What exactly is the reactive centrifugal force (split)


by A.T.
Tags: centrifugal, force, reactive, split
Andrew Mason
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Jan31-13, 06:16 PM
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Quote Quote by stevendaryl View Post
Hmm. Let's see if that's true. Suppose you have three objects 1, 2, 3. There are correspondingly 6 forces:
  1. [itex]\stackrel{\rightarrow}{F_{12}}[/itex] = force of 1 on 2
  2. [itex]\stackrel{\rightarrow}{F_{21}}[/itex] = force of 2 on 1
  3. [itex]\stackrel{\rightarrow}{F_{13}}[/itex] = force of 1 on 3
  4. [itex]\stackrel{\rightarrow}{F_{31}}[/itex] = force of 3 on 1
  5. [itex]\stackrel{\rightarrow}{F_{23}}[/itex] = force of 2 on 3
  6. [itex]\stackrel{\rightarrow}{F_{32}}[/itex] = force of 3 on 2

The fact that the situation is static means that the force on any object is zero. So we have:
  1. [itex]\stackrel{\rightarrow}{F_{12}} + \stackrel{\rightarrow}{F_{32}} = 0[/itex]
  2. [itex]\stackrel{\rightarrow}{F_{13}} + \stackrel{\rightarrow}{F_{23}} = 0[/itex]
  3. [itex]\stackrel{\rightarrow}{F_{21}} + \stackrel{\rightarrow}{F_{31}} = 0[/itex]

That's three constraints. I don't see how you can derive that
[itex]\stackrel{\rightarrow}{F_{12}} + \stackrel{\rightarrow}{F_{21}} = 0[/itex]
If it is static, all the forces on each object must sum to 0. So:

[itex]\vec{F}_{12} + \vec{F}_{32} + \vec{F}_{13} + \vec{F}_{23} + \vec{F}_{21} + \vec{F}_{31} = 0[/itex]

You get that from the first law.

If [itex]\vec{F}_{12} + \vec{F}_{32} = 0[/itex] and if 2 is not accelerating then either [itex]\vec{F}_{12} = \vec{F}_{32} = 0[/itex] or there are other forces acting on 2.

In that case you would have to write:

[itex]\vec{F}_{12} + \vec{F}_{32} + \vec{F}_{13} + \vec{F}_{23} + \vec{F}_{21} + \vec{F}_{31} + \sum F_{i(ext)} = 0[/itex], which is again just the first law.

Take a simpler case of a magnet (body 1) and an iron bar (body 2) stuck to it by the magnet's magnetic force. The total force of the magnet on the bar is 0 and the total force of the bar on the magnet is 0. If it were otherwise, there would be acceleration.

AM
Andrew Mason
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Jan31-13, 06:33 PM
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Quote Quote by A.T. View Post
It is the interpretation of Newton's third that you will find in every modern physics book. It doesn't matter how you call this Law ("3rd law" or "generalized 3rd law"). All your hand waving cannot change the fact that:
- It is a law of physics.
- It states two equal but opposite forces.
- It applies to static cases too.
- It means that my astronauts exert a centrifugal reaction force on the station
We agree that there is a force exerted on the space station by the astronaut that is a reaction to the force of the space station causing the astronaut to experience centripetal acceleration. All I am saying is that that reaction force causes the space station centre of mass to undergo centripetal acceleration toward the centre of rotation. So the reaction to the centripetal change of motion of the astronaut is a centripetal change of motion of the thing that is acting on the astronaut. And that is what the third law says must occur.

AM
Andrew Mason
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Jan31-13, 06:38 PM
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Quote Quote by stevendaryl View Post
I agree with you that thinking of equal and opposite forces works better than thinking of equal and opposite changes of motion, but in the case you are talking about, the change in momentum of one of the masses is equal and opposite to the vectorial sum of the changes of momentum of the other two masses. That might be what AM meant.
Exactly. See my post #135 above.

AM
Andrew Mason
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Jan31-13, 06:45 PM
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Quote Quote by DaleSpam View Post
Yes, this is correct. Note that the gravitational force (mg) is not equal to the change in momentum (ma=0). So your correct statement here contradicts your own incorrect statement elsewhere.
The reason there is no change in momentum is because the body and the earth are each exerting a total force of 0 on each other. If it were otherwise, there would be an acceleration.

Consider three identical masses, each connected by identical springs of neglible mass to both of the others and starting at rest from a position where the springs are all equally stretched. Each object is experiencing a change in motion and there is no other body experiencing an equal and opposite change in motion.
True, but there is a change in motion of the other two that combines to be equal and opposite to the change in motion of the first. The third law says that you cannot have just one body changing its motion. See my post #135 above.

AM
DaleSpam
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Jan31-13, 09:14 PM
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Quote Quote by Andrew Mason View Post
It is interesting that Newton speaks of "bodies ... not hindered by any other impediments". One can argue whether he was equating action with "force on an otherwise unimpeded body", but that is not an unreasonable interpretation.
That phrase seems to contradict your position and support the mainstream interpretation. I do not think that this reference supports your interpretation.

Also, this may be a translation error. Other translations are pretty clear on the standard interpretation. "Whatever draws or presses another is as much drawn or pressed by that other. If you press a stone with your finger, the finger is also pressed by the stone. "
http://www.marxists.org/reference/su.../en/newton.htm

Do you have any mainstream scientific reference that clearly supports your unorthodox stance? To ensure clarity, it would be best if it expressed the law mathematically. I.e. the usual version is something like [itex]F_{12}=-F_{21}[/itex] but your version would be something like [itex]dp_1/dt = - dp_2/dt[/itex]. Can you find a reference that actually expresses the third law that way?

Here are some clear references for the usual way:
http://en.wikipedia.org/wiki/Newton%....27s_third_law
http://www.phy6.org/stargaze/Snewton3.htm
http://www.ck12.org/concept/Newtons-...s-Third-Law%2F
DaleSpam
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Jan31-13, 09:20 PM
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Quote Quote by Andrew Mason View Post
The reason there is no change in momentum is because the body and the earth are each exerting a total force of 0 on each other. If it were otherwise, there would be an acceleration.
Agreed. Since the total force is zero then, according to your interpretation, there is no third law pair for either the normal force or the gravitational force.

Quote Quote by Andrew Mason View Post
True, but there is a change in motion of the other two that combines to be equal and opposite to the change in motion of the first.
But that isn't what the third law says, even in your interpretation as expressed above. The Newton quote certainly doesn't lend support to this weird three-body version of the third law. You are just making things up as you go along, and contradicting yourself and your sources.
A.T.
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Feb1-13, 01:18 AM
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Quote Quote by Andrew Mason View Post
All I am saying is that that reaction force causes the space station centre of mass to undergo centripetal acceleration toward the centre of rotation.
And you are wrong as usual. The space station centre of mass has no acceleration toward the centre of rotation. The space station centre of mass is static at the centre of rotation.



Needless to say that even if you were right, it would be still irrelevant. It would not change the fact that the contact force Frcf acts centrifugally on the wall.
A.T.
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Feb1-13, 01:33 AM
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Quote Quote by Andrew Mason View Post
Do you consider Newton to be mainstream?
Your interpretation of Newton is not mainstream.

But even if you were right about what he initially meant centuries ago in his Latin prose, it would still be irrelevant. Laws of physics are often generalized over the course of time. Today we have a 3rd law that applies to forces regardless of accelerations.
A.T.
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Feb1-13, 03:18 AM
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Quote Quote by DaleSpam View Post
your version would be something like [itex]dp_1/dt = - dp_2/dt[/itex].
Expressed that way, it would be a rather useless law, that applies only in a special case where no other forces are acting (or cancel exactly). It would not be a law, but rather a case specific rule. A law in physics is usually a generally applicable rule.
Quote Quote by DaleSpam View Post
Can you find a reference that actually expresses the third law that way?
I'm sure many school books use the simplest possible example (two bodies, one interaction) to introduce the 3rd Law. But if Andrew goes a few pages further, he should find examples where there are more forces acting.
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Quote Quote by A.T. View Post
And you are wrong as usual. The space station centre of mass has no acceleration toward the centre of rotation. The space station centre of mass is static at the centre of rotation.
Are you including both astronauts in the space station? The space station I refer to includes only the other astronaut. Read my #136 post:
"With respect to the astronaut space-station example, the third law says that a body undergoing centripetal change in motion due to contact with another unimpeded body causes that other body to have an equal change of motion in the opposite direction. So if we are talking about the astronaut undergoing a centripetal change in motion due to contact with the space station/other astronaut (taken as one body), the third law "reaction" is a change in motion of the space station/other astronaut in the opposite direction. That change in motion always points to the centre of rotation."
Here is a simple question: If the reaction force to the centripetal acceleration of astronaut 1 is a centrifugal static force (ie it does not result in any contrary change in motion of the thing that is accelerating astronaut 1) what is the force that causes the centripetal acceleration of the centre of mass of the space station/other astronaut?

Needless to say that even if you were right, it would be still irrelevant. It would not change the fact that the contact force Frcf acts centrifugally on the wall.
We agree on the force and we agree on the direction. And all I am saying is that since this force causes the rest of the mass to which the astronaut is connected to undergo acceleration toward the centre of rotation, it is a true centripetal force. Calling it centrifugal is a misnomer.

AM
Andrew Mason
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Feb1-13, 06:40 AM
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Quote Quote by DaleSpam View Post
Agreed. Since the total force is zero then, according to your interpretation, there is no third law pair for either the normal force or the gravitational force.

But that isn't what the third law says, even in your interpretation as expressed above. The Newton quote certainly doesn't lend support to this weird three-body version of the third law. You are just making things up as you go along, and contradicting yourself and your sources.
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.

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.

AM
stevendaryl
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Feb1-13, 06:40 AM
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Quote Quote by Andrew Mason View Post
You get that from the first law.

If [itex]\vec{F}_{12} + \vec{F}_{32} = 0[/itex] and if 2 is not accelerating then either [itex]\vec{F}_{12} = \vec{F}_{32} = 0[/itex] or there are other forces acting on 2.
I don't get that, at all. From [itex]\vec{F}_{12} + \vec{F}_{32} = 0[/itex], you get [itex]\vec{F}_{12} = - \vec{F}_{32}[/itex], not that
[itex]\vec{F}_{12} = \vec{F}_{32} = 0[/itex]

The fact that 2 is not accelerating is what tells us that [itex]\vec{F}_{12} + \vec{F}_{32} = 0[/itex].
A.T.
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Feb1-13, 06:42 AM
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Quote Quote by Andrew Mason View Post
The space station I refer to includes only the other astronaut.
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.
Quote Quote by Andrew Mason View Post
...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.
stevendaryl
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Feb1-13, 07:00 AM
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Quote Quote by Andrew Mason View Post
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.
But with the "force" interpretation of the third law, it is easy to analyze the interaction, at least in the special case (which is most relevant in most applications of Newton's laws) in which all forces are either (1) instantaneous two-body interactions, or (2) contact forces.

As I said, in a three-body problem, there are 6 two-body interaction forces:

[itex]\vec{F_{12}}, \vec{F_{21}}, \vec{F_{13}}, \vec{F_{31}}, \vec{F_{23}}, \vec{F_{32}}[/itex]

where [itex]\vec{F_{IJ}}[/itex] means the force of object [itex]I[/itex] on object [itex]J[/itex]. If it's a static situation, the sum of the forces on each object is zero. So we conclude:

[itex]\vec{F_{12}} = -\vec{F_{32}}[/itex]
[itex]\vec{F_{13}} = -\vec{F_{23}}[/itex]
[itex]\vec{F_{21}} = -\vec{F_{31}}[/itex]

Then Newton's third law (interpreted in terms of equal and opposite forces) implies
[itex]\vec{F_{12}} = -\vec{F_{21}}[/itex]
[itex]\vec{F_{13}} = -\vec{F_{31}}[/itex]
[itex]\vec{F_{23}} = -\vec{F_{32}}[/itex]

These 6 equations have the solution
[itex]\vec{F_{12}} = \vec{F_{21}} = \vec{F_{13}} = \vec{F_{31}}= \vec{F_{23}} = \vec{F_{32}} = 0[/itex]

But I don't see how you can get that conclusion without invoking the force version of the 3rd law.
Andrew Mason
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Feb1-13, 07:03 AM
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Quote Quote by A.T. View Post
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
DaleSpam
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Feb1-13, 07:10 AM
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Quote Quote by Andrew Mason View Post
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 Quote by Andrew Mason View Post
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 [itex]dp_i/dt=\Sigma dp_{j \ne i}/dt[/itex]. You need to find somewhere that states not just that this expression is true, but that this is Newton's 3rd law.
A.T.
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Feb1-13, 07:29 AM
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Quote Quote by Andrew Mason View Post
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 Quote by Andrew Mason View Post
There is an apparent force in the rotating frame.
And a real centrifugal force on the wall Frcf. It exists in every frame.
Quote Quote by Andrew Mason View Post
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.
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Feb1-13, 09:04 AM
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Quote Quote by A.T. View Post
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


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