Mentor

## What exactly is the reactive centrifugal force (split)

 Quote by Andrew Mason It is not a weird three-body version of the third law. Body A exerts forces on both Body B and Body C. Body C and Body B exert equal and opposite forces on Body A. But in order to analyse the interaction using forces you need to know the instantaneous direction and duration of these forces and whether the bodies are rigid or deform during the interaction etc. It is not easy to analyse using forces.
It is clear that you know that the usual formulation of the 3rd law is correct. Whether or not it is easy is irrelevant. There are lots of laws of physics that are not easy.

 Quote by Andrew Mason The third law says that the action (which we can take to be A's action on B and C) is accompanied by an equal and opposite reaction (the action of B and C on A). So we can say that the change in motion of A is accompanied by equal and opposite changes in motion of B and C. This is true at all times before during and after the interaction.
This isn't going to work. Not only are you trying to peddle the idea that in "to every action there is an equal and opposite reaction" the word "action" refers to changes in momentum; you are now trying to say that the third law is "to every action there is an equal and opposite sum of reactions". Good luck finding some authoritative support for that. It certanily isn't what Newton said above, and I have never seen it expressed that way.

Please find a clear reference. Now the mathematical expression you are looking for is $dp_i/dt=\Sigma dp_{j \ne i}/dt$. You need to find somewhere that states not just that this expression is true, but that this is Newton's 3rd law.

 Quote by Andrew Mason I am not changing anything.
Yes you are. The definition of the objects is part of the example, to demonstrate reactive centrifugal forces. You don’t get to change my example, just because it proves you wrong.
 Quote by Andrew Mason There is an apparent force in the rotating frame.
And a real centrifugal force on the wall Frcf. It exists in every frame.
 Quote by Andrew Mason The real forces are causing accelerations toward the centre.
There is no acceleration in the rotating frame. It is all static. But there still is a real centrifugal force on the wall, exerted by the astronaut. Accelerations are irrelevant to this.

Recognitions:
Homework Help
 Quote by A.T. Yes you are. The definition of the objects is part of the example, to demonstrate reactive centrifugal forces. You don’t get to change my example, just because it proves you wrong. And a real centrifugal force on the wall Frcf. It exists in every frame.
Ok. It is a centrifugal force that has as its only physical effect the acceleration of the centre of mass of the rest of the system toward the centre of rotation. Is that how you want to explain this to students?

 There is no acceleration in the rotating frame. It is all static. But there still is a real centrifugal force on the wall, exerted by the astronaut. Accelerations are irrelevant to this.
It is rotating. Everything is accelerating. We use the inertial frame to analyse forces. The absence of apparent acceleration in the non-inertial frame is not real. Why are you even referring to this? This makes it very difficult for the student to distinguish between the apparent centrifugal force and the real reaction force (which you say is a static centrifugal force and I say is a non-static centripetal force).

AM

 Quote by Andrew Mason Ok. It is a centrifugal force that has as its only physical effect the acceleration of the centre of mass of the rest of the system toward the centre of rotation.
Wrong and irrelevant. It can have other physical effects, like creating a dent in the wall. But the physical effects of the force are irrelevant to the fact that it acts centrifugally.
 Quote by Andrew Mason Static?!!! You can't be serious.
Yes, static. Nothing moves in the rotating frame of my scenario.
 Quote by Andrew Mason It is rotating.
Not in the rotating frame.

 Quote by Andrew Mason Take the simplest example of the three masses in a straight line: 1, 2, 3. None are accelerating. So: (1)$\vec{F}_{12} + \vec{F}_{32} = 0$ and (2)$\vec{F}_{21} = 0$ and (3)$\vec{F}_{23} = 0$. But $\vec{F}_{23} = -\vec{F}_{32}$ and $\vec{F}_{21} = -\vec{F}_{12}$ So that means: $\vec{F}_{12} = \vec{F}_{32} = 0$ The same thing applies if they are arranged in a triangle or any other shape. AM
No, it doesn't. Dale gave a counterexample. Here's a different one: You have three particles 1, 2, and 3. For simplicity, let them be in a straight line, so we only need consider the x-component of the forces. Let

$F_{12} = 0$
$F_{32} = 0$
$F_{13} = 1$
$F_{23} = -1$
$F_{31} = 2$
$F_{21} = -2$

That satisfies Newton's first law, but not the third.

Recognitions:
Homework Help
 Quote by A.T. Wrong and irrelevant. It can have other physical effects, like creating a dent in the wall. But the physical effects of the force are irrelevant to the fact that it acts centrifugally.
A dent in the wall would require a momentary reduction in the centripetal force being applied by the space station to the astronaut (at the given rotational speed) and a corresponding reduction in the astronaut's reaction force. If you say this reaction force is a centrifugal force it is very strange that it only causes an outward effect when it is reduced.

 Yes, static. Nothing moves in the rotating frame of my scenario.
You were posting as I was editing. I added:
It is rotating. Everything is accelerating. We use the inertial frame to analyse forces. The absence of apparent acceleration in the non-inertial frame is not real. Why are you even referring to this? This makes it very difficult for the student to distinguish between the apparent centrifugal force and the real reaction force (which you say is a static centrifugal force and I say is a non-static centripetal force).
 Not in the rotating frame.
We don't analyse real forces in the rotating frame.

AM

Mentor
 Quote by Andrew Mason A dent in the wall would require a momentary reduction in the centripetal force being applied by the space station to the astronaut (at the given rotational speed) and a corresponding reduction in the astronaut's reaction force.
There is no need for a reduction in the centripetal force in order to cause a dent. You are again taking one specific scenario and making a mistaken generalization.

 Quote by Andrew Mason We don't analyse real forces in the rotating frame.
Yes, we do. We also analyze fictitious forces there, but real forces are not ignored.

I note that you have not found a reference that writes a clear mathematical expression corresponding to your version and calls that the third law. All you have is an ambiguous translation that cannot be said to clearly contradict your approach.

 Quote by Andrew Mason If you say this reaction force is a centrifugal force it is very strange that it only causes an outward effect when it is reduced.

When you sand on a table, you exert a downwards force on the table. When that downwards force starts deforming the table downwards, there eventually will be a momentary reduction in that downwards force. Do you find it strange to call it a downwards force, because it creates a downwards effect when it is reduced?

 Quote by Andrew Mason We use the inertial frame to analyse forces.
Who is "we"? I show the forces in both frames in the diagram.

Recognitions:
Homework Help
 Quote by DaleSpam There is no need for a reduction in the centripetal force in order to cause a dent. You are again taking one specific scenario and making a mistaken generalization.
I suppose someone could soften the floor so the astronaut made a dent. Is that what you mean? The process of making the dent still involves an increase of the distance from the centre of rotation, which reduces the centripetal force and the reaction to that force.

 Yes, we do. We also analyze fictitious forces there, but real forces are not ignored. I note that you have not found a reference that writes a clear mathematical expression corresponding to your version and calls that the third law. All you have is an ambiguous translation that cannot be said to clearly contradict your approach.
I gave you Newton's explanation of the third law. You don't accept that? The translations from the latin are correct. Any ambiguity is Newton's not the translator's.

Acceleration is the second time derivative of the body's displacement vector relative to an inertial point. If you use as your reference a non-inertial point whose acceleration is constantly changing, how would you measure real acceleration?

AM

Recognitions:
Homework Help
 Quote by A.T. What is strange about this? When you sand on a table, you exert a downwards force on the table. When that downwards force starts deforming the table downwards, there eventually will be a momentary reduction in that downwards force. Do you find it strange to call it a downwards force, because it creates a downwards effect when it is reduced?
Suppose the top 10 cm of the floor of the space station was softened so that the astronaut's feet and centre of mass move 10 cm farther from the centre and cause a 10 cm dent in the floor. Is this dent caused by the reaction force to centripetal acceleration or is it the inertial effect of the reduction in centripetal force due to the softening of the floor? There is a big difference: one is a real force and the other is not. If one calls both forces (the real reaction force of the astronaut on the space station and the fictitious centrifugal force observed only in the rotating frame) centrifugal it gets very confusing. Even Newton was confused by centrifugal force.

AM

 Quote by Andrew Mason Any ambiguity is Newton's
There is no ambiguity in physics books today. The law deals with forces and accelerations are irrelevant.

 Quote by Andrew Mason Is this dent caused by the reaction force to centripetal acceleration force
Yes. But don't confuse forces and accelerations.

Recognitions:
Homework Help
 Quote by A.T. Yes. But don't confuse forces and accelerations.
Why is it not caused simply by the astronaut's inertia due to the inability of the floor to provide the needed centripetal acceleration? Why is it so fundamentally different if that 10 cm of the floor just momentarily dissolved?

By the way, I put it that way because you insist that one can have a force that produces no acceleration as a third law reaction to one that does produce acceleration. There is nothing wrong or confused in saying the force is related to an acceleration.

AM

Mentor
 Quote by Andrew Mason I suppose someone could soften the floor so the astronaut made a dent. Is that what you mean? The process of making the dent still involves an increase of the distance from the centre of rotation, which reduces the centripetal force and the reaction to that force.
No, the centripetal force is $mr\omega^2$, it increases as the distance from the center increases.

 Quote by Andrew Mason I gave you Newton's explanation of the third law. You don't accept that? The translations from the latin are correct. Any ambiguity is Newton's not the translator's.
Then Newton was ambiguous. I read it and to me it seems to be talking about forces. You read it and to you to you it seems to be talking about changes in momentum. It does not clearly support your position (nor does it clearly support mine).
EDIT: Actually, after having read the entire quote in context it seems quite clear that Newton is not supporting your view. The part of the quote you cited is not the law, but the last of three examples. First he gives the "action-reaction" formulation of the law. Then he clearly talks about forces, not accelerations, in an example of a finger pressing a stone and second example of a horse drawing a stone. Then he introduces a third example, this one clearly indicating the special case of two bodies with no other forces acting so that the force is equal to the change in momentum (by his second law). In that case your formulation is equivalent to the usual formulation, but in all of the other examples he clearly intended the usual formulation, not yours.
That is why I prefer sources that use math which is clear and unambiguous. All of those that I have seen clearly contradict your position. Can you provide any unambiguous (with math) support for your position? I provided unambiguous (with math) support for mine, so I don't think it is unreasonable to require the same from you.

 Quote by Andrew Mason Acceleration is the second time derivative of the body's displacement vector relative to an inertial point. If you use as your reference a non-inertial point whose acceleration is constantly changing, how would you measure real acceleration?
What is "real acceleration"? I would measure "coordinate acceleration" as the second time derivative of my coordinate location. I would measure "proper acceleration" with an accelerometer. I don't know what "real acceleration" is.

Mentor
 Quote by Andrew Mason Suppose the top 10 cm of the floor of the space station was softened so that the astronaut's feet and centre of mass move 10 cm farther from the centre and cause a 10 cm dent in the floor. Is this dent caused by the reaction force to centripetal acceleration or is it the inertial effect of the reduction in centripetal force due to the softening of the floor?
It is caused by the centrifugal reaction force. It is certainly not caused by anything related to "the reduction in centripetal force" since the centripetal force does not reduce but instead increases.

Recognitions:
Homework Help
 Quote by DaleSpam It is caused by the centrifugal reaction force. It is certainly not caused by anything related to "the reduction in centripetal force" since the centripetal force does not reduce but instead increases.
??? The reaction force completely disappears if the centripetal force disappears, which is what happens when the floor completely dissolves. So how can the reaction force cause the outward motion?

I know it seems counter-intuitive, but with no external torque the centripetal force decreases as r increases. We see this from the fact that gravity, which supplies the centripetal force for orbiting bodies, varies as 1/r^2.

AM

Recognitions:
Homework Help
 Quote by DaleSpam No, the centripetal force is $mr\omega^2$, it increases as the distance from the center increases.
Yes, but ω decreases as 1/r (L = constant) so the centripetal force decreases as 1/r: $F_c =mr\omega^2 = L^2/mr$

 Then Newton was ambiguous. I read it and to me it seems to be talking about forces. You read it and to you to you it seems to be talking about changes in momentum. It does not clearly support your position (nor does it clearly support mine).
It is an interesting question, I'll grant you that.

I found this commentary which wrestles with this very issue.

I know what you are saying but the real problem I have is that Newton talks about force in his first two laws but talks about action and reaction in his third and does not mention force. The first two laws are quite straight forward and unambiguous. But the third is not, due to his use of undefined terms. That said, we should not be too critical of any confusion Newton may have had or caused by his statement of this law. Newton's laws are a truly remarkable achievement when one considers the state of physical science at the time.

 That is why I prefer sources that use math which is clear and unambiguous. All of those that I have seen clearly contradict your position. Can you provide any unambiguous (with math) support for your position? I provided unambiguous (with math) support for mine, so I don't think it is unreasonable to require the same from you.
I don't think it is a matter of contradicting "my position" because I am just saying that Newton's third law is a statement about conservation of motion (momentum) and I don't think conservation of momentum can be contradicted.

 What is "real acceleration"? I would measure "coordinate acceleration" as the second time derivative of my coordinate location. I would measure "proper acceleration" with an accelerometer. I don't know what "real acceleration" is.
"Real acceleration" is a non-zero second derivative with respect to time of a body's displacement from a fixed point in an inertial reference frame. It does not matter what inertial reference frame you select. It is the same in all (in Galilean relativity).

AM