Is the concept of reactive centrifugal force valid?

[A.T.]

I say simply that the third law reaction to a centripetal force causing centripetal acceleration is always an equal and opposite centripetal force causing centripetal acceleration as measured in an inertial frame of reference.

Yes, we got that. It is still not true in general, based on the common definition of "centripetal" & "centrifugal" (see post #196). Changing definitions, calling forces "effects" and pointing out special cases proves nothing.

 

[Dale]

I say simply that the third law reaction to a centripetal force causing centripetal acceleration is always an equal and opposite centripetal force causing centripetal acceleration as measured in an inertial frame of reference.

Which has been demonstrated to be incorrect by many counter-examples where the 3rd law reaction was an equal and opposite centrifugal force opposing the centripetal acceleration.

 

[Dale]

The whole point of this exercise is to determine the magnitude and direction of the reaction force to the centripetal force. The reaction to the centripetal force is not on that rope. It is on what is connected to the "fixed" end which, by the way, cannot possibly be "fixed" to an inertial reference frame. If you refuse to tell us what it is fixed to so we can choose an inertial reference frame you cannot analyse the problem. End of story.

And yet, I analyzed it fine without any additional information. Study my analysis and show me any point which violates any of Newton's laws. For each object Newton's 2nd law is satisfied because the net force acting on that object is equal to ma, and for each interaction Newton's 3rd law is satisfied because the forces are equal and opposite. Show me any contradiction where Newton's laws are not satisfied. There is no requirement that Newton's laws only be applied to isolated systems where all of the forces are internal.

So let's do it right. We have two such ropes and masses and we let them rotate about the centre of mass.

I will be glad to analyze your new problem after you have analyzed mine.

Without reference to your new scenario, can you point to any of the 4 objects in my analysis for which Newton's 2nd law is violated or any pair of those objects for which Newton's 3rd law is violated? If not, then the analysis is correct.

 

[A.T.]

It is on what is connected to the "fixed" end which, by the way, cannot possibly be "fixed" to an inertial reference frame.

The point on the rope 1m left from mass A is at rest in an inertial frame. You can assume that as given. This is trivial to achieve but the details of it are irrelevant for the questions.

 

[A.T.]

So let's do it right. We have two such ropes and masses and we let them rotate about the centre of mass. We measure all forces in the non-rotating inertial reference frame of the centre of mass/rotation. We will call the balls A1, B1 and A2, B2 and the respective ropes the same. Here is my analysis:

1)Accelerations:
a_{A1} = (1 m)(6.28 s^{-1})^2 = 39.5 m/s^2 \text{centripetal}
a_{B1} = (2 m)(6.28 s^{-1})^2 = 79.0 m/s^2 \text{centripetal}
a_{A2} = (1 m)(6.28 s^{-1})^2 = 39.5 m/s^2 \text{centripetal}
a_{B2} = (2 m)(6.28 s^{-1})^2 = 79.0 m/s^2 \text{centripetal}


2) Forces:
F_{A1}=(1 kg)(39.5 m/s^2)=39.5 N \text{centripetal}
F_{B1}=(1 kg)(79.0 m/s^2)=79.0 N \text{centripetal}
F_{A2}=(1 kg)(39.5 m/s^2)=39.5 N \text{centripetal}
F_{B2}=(1 kg)(79.0 m/s^2)=79.0 N \text{centripetal}

3) The rope tensions supply the centripetal forces. So the tension in ropes A1 and in A2 is the sum of these centripetal forces on masses A1 and B1: 118.5 N. The tension in ropes B1 and B2 provide only the centripetal force on masses B1 and B2 respectively: 39.5N.

The force acting on A1 is equal and opposite to the force acting on A2 and is the difference in rope tensions: 79N. Both are centripetal.

Similarly the force acting on B1 is equal and opposite to the force on B2 and is 39.5 N. Both are centripetal.

Those are the action and reaction pairs and they are the only forces.

AM

Where is the answer to point 4) for your scenario, listing ALL the forces (not only net forces) acting on each mass?

 

[Andrew Mason]

There is no requirement that Newton's laws only be applied to isolated systems where all of the forces are internal.

There is if you want to analyse third law pairs of forces. If you disagree, try explaining the third law reaction force to the force on me when I fall from a tree accelerating at 9.8 m/sec^2 toward the ground without viewing the earth/me as an isolated system.

I will be glad to analyze your new problem after you have analyzed mine.

I will when you give me all the information needed to analyse it properly.

Without reference to your new scenario, can you point to any of the 4 objects in my analysis for which Newton's 2nd law is violated or any pair of those objects for which Newton's 3rd law is violated? If not, then the analysis is correct.

Certainly.

You say:

"Since the rope is massless, by Newton's 2nd law F_{A2}+F_{B2}=ma=0 so by Newton's 3rd law F_{2A}=-F_{A2}=F_{B2}=-F_{2B}=79.0 N \text{outwards}"

The problem is that you are measuring a pseudo force. You are applying Newton's laws in a rotating frame of reference. Of course, that is how it appears in the rotating reference frame. That is why the concept of centrifugal force is useful. B appears to be pulling on A but, in fact, it is not. This is because A is not pulling on B. Both are pulling on whatever it is the other end of the rope is connected to and the reaction force is on whatever it is connected to. You can prove this by cutting the rope that connects A and B. If you cut the rope between B and A, A's circular rotation motion does not change at all.

AM

 

[Andrew Mason]

Where is the answer to point 4) for your scenario, listing ALL the forces (not only net forces) acting on each mass?

I said that those are ALL the forces. Those are the ONLY forces.

AM

 

[A.T.]

I said that those are ALL the forces. Those are the ONLY forces.

No, you gave only the net forces. You computed them by taking the "difference in rope tensions" at A1 & A2. This is a cheap semantic trick to hide that fact that you are adding the individual force vectors to get the net force.

In point 4) you were supposed to list all the individual forces acting on A1. There are two ropes at A1. So there are two individual forces acting on A1. Please list those two individual interaction forces and their directions.

 

[Andrew Mason]

No, you gave only the net forces. You computed them by taking the "difference in rope tensions" at A1 & A2. This is a cheap semantic trick to hide that fact that you are adding the individual force vectors to get the net force.

This is why centrifugal forces create confusion!

If you add into this configuration centrifugal forces on A1 (and A2) you end up with the following tension on the rope between A1 and the centre of mass (and A2, its the same rope):

Centripetal force on A1: 39.5N plus
Centripetal force on B1: 79 N plus
Centrifugal force on A1: 79N

Total tension: 197.5 N

Since the rope provides the force required to accelerate BOTH A1 and B1, the tension includes the centripetal forces required for A1 and B1.If you add to that the centrifugal forces you get 197.5 N which is certainly not the tension that is measured.

In point 4) you were supposed to list all the individual forces acting on A1. There are two ropes at A1. So there are two individual forces acting on A1.

No there is not. There is a real centripetal force and a pseudo centrifugal force. I listed only the REAL forces.

AM

 

[A.T.]

There are two ropes at A1. So there are two individual forces acting on A1.

No there is not. There is a real centripetal force and a pseudo centrifugal force. I listed only the REAL forces.

Nonsense. Forces from attached ropes are electro-magnetic interaction forces, and act in every frame. Inertial forces in non-inertial frames are not acting trough attached ropes, but directly on every piece of mass.

Two ropes under tension attached to A1 means two individual interaction forces acting at A1. Please list those two forces and their directions.

 

[Andrew Mason]

Nonsense. Forces from attached ropes are electro-magnetic interaction forces, and act in every frame. Inertial forces in non-inertial frames are not acting trough attached ropes, but directly on every piece of mass.

Two ropes under tension attached to A1 means two individual interaction forces acting at A1. Please list those two forces and their directions.

I have listed them. If you keep asking the same question you get the same answer.

AM

 

[A.T.]

Two ropes under tension attached to A1 means two individual interaction forces acting at A1. Please list those two forces and their directions.

I have listed them.

You listed two forces on A1:

Centripetal force on A1: 39.5N
Centrifugal force on A1: 79N

Then you claimed that one of them is an inertial force (pseudo force as you call it). But I asked you about the two interaction forces (real forces as you call them) acting on A1. I'm not interested in inertial forces, because we analyze an inertial frame.

So do you now say that the two forces listed above are in fact interaction forces (real forces)?

 

[Andrew Mason]

Then you claimed that one of them is an inertial force (pseudo force as you call it). But I asked you about the two interaction forces (real forces as you call them) acting on A1. I'm not interested in inertial forces, because we analyze an inertial frame.

So do you now say that the two forces listed above are in fact interaction forces (real forces)?

I am not sure what you mean by interacting. There is no force applied by B to A or by A to B in Dale's scenario. If I cut the rope between B and A there will be no change in the way A is rotating assuming that the rope is tethered to some (infinitely massive) fixed point.

But in my scenario they are all interacting in the sense that if I cut any rope, the rotations and motions of all of them will change. If I cut an outer rope, the three tethered masses will, through inertial reconfiguration, rotate about a different centre of mass and that centre of mass will move away from the untethered mass in uniform motion. If I cut a middle rope, the two tethered masses will each rotate about their respective centres of mass and those centres of mass will move away from each other in uniform motion.

AM

 

[Doc Al]

I am not sure what you mean by interacting. There is no force applied by B to A or by A to B in Dale's scenario.

So you claim the tension in the connecting rope between them is zero and that if you cut the rope, B will miraculously continue to travel in a circle?

 

[Andrew Mason]

So you claim the tension in the connecting rope between them is zero

When did I say that? I gave you the tension as 79N.

and that if you cut the rope, B will miraculously continue to travel in a circle?

Of course not. That would require a centripetal force. When the rope is cut there is none. B will just continue in the same motion that it had when the rope was cut.

AM

 

[Doc Al]

So you claim the tension in the connecting rope between them is zero

When did I say that? I gave you the tension as 79N.

Right in your last post:

There is no force applied by B to A or by A to B in Dale's scenario.

Or did you mean something other than rope tension?

 

[A.T.]

I asked you about the two interaction forces (real forces as you call them) acting on A1.

I am not sure what you mean by interacting.

I was not talking about "interacting". I was talking about "interaction forces". And I translated it into the term that you tend to use: "real forces".

So I think you know exactly what I mean. You are just playing dumb and derailing:

...blah...rotations ...blah...motions ...If I cut the rope...blah...

Not what I asked. Simple questions:

Two ropes under tension are attached to A1.

Yes or No?

Two individual interaction forces (real forces as you call them) are acting at A1.

Yes or No?

You listed two forces on A1:
Centripetal force on A1: 39.5N
Centrifugal force on A1: 79N
They are both interaction forces (real forces as you call them)

Yes or No?

 

[Andrew Mason]

Right in your last post:

Or did you mean something other than rope tension?

The tension between A1 and B1 provides the acceleration force on B1 (in my scenario) and the reaction force from B2 to B1. That is what the tension force does. The tension provides both centripetal forces. That accounts fully for its tension.

In Dale's scenario you cannot see the reaction force because he wants to hide from us the body to which the left end of the rope is tied.

AM

 

[Doc Al]

In Dale's scenario you cannot see the reaction force because he wants to hide from us the body to which the left end of the rope is tied.

On the contrary, Dale's scenario is crystal clear. The rope between A and B exerts a real force on both. Do you deny this?

 

[Dadface]

I would like to make a comment about the mass on a rope problem.The problem,as presented,describes a thought experiment and a more detailed analysis requires extra knowledge about the relevant parts of the rest of the system.
As an example is the rope rotating untethered in space?If so how can it do that? Could it be tethered and rotating on earth? if so in what sort of circle,horizontal or vertical or somewhere between?For earthbound rotations the two relevant forces on the mass are its weight and the tension in the rope.
Consider the simpler example where there is a single mass.Let this be rotating in a vertical circle and with a speed such at the topmost point of its rotation all of the centripetal force is provided by the weight of the mass,the tension being zero.At this point the action reaction bodies are the mass and the Earth one being pulled down the other being pulled up.Where is the centrifugal force?

 

[olivermsun]

In your example, the "reaction" centrifugal force would be the upward gravitational pull that the mass exerts on the earth.

 

[Andrew Mason]

I was not talking about "interacting". I was talking about "interaction forces". And I translated it into the term that you tend to use: "real forces".

There are no interaction forces between A1 and B1 in the sense that the tension in the rope between A1 and B1 can only be accounted for by the centripetal forces on A1 and B1 and their respective reaction forces on B2 and A2. The tension is fully accounted for with those forces. You would have to ignore those forces in order to add this mysterious centrifugal force. You can't do that.

So this "centrifugal force" does not add any tension to the rope. Do the math. If you add the centrifugal "forces" you get too much tension. This is hidden from you in Dale's scenario because he won't let you see what is on the other side of the rotating system, that's all.

So I think you know exactly what I mean. You are just playing dumb and derailing:

I know what you mean and why you think there is a centrifugal force. That is not difficult to see.

You have to read carefully what I am saying. Rotational motion is a very difficult area of physics. Don't assume that someone is acting in bad faith because they disagree with you. Not only is that not a very convincing way to persuade the other side, it makes it more difficult for you to want to see and fully understand the contrary view. I argue every day for a living. I can assure you my position is taken in utmost good faith.

Not what I asked. Simple questions:

Two ropes under tension are attached to A1.

Yes or No?

Yes.

Two individual interaction forces (real forces as you call them) are acting at A1.

Yes or No?

Well there is the rope tension between A1 and A2 and between B1 and B2. That means there is rope tension between B1 and A1 but that is not because B1 is exerting a force on A1. There is no such force. The tensions between B1 and A1, B2 and A2 and between A1 and A2 are fully accounted for by the centripetal forces acting on the masses.

Why not ask why the centrifugal force you say is acting on A1 does not show up in the analysis that I gave of the tensions? If the centrifugal force was real the tension in A1-A2 would have to be 197.5 N. Why is it only 118.5 N?

You listed two forces on A1:
Centripetal force on A1: 39.5N
Centrifugal force on A1: 79N
They are both interaction forces (real forces as you call them)

Yes or No?

The centripetal force is a real force and it produces real acceleration. The centrifugal force on A1 is not a real force.

AM

 

[Dadface]

In your example, the "reaction" centrifugal force would be the upward gravitational pull that the mass exerts on the earth.

Agreed that it is a "reaction" force .Now consider the mass moving to other positions. 23.36 in the UK and time for beddy byes.Night night sleep tight.

 

[Andrew Mason]

On the contrary, Dale's scenario is crystal clear. The rope between A and B exerts a real force on both. Do you deny this?

Yes. I say that is not correct. The rope exerts a real force on B and a real force on the body to which the post forms a rigid part. It does not exert a force on A. If it exerted a force on A then there would be additional tension in the rope: the centripetal force required to accelerate A and B plus the centrifugal force of B on A. But the tension is only 118.5 N, which is the centripetal force required to accelerate B and A. Why does the centrifugal force not show up in the tension?

I am having difficulty understanding why you are arguing in favour of centrifugal force as a real force. I am certainly not the first person to take the position that there is no such thing as a centrifugal force. Until I read this Wikipedia article on "reactive centrifugal force" I had never heard of the term. I have never found a physics text that talks about centrifugal force as a real force or that makes any distinction between centrifugal reaction force and centrifugal force. Engineering books tend to gloss over the physics and don't count.

As has been pointed out, there is a good discussion of this in "[URL discussion page for the Wikipedia article.
[/URL]
AM

 

[Dale]

There is if you want to analyse third law pairs of forces.

This is not correct. Here is some course material for you to read on Newton's 3rd law:
http://s3.amazonaws.com/cramster-resource/8637_n_21740.pdf

Example 4.5 (on p2) is a non-isolated system, see the note about the external force. Page 8 also has a second example using Newton's 3rd on a non-isolated system.

Can you find any reference that supports your position that you are required to only analyze 3rd law pairs of forces in isolated systems? Of course, you can find examples of Newton's 3rd law applied to isolated systems, but can you find any reference that says explicitly that it is not possible to use Newton's 3rd law in non-isolated systems?

I will when you give me all the information needed to analyse it properly.

Already done, as you can see by the fact that I did analyze it properly.

You say:

"Since the rope is massless, by Newton's 2nd law F_{A2}+F_{B2}=ma=0 so by Newton's 3rd law F_{2A}=-F_{A2}=F_{B2}=-F_{2B}=79.0 N \text{outwards}"

Yes, that is all correct.
There are two forces on the rope
F_{A2}+F_{B2} = 79.0 N \text{inwards} + 79.0 N \text{outwards} = 0
And because the rope is massless
ma=0a=0
So Newton's 2nd law is satisfied since F_{net}=ma.

The reaction force to F_{A2} is
F_{2A}=79.0 N \text{outwards}
And the reaction force to F_{B2} is
F_{2B}=79.0 N \text{inwards}
So Newton's 3rd law is also satisfied since F_{A2}=-F_{2A} and F_{B2}=-F_{2B}

So that does not in any way violate Newton's laws.

Now that I have demonstrated that Newton's laws are satisfied, do you now recognize that the analysis is correct and that no additional information is needed to completely characterize all of the forces in the system? If not, then you need to specifically identify which force(s) violate which of Newton's laws. This will take the form of specifically identifying an object where the net force on that object is different from ma or identifying a pair of objects where the 3rd law pair is not equal and opposite. Otherwise you need to admit that the analysis is correct.

The problem is that you are measuring a pseudo force. You are applying Newton's laws in a rotating frame of reference.

No, this is all from an inertial frame. I don't know what would make you believe that I was doing the analysis in the rotating frame. The fictitious forces in a non-inertial frame violate Newton's 3rd law, and all of the forces in my analysis obey Newton's 3rd law. Also, fictitious forces in a non-inertial frame are always proportional to the mass, so since the mass of the rope is 0 the inertial force would also be 0 even if I had made that mistake.

 

[Doc Al]

Yes. I say that is not correct. The rope exerts a real force on B and a real force on the body to which the post forms a rigid part. It does not exert a force on A. If it exerted a force on A then there would be additional tension in the rope: the centripetal force required to accelerate A and B plus the centrifugal force of B on A. But the tension is only 118.5 N, which is the centripetal force required to accelerate B and A. Why does the centrifugal force not show up in the tension?

If you look at A and B as a single system, then the tension between them is an internal force and thus irrelevant. When you look at them separately, then you must include the tension in your calculations. There are two forces on A: The inward force from one rope and the outward force from the other. (That outward force is what is being called a 'centrifugal' force.)

I am having difficulty understanding why you are arguing in favour of centrifugal force as a real force. I am certainly not the first person to take the position that there is no such thing as a centrifugal force. Until I read this Wikipedia article on "reactive centrifugal force" I had never heard of the term. I have never found a physics text that talks about centrifugal force as a real force. Engineering books gloss over the physics and don't count.

As far as the ridiculous term 'reactive centrifugal force' goes, I've written several posts objecting to its use. Of course, they are using 'centrifugal' in the trivial sense of 'away from the center', so it's really not a big deal.

Personally, except in the context of this thread, I never use the term 'centrifugal force' except when referring to the pseudoforce that appears when analyzing things from a rotating frame. In fact, I'm not even too fond of the term 'centripetal force', since that is often misunderstood. But at least the individual forces that contribute to the 'centripetal force' are 'real', noninertial forces that have actors.

So I am not arguing in favor of the term 'centrifugal force' as a real force, though the 'centrifugal' forces in this thread are real forces. That's a trivial semantic issue. What I am arguing for is the correct application and understanding of Newton's 3rd law and even the meaning of 'centripetal force'; these are concepts that you've repeatedly abused in this thread.

 

[Dale]

The problem,as presented,describes a thought experiment and a more detailed analysis requires extra knowledge about the relevant parts of the rest of the system.

No it doesn't. See post 197 where I did a complete analysis without any extra knowledge. The system is not isolated, but I specified that gravity was absent. So all necessary information is provided to answer the questions.

 

[Dale]

The rope exerts a real force on B

If you admit that the rope exerts a real force on B then by Newton's 3rd law B exerts a real force on the rope and this real force is in the opposite direction. Therefore one of those forces is pulling inwards (centripetal) and the other is pulling outwards (centrifugal).

 

[Dale]

Personally, except in the context of this thread, I never use the term 'centrifugal force' except when referring to the pseudoforce that appears when analyzing things from a rotating frame. In fact, I'm not even too fond of the term 'centripetal force', since that is often misunderstood.

Agreed, completely.

 

[Andrew Mason]

No it doesn't. See post 197 where I did a complete analysis without any extra knowledge. The system is not isolated, but I specified that gravity was absent. So all necessary information is provided to answer the questions.

Dadface is correct. You are trying to avoid the essence of the problem by postulating an unbalanced rotating mass about a perfectly "fixed" point of rotation and treating it as inertial frame of reference. That is IMPOSSIBLE in our present universe according to Newton's laws. You can't just make stuff up. It has to correspond to the real universe.

AM

 

[Andrew Mason]

If you admit that the rope exerts a real force on B then by Newton's 3rd law B exerts a real force on the rope and this real force is in the opposite direction. Therefore one of those forces is pulling inwards (centripetal) and the other is pulling outwards (centrifugal).

If I stretch an elastic, the tension force in the elastic is a force that is directed toward the centre of the elastic. The direction of the force depends on which side of the centre of mass of the elastic you are looking at. Ergo, the tension is centripetal. Same thing with the rope, just not as obvious.

Just add up the tensions on the ropes. Add up the centripetal forces on the masses. They add up to the tensions. How can the centrifugal force contribute more than this to tension? In the non-inertial frame the centrifugal "force" is needed to explain the tension because you are not observing any centripetal acceleration. In the inertial frame, if you add the centripetal force you get too much tension.

Do you disagree with my figures for tension (ie. 118.5N and 79 N)?

AM

 

[Andrew Mason]

If you admit that the rope exerts a real force on B then by Newton's 3rd law B exerts a real force on the rope and this real force is in the opposite direction.

The rope can't exert a reaction force unless it has mass. (I thought you said these were massless ropes). You seem to be ignoring the fact that the left end of the rope is connected to another mass. It is the mass that the rope is connected to that exerts the reaction to the pull on the rope.

AM

 

[rcgldr]

If you admit that the rope exerts a real force on B then by Newton's 3rd law B exerts a real force on the rope and this real force is in the opposite direction. Therefore one of those forces is pulling inwards (centripetal) and the other is pulling outwards (centrifugal).

The rope can't exert a reaction force unless it has mass.

I don't see any claim that the force the rope exerts onto object B is a reaction force. It's object B that exerts an equal and opposing reaction force to the end of the rope (due to the centripetal acceleration).

Going back to what I think is a simpler example is a rocket in space free of any external forces, and using it's engine to maintain a circular path. At the engine, there is a centripetal force exerted on the rocket and a centrifugal force exerted on the ejected fuel. In this case, both forces could be considered reactive.

Trying to get back on what I thought was the topic (and repeating myself), wasn't this thread supposed to be about the validity of the term "reactive centrigufal force"? The wiki articles cite a few references where that term is used, and after discussion at wiki, those involved agreed it was a valid term. The wiki article does need a correction to note that sometimes centripetal forces coexist with reactive centrifugal forces and sometimes they don't, but this doesn't invalidate the terminology.

 

[Andrew Mason]

If you look at A and B as a single system, then the tension between them is an internal force and thus irrelevant. When you look at them separately, then you must include the tension in your calculations. There are two forces on A: The inward force from one rope and the outward force from the other. (That outward force is what is being called a 'centrifugal' force.)

When you look at A and B like that without reference to the centre of rotation or some other inertial frame, you are in a non-inertial frame of reference. In an inertial frame of reference (which may or may not be very close to the post - Dale will not provide that information), you can see that the tension between A and B is entirely a function of B's acceleration and mass. It has nothing to do with A. In other words, if B was pulling on A as a reaction force to the centripetal force on it, A's mass would matter. Since it does not, the reaction force to the centripetal force on B has some other origin (ie. the mass that the post is connected to, which Dale wants to prevent us from knowing anything about).

AM

 

[A.T.]

I argue every day for a living

I'm delighted to hear that you don't work in math/physics. The reason why you switched to law is also obvious from this thread.

I know what you mean and why you think there is a centrifugal force.

It is the result of your analysis: Centrifugal force on A1: 79N

The centrifugal force on A1 is not a real force.

It must me a real force, because you analyzed an inertial frame of reference.

 

[A.T.]

I would like to make a comment about the mass on a rope problem.The problem,as presented,describes a thought experiment and a more detailed analysis requires extra knowledge about the relevant parts of the rest of the system.

Nobody asked for a "more detailed analysis". The specific questions posed by DaleSpam can all be answered without any extra knowledge.

 

[A.T.]

Until I read this Wikipedia article on "reactive centrifugal force" I had never heard of the term.

Who cares? If you want to question if the term is used widely enough to grant a Wikipedia entry, do this on the discussion page of Wikipedia.

Engineering books tend to gloss over the physics and don't count.

Tell this to the Wiki authors. They will tell you that it is perfectly valid to have Wikipedia articles about terms which are used solely/mainly in engineering.

 

[Doc Al]

When you look at A and B like that without reference to the centre of rotation or some other inertial frame, you are in a non-inertial frame of reference. In an inertial frame of reference (which may or may not be very close to the post - Dale will not provide that information), you can see that the tension between A and B is entirely a function of B's acceleration and mass. It has nothing to do with A. In other words, if B was pulling on A as a reaction force to the centripetal force on it, A's mass would matter. Since it does not, the reaction force to the centripetal force on B has some other origin (ie. the mass that the post is connected to, which Dale wants to prevent us from knowing anything about).

Utter nonsense. The frame in which we are analyzing the problem is clearly an inertial frame. A and B are rotating! There are no inertial forces, only real forces.

What forces act on A? The tensions from the two ropes.
What forces act on B? The tension from the second rope.

Since we know the acceleration, we can apply Newton's 2nd law and figure out what those forces are.

And sure, since the ropes are massless, you can consider them as just transmitting the force between bodies. The force that A and B exert on each other is a real force. And of course those forces comply with Newton's 3rd law.

 

[Dadface]

Nobody asked for a "more detailed analysis". The specific questions posed by DaleSpam can all be answered without any extra knowledge.

Well I think Andrew Mason wants extra knowledge and in order to make a start on the problem I would want extra knowledge.Look at what we have,a rope assembley rotating about one end of the rope and in the absence of gravity and anything else.How can this happen? Is the assembley in deep space,perhaps it is in free fall?There are many questions but just for now perhaps you can tell me how this set up,which is based on a thought experiment,can be brought into the real world.

 

[A.T.]

Look at what we have,a rope assembley rotating about one end of the rope and in the absence of gravity and anything else.How can this happen?

Irrelevant for the questions asked.

Is the assembley in deep space,perhaps it is in free fall?

It is rotating around the left end in an inertial reference frame.

There are many questions but just for now perhaps you can tell me how this set up which is based on a thought experiment can be brought into the real world.

Irrelevant for the questions asked.

 

[Dadface]

Irrelevant for the questions asked.

It is rotating around the left end in an inertial reference frame.

Irrelevant for the questions asked.

So are you preferring to keep the question unchanged being based on a thought experiment which has no relevance to the real world?

 

[Dale]

Well I think Andrew Mason wants extra knowledge and in order to make a start on the problem I would want extra knowledge.

Then you two would both fail freshman physics. No extra knowledge is required to answer the questions asked. If this question were on a freshman physics test and you didn't even start the problem because you asked for extra knowledge then you would not even get partial credit for the question.

Look at what we have,a rope assembley rotating about one end of the rope and in the absence of gravity and anything else.How can this happen?

By applying an external force to the left end of the rope. The system is not isolated as has been mentioned many times. The details of how you apply that external force do not matter.

So are you preferring to keep the question unchanged being based on a thought experiment which has no relevance to the real world?

It is completely relevant to the real world. Non-isolated systems are analyzed all the time. The only unrealistic part is the "massless rope" approximation.

 

[A.T.]

So are you preferring to keep the question unchanged being based on a thought experiment which has no relevance to the real world?

The questions you have asked have no relevance to the solution of the problem within Newtonian mechanics. As for "relevance to the real world" : You could just as well demand information on the color of the mass spheres, because in the real world, they always have some color.

 

[Dale]

If I stretch an elastic, the tension force in the elastic is a force that is directed toward the centre of the elastic. The direction of the force depends on which side of the centre of mass of the elastic you are looking at.

Yes, as shown in my analysis. And since the center of mass of the elastic is not the center of rotation then one of those forces is inwards and one is outwards.

The rope can't exert a reaction force unless it has mass. (I thought you said these were massless ropes).

Interesting. So you believe that Newton's 3rd law does not apply to massless objects also. Ropes can clearly exert action forces, and if it couldn't exert a reaction force then you could build a reactionless drive. This is clearly absurd, and is just another basic misunderstanding of Newton's laws on your part.

I am still waiting for a reference that Newton's 3rd law does not apply inside non-isolated systems, so now please also provide a reference that Newton's 3rd law does not apply when the traditional massless-rope approximation is used. Because you have a flawed understanding of Newton's laws you are getting a flawed understanding of the problem posed.

 

[Andrew Mason]

I'm delighted to hear that you don't work in math/physics. The reason why you switched to law is also obvious from this thread.

Law teaches us that ad hominen arguments are not persuasive. You should take that to heart in any debate.

Besides, what does it matter who is making the case here? I am saying that the alleged centrifugal force in either my or Dale's scenario does not appear in the measurements of the tensions. So if it can't be measured it does not agree with reality. So you are talking about a fictitious concept. "It does not make any difference how smart you are, who made the guess, or what his name is – if it disagrees with experiment it is wrong." Richard Feynman.

AM

 

[Doc Al]

I am saying that the alleged centrifugal force in either my or Dale's scenario does not appear in the measurements of the tensions.

The tension is the 'centrifugal' force!

How about this: Define 'centrifugal force'.

So if it can't be measured it does not agree with reality. So you are talking about a fictitious concept. "It does not make any difference how smart you are, who made the guess, or what his name is – if it disagrees with experiment it is wrong." Richard Feynman.

 

[Dadface]

Then you two would both fail freshman physics. No extra knowledge is required to answer the questions asked. If this question were on a freshman physics test and you didn't even start the problem because you asked for extra knowledge then you would not even get partial credit for the question.


(It is most likely that the question as it has been stated would not meet the criteria to be included in an exam.)


By applying an external force to the left end of the rope. The system is not isolated as has been mentioned many times. The details of how you apply that external force do not matter.


(In the original question it was stated that the whole assembley is swing about the left side of the rope and above it is stated by applying an external force to the left end of the rope.Thats two different ways of saying the same thing I suppose and that seems fair enough for an earthbound or similar problem where there is gravity and suitable fixing points.The difficulty here is that there is no gravity so where is the assembley and how is it fixed or swung about.The details of fixing do matter and if these details are given the solutions to the problem change.)


It is completely relevant to the real world. Non-isolated systems are analyzed all the time. The only unrealistic part is the "massless rope" approximation.

(They may well be analysed all the time but the limitations of any analysis should always be considered during the analysis.)

My comments replying to DaleSpam are in brackets.Sorry I do not know how to improve the presentation

 

[Andrew Mason]

What forces act on A? The tensions from the two ropes.

No. No. No. This is simply wrong. This would be the case if B was pulling on A because A was moving the same as B. You seem to be treating A and B as being in the same reference frame. They are not.

You say you are analysing this in an inertial frame but in an inertial frame A is not accelerating at the same rate as B. You have to take that fact into account. If I just took Dale's rope and pulled it to the left A and B would accelerate at the same rate and of course there would be a force between A and B and a third law reaction force in the opposite direction back toward A. But that is not occurring here. Rotation is very, very subtle and confusing that way.

Suppose in Dale's scenario that the rope between the centre post and A was a hollow cylindrical rope that just went to A. A smaller diameter rope sits inside and goes through A, without connecting to A, all the way to B. Would there be any difference to the rotation of masses A and B if I picked up the left end of both ropes and swung them?

I can't see any. The tensions would be the same in the ropes between the post and A. They would sum to 118.5 N. (39.5 for the hollow rope and 79 for the one inside). You can see that the force on B does not result in any pull on A and yet the motions of both A and B are exactly the same as in Dale's senario. If you disagree, where is the different force on A or B and where is there different tension?

AM

 

[Dadface]

The questions you have asked have no relevance to the solution of the problem within Newtonian mechanics. As for "relevance to the real world" : You could just as well demand information on the color of the mass spheres, because in the real world, they always have some color.

It has every relevance.I would prefer to see a well defined problem that applies to the real world.Demand to see the colour!Are you taking the Mick?
Anyway,it's all right for you reprobates I now have the wife demanding that I take her shopping.

 

[Doc Al]

No. No. No. This is simply wrong. This would be the case if B was pulling on A because A was moving the same as B. You seem to be treating A and B as being in the same reference frame. They are not.

Looks like we can add 'reference frames' to the list of concepts you do not understand. We are viewing the motion of A and B from an inertial reference frame. A and B do not 'belong' to a reference frame.

You say you are analysing this in an inertial frame but in an inertial frame A is not accelerating at the same rate as B. You have to take that fact into account.

Of course their accelerations are different and of course that fact has been taken into account.

If I just took Dale's rope and pulled it to the left A and B would accelerate at the same rate and of course there would be a force between A and B and a third law reaction force in the opposite direction back toward A. But that is not occurring here.

You think that somehow being in rotation invalidates Newton's 3rd law.

Rotation is very, very subtle and confusing that way.

Apparently so, to some.

Suppose in Dale's scenario that the rope between the centre post and A was a hollow cylindrical rope that just went to A. A smaller diameter rope sits inside and goes through A, without connecting to A, all the way to B. Would there be any difference to the rotation of masses A and B if I picked up the left end of both ropes and swung them?

Irrelevant to the problem at hand.

I can't see any. The tensions would be the same in the ropes between the post and A. They would sum to 118.5 N. (39.5 for the hollow rope and 79 for the one inside). You can see that the force on B does not result in any pull on A and yet the motions of both A and B are exactly the same as in Dale's senario. If you disagree, where is the different force on A or B and where is there different tension?

If B didn't exist the tension on the first rope would be different. The only way that B can influence A is via the tension it exerts on A via the connecting rope.