What exactly is the reactive centrifugal force (split)

[Andrew Mason]

Nice try. Why not include both astronauts in the space station, so we have just one object, no interactions and finally no reactive centrifugal force.

Sorry, you don't get to decide how people divide the system into parts. In my example there are 3 objects: station & 2 astronauts. This is a perfectly valid way to analyze the scenario. So deal with it, and don't try to change it.

I am not changing anything. Are you saying that the presence of the other astronaut does not affect the force that the space station applies to the first? Since it does, then you have to include it as part of the "other body" that is interacting with the first astronaut.

Every atom in that space station/2 astronaut example is accelerating. There are a gazillion tension forces. The total force on each atom is the atom's mass multiplied by its centripetal acceleration.

Irrelevant. The force is there, regardless of acceleration. In the rotating frame there is no acceleration at all, but there still is a force on the wall pointing outwards: centrifugal.

There is an apparent force in the rotating frame. The real forces are causing accelerations toward the centre.

AM

 

[Dale]

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.

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.

 

[A.T.]

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.

There is an apparent force in the rotating frame.

And a real centrifugal force on the wall F_rcf. It exists in every frame.

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.

 

[Andrew Mason]

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 F_rcf. 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

 

[A.T.]

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.

Static?! You can't be serious.

Yes, static. Nothing moves in the rotating frame of my scenario.

It is rotating.

Not in the rotating frame.

 

[stevendaryl]

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.

 

[Andrew Mason]

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

 

[Dale]

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.

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.

 

[A.T.]

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.

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?

We use the inertial frame to analyse forces.

Who is "we"? I show the forces in both frames in the diagram.

 

[Andrew Mason]

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

 

[Andrew Mason]

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

 

[A.T.]

Any ambiguity is Newton's

There is no ambiguity in physics books today. The law deals with forces and accelerations are irrelevant.

 

[A.T.]

Is this dent caused by the reaction force to centripetal [STRIKE]acceleration[/STRIKE] force

Yes. But don't confuse forces and accelerations.

 

[Andrew Mason]

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

 

[Dale]

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.

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.

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.

 

[Dale]

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.

 

[Andrew Mason]

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

 

[Andrew Mason]

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

 

[A.T.]

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 not caused by the Big Bang due to the fact that without the Big Bang there would be no astronaut and space station in the first place? You can always construct endless cause-effect chains. But usually we consider forces acting on the object as the direct cause of the objects deformation. Therefore only a force acting on the wall can be a direct cause of the wall's deformation.

But keep in mind that the whole causation reasoning about forces is often more fuzzy intuitive mythology, than physics. It doesn't affect the quantitative results. The action doesn't cause the reaction in the 3rd law. They both happen simultaneously, and physics just says they are equal but opposite. The same is true for net force and acceleration in the 2nd law. Although we often say that the net force causes the acceleration, they both happen simultaneously, and physics just says they are proportional.

 

[A.T.]

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.

By insisting on your interpretation of Newtons 3rd you are contradicting modern mainstream physics. That is because the mainstream interpretation is more general than your interpretation. So you are basically saying that Newtons 3rd doesn't apply to all those cases, where mainstream physics says it does apply. See also my remarks in post #145.

 

[A.T.]

L = constant

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

 

[Dale]

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

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

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

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

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

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

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

 

[Andrew Mason]

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

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

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

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

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

AM

 

[Andrew Mason]

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

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

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

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

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

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

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

AM

 

[Dale]

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

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

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

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

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

Thank you!

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

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

 

[A.T.]

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

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

 

[A.T.]

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

Read post #69.

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

 

[Dale]

Andrew Mason,

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

 

[A.T.]

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

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

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

 

[Andrew Mason]

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

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

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

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

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

AM

 

[Andrew Mason]

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

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

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

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

AM

 

[A.T.]

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

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

 

[A.T.]

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

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

No. Read it again.

 

[Dale]

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

Sure, Newton's 2nd law.

That can never happen here.

Why not?

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

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

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

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

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

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

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

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

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

 

[A.T.]

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

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

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

 

[Andrew Mason]

No. Read it again.

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

AM

 

[A.T.]

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

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

 

[Andrew Mason]

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

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

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

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

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

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

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

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

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

AM

 

[A.T.]

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

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

 

[Dale]

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

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

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

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

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

 

[A.T.]

for extended bodies a centrifugal reaction force can cause centripetal acceleration

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

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

 

[Dale]

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

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

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

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

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

 

[Andrew Mason]

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

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

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

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

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

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

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

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

AM

 

[A.T.]

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

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

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

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

 

[A.T.]

It does not matter how you cut it

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

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

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

We can only look at the forces that cause acceleration.

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

Whose "naming convention"?

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

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

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

 

[Andrew Mason]

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

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

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

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

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

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

AM

 

[Andrew Mason]

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

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

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

A cite would be helpful.

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

AM

 

[A.T.]

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

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

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

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

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

 

[Dale]

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

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

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

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

 

[Dale]

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

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

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

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

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

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