Does it really work? [Answered - No]

[J.C.MacSwell]

Let's accelerate thin ring inside u-track without external forces. In this case thin ring gains momentum to track. Correct?

Thin ring will cause centripetal force on curve part of track. Track gains another momentum with opposite on this period of time. Correct?

During deceleration time the thin ring gains momentum to the track again. Direction for acceleration/deceleration momentums on u-track will be identical. Correct?

For thin ring Net momentums curve part of track will be less than sum of acceleration/deceleration momentums. For solid disk they will be equal to each other. Why? Because the thin ring has more energy than solid disk has. And during deceleration time (let's talk about that part. I didn't describe how to accelerate thin ring) thin ring gains bigger momentum to track than solid disk can do.

Inside Isolated system law of conservation will work. But for external viewers Isolated System will oscillate and move into one direction on each circle.

 

BTW.

I updated site. I did a lot of changes. Please take a look.

 

So, again, how do you accelerate the thin ring without external forces on the straight part of the track?

 

Just a note I have to add: You are applying Newtons Laws to something you have drawn, some geometry, some representation of a physical system. The result you are getting is an error in logic or in your assumptions. How can it be otherwise when you are basing it on Newton's laws alone and nothing else? You are not doing an experiment.

Edited May 28, 2009 by J.C.MacSwell

[ABV]

So, again, how do you accelerate the thin ring without external forces on the straight part of the track?

Some implementation things?

Magnet gun, percussion device, turned wheel, and much, much more. All these devices should stand on oval track.

 

 

Just a note I have to add: You are applying Newtons Laws to something you have drawn, some geometry, some representation of a physical system. The result you are getting is an error in logic or in your assumptions. How can it be otherwise when you are basing it on Newton's laws alone and nothing else? You are not doing an experiment.

 

 

There is no error. This is paradox. I you see some errors on force directions or any kind of things, let me know. If it’s real error – I fix it.

But again. This is not logical error -This is paradox. We can discuss about each part of this presentation.

I agree. Isolated System should stand on one position. At least, modern physics say about it. The law of momentum conservation is working there, inside Isolated System. But I see equation conflict between momentums from curve and straight parts of track.

 

A few questions.

Will extra energy allow to rolling body moves longer?

Which rolling body, thin ring or solid disk, with same mass and velocity moves longer? (Let’s use same rolling resistant in both cases)

Are Newton’s laws will change from track geometry?

 

 

P.S.

The experiments may be not so clear, but shows whole test dynamic in comparing view.

Edited May 29, 2009 by ABV

[J.C.MacSwell]

Some implementation things?

Magnet gun, percussion device, turned wheel, and much, much more. All these devices should stand on oval track.

 

 

 

 

 

There is no error. This is paradox. I you see some errors on force directions or any kind of things, let me know. If it’s real error – I fix it.

But again. This is not logical error -This is paradox. We can discuss about each part of this presentation.

I agree. Isolated System should stand on one position. At least, modern physics say about it. The law of momentum conservation is working there, inside Isolated System. But I see equation conflict between momentums from curve and straight parts of track.

 

A few questions.

Will extra energy allow to rolling body moves longer?

Which rolling body, thin ring or solid disk, with same mass and velocity moves longer? (Let’s use same rolling resistant in both cases)

Are Newton’s laws will change from track geometry?

 

 

P.S.

The experiments may be not so clear, but shows whole test dynamic in comparing view.

Show an isolated system with no external forces.

 

Show all the forces at one, and only one, point in time.

 

Use arrows to show the direction of each force.

 

Let the length of the arrow represent the magnitude of the force.

 

If you haven't misrepresented any forces: You will see that the net force on the system is exactly zero.

This will work for every point in time you consider.

 

If you consider many points at once, it will be too easy to fool yourself, by 1. misrepresenting some of the forces relative to others, and 2. leaving some out, both of which you are doing.

Edited May 29, 2009 by J.C.MacSwell

[ABV]

Show an isolated system with no external forces.

Are any external forces present there?

 

Show all the forces at one, and only one, point in time.

This Isolated System has 4 stages. I may be devide image to all of them, But it won't be helpful. (But I'll think about it)

 

Use arrows to show the direction of each force.

All forces have a direction. Did I miss the direction for any foce there?

 

Let the length of the arrow represent the magnitude of the force.

Please read text. "Net momentum which rolling body gains to track from these forces will be a parallel to acceleration/deceleration momentums too"

 

If you haven't misrepresented any forces: You will see that the net force on the system is exactly zero.

This will work for every point in time you consider.

I disagree. No any law of forces conservation is present on Classical Mechanics. If we talk about law of linear momentum conservation, I showed may examples on previous posts and explained how it works.

 

If you consider many points at once, it will be too easy to fool yourself, by 1. misrepresenting some of the forces relative to others, and 2. leaving some out, both of which you are doing.

no comments

[J.C.MacSwell]

Are any external forces present there?

 

Yes. There is no reaction force shown when the ring is accelerated and decelerated. The traction force is a lesser component.

 

This Isolated System has 4 stages. I may be devide image to all of them, But it won't be helpful. (But I'll think about it)

You are adding the four stages up, as if they represent a full circuit

 

All forces have a direction. Did I miss the direction for any foce there?

Other than leaving some out, they are almost correct, except the end ones represent the full curve where the centripetal force direction will vary. You can claim it represents the average direction only in the case of a static track, so this is incorrect for an isolated system where the track will be displaced, and you may also misrepresent the average magnitude. Very easy to make mistakes (wrong assumptions) as the speed will vary around the curve.

 

Please read text. "Net momentum which rolling body gains to track from these forces will be a parallel to acceleration/deceleration momentums too"

 

Something is lost to me in the translation. Do you believe the forces are proportional to the length of the arrows shown?

 

I disagree. No any law of forces conservation is present on Classical Mechanics. If we talk about law of linear momentum conservation, I showed may examples on previous posts and explained how it works.

 

If there are no external forces on an isolated system, the net forces will be zero. If the center of gravity of an isolated system is at rest, it will remain at rest. It cannot move, not even wobble. The system cannot gain, even temporarily, any net momentum it didn't already have. None of your explanations seemed to recognize this, and you added and subtracted momentums that you believed you gain and lost. Making more errors to one side than the other, and you magically end up breaking physical laws.

 

no comments

Edited May 29, 2009 by J.C.MacSwell

[ABV]

Which comes back to:

 

No, it doesn't work as described. If it did then congratulations, you are breaking known laws of physics, as what is described is impossible.

 

I didn’t break any physics laws. I’m using physics laws for explanation.

Please do not mix Newton's third law and law of momentum conservation.

I’m using both of them there.

[J.C.MacSwell]

I didn’t break any physics laws. I’m using physics laws for explanation.

Please do not mix Newton's third law and law of momentum conservation.

I’m using both of them there.

 

You are claiming you have a design whereas an isolated physical system can change it's momentum. Are you not?

[ABV]

You are claiming you have a design whereas an isolated physical system can change it's momentum. Are you not?

 

The law of momentum conservation works for Isolated System Internally. But to make it work, Isolated System must move into one direction.

 

Please pay attention to equation between thin ring and solid disk momentums on curve part of track. And momentum difference on deceleration part of track (has better explanation than acceleration part). If you agree with that, what will happen with this Isolated System on end of rolling body moving circle?

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Actually one thing will be different. The thin ring Acceleration/deceleration acting time will be longer than solid disk acting time for same parts of track. It means Full circle time for rolling thin ring and solid disk with same mass and velocities on curve part of track will be different.

[J.C.MacSwell]

The law of momentum conservation works for Isolated System Internally. But to make it work, Isolated System must move into one direction.

 

Please pay attention to equation between thin ring and solid disk momentums on curve part of track. And momentum difference on deceleration part of track (has better explanation than acceleration part). If you agree with that, what will happen with this Isolated System on end of rolling body moving circle?

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Actually one thing will be different. The thin ring Acceleration/deceleration acting time will be longer than solid disk acting time for same parts of track. It means Full circle time for rolling thin ring and solid disk with same mass and velocities on curve part of track will be different.

 

If they are at the same velocity there is no difference in translational momentum between objects of equal mass. Neither the ring nor the disc can magically gain any momentum to impart to the track, without taking it from the track, unless an outside force is involved.

[ABV]

If they are at the same velocity there is no difference in translational momentum between objects of equal mass. Neither the ring nor the disc can magically gain any momentum to impart to the track, without taking it from the track, unless an outside force is involved.

 

There is no magic – difference capacity of kinetic energy.

Rolling bogy kinetic energy depend o 2 parts - linear and angular.

The thin ring has angular part of kinetic energy bigger than the same part of energy for solid disk. Rolling body with highest energy gets longer ride than same body with lower energy. This longer ride with same rolling resistance gains higher momentum to the track.

[J.C.MacSwell]

There is no magic – difference capacity of kinetic energy.

Rolling bogy kinetic energy depend o 2 parts - linear and angular.

The thin ring has angular part of kinetic energy bigger than the same part of energy for solid disk. Rolling body with highest energy gets longer ride than same body with lower energy. This longer ride with same rolling resistance gains higher momentum to the track.

 

No amount of energy can overcome the law of conservation of momentum. Balance the momentum equations. Balance the energy equations. Just because it has more kinetic energy does not mean it has more momentum.

 

I understand what you think is going on, but it doesn't work that way.

Edited May 29, 2009 by J.C.MacSwell

[ABV]

No amount of energy can overcome the law of conservation of momentum. Balance the momentum equations. Balance the energy equations.

.

All equations are present inside Isolated System.

 

The momentum is depending from value of energy. All kinetic energy would discharge and convert to work, with 2 values - resistance force and acting time. Momentum is just multiplication of force and acting time values. Correct?

 

I understand what you think is going on, but it doesn't work that way.

 

Which way it should? Where extra kinetic energy goes?

[J.C.MacSwell]

All equations are present inside Isolated System.

 

The momentum is depending from value of energy. All kinetic energy would discharge and convert to work, with 2 values - resistance force and acting time.

 

This is not correct. Misleading at best.

 

 

Momentum is just multiplication of force and acting time values. Correct?

 

Correct. Just keep it mind that the force and momentum are vector quantities, unlike energy.

[ABV]

This is not correct. Misleading at best.

What do you mean. Could you explain in details please.

 

 

 

Correct. Just keep it mind that the force and momentum are vector quantities, unlike energy.

 

And what? You mean momentum and energy does not relate to each other?

 

P.S.

I didn't get answer about extra energy. Where it goes?

[J.C.MacSwell]

What do you mean. Could you explain in details please.

 

 

 

 

 

And what? You mean momentum and energy does not relate to each other?

 

P.S.

I didn't get answer about extra energy. Where it goes?

 

You put it and more in on one side of the track, and remove it on the other, along with other "non-extra" inputs and extractions. It just takes more to input to accelerate the ring than the disc, and also more to extract on the other side.

Edited May 29, 2009 by J.C.MacSwell

[ABV]

You put it and more in on one side of the track, and remove it on the other, along with other "non-extra" inputs and extractions. It just takes more to input to accelerate the ring than the disc, and also more to extract on the other side.

 

First. You try to reverse calculation. This is incorrect.

Second. thin ring has more energy than solid disk at same velocity. And this thin ring energy gains more momentum to the track than solid disk does.

 

The rolling movement is more complicated than linear, but try to understand rolling body convert energy to work by rolling friction ONLY. Base on spent energy value, rolling body gains momentum to track. Energy cannot disappear without any reason, same cannot come from nowhere.

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You put it and more in on one side of the track, and remove it on the other, along with other "non-extra" inputs and extractions. It just takes more to input to accelerate the ring than the disc, and also more to extract on the other side.

 

One more question. Will thin ring get longer ride than solid disk with same mass and velocity? Yes or No?

(please use same rolling resistance on a straight track)

[J.C.MacSwell]

First. You try to reverse calculation. This is incorrect.

Second. thin ring has more energy than solid disk at same velocity. And this thin ring energy gains more momentum to the track than solid disk does.

 

The rolling movement is more complicated than linear, but try to understand rolling body convert energy to work by rolling friction ONLY. Base on spent energy value, rolling body gains momentum to track. Energy cannot disappear without any reason, same cannot come from nowhere.

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One more question. Will thin ring get longer ride than solid disk with same mass and velocity? Yes or No?

(please use same rolling resistance on a straight track)

 

Yes.

[ABV]

Yes.

 

Thank you for answer.

 

But if look on this Isolated System, we can see same straight tracks inside. Newton laws should not be change. And law of conservation mast work for Isolated System also.

 

Paradox.

 

This is I try to explain.

[J.C.MacSwell]

Thank you for answer.

 

But if look on this Isolated System, we can see same straight tracks inside. Newton laws should not be change. And law of conservation mast work for Isolated System also.

 

Paradox.

 

This is I try to explain.

 

Start with a free body diagram.

 

Draw the forces on ring or disc showing rolling resistance.

[ABV]

Start with a free body diagram.

 

Draw the forces on ring or disc showing rolling resistance.

 

Will it helps? These diagrams will be the same for thin ring and solid disk.

But thin ring acting time for deceleration period will be higher than same deceleration acting time for solid disk.

[J.C.MacSwell]

Will it helps? These diagrams will be the same for thin ring and solid disk.

But thin ring acting time for deceleration period will be higher than same deceleration acting time for solid disk.

 

Draw it and see.

[ABV]

Draw it and see.

 

This is link to wikipedia.

http://en.wikipedia.org/wiki/Rolling_resistance

[J.C.MacSwell]

This is link to wikipedia.

http://en.wikipedia.org/wiki/Rolling_resistance

 

The drawing shows a wheel being pulled with constant velocity against rolling resistance. Deceleration due to rolling resistance will be different from that.

[ABV]

The drawing shows a wheel being pulled with constant velocity against rolling resistance. Deceleration due to rolling resistance will be different from that.

 

2 questions.

Is rolling resistance depends on rolling body shape?(thin ring and solid disk only)

Is rolling resistance depends on acting time?

[J.C.MacSwell]

2 questions.

Is rolling resistance depends on rolling body shape?(thin ring and solid disk only)

Is rolling resistance depends on acting time?

 

It depends on a lot of factors. I was hoping you would draw the force vectors and see the difference between the disc and ring and to see what it meant to have the "same" rolling resistance for decelerating discs or rings. The resultant force vectors cannot have the same magnitude, direction and application point in each case (unless at least one of them is skidding to some extent)