Is the concept of reactive centrifugal force valid?

[Andrew Mason]

Sorry, but "is capable of producing acceleration" is just vague gibberish again.

When I lean against a wall, I exert a real force on the wall, but it doesn't result in any acceleration. Equally the astronaut exerts a real centrifugal force on the wall which doesn't result in centrifugal acceleration.

What if the wall, after holding him in broke down just a little and gave way a little bit. Would the astronaut experience a little bit of centrifugal acceleration until the wall eventually held him again?

AM

 

[Dale]

1.If only this paragraph had been sent I would have replied that in order to get any answer at all it would be necessary to change at least one feature of the problem as it was originally set.

Do you understand now how that is not the case?

2.When I read this comment things started to drop into place including why there were some seemingly rather strange replies to my posts.In the first paragraph I was told not to change the problem but in the second paragraph it was suggested that I can in fact change the problem by using things such as extra rope.

The extra rope doesn't change the problem, it simply specifies some of the irrelevant details. In fact, for your design the extra rope is required in order to avoid changing the problem. As long as the external system supplies the right force to the center it is a legitimate external system for the problem and doesn't change any of the givens.

Do you feel that you understand now what it means to change the problem and why the details are irrelevant?

 

[rcgldr]

What if the wall, after holding him in broke down just a little and gave way a little bit. Would the astronaut experience a little bit of centrifugal acceleration until the wall eventually held him again?

From an inertial frame of reference, if the astronaut is not touching any surface of the rotating space station, then the astronaut moves with constant velocity without any acceleration. It's the surfaces of the space station that accelerate and eventually collide with the astronaut. From the rotating space station frame of reference, it appears that the astronaut is accelerating due to fictitious forces (centrifugal, corilios), and not the space station.

 

[A.T.]

What if the wall, after holding him in broke down ...

Changing the scenario will not prove that the real reactive centrifugal force doesn't exist in my scenario.

It is trivial to see that if you remove the centripetal force on the astronaut by the wall, then you also remove the centrifugal reaction force by the astronaut on the wall. It is a 3rd law force pair, that always acts together. The key thing about Newtons 3d is:

Both forces in Newtons 3d are always real forces (interaction forces). They both act in every reference frame. They both are exerted by some object on another object.

This is what differentiates the real reactive centrifugal force on the wall, from the inertial centrifugal force (pseudo / fictitious force) on the astronaut. The inertial force exists only in the rotating frame. The inertial force acts directly on the object, regardless of any interactions with other objects.

 

[Andrew Mason]

In Dale's scenario what would happen to the tension in the rope between A and B if the rope from A to the centre was cut? Would the tension between A and B relax?

In linear acceleration (eg. pulling the left end of Dale's rope to the left with constant force) there would be a tension in the rope between A and B while the acceleration was occurring but the tension would start to relax as soon as the rope to the left of A was cut. The two masses would accelerate toward each other for a brief moment after the rope was cut as the rope between them contracted. But in the case of rotation I am not sure it would. I am still thinking about it but I would appreciate your comments.

AM

 

[Andrew Mason]

Changing the scenario will not prove that the real reactive centrifugal force doesn't exist in my scenario.

It is trivial to see that if you remove the centripetal force on the astronaut by the wall, then you also remove the centrifugal reaction force by the astronaut on the wall. It is a 3rd law force pair, that always acts together. The key thing about Newtons 3d is:

Both forces in Newtons 3d are always real forces (interaction forces). They both act in every reference frame. They both are exerted by some object on another object.

This is what differentiates the real reactive centrifugal force on the wall, from the inertial centrifugal force (pseudo / fictitious force) on the astronaut. The inertial force exists only in the rotating frame. The inertial force acts directly on the object, regardless of any interactions with other objects.

I am confused as to what the essential difference is, as you understand it, between a pseudo centrifugal force and a real centrifugal force. Can you give me an example of this pseudo/fictitious centrifugal force on the astronaut? The only force on the astronaut is a centripetal force, which seems to me to be real.

AM

 

[A.T.]

Both forces in Newtons 3d are always real forces (interaction forces). They both act in every reference frame. They both are exerted by some object on another object.

This is what differentiates the real reactive centrifugal force on the wall, from the inertial centrifugal force (pseudo / fictitious force) on the astronaut. The inertial force exists only in the rotating frame. The inertial force acts directly on the object, regardless of any interactions with other objects.

I am confused as to what the essential difference is, as you understand it, between a pseudo centrifugal force and a real centrifugal force.

I just listed the differences above. In fact I listed them pages ago in my diagram:

Can you give me an example of this pseudo/fictitious centrifugal force on the astronaut?

See F_icf in the diagram for the co-rotating frame.

The only force on the astronaut is a centripetal force...

This is only true in the inertial frame. In the rotating frame there is also an inertial centrifugal force on the astronaut.

...which seems to me to be real.

The centripetal force in this case is real because it is an interaction force, not because it happens to be the net force in the inertial frame and thus also determines the acceleration in that frame. So don't confuse "real force" and "net force" again.

 

[rcgldr]

Can you give me an example of this pseudo/fictitious centrifugal force on the astronaut?

I already mentioned an example. An astronaut jumps inwards from the wall of the space station. During the period of time that the astronaut is not in contact with the surfaces of the rotating space station, from the perspective of the rotating space station frame of reference, the astronaut appears to be accelerated outwards by a fictitious centrifugal force (there's also a fictitious coriolis force, and I'm not sure which is considered responsible for the apparent outwards acceleration), when in fact from an inertial frame of reference the astronaut is moving at constant velocity (no acceleration).

 

[A.T.]

I already mentioned an example. An astronaut jumps inwards from the wall of the space station. During the period of time that the astronaut is not in contact with the surfaces of the rotating space station, from the perspective of the rotating space station frame of reference, the astronaut appears to be accelerated outwards by a fictitious centrifugal force (there's also a fictitious coriolis force, and I'm not sure which is considered responsible for the apparent outwards acceleration),

The coordinate acceleration in the rotating frame is determined by the net force in the rotating frame. If the astronaut is not interacting with the station, the net force in the rotating frame is the vector sum of the inertial forces acting on him:
- inertial centrifugal force
- Coriolis force

The direction of the inertial centrifugal force is always outwards. The direction of the Coriolis force depends on his movement direction in the rotating frame, which depends on how he jumped and the phase of his jump. The Coriolis force can be zero, or have a centrifugal component, or have a centripetal component or have no radial component at all, just a tangential one.

 

[sankalpmittal]

Sorry, you were not close enough.

Doc Al's analysis on tension of string is precise and correct .

Centripetal force is real. It is just one of ordinary forces acting on a whirling body by a string as string has restoring force towards center which is centripetal force and hence is directed towards the center of whirling system. Centripetal force causes centripetal acceleration. Centripetal force obeys 3rd law of Newton - that is there is another body which experiences equal and opposite reaction. But we might not confuse it with centrifugal force .

Centrifugal force is not quite real. It does not obey 3rd law of Newton - there is no source of centrifugal force. Centrifugal force exists only in rotating frame of reference.

If you stand on rotating carousel, you are at rest with respect to carousel. The force of friction acting on the soles of your shoes pushes you inward. Why you remain at rest with respect to the carousel? Because in rotating frame of reference of carousel there is imaginary centrifugal force acting on you outward, which has no source. This force exactly compensates centripetal force of friction, and you have zero acceleration and remain at rest.

 

[A.T.]

Centripetal force is real.

Not in general, but in inertial frames yes.

Centripetal force obeys 3rd law of Newton - that is there is another body which experiences equal and opposite reaction.

And that equal and opposite reaction is sometimes "centrifugal", which means only "away from the center".

But we might not confuse it with centrifugal force .

Why we must not confuse is:
- reactive centrifugal force
- inertial centrifugal force

Centrifugal force is not quite real. It does not obey 3rd law of Newton - there is no source of centrifugal force. Centrifugal force exists only in rotating frame of reference.

That is correct for the inertial centrifugal force.

If you stand on rotating carousel, you are at rest with respect to carousel. The force of friction acting on the soles of your shoes pushes you inward. Why you remain at rest with respect to the carousel? Because in rotating frame of reference of carousel there is imaginary centrifugal force acting on you outward, which has no source. This force exactly compensates centripetal force of friction, and you have zero acceleration and remain at rest.

Correct description of the inertial centrifugal force. I bolded the key differences to the reactive centrifugal force exerted on the carousel by your feet, which exists in every frame and has a second object as source (your feet).

 

[Lobezno]

They should really give the real, "reactive" centrifugal force a new name so as to avoid confusion.

The second definition of centrifugal force, the force on the body that exerts the centripetal force IS REAL and is required by Newton's Third Law. It's most unfortunate that is has the same name as the fictitious force that, in a non-inertial frame of reference, appears on the body undergoing circular motion.

 

[Andrew Mason]

I have come to the somewhat surprising conclusion that, in Dale’s scenario, the tension in the rope between A and B will decrease but not go to 0 if the rope from A to the centre is cut. It will reduce to 39.5N. This is because A and B are actually rotating about each other as they both rotate about the pivot. As I have said before, the physics of rotation is subtle and not always obvious.

If you follow the centre of mass of A-B (ie. 1.5 m from the centre pivot) and plot the movement of A and B relative to that centre of mass you will see that they rotate about that A-B centre of mass at the same angular speed they rotate around the pivot: 1 rotation per second. The two rotations are perfectly synchronized so it looks like only one rotation.

One could view this is a compound rotation. It is similar to the moon’s rotations: always showing the same face toward the Earth because the moon is rotating about its own centre of mass at exactly the same angular speed as it is rotating about the earth-moon centre of mass.

When the rope from the centre to A is cut the rotation of A and B does not change. This means that A and B will keep rotating relative to each other about their centre of mass so the tension created by that rotation will remain unchanged. What will change is the tension due to the lack of acceleration of their centre of mass toward the pivot.

Dale’s scenario is equivalent to having a 1.5 m rope connecting the pivot to the centre of mass of A and B (ie. half way between A and B at 1.5 m from the pivot) and rotating everything about the pivot at 1 rev/sec while A and B rotate about their centre of mass at 1 rev./sec.

Here are the accelerations, forces and rope tensions in such a scenario:

1)Accelerations:

Acceleration of the AB centre of mass:
[itex]a_{ABcm} = (1.5 m)(6.28 s^{-1})^2 = 59.25 m/s^2 \text{centripetal}[/itex]

Acceleration of A and B relative to the AB cm:
[itex]a_{B_{ABcm}} = (.5 m)(6.28 s^{-1})^2 = 19.75 m/s^2 \text{centripetal}[/itex]
[itex]a_{A_{ABcm}} = (.5 m)(6.28 s^{-1})^2 = 19.75 m/s^2 \text{centripetal}[/itex]

Acceleration of A and B relative to the central pivot:
[itex]a_{A_{cp}} = a_{ABcm} - a_{A_{ABcm}} = 59.25 – 19.75 = 39.5 m/s^2
\text{centripetal}[/itex]
[itex]a_{B_{cp}} = a_{ABcm} + a_{B_{ABcm}} = 59.25 + 19.75 = 79.0 m/s^2 \text{centripetal}[/itex]

2) Forces:

[itex]F_{A}= m_Aa_{A_{cp}} = (1 kg)(39.5 m/s^2)=39.5 N \text{centripetal}[/itex]
[itex]F_{B}= m_Ba_{B_{cp}} (1 kg)(79.0 m/s^2)=79.0 N \text{centripetal}[/itex]

3) Rope tensions:

In the rope between the central pivot and the AB centre of mass:
[itex]T_{rope1}= m_{ABcm}a_{ABcm} = 2 kg \times 59.25 m/sec^2 = 118.5N[/itex]

In the rope between the AB centre of mass and A:
[itex] T_{rope2}= m_Aa_A = m_A(a_{ABcm} - a_{A_{ABcm}}) = 59.25 - 19.75 N = 39.5N [/itex]

In the rope between the AB centre of mass and B:
[itex]T_{rope2}= m_Ba_B = m_B(a_{ABcm} + a_{B_{ABcm}}) = 59.25 + 19.75 N = 79N [/itex]From the above one can see that the positions of A and B, their accelerations, forces and the AB rope tension are all identical to those in Dale’s scenario.

So, I can now agree (I apologize for my earlier intransigence) that there is a real force on A from B. And I now admit that a force on A of 39.5 N is directed outward in relation to the central pivot but which is centripetal in relation to the centre of mass of AB.

So, Doc Al and Dalespam and the others have been right that A and B are pulling on each other (again I apologize for being so stubborn about that). I now see why that is. It has to do with the relative rotation of A and B.

There remains only a debate about whether the compounding of centripetal forces like this results in a centrifugal force.

If A and B are rotating about their centre of mass then the forces between A and B are centripetal as between A and B. The forces on A and B are directed to the centre of mass of A and B.

So all forces here could be correctly viewed as centripetal in the sense that they are all toward a centre of rotation. They are just not all toward the same centre of rotation.

Whether this makes it centrifugal or centripetal depends on which centre you are referring to and the precise definition of centripetal and centrifugal force. A compound rotation like this highlights the need for a clear and precise definition. All forces could be correctly viewed as centripetal since they are all directed toward, and result entirely from the rotation of A and B around a centre of rotation. In any event, all the forces are real.

AM

 

[Dale]

I am confused as to what the essential difference is, as you understand it, between a pseudo centrifugal force and a real centrifugal force.

As mentioned above, there are a few key differences between inertial forces (aka fictitious forces, or pseudo forces). First and most importantly, they violate Newton's laws. Specifically, they violate Newton's 3rd law, i.e. there is no equal and opposite reaction to an inertial force. Second, they are frame varying, i.e. they exist in some reference frames but not in all reference frames. Third, they are always proportional to the mass of the object (this feature is one of the hints that led Einstein to the equivalence principle).

 

[Dale]

Doc Al and Dalespam and the others have been right that A and B are pulling on each other

I think that Doc Al might say it that way, but I wouldn't. I would say that A and B are each pulling on the rope.

 

[Dadface]

Do you understand now how that is not the case?

The extra rope doesn't change the problem, it simply specifies some of the irrelevant details. In fact, for your design the extra rope is required in order to avoid changing the problem. As long as the external system supplies the right force to the center it is a legitimate external system for the problem and doesn't change any of the givens.

Do you feel that you understand now what it means to change the problem and why the details are irrelevant?

Dalespam,at the time of asking it was most relevant for me to ask for the details because the question was unclear and seemed to be describing an impossible situation.Looking at it in retrospect it can be seen that the question is ambiguous.You may be able to see this ambiguity if you look at your question(post 190 page 12).
As I understood the question at the time I thought that the word assembley referred to the assembley you described and since you called this the whole assembley I though that there was nothing else.
The mental pictures I formed of the event included the assembley somehow rotating in space about the end of the rope but with nothing else attached to the rope.Of course I had to ask for details.You can see something about how i interpreted your question if you read again some of my earlier posts(220 page 19 227 page14 etc)
In a later post you said that the system was not isolated.That should have given me a clue but it just washed over me at the time and I just did not spot the significance of it.Hands up to that.
In post 242 you referred to "applying an external force" to the end of the rope.At last I had been given a detail and the question made some sense,but there was another problem.I took the part description "the end of the rope" literally and by that I understood it to mean that whatever else is attached to the assembley has to be at the end of the rope and not beyond it.In other words in order to not change the linear dimensions of the new "whole assembley" the total length had to be at its original defined length of 2m.
I was puzzled by that requirement but I went along with it anyway as evidenced by my following posts.Did you not find it odd that my analysis,for example,had the mass M being positioned exactly at the end of the rope?
Eventually the relevant extra details I needed for the question to make sense came out,but sort of indirectly.It was a long time getting there.

 

[Doc Al]

I think that Doc Al might say it that way, but I wouldn't. I would say that A and B are each pulling on the rope.

I also prefer the more accurate statement that A and B each pull the rope (and, of course, that the rope pulls back on both A and B). That's the most complete statement. For some purposes it's OK to be a bit sloppy and say that "A and B pull on each other", treating the rope as merely transmitting the force. But I don't think that's a good idea for this thread, as the whole issue is to identify the forces involved most accurately.

 

[rcgldr]

The wiki article on centrifugal force states that it exists for all objects observed from a rotating frame and is equal and opposing to what would be the mass of the observed object times the centripetal acceleration of a point on the rotating frame at the same distance from the center of rotation of the rotating frame as the observed object, except the point is not moving wrt the rotating frame. Centripetal acceleration of that point is v² / r wrt to an inertial frame and centrifugal force = m (mass of object) v² / r. The corilios force exists for all objects that appear to be moving when observed from a rotating frame. For an object at "rest" in an intertial frame, the corilios force will have double the magnitude and oppose centrifugal force, resulting an apparent but fictitious centripetal force of the object as observed from a rotating frame.

My point here is that in the case of an object not moving wrt to a rotating frame, then the centrifugal force wrt the rotating frame is the same as the reactive centrifugal force wrt an inertial frame. In this case the centrifugal force is real and the same in both frames, resulting in an outwards force exerted on whatever surface of the rotating frame that causes the object to travel in a circular path wrt inertial frame.

 

[Dale]

Looking at it in retrospect it can be seen that the question is ambiguous.You may be able to see this ambiguity if you look at your question(post 190 page 12).
...
In a later post you said that the system was not isolated...
In post 242 you referred to "applying an external force" to the end of the rope...
Eventually the relevant extra details I needed for the question to make sense came out,but sort of indirectly.It was a long time getting there.

I don't know if you have taken a basic physics class, but this type of problem is very common on homework or tests. Usually you are given enough information to come to a unique solution for the question asked, and no extra information.

Systems are usually not explicitly identified as isolated or non-isolated, it is inferred from the analysis itself. You are usually not told that if you want to envision things outside of the problem you are free to add a length of rope if you happen to be imagining some specific external mechanism that requires it.

The question is unambiguous because it has enough information to come to a unique solution. These other details that came out later are not things that were missing from the question, they are things that should have been part of the analysis (the non-isolated system) or only incidental to your particular specification of the external system (the extra length of rope).

 

[Andrew Mason]

I also prefer the more accurate statement that A and B each pull the rope (and, of course, that the rope pulls back on both A and B). That's the most complete statement. For some purposes it's OK to be a bit sloppy and say that "A and B pull on each other", treating the rope as merely transmitting the force. But I don't think that's a good idea for this thread, as the whole issue is to identify the forces involved most accurately.

I understand what you are saying. But you can't apply a force to something that has no mass. If the rope is massless, then the force that A applies to the rope is zero and all the forces between A and B are exerted by A on B and vice-versa.

If there is no force applied to the rope, it would effectively be force at a distance. Which is why real ropes cannot be massless. But for this problem as stated, it really is force at a distance.

AM

 

[Dale]

I thought you changed your mind on that earlier and agreed that Newtons laws do apply to massless ropes. If not, please find a mainstream scientific reference. I don't think there is any support for the claim that you cannot apply a force to something massless. Certainly an unbalanced force can't be applied, but we didn't do that here.

In any case, massless ropes, like frictionless ramps, and other such idealized devices, are understood as merely being reasonable approximations for these kinds of problems.

 

[Andrew Mason]

I thought you changed your mind on that earlier and agreed that Newtons laws do apply to massless ropes. If not, please find a mainstream scientific reference. I don't think there is any support for the claim that you cannot apply a force to something massless. Certainly an unbalanced force can't be applied, but we didn't do that here.

Well Newton's laws apply to massless ropes. It is just that if they have 0 mass any force applied to a massless rope results in zero force exerted on the rope and zero reaction force from the rope. F = ma. That's all I am saying. So we treat the rope as simply allowing force to be exerted by and on the masses that are connected to it. If B applies a pulling force on one end of a massless rope and A is the only other mass connected to the rope, the reaction of A is entirely on B. That would not be the case if the rope had mass.

In any case, massless ropes, like frictionless ramps, and other such idealized devices, are understood as merely being reasonable approximations for these kinds of problems.

Massless=negligible for the purposes of the problem at hand. I understand.

AM

 

[Dadface]

I don't know if you have taken a basic physics class, but this type of problem is very common on homework or tests. Usually you are given enough information to come to a unique solution for the question asked, and no extra information.

Systems are usually not explicitly identified as isolated or non-isolated, it is inferred from the analysis itself. You are usually not told that if you want to envision things outside of the problem you are free to add a length of rope if you happen to be imagining some specific external mechanism that requires it.

The question is unambiguous because it has enough information to come to a unique solution. These other details that came out later are not things that were missing from the question, they are things that should have been part of the analysis (the non-isolated system) or only incidental to your particular specification of the external system (the extra length of rope).

1.In the UK these types of problems are very common indeed.Circular motion is treated quantitatively in S and A level physics courses(for 17 and 18 year olds).I can't recall an actual exam question on this topic where enough information wasn't given.

2.I concur with your second point and would add that,depending on the problem,any additional information can be irrelevant,misleading and can result in the question exceeding any desired length limits.

3.The question is ambiguous in that it can be interpreted as describing an impossible event and one that cannot be analysed.Let me illustrate this by comparing your question to one I just selected.The question is from a past university of Cambridge A level exam and I will quote just a part of it:
After giving numerical information about a stone on a string the question continues..."The stone is whirled in a vertical circle the axis of rotation being at a height of 100 cm above the ground".It is easy to see similarities between this question and yours when you wrote "The whole assembley is swung about the left end of the rope"
The relevant difference between the questions is that in the first question the implication is that the structure is not just the stone and the string but that there is an additional external structure,the details of which are not necessary to answer the question,but the prescence of which is necessary for the event to occur.
In your question the event happened in zero gravity.Was it in space? Was there something else to provide the necessary external structure?If there was something else then fine,but the implication I read into your question was that the rope with masses attached was the "whole" assembley in other words there was nothing else.

 

[Doc Al]

Well Newton's laws apply to massless ropes.

Sure they do.

It is just that if they have 0 mass any force applied to a massless rope results in zero force exerted on the rope and zero reaction force from the rope. F = ma. That's all I am saying.

That's wrong. Any unbalanced force applied to a massless object would result in infinite acceleration--that just means that the net force on the massless rope must always be zero. Not that you can't exert a force on it!

The massless rope approximation just means that the tension in the rope is uniform. You can certainly pull the rope and it certainly pulls back.

If completely massless bothers you, give the rope some very small mass. See what changes.

 

[Dale]

any force applied to a massless rope results in zero force exerted on the rope

This is a self-contradiction.

 

[Doc Al]

I would say that you can apply a force but you cannot exert a force.

LOL! Get a grip, man!

You are attributing magical powers to this massless rope. Somehow it can transmit a force without exert a force.

I would prefer to say that you cannot exert a force at all on a massless rope because a massless rope cannot stretch. If it stretched, there would be a difference in force between ends while it stretched, and that can never happen with a massless rope. So the force applied to an end of the rope does not exert a force, net or otherwise, on the rope at all. For the purpose of the physics here we really are transmitting force at a distance through the concept of mechanical tension.

You are simply defining a massless rope out of existence, even though such an idealization is used all the time. The idealization just makes the analysis easier and eliminates non-essentials.

Does a real rope stretch? Sure. But our idealized massless rope is inextensible. Just for convenience. If that bothers you, let it stretch a bit. So what?

Does a real rope have mass? Sure. But our idealized rope does not. Again, just for convenience. If you like, use a real, massive rope!

You are massively (get it?) missing the point here.

 

[A.T.]

They should really give the real, "reactive" centrifugal force a new name so as to avoid confusion.

Maybe, but this is not the task of Wikipedia. On the contrary: it is against their rules to invent new terms, and put them there. You have to provide references showing that a term is already in use, so I assume there are such references for the real centrifugal force.

Personally I would prefer that the inertial centrifugal force gets a new name. "Centrifugal" is simply too general to describe this specific concept. The other inertial forces in rotating frames have specific names: Coriolis force & Euler force. Bernoulli and Lagrange seem to be the ones who identified inertial centrifugal force as a pseudo force. But this is not going to stick anyway.

 

[Andrew Mason]

That's wrong. Any unbalanced force applied to a massless object would result in infinite acceleration--that just means that the net force on the massless rope must always be zero. Not that you can't exert a force on it!

We are both saying that the forces exerted on each end MUST BE equal and opposite at all times.

I understand what you are saying. You are saying you can exert a force but you cannot exert a net force on a massless rope. I would say that you can apply a force but you cannot exert a force. It is an interesting, although for this problem immaterial, difference in the way of looking at the same thing. We are quibbling about how many angels can dance on the head of a pin.

I would prefer to say that you cannot exert a force at all on a massless rope because a massless rope cannot stretch. If it stretched there would have to be non-equal, opposing forces on the ends on the rope (ie a net force on the rope), and that cannot happen with a massless rope. So the force applied to an end of the rope does not exert a force, net or otherwise, on the rope at all. For the purpose of the physics here we really are transmitting force at a distance through the concept of mechanical tension.

Now Dale says it is a contradiction to say that one can apply a force but not exert a force on something. You are saying that "apply" and "exert" a force are the same thing. I make a distinction.

Consider the interaction between a photon and an atom when an atom releases a photon. The photon (a massless particle) carries with it momentum [itex]h/\lambda[/itex] and causes a change in momentum to the atom in the opposite direction of [itex]\Delta p_e = h/\lambda[/itex]. If we say that the photon exerts a force on the atom, then for Newton's third law (which is fundamental to all of physics, quantum mechanics included) to hold we would have to say that there is a force exerted on the photon. But, for reasons we both understand, it is meaningless to talk about a force being exerted on a photon.

When the photon is absorbed by another atom, a distance s away, there is a change in momentum of that atom that is equal and opposite to the change in momentum of the emitting atom. It took a finite time to accomplish that change in momentum ([itex]\Delta t =s/c[/itex]). So we can say that the emitting atom exerts a force on the absorbing atom: [itex]F = \Delta p/\Delta t = hc/\lambda s[/itex]. We could say that the force is applied by means of the exchange of a massless photon. We just can't say that the force is exerted on the photon.

AM

 

[Doc Al]

We are both saying that the forces exerted on each end MUST BE equal and opposite at all times.

OK.

I understand what you are saying. You are saying you can exert a force but you cannot exert a net force on a massless rope. I would say that you can apply a force but you cannot exert a force.

Sorry, but that sounds like gibberish.

I would prefer to say that you cannot exert a force at all on a massless rope because a massless rope cannot stretch. If it stretched there would have to be non-equal, opposing forces on the ends on the rope (ie a net force on the rope), and that cannot happen with a massless rope.

(1) We are talking about an idealized rope here--who cares about the details about how it stretches or doesn't stretch. (Assume an inextensible rope.)
(2) Why in the world do you think you need non-equal forces on the ends of the rope to stretch it? All you need is tension.

So the force applied to an end of the rope does not exert a force, net or otherwise, on the rope at all. For the purpose of the physics here we really are transmitting force at a distance through the concept of mechanical tension.

More self-contradictory statements. Mechanical tension is distinctly local acting. Each part of the rope exerts a force on the neighboring part.

If you're having trouble with a massless rope just use a massive one! This is really beside the point to the issues raised in this thread.

Now you say it is a contradiction to say that one can apply a force but not exert a force on something. You are saying that "apply" and "exert" a force are the same thing. I make a distinction.

Consider the interaction between a photon and an atom when an atom releases a photon. The photon (a massless particle) carries with it momentum [itex]h/\lambda[/itex] and causes a change in momentum to the atom in the opposite direction of [itex]\Delta p_e = h/\lambda[/itex]. If we say that the photon exerts a force on the atom, then for Newton's third law (which is fundamental to all of physics, quantum mechanics included) to hold we would have to say that there is a force exerted on the photon. But, for reasons we both understand, it is meaningless to talk about a force being exerted on a photon.

When the photon is absorbed by another atom, a distance s away, there is a change in momentum of that atom that is equal and opposite to the change in momentum of the emitting atom. It took a finite time to accomplish that change in momentum ([itex]\Delta t =s/c[/itex]. So we can say that the emitting atom exerts a force on the absorbing atom. We could say that the force is applied by means of the exchange of a massless photon. We just can't say that the force is exerted on the photon.

An irrelevant digression. We are talking about classical, macroscopic contact forces here, not interactions at the atomic level.

 

[A.T.]

If you're having trouble with a massless rope just use a massive one!

Or simply consider the rotating space station scenario, without any ropes.

 

[Doc Al]

Or simply consider the rotating space station scenario, without any ropes.

Exactly. Or the merry-go-round. Nothing 'massless' in those scenarios, so no temptation to invoke 'action at a distance'.

 

[Dale]

If it stretched there would have to be non-equal, opposing forces on the ends on the rope (ie a net force on the rope)

This is not correct.

You are saying that "apply" and "exert" a force are the same thing. I make a distinction.

Do you have any reference for that distinction?

In any case, are you finally in agreement with my analysis, or do you wish to propose your own? After we have pinned that down, then I am glad to analyze any other scenario of your choosing.

 

[A.T.]

My point here is that in the case of an object not moving wrt to a rotating frame, then the centrifugal force wrt the rotating frame is the same as the reactive centrifugal force wrt an inertial frame.

Just to clarify: In this specific case they happen to have the same direction and magnitude, but this doesn't make it "the same force". They act on different objects (see https://www.physicsforums.com/attachment.php?attachmentid=38327&stc=1&d=1314480216").

 

[Andrew Mason]

This is not correct.

Do you have any reference for that distinction?

In any case, are you finally in agreement with my analysis, or do you wish to propose your own? After we have pinned that down, then I am glad to analyze any other scenario of your choosing.

Let's just say that you can apply a force through the rope to the mass on the other end an not quibble about whether you can apply or exert a real force to a non-real massless rope.

AM

 

[Dale]

But that implies action at a distance.

Anyway, do you now accept my analysis or have your own?