Is the concept of reactive centrifugal force valid?

[Doc Al]

Who said you can have tension in a rope without two masses pulling on it? Not me. Never. .Of course something is pulling on it. Just not A. The mass at the other end of the rope is pulling on it. A is not pulling on the rope between A and B. That is what I am saying.

I hear what you're saying and it doesn't make sense. The only things touching the rope between A and B--and thus able to exert a force on it--are A and B. Of course both A and B pull on the rope.

The 'mass at the other end of the rope' (in your scenario, I assume) cannot directly exert a force on the rope between A and B. There is no mysterious action at a distance happening here!

 

[Andrew Mason]

Although this is true it is not what Newton's 3rd law says. In any case, the system is not isolated, so it can accelerate and the center of rotation does not need to be the center of mass.

Ok. So the centre of rotation accelerates! Finally.

Now you have a non-inertial frame of reference for the pivot. How are we supposed to determine forces and accelerations from that? Of course you are going to have centrifugal "forces" appearing if you treat the pivot as an inertial frame. The only 3d law force that you can have on the pivot will appear to be OUTWARD. But one is fooled by the fact that the pivot is accelerating. In fact, it is accelerating toward the centre of mass (which defines the origin of an inertial frame of reference).

AM

 

[Dale]

Ok. So the centre of rotation accelerates! Finally.

No, the center of mass of the non-isolated system accelerates (as required by Newtons 2nd). The center of rotation does not. They are not the same for non-isolated systems.

Now you have a non-inertial frame of reference for the pivot.

If the reference frame were non-inertial then by definition Newtons laws would not hold in my analysis. Since Newtons laws hold in my analysis, the reference frame is inertial.

 

[A.T.]

I will keep as close to the original problem as is possible.

Sorry, you were not close enough.

 

[Dale]

It is pulling on something that Dale will not provide us information on. It has to be an infinite mass.

The information is irrelevant and unnecessary for the question posed, and it does not have to be an infinite mass as explained to Dadface.

 

[Dale]

There is no mysterious action at a distance happening here!

Ooh, action at a distance rope, that would be a great product. I bet rock climbers would love it.

 

[A.T.]

It is pulling on something that Dale will not provide us information on. It has to be an infinite mass.

Really? In your own scenario, where you doubled & mirrored Dales arrangement you didn't have an infinite mass to keep the center of rotation 1m from A1 inertial. And you got the same 79N centrifugal reaction force on A1 that Dale got. So the extra information and your modification where completely irrelevant for the forces acting at the masses.

 

[A.T.]

Are you seriously suggesting that the surface of the rotating ship doesn't exert a real force against the feet of the passenger?

I think you should ask Andrew to provide his definition of "real force". It is obviously not the commonly used one. Andrew thinks that forces that are present in inertial frames and are exerted by one object on another object are not real forces.

 

[Andrew Mason]

Really? In your own scenario, where you doubled & mirrored Dales arrangement you didn't have an infinite mass to keep the center of rotation 1m from A1 inertial. And you got the same 79N centrifugal reaction force on A1 that Dale got. So the extra information and your modification where completely irrelevant for the forces acting at the masses.

It has to be an infinite mass if the centre of rotation is that pivot with nothing else rotating about that pivot. That was Dale's scenario. That is why I added the oppositely rotating tethered masses.

AM

 

[Andrew Mason]

I think you should ask Andrew to provide his definition of "real force". It is obviously not the commonly used one. Andrew thinks that forces that are present in inertial frames and are exerted by one object on another object are not real forces.

A real force f is something that is capable of producing a real acceleration (ie. as measured in an inertial frame of reference) a on a mass m such that a = f/m

AM

 

[Dale]

It has to be an infinite mass if the centre of rotation is that pivot with nothing else rotating about that pivot. That was Dale's scenario.

That is not my scenario. If there were nothing else then the system would be isolated, which it is not. I have said that the details of what is there are irrelevant, not that there is nothing there.

 

[A.T.]

That is why I added the oppositely rotating tethered masses.

And your addition changed nothing about the 79N centrifugal reaction force exerted on A by the outer rope in an inertial reference frame. So you wasted your time.

 

[A.T.]

A real force f is something that is capable of producing a real acceleration (ie. as measured in an inertial frame of reference)...

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.

...a on a mass m such that a = f/m

I see. So by "real force" you actually mean "net force" (that is the name everone else uses for the "f" in a=f/m)

Look, why don't you just learn and use the terms everyone else uses. Then you will not be confused by Wikipedia articles, and your posts here will not sound like the chewbacca defense.

 

[Dadface]

Here is where you are changing the problem. The problem isn't broad, it is specific. The location of the center of rotation isn't something you have to assume, it is something that is given. If you change that then you are changing the problem. Don't do that.

For any mass M there is some length L of extra rope which you can attach to the left end of the existing rope to get the entire assembly to rotate about the given point in isolation, but you must satisfy the constraint of rotation about the given point. Alternatively, you can have whatever is to the left be non-isolated also, or more complicated shape than a simple mass and rope, or actively generating thrust via rocket exhaust or ...

There are an infinite number of such assemblies satisfying the problem. All of them will result in the same answers as given above. The details are irrelevant.

Allow me to respond to each of the three paragraphs in turn:

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.Having the axis of rotation at the end of the rope requires that the geometry of the system be changed but maintaining the geometry of the system requires that the axis of rotation be changed.Catch 22 comes to mind.

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.
Before reading this comment my understanding of the problem ,as evidenced by several posts (see for example 242,264 and 267) was that any fixings had to be at the end of the rope and not beyond it.The problem was unclear but previously when clarification was asked for a typical answer was something like "the details are irrelevant".I think I will use that answer the next time my friend asks for instructions on how to defuse a bomb.

3.I just love infinity.(no I don't sometimes I think infinity is a total scumbag)

 

[Dadface]

Sorry, you were not close enough.

Hey,its great to see that you are taking the opportunity to have a lttle dig.

 

[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:
a_{ABcm} = (1.5 m)(6.28 s^{-1})^2 = 59.25 m/s^2 \text{centripetal}

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

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

2) Forces:

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

3) Rope tensions:

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

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

In the rope between the AB centre of mass and B:
T_{rope2}= m_Ba_B = m_B(a_{ABcm} + a_{B_{ABcm}}) = 59.25 + 19.75 N = 79NFrom 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.