Is the concept of reactive centrifugal force valid?

[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.

 

[Andrew Mason]

Then you two would both fail freshman physics.

Please don't resort to ad hominem and condescending comments. No one here would fail freshman physics. This is an issue that has fooled many good physicists. It seems simple but it isn't. Dadface is making a reasonable point that you refused to answer for some unfathomable reason.

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.

I would give him marks for asking the question. But you are right that you can answer the questions as to the forces on A and B and the rope tensions. You just cannot explain all the physics because the reaction forces to A and B's rotation are not on A or B. They are on whatever it is that the left side is connected to.

AM

 

[Andrew Mason]

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.

There is no force between A and B in the hollow rope scenario.

The tension on the first rope would be the same as in Dale's scenario if B didn't exist in Dale's scenario. It would be 39N + 0.

I asked how Dale's scenario differed from this one in terms of the motions and forces. You have not provided an answer.

AM

 

[A.T.]

Law teaches us that ...

Nothing that you learned in law school will help you in a physics debate. You could just as well try this:

So if it can't be measured ...

It can be measured, just like any other real force acting trough a rope on a mass.

 

[Andrew Mason]

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.

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

But you are treating the forces between A and B exactly the same way as if they were in the same reference frame. If you disagree, please explain how the tension between A and B here differs from a scenario in which A undergoes linear acceleration with B tethered to it by that rope.

AM

 

[Doc Al]

I asked how Dale's scenario differed from this one in terms of the motions and forces. You have not provided an answer.

Your scenario is irrelevant to Dale's. By your own calculation, the tensions are different.

How about answering all the question you've been asked instead of dodging them with red herrings?

Among many other questions we are waiting for you to answer is this:

Please define 'centrifugal force' as used in this thread.

And while you're at it, please define 'centripetal force' as well.

 

[Doc Al]

But you are treating the forces between A and B exactly the same way as if they were in the same reference frame.

I have no idea what you're talking about. We are viewing things from an inertial frame, so we can apply Newton's 2nd law without modification.

If you disagree, please explain how the tension between A and B here differs from a scenario in which A undergoes linear acceleration with B tethered to it by that rope.

One difference: In Dale's scenario the two masses have different accelerations, but in yours they would have the same acceleration. No problem: Just apply Newton's 2nd law.

 

[A.T.]

I have no idea what you're talking about.

Chewbacca!

 

[Dadface]

Let me rephrase an earlier question.We have been told that the assembley is rotating about one end of the rope by applying an external force to that end ,the details of how that force is applied being irrelevant.If the details are irrelevant then the answers we obtain should be the same regardless of the value of the effective mass of the fixing system.
Let the effective mass equal M and if so according to the calculations and statements made earlier this mass will have a resultant force of 118.5N acting on it,this force acting towards the other two masses.Do we ignore the acceleration that results from this resultant force and still take it that the rotation is about the end of the rope?
Play around with the equation and you can calculate the acceleration of M so ,for example,as M approaches zero the acceleration approaches infinity,in other words masses A and B will be rotating about a rope end which is accelerating towards them at an incredibly high value.
Is the centre of rotation still at the end of the rope when M is Ikg or any other value?Of course not,when we treat it as a real problem the answers change and the analysis becomes more complex.The axis of rotation does get closer to the rope end as M approaches infinity but then gravity comes into the problem.

 

[A.T.]

Let me rephrase an earlier question.We have been told that the assembley is rotating about one end of the rope by applying an external force to that end ,the details of how that force is applied being irrelevant.

They obviously are. Andrew made the whole thing symmetrical and balanced, and still got the same 79N centrifugal force acting on the inner mass(es) trough the outer rope.

If the details are irrelevant then the answers we obtain should be the same regardless of the value of the effective mass of the fixing system.

As long as your "fixing system" satisfies the condition: The left end of the rope (1m from mass A) is at rest in an inertial frame the details of the "fixing system" are irrelevant. Simply attaching some mass at some distance will not in general satisfy the condition, so you are changing the scenario.

 

[RonL]

Maybe the best solution is a new scenario, we can also give color for Dadface

Make everything liquid,

A translucent red resist, dropped on a slowly spinning silicon wafer, then speed is ramped up.

The liquid rope? spreads across the wafer and the mass of liquid in contact with the thinning and spreading resist rope, would appear much like a liquid ring, sliding and rolling toward the edge of the wafer where it breakes contact and flies off into the vacuum waste chamber.

I think the liquid can show a transformation between fictious and real forces.
All actions of this process are real and should fall into all the laws and principles being discussed.

I would hope someone good with words and familiar with this process could make the connection of all the actions taking place and help clear any misconceptions.

 

[Andrew Mason]

Your scenario is irrelevant to Dale's. By your own calculation, the tensions are different.

! My tensions are the same. So how does my scenario differ than Dale's different? That's all I asked.

Please define 'centrifugal force' as used in this thread.

And while you're at it, please define 'centripetal force' as well.

I don't know how you are defining centrifugal force as it is used in this thread. I agree with the following definitions:

http://science.howstuffworks.com/centrifugal-force-info.htm": Centrifugal Force, in physics, the tendency of an object following a curved path to fly away from the center of curvature. Centrifugal force is not a true force; it is a form of inertia (the tendency of objects that are moving in a straight line to continue moving in a straight line). Centrifugal force is referred to as a force for convenience—because it balances centripetal force, which is a true force. If a ball is swung on the end of a string, the string exerts centripetal force on the ball and causes it to follow a curved path. The ball is said to exert centrifugal force on the string, tending to break the string and fly off on a tangent.

"[URL
Centrifugal Force:[/URL] An object traveling in a circle behaves as if it is experiencing an outward force. This force, known as the centrifugal force, depends on the mass of the object, the speed of rotation, and the distance from the center. The more massive the object, the greater the force; the greater the speed of the object, the greater the force; and the greater the distance from the center, the greater the force. It is important to note that the centrifugal force does not actually exist. We feel it, because we are in a non-inertial coordinate system. Nevertheless, it appears quite real to the object being rotated.


http://hyperphysics.phy-astr.gsu.edu/hbase/corf.html#cent" Whereas the centripetal force is seen as a force which must be applied by an external agent to force an object to move in a curved path, the centrifugal and coriolis forces are "effective forces" which are invoked to explain the behavior of objects from a frame of reference which is rotating.


As far as centripetal force is concerned, it is pretty straightforward as well:

http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html#cf"Any motion in a curved path represents accelerated motion, and requires a force directed toward the center of curvature of the path. This force is called the centripetal force which means "center seeking" force.

AM

 

[Andrew Mason]

To Doc Al and Dalespam: Two simple questions:

1. Suppose in Dale's scenario I simply move A toward the centre of rotation. Does the centrifugal force of B on A change? My thinking is that there would still be a tension of 79 N on the rope running between A and B, so I assume that you would say that there is a reactive centrifugal force of 79N exerted by B on A. Do I understand you correctly?

2. Suppose I just have B rotating on a rope about a central pivot. Do you say there a centrifugal force between the center and B? There is certainly tension? So why would there not be a centrifugal force? Does the mass of the pivot (not what it is connnected to, just the pivot) affect this?

AM

 

[Andrew Mason]

Maybe the best solution is a new scenario, we can also give color for Dadface

Make everything liquid,

A translucent red resist, dropped on a slowly spinning silicon wafer, then speed is ramped up.

The liquid rope? spreads across the wafer and the mass of liquid in contact with the thinning and spreading resist rope, would appear much like a liquid ring, sliding and rolling toward the edge of the wafer where it breakes contact and flies off into the vacuum waste chamber.

I think the liquid can show a transformation between fictitious and real forces.
All actions of this process are real and should fall into all the laws and principles being discussed.

This is similar to Newton's bucket: the surface of a liquid in a bucket will change (it will form a parabaloid shape) if you hang the bucket from a rope, twist the rope and then let it spin.

What happens in your example, and in Newton's bucket, is that the forces between the molecules of liquid are not strong enough to provide the centripetal acceleration needed to keep the water moving in a circle as the wafer's/bucket's spin increases. In the wafer's frame of reference, the liquid will move outward as if there is a force pushing it out. But it is just the tendency of the water molecules to move in a straight line due to inertia. In the bucket case, water molecules build up on the inside surface of the bucket due to this inertial effect.

AM

 

[Dadface]

They obviously are. Andrew made the whole thing symmetrical and balanced, and still got the same 79N centrifugal force acting on the inner mass(es) trough the outer rope.


As long as your "fixing system" satisfies the condition: The left end of the rope (1m from mass A) is at rest in an inertial frame the details of the "fixing system" are irrelevant. Simply attaching some mass at some distance will not in general satisfy the condition, so you are changing the scenario.

(By fixing system I mean what,if anything,is at the end of the rope)

You chose to restate two conditions of the fixing system but chose to ignore other conditions the main one being a relevant property of the system which is its mass.It is easy to just state that the left end of the rope is at a certain distance from A and "at rest in an inertial frame" but how can this be achieved?The scenario is fine but the system needs to be defined in greater detail such that the conditions can be realized or approximated to.
The analysis that has been carried out with this problem applied Newtons laws to each of the two masses but was not applied to the fixing system this being a necessary part of the assembley but being ignored.The extended analysis,which should take into account the whole assembley and not just arbitarily chosen parts of it is easy to carry out.Just assume that the fixing system has a mass,call it M for example and take it from there.
I think the main difficulties here arise from the fact that the problem is based on a thought experiment.Thought experiments have proved their worth many times but they should always be used with care and the consequences of any simplifying assumptions made should always be considered.

Gosh it's 21.55.Me thinks it's time for a pear cider and perhaps a glass of Chablis.

 

[Doc Al]

To Doc Al and Dalespam: Two simple questions:

1. Suppose in Dale's scenario I simply move A toward the centre of rotation. Does the centrifugal force of B on A change? My thinking is that there would still be a tension of 79 N on the rope running between A and B, so I assume that you would say that there is a reactive centrifugal force of 79N exerted by B on A. Do I understand you correctly?

2. Suppose I just have B rotating on a rope about a central pivot. Do you say there a centrifugal force between the center and B? There is certainly tension? So why would there not be a centrifugal force? Does the mass of the pivot (not what it is connnected to, just the pivot) affect this?

I'd say that B (via the rope) will exert an outward (thus 'centrifugal') force on whatever the rope attaches to. Why would you think the two scenarios would be any different in that regard?

 

[A.T.]

I don't know how you are defining centrifugal force as it is used in this thread.

I gave you the defintions consistent with the Wikipedia articles:

Centripetal : Force direction is the same as the direction from the point of force application to the center of rotation.
Centrifugal : Force direction is the same as the direction from the center of rotation to the point of force application, or center of rotation and point of force application are the same point.

I agree with the following definitions:

- If you think that your definition is more common than the one above and should therefore be used in Wikipedia, argue about it on the Wikipedia discussion page.

- If you want to discuss the physics in the Wikipedia articles, then you have to stick to the definitions they currently use. Otherwise you are not discussing the articles, just your misinterpretations of them.

 

[Dale]

The details of fixing do matter and if these details are given the solutions to the problem change.

No, the details of the fixing do not matter. As long as you don't change any of the specified conditions the solution to the problem will not change.

Let me rephrase an earlier question.We have been told that the assembley is rotating about one end of the rope by applying an external force to that end ,the details of how that force is applied being irrelevant.If the details are irrelevant then the answers we obtain should be the same regardless of the value of the effective mass of the fixing system.

That is correct.

Let the effective mass equal M and if so according to the calculations and statements made earlier this mass will have a resultant force of 118.5N acting on it,this force acting towards the other two masses.Do we ignore the acceleration that results from this resultant force and still take it that the rotation is about the end of the rope?

The rotation is about the end of the rope. That was specified in the problem. However you want to apply the force to the end of the rope is up to you, but you cannot change the problem without, uh, changing the problem. If you do not change the problem then you will not alter any of the answers that I posted, regardless of the nature of the set up "to the left" of the end of the rope.

Play around with the equation and you can calculate the acceleration of M so ,for example,as M approaches zero the acceleration approaches infinity,in other words masses A and B will be rotating about a rope end which is accelerating towards them at an incredibly high value.

Please feel free to demonstrate that equation. I do not think that your claim here is correct unless you change the problem.

Is the centre of rotation still at the end of the rope when M is Ikg or any other value?Of course not

If the center of rotation isn't at the end of the rope then you have changed the problem. There are an infinite number of ways to achieve that, and which one you pick is not important.

 

[Doc Al]

! My tensions are the same. So how does my scenario differ than Dale's different? That's all I asked.

Since you keep changing the scenario instead of analyzing the one given, perhaps I mixed up which one you were talking about. Are you talking about your hollow rope scenario? Or your four mass version? Link to the post you are asking about.

I don't know how you are defining centrifugal force as it is used in this thread.

It's been defined over and over in this thread. No excuse.

I agree with the following definitions:

http://science.howstuffworks.com/centrifugal-force-info.htm": Centrifugal Force, in physics, the tendency of an object following a curved path to fly away from the center of curvature. Centrifugal force is not a true force; it is a form of inertia (the tendency of objects that are moving in a straight line to continue moving in a straight line). Centrifugal force is referred to as a force for convenience—because it balances centripetal force, which is a true force. If a ball is swung on the end of a string, the string exerts centripetal force on the ball and causes it to follow a curved path. The ball is said to exert centrifugal force on the string, tending to break the string and fly off on a tangent.

This a ludicrously confused 'definition'. In the first part it seems to treat centrifugal force as a pseudoforce (what else can they mean by it saying it is 'not a true force'). Then they go on to state how the 'ball is said to exert a centrifugal force on the string'. Laughable.

"[URL
Centrifugal Force:[/URL] An object traveling in a circle behaves as if it is experiencing an outward force. This force, known as the centrifugal force, depends on the mass of the object, the speed of rotation, and the distance from the center. The more massive the object, the greater the force; the greater the speed of the object, the greater the force; and the greater the distance from the center, the greater the force. It is important to note that the centrifugal force does not actually exist. We feel it, because we are in a non-inertial coordinate system. Nevertheless, it appears quite real to the object being rotated.

This is better, but is not the meaning of 'centrifugal force' used in this thread.

http://hyperphysics.phy-astr.gsu.edu/hbase/corf.html#cent" Whereas the centripetal force is seen as a force which must be applied by an external agent to force an object to move in a curved path, the centrifugal and coriolis forces are "effective forces" which are invoked to explain the behavior of objects from a frame of reference which is rotating.

Again, another definition of it as a pseudoforce; not relevant for this thread.

As used in this thread, as has been repeated ad nauseam, 'centrifugal' merely means 'acting outward from the center'.

As far as centripetal force is concerned, it is pretty straightforward as well:

http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html#cf"Any motion in a curved path represents accelerated motion, and requires a force directed toward the center of curvature of the path. This force is called the centripetal force which means "center seeking" force.

A bit sloppy, but workable. The 'centripetal force' is the net force acting on object that is centripetally accelerating.

 

[Andrew Mason]

I'd say that B (via the rope) will exert an outward (thus 'centrifugal') force on whatever the rope attaches to. Why would you think the two scenarios would be any different in that regard?

I didn't and I don't. I just wanted to be sure I understood you.

Then you admit that the mass of the object that the rope attaches is relevant, because the acceleration of that object due to this force is determined by its mass. a = f/m. Second law. Good. This is progress.

If the rope is tied to a 10 kg mass, let's call it C, would you not agree that C would rotate about the centre of mass of this system? Would you not agree then that the force on on C would be a centripetal one? Where is the centrifugal force?

AM

 

[Dale]

Please don't resort to ad hominem and condescending comments. No one here would fail freshman physics.

People who don't know how to apply Newton's laws to homework-style questions fail freshman physics. You don't know how to apply Newton's laws to homework-style questions.

You just cannot explain all the physics because the reaction forces to A and B's rotation are not on A or B.

Here is another example of your misunderstanding of Newton's laws. If X exerts a force on Y then the reaction is the force that Y exerts on X. The only force on B is exerted by rope 2, so the reaction force from B is the force on rope 2 exerted by B.

But you are treating the forces between A and B exactly the same way as if they were in the same reference frame.

I was working in an inertial frame. All of the forces are real, so they exist in all reference frames.

To Doc Al and Dalespam: Two simple questions:

Again, I would be glad to entertain other scenarios once you analyze the scenario I gave. I think it is only fair for scenarios to be analyzed in the order presented. You can either do your own analysis of my scenario, or confirm that you agree with mine.

Btw, I am also still waiting for the two mainstream scientific references. One that supports your claim that Newton's 3rd law doesn't apply to non-isolated systems, and another that supports your claim that Newton's 3rd law doesn't apply to "massless" ropes.

 

[Doc Al]

Then you admit that the mass of the object that the rope attaches is relevant, because the acceleration of that object due to this force is determined by its mass. a = f/m. Second law. Good. This is progress.

The mass is irrelevant in Dale's scenario since the end of the rope attached to the center of rotation.

If the rope is tied to a 10 kg mass, let's call it C, would you not agree that C would rotate about the centre of mass of this system?

What happened to the fixed pivot? You're changing the scenario once again. Come on.

Would you not agree then that the force on on C would be a centripetal one? Where is the centrifugal force?

1. You are probably (who knows?) using a different definition of 'centrifugal' than others are in this thread. (I just read your definitions.)
2. I'll be glad to consider alternative scenarios once the given ones have been adequately agreed upon.

 

[A.T.]

This a ludicrously confused 'definition'. In the first part it seems to treat centrifugal force as a pseudoforce (what else can they mean by it saying it is 'not a true force'). Then they go on to state how the 'ball is said to exert a centrifugal force on the string'. Laughable.

Yeah: It acts on the string but balances the centripetal force on the ball. But hey, it's from a howstuffworks.com! Much better than engineering books that don't count as references for Andrew.

This is better, but is not the meaning of 'centrifugal force' used in this thread.

Better, but not much better: "An object traveling in a circle behaves as if it is experiencing an outward force." So the centrifugal pseudo force acts in the inertial frame, where the object is traveling in a circle?

 

[A.T.]

Just for fun

Astronauts exerting centrifugal reaction forces on their space ship:

http://www.youtube.com/watch?v=UpiC-KbtLYI#t=12s

It seems that it would have to rotate faster than that (see reflection at 0:20) in order to generate 1g. But by running with the rotation you can increase the centrifugal reaction forces.

 

[Andrew Mason]

People who don't know how to apply Newton's laws to homework-style questions fail freshman physics. You don't know how to apply Newton's laws to homework-style questions.

I am trying to get a physics professor who has written a leading text on classical mechanics to opine on this. I will let him grade my analysis, but thanks. No one can argue that you mince words.

Here is another example of your misunderstanding of Newton's laws. If X exerts a force on Y then the reaction is the force that Y exerts on X. The only force on B is exerted by rope 2, so the reaction force from B is the force on rope 2 exerted by B.

You are assuming that A is pulling on the rope. It isn't. How can it possibly pull on the rope if it is accelerating away from B more slowly than B is accelerating toward A? Simple question? How is that possible?

Btw, I am also still waiting for the two mainstream scientific references. One that supports your claim that Newton's 3rd law doesn't apply to non-isolated systems, and another that supports your claim that Newton's 3rd law doesn't apply to "massless" ropes.

I have never made either claim. Such claims would be false, of course. Newton's 3d law applies to all interactions. Otherwise, the most fundamental law we know about, the conservation of momentum, would be violated. No one is suggesting that.

Newton's law applies to non-isolated systems. But unless you know the motion of the centre of mass how do you determine an inertial frame of reference in which to measure the forces?

Newton's third law applies to massless ropes provided the frame of reference you are measuring the forces in is the same for the forces at both ends of the rope. You are not doing that. You think you are, but you really aren't. That issue will be cleared up shortly.

AM

 

[Andrew Mason]

Astronauts exerting centrifugal reaction forces on their space ship:

http://www.youtube.com/watch?v=UpiC-KbtLYI#t=12s

It seems that it would have to rotate faster than that (see reflection at 0:20) in order to generate 1g. But by running with the rotation you can increase the centrifugal reaction forces.

Are you seriously suggesting that this "centrifugal force" is a real force? (Kubrick took a lot of artistic license in portraying space travel but he did not change physics that much).

By running against the rotation you could become weightless. Is that because the centrifugal and centripetal forces balance or is it because there is no more rotation?

AM

 

[Doc Al]

Are you seriously suggesting that this "centrifugal force" is a real force?

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

 

[RonL]

This is similar to Newton's bucket: the surface of a liquid in a bucket will change (it will form a parabaloid shape) if you hang the bucket from a rope, twist the rope and then let it spin.

What happens in your example, and in Newton's bucket, is that the forces between the molecules of liquid are not strong enough to provide the centripetal acceleration needed to keep the water moving in a circle as the wafer's/bucket's spin increases. In the wafer's frame of reference, the liquid will move outward as if there is a force pushing it out. But it is just the tendency of the water molecules to move in a straight line due to inertia. In the bucket case, water molecules build up on the inside surface of the bucket due to this inertial effect.

AM

Thanks for your answer Andrew,
There is much more going on than your answer would suggest, but I feel the example is a bit of an intrusion to the thread, so I'll leave it alone.
Should anyone up to speed on wafer coating physics see anything useful in the example, maybe the action/reaction can represent a clear and intuitive example of what pushes and pulls the resist (in my mind it has motion in four directions, all as it moves to the edge of the wafer).

I'll get back to my observation post and see if I can learn some more.:shy:

Ron

 

[Andrew Mason]

Thanks for your answer Andrew,
There is much more going on than your answer would suggest, but I feel the example is a bit of an intrusion to the thread, so I'll leave it alone.
Should anyone up to speed on wafer coating physics see anything useful in the example, maybe the action/reaction can represent a clear and intuitive example of what pushes and pulls the resist (in my mind it has motion in four directions, all as it moves to the edge of the wafer).

I'll get back to my observation post and see if I can learn some more.:shy:

Ron

It is a bit complicated. What pushes or pulls the resist (which I gather is a liquid) are the friction forces between the resist and the wafer as well as the forces between the liquid molecules.

These friction forces apply to the liquid at the point where the resist first contacts the wafer and tend to drag the molecules with the wafer in a tangential path. But as soon as that happens, the wafer accelerates away from that tangential path (ie. toward the centre) and tries to pull the resist with it. As the rotational speed of the wafer increases, the tangential speed of the resist increases. But, more important, the friction forces are not strong enough to provide the needed centripetal acceleration. The molecules' inertia keeps them moving in a tangential direction which means that they keep going in whatever direction they were traveling when the friction forces failed, and they leave the wafer. In effect, they are thrown off the wafer. It is not a force that causes them to leave. It is the insufficiency of the force that is required to make them stay that does it.

AM

 

[Andrew Mason]

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

No. Of course it does. The passenger is undergoing centripetal acceleration. We went through this about 200 posts ago. That does not mean that he exerts a centrifugal reaction force on the space station. The reaction force causes the centre of mass of the space station to accelerate toward the centre of rotation. So the reaction force is a centripetal force as well.

AM

 

[Dale]

You are assuming that A is pulling on the rope.

If the rope is pulling on B then B is pulling on the rope in the opposite direction, that is required by Newton's 3rd law. No knowledge of A is necessary although in this case it is true that A is pulling on the rope.

It isn't. How can it possibly pull on the rope if it is accelerating away from B more slowly than B is accelerating toward A? Simple question? How is that possible?

The distance is what is important for determining if a rope is under tension, not the relative acceleration.

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.

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

Btw, I am also still waiting for the two mainstream scientific references. One that supports your claim that Newton's 3rd law doesn't apply to non-isolated systems, and another that supports your claim that Newton's 3rd law doesn't apply to "massless" ropes.

I have never made either claim.

Such claims would be false, of course. Newton's 3d law applies to all interactions. Otherwise, the most fundamental law we know about, the conservation of momentum, would be violated. No one is suggesting that.

Newton's law applies to non-isolated systems. ...

Newton's third law applies to massless ropes ...

I am glad that you have changed your mind about these important issues. Given that, do you now agree that my analysis is correct?