shear wave where the floor material transitions from stress to stress-free. During the time between when the bolts are suddenly cut and when that shear wave reaches the feet of the astronaut the centrifugal reaction force still exists, the feet and floor are still in contact, and the floor is accelerating in a direction away from the center. It may help to think of the floor as being made of a stretchy rubber material.
You appear to be saying that the relaxation of the stress forces within the floor will cause the floor to expand against the astronaut (and, presumably the relaxation of similar tensions in the astronaut will cause the astronaut to expand against the floor section that has been liberated from the space station). But once the bolts are cut the centre of mass of the astronaut and floor section (taken together so long as they are exerting forces on each other) defines an inertial reference frame - no external forces are acting on them. So there is no acceleration of the centre of mass of the floor/astronaut. The astronaut and floor section will briefly (very briefly) push off against each other as the tensions are relaxed, but that is not a reaction to the centripetal force which is zero at that moment.

Finally, the third paragraph:
"The centrifugal force is every bit as "centrifugal" as the centripetal force is "centripetal". The centrifugal force points away from the center, the centripetal points towards the center. If either is unbalanced then it will result in acceleration in the corresponding direction. If there are other forces involved then the actual acceleration depends on the net force, per Newton's 2nd law."
Again, the centripetal force is always unbalanced because it is defined as centripetal acceleration x mass. There is never centrifugal acceleration (unless you are in an accelerating frame of reference) even if one were to accept the idea of calling the reaction force to the force causing centripetal acceleration "centrifugal".

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A.T.
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Feb8-13, 03:00 AM
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Quote Quote by Andrew Mason View Post
The rocket is quite a bit more complicated.
It is a dead simple, minimal scenario. If your proposed naming reasoning gets complicated here already, then I doubt it will catch on.

Quote Quote by Andrew Mason View Post
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.
There is no inertial space station in the rocket scenario. Here it is again:

A space ship is moving on a circular path as seen from an inertial frame, by firing its engine 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.

Quote Quote by Andrew Mason View Post
Again, the centripetal force is always unbalanced because it is defined as centripetal acceleration x mass.
That is the definition of "net force", not of "centripetal force". The net force can be centripetal but so can individual forces, which are different from the net force.


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