What exactly is the reactive centrifugal force (split)

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SUMMARY

The discussion centers on the concept of reactive centrifugal force, particularly in the context of astronauts interacting within a space station. Participants clarify that the only force acting on each astronaut is the centripetal force from the space station's structure, while the force each astronaut exerts on the station creates tensions that result in a centripetal force on the other astronaut. The debate emphasizes the distinction between centripetal and centrifugal forces, asserting that centrifugal forces point outward from the center of rotation. The conversation also touches on Newton's Third Law, highlighting the importance of understanding action-reaction pairs in the context of forces and motion.

PREREQUISITES
  • Understanding of Newton's Third Law of Motion
  • Familiarity with centripetal and centrifugal forces
  • Basic knowledge of inertial and interaction forces
  • Concept of reference frames in physics
NEXT STEPS
  • Research the differences between centripetal and centrifugal forces in rotating systems
  • Study Newton's Third Law in various physical contexts
  • Explore the implications of inertial forces in non-inertial reference frames
  • Examine case studies involving forces in space environments, particularly in orbital mechanics
USEFUL FOR

Physics students, educators, and professionals in aerospace engineering or any field involving dynamics and forces in rotating systems will benefit from this discussion.

  • #91
DaleSpam said:
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.
 
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  • #92
A.T. said:
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.
 
  • #93
Code:
A.T. said:
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.

A.T. said:
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
 
  • #94
DaleSpam said:
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.
DaleSpam said:
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.
 
  • #95
Andrew Mason said:
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.
Andrew Mason said:
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.
Andrew Mason said:
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.
Andrew Mason said:
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.
Andrew Mason said:
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
 
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  • #96


DaleSpam said:
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
 
  • #97
DaleSpam said:
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
 
  • #98
Andrew Mason said:
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.

Andrew Mason said:
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.
 
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  • #99


Andrew Mason said:
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.

Andrew Mason said:
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.
 
  • #100
Andrew Mason said:
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).

Andrew Mason said:
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.
 
  • #101
Andrew Mason said:
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

Andrew Mason said:
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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  • #102
DaleSpam said:
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
 
  • #103
Andrew Mason said:
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.
Andrew Mason said:
That is the fictitious centrifugal force.
No, that is nonsense in any case. A fictitious force is never part of a 3rd law pair.
 
  • #104
Andrew Mason said:
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.

Andrew Mason said:
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.

Andrew Mason said:
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.

Andrew Mason said:
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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  • #105
DaleSpam said:

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
 
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  • #106
Andrew Mason said:
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.

Andrew Mason said:
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.

Andrew Mason said:
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.

Andrew Mason said:
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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  • #107
DaleSpam said:
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".

AM
 
  • #108
Andrew Mason said:
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.

Andrew Mason said:
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 spaceship 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.

Andrew Mason said:
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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  • #109
Andrew Mason said:
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?
I don't know why you would say "the centripetal force (presumably by the force exerted by the floor on the astronaut)" when I explicitly said "the floor is under stress with centripetal forces from the bolts". Please draw a free-body diagram for the floor. There are two sets of forces on the floor, one from the bolts, one from the astronaut. The forces from the bolts are directed centripetally. The forces from the astronaut are directed centrifugally. The centripetal force from the bolts is greater than the centrifugal reaction force from the astronaut, giving a net centripetal force as required by Newton's 2nd law. They are not a third law pair since (1) they act on the same object (the floor) and (2) they are between two different pairs of objects.

Andrew Mason said:
"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.
But I am not taking them together. I am analyzing the section of floor separately, which is perfectly legitimate. Taking it separately, there remains a "reaction" force on the floor from the astronaut and that force causes the floor to accelerate centrifugally. Likewise, there remains an "action" force on the astronaut from the floor which causes the astronaut to accelerate centripetally. There is simply no avoiding that.

Andrew Mason said:
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".
Can you provide a reference for this? I.e. a reference which states that, in a situation with multiple forces acting on an object, the centripetal force is the net force? I had always taken the centripetal force to be the individual force which was acting centripetally, not the net force, but perhaps I was wrong in this.

In any case, that doesn't change the analysis for the astronaut and floor during the shear wave propagation. The floor is exerting an unbalanced centripetal force on the astronaut which is accelerating centripetally. The reaction to that force is the centrifugal force on the floor which is also unbalanced and therefore the floor is accelerating centrifugally.
 
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  • #110


Andrew Mason said:
This is not a good example because the gravitational force and normal force are not third law pairs. They are not equal and opposite, for one thing. The normal force is always a bit less than the gravitational force except at the poles.
AM


Why does the normal force is always a bit less than the gravitational force?
 
  • #111


GT1 said:
Why does the normal force is always a bit less than the gravitational force?

The Earth is rotating. If you are "standing still" on your bathroom scale, you are actually traveling in a circular path at a rate of about one rotation per 24 hours.

If you add the inward gravitational force and the outward normal force together, the resultant is the small centripetal force required to keep you on this circular path.

If normal force were equal to gravitational force then you would not be accelerating. You would find yourself following a straight line path tangent to the rotating Earth and would quickly find yourself in outer space.
 
  • #112
DaleSpam said:
I don't know why you would say "the centripetal force (presumably by the force exerted by the floor on the astronaut)" when I explicitly said "the floor is under stress with centripetal forces from the bolts". Please draw a free-body diagram for the floor. There are two sets of forces on the floor, one from the bolts, one from the astronaut. The forces from the bolts are directed centripetally. The forces from the astronaut are directed centrifugally. The centripetal force from the bolts is greater than the centrifugal reaction force from the astronaut, giving a net centripetal force as required by Newton's 2nd law. They are not a third law pair since (1) they act on the same object (the floor) and (2) they are between two different pairs of objects.
Ok. Fair enough.

But I am not taking them together. I am analyzing the section of floor separately, which is perfectly legitimate. Taking it separately, there remains a "reaction" force on the floor from the astronaut and that force causes the floor to accelerate centrifugally. Likewise, there remains an "action" force on the astronaut from the floor which causes the astronaut to accelerate centripetally. There is simply no avoiding that.
Once the bolts are cut, the centre of mass of the astronaut/floor stops accelerating. While the astronaut and floor are still in contact, the centres of mass of bodies rotate about their common centre of mass. The relaxation of tensions just causes them to accelerate away from each other (briefly) ie. they push away on each other. That force is not a centripetal force because it is not constantly directed to a central point.

Can you provide a reference for this? I.e. a reference which states that, in a situation with multiple forces acting on an object, the centripetal force is the net force? I had always taken the centripetal force to be the individual force which was acting centripetally, not the net force, but perhaps I was wrong in this.
Centripetal just means always pointing to the centre, of course. And it is created naturally by rotation of a rigid or constrained body.

You could increase the radial tension artificially. For example in a bicycle wheel, you could tighten the spokes. This would have no bearing on the centripetal acceleration experienced by the rotating bicycle wheel. You would then have a centripetal force due to rotation plus a static radial tension.

In any case, that doesn't change the analysis for the astronaut and floor during the shear wave propagation. The floor is exerting an unbalanced centripetal force on the astronaut which is accelerating centripetally. The reaction to that force is the centrifugal force on the floor which is also unbalanced and therefore the floor is accelerating centrifugally.
It is not directed constantly toward a central point. The astronaut and floor are no longer rotating with the space station when the bolts are cut. They are only rotating about their centre of mass which is inertial (not accelerating). The push against each other is a linear force separating the two parts. It is not directed to the same point. I don't see how it could be centripetal.

AM
 
  • #113
Andrew Mason said:
That force is not a centripetal force because it is not constantly directed to a central point.
Yes, it is. Remember, the floor continues to rotate and the normal force continues to be perpendicular to the floor. For the parts that have not been reached by the shear wave, everything continues as normal.

Andrew Mason said:
Centripetal just means always pointing to the centre, of course. And it is created naturally by rotation of a rigid or constrained body.
I note that yet again you are unable to produce any reference supporting one of your claims when asked to do so.
 
  • #114
DaleSpam said:
Yes, it is. Remember, the floor continues to rotate and the normal force continues to be perpendicular to the floor. For the parts that have not been reached by the shear wave, everything continues as normal.
What is the source of the force that causes centripetal acceleration of the centre of mass of the floor after the bolts are cut?

After the bolts are cut, the floor continues to rotate about its centre of mass, not the centre of mass of the space station. When it is rotating with the space station, its centre of mass is accelerating but when it is freed (bolts cut) its centre of mass stops accelerating.

I note that yet again you are unable to produce any reference supporting one of your claims when asked to do so.
See any good physics text on the derivation of centripetal force: Alonso Finn, Physics, Addison Wesley p. 120-121;Barger and Olssen, Classical Mechanics, p 112, you will see the derivation of central force, defined as the force on a rotating body where dL/dt = 0:

\vec{L} = \vec{r}\times\vec{p}

\frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt}\times\vec{p} + \vec{r}\times\frac{d\vec{p}}{dt}

\frac{d\vec{L}}{dt} = \vec{v}\times m\vec{v} + \vec{r}\times\frac{d\vec{p}}{dt}

\frac{d\vec{L}}{dt} = 0 + \vec{r}\times\frac{d\vec{p}}{dt} = \vec{r}\times\vec{F}

If \frac{d\vec{L}}{dt}=0 then \vec{F} is entirely radial: \vec{r}\times\vec{F}=0 and this is called the centripetal force.

AM
 
  • #115
Andrew Mason said:
What is the source of the force that causes centripetal acceleration of the centre of mass of the floor after the bolts are cut?
The center of mass of the floor accelerates centrifugally immediately after the bolts are cut. The source of the force is the centrifugal reaction force from the astronaut.

Andrew Mason said:
After the bolts are cut, the floor continues to rotate about its centre of mass, not the centre of mass of the space station. When it is rotating with the space station, its centre of mass is accelerating but when it is freed (bolts cut) its centre of mass stops accelerating.
It isn't "freed" when the bolts are cut. It is "freed" when the shear wave reaches the astronaut's feet and they lose contact. During the time between the bolts being cut and the shear wave reaching the astronaut's feet the floor is in non-rigid body motion. Overall, the COM is accelerating centrifugally under the influence of the centrifugal force, outside the wavefront the floor is accelerating centrifugally, and inside the wavefront the floor continues to accelerate centripetally.

If you are not so familiar with stress and strain then perhaps it will help to think of the astronaut as standing on a spring. While the spring is decompressing the "top" continues to accelerate centripetally while the "bottom" and the COM accelerate centrifugally. It isn't quite the same since the shear wave is progressing laterally while the spring is expanding radially, but both are governed by Hooke's law, so they are closely related.

Again, the term "reactive centrifugal force" is common and you should be familiar with it. You have valid reasons for disliking it, but those reasons don't make the term go away. I sympathize with your position since I have similar objections to other terms that just won't go away, no matter how I wish they would (e.g. "relativistic mass").
 
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  • #116
DaleSpam said:
The center of mass of the floor accelerates centrifugally immediately after the bolts are cut. The source of the force is the centrifugal reaction force from the astronaut.
That reaction force ends as soon the centripetal acceleration of the astronaut ends.

It seems to me that the source of the "force" is the relaxation of the tension, so there is no net force on the centre of mass of either the astronaut or the floor. The relaxation of the tension creates a compression wave that propagates through the floor material much the same way that a stretched spring would oscillate if you were swinging it around and then let it go. The duration of those oscillations would depend on the damping forces within the spring. Are you suggesting that the centre of mass of the spring would still be undergoing centripetal acceleration while the oscillations continued?

If you are not so familiar with stress and strain then perhaps it will help to think of the astronaut as standing on a spring. While the spring is decompressing the "top" continues to accelerate centripetally while the "bottom" and the COM accelerate centrifugally. It isn't quite the same since the shear wave is progressing laterally while the spring is expanding radially, but both are governed by Hooke's law, so they are closely related.
If it is propagating laterally, how can it be the reaction to centripetal acceleration, which is entirely radial?

Again, the term "reactive centrifugal force" is common and you should be familiar with it. You have valid reasons for disliking it, but those reasons don't make the term go away. I sympathize with your position since I have similar objections to other terms that just won't go away, no matter how I wish they would (e.g. "relativistic mass").
Within any extended body rotating about a central body and whose orientation with respect to the radial vector from the centre of rotation does not change, there are necessarily tension forces that will arise within the body. This is due to the fact that different parts of the body are at a different distances from the centre of rotation (ie. centripetal acceleration of the part is different than ω^2rcom). These tensions are all directed radially from the com of the extended body. When the com of the extended body stops accelerating, these tensions continue because the extended body continues to rotate about its centre of mass. So there is no relaxation of these tensions.

AM
 
  • #117
Andrew Mason said:
That reaction force ends as soon the centripetal acceleration of the astronaut ends.
Yes, clearly.

Andrew Mason said:
It seems to me that the source of the "force" is the relaxation of the tension, so there is no net force on the centre of mass of either the astronaut or the floor.
Why wouldn't there be a net force? There is the force between the floor and the astronaut and there are no other forces on either to balance that force, so clearly there is a net force on each.

Andrew Mason said:
If it is propagating laterally, how can it be the reaction to centripetal acceleration, which is entirely radial?
Shear waves are transverse waves. The wave propagates laterally and the forces are transverse to the wave, i.e. the forces are radial.

That is the key difference between the floor example and the spring example. They both follow Hooke's law, but for the spring example, the wave would be a compression wave propagating radially rather than a shear wave propagating laterally. In both cases, the forces are radial. If it is easier for you to grasp the spring example then we can do that.
 
  • #118
DaleSpam said:
Yes, clearly.

Why wouldn't there be a net force? There is the force between the floor and the astronaut and there are no other forces on either to balance that force, so clearly there is a net force on each.
If they are in contact with each other they can have equal and opposite forces on each other. I meant there would be no centripetal acceleration of the centre of mass of either the astronaut or the liberated floor section toward the centre of rotation of the space station.

Shear waves are transverse waves. The wave propagates laterally and the forces are transverse to the wave, i.e. the forces are radial.

That is the key difference between the floor example and the spring example. They both follow Hooke's law, but for the spring example, the wave would be a compression wave propagating radially rather than a shear wave propagating laterally. In both cases, the forces are radial. If it is easier for you to grasp the spring example then we can do that.
The physics here cannot depend on whether the wave is a shear wave or a compression wave. I don't really see much of a shear wave here because the difference in centripetal force across the thickness of the floor is very small, assuming the thickness of the floor is small compared to its distance from the centre of rotation.

It is very easy to create an acceleration that is outward from the centre of rotation at a given moment. You would just have to have an extended object rotating about a central point at speed ω and also rotating about its centre of mass at speed ω+Δω. This is what happens in some amusement rides (eg. Tilt A Whirl). You can set up compressed springs and then release them, sending mass outward. Of course, these are not third law pairs to the centripetal force.

AM
 
  • #119
Andrew Mason said:
The physics here cannot depend on whether the wave is a shear wave or a compression wave
This is a very weird comment. Shear waves and compression waves are physically different waves. As I have described, we can set up a scenario with either kind of wave, but they are physically different scenarios.

So if you have trouble understanding shear waves we can do compression waves. But I don't know why you would say that the physics cannot depend on the nature of the wave.

Andrew Mason said:
I don't really see much of a shear wave here because the difference in centripetal force across the thickness of the floor is very small
What does the difference in centripetal force across the thickness of the floor have to do with shear? I don't think you understand shear stress at all.

Do you have access to a statics textbook? I am glad to help you with shear, but if you prefer we can switch to the compression example. Just let me know your preference.
 
  • #120
DaleSpam said:
This is a very weird comment. Shear waves and compression waves are physically different waves. As I have described, we can set up a scenario with either kind of wave, but they are physically different scenarios.

So if you have trouble understanding shear waves we can do compression waves. But I don't know why you would say that the physics cannot depend on the nature of the wave.

What does the difference in centripetal force across the thickness of the floor have to do with shear? I don't think you understand shear stress at all.
Well, I was assuming you were talking about a stress that is the result of two opposing forces that act on the same body in a way that tends to shear - make parts of the body that are close together move in opposite directions. Not sure how that applies here so if I am missing something, feel free to enlighten me.

Do you have access to a statics textbook? I am glad to help you with shear, but if you prefer we can switch to the compression example. Just let me know your preference.
No mention of shear stress in any of my physics texts. Sounds like engineering.

AM
 

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