Question about odd and even sets

[J.C.MacSwell]

I thought you'd like to know the answer for this question.

How many times the translation velocity for case 2 is lower than translation velocity for case 1?

 

No?

 

As defined in your original cases 1 and 2 in post 30 they are the same.

 

However, having said that, I'm pretty sure you meant something different from what you wrote.

Equal linear momentum inputs to equal mass systems originally at rest means that their final velocities will be the same.

Edited July 28, 2009 by J.C.MacSwell

[ABV]

Equal linear momentum inputs to equal mass systems originally at rest means that their final velocities will be the same.

 

Yes - for translation movement - only linear momentum - the natural phenomenon

Yes - for rotation movement - only angular momentum. -the natural phenomenon

Yes/No/Maybe? - for translation with rotation movement - there both linear and angular momentums. - the natural phenomenon also.

 

The natural phenomenon the translation with rotation movement is multidimention. Cannot be described like a sum of two movements. Because laws of angular and linear momentums are completelly different.

 

You would try to describe situation just for linear to linear movement momentum transfer action.

 

Here is linear to linear with rotation movement momentum transfer action.

It can't be covered by linear to linear momentum transfer law. This is the same if you try to describe rotation to rotation angular momentum transfer by linear law.

 

This is the mistake.

[J.C.MacSwell]

Yes - for translation movement - only linear momentum - the natural phenomenon

Yes - for rotation movement - only angular momentum. -the natural phenomenon

Yes/No/Maybe? - for translation with rotation movement - there both linear and angular momentums. - the natural phenomenon also.

 

The natural phenomenon the translation with rotation movement is multidimention. Cannot be described like a sum of two movements. Because laws of angular and linear momentums are completelly different.

 

You would try to describe situation just for linear to linear movement momentum transfer action.

 

Here is linear to linear with rotation movement momentum transfer action.

It can't be covered by linear to linear momentum transfer law. This is the same if you try to describe rotation to rotation angular momentum transfer by linear law.

 

This is the mistake.

 

This is due to the angular moment input due to the "off center" linear momentum input. It has no affect on the final linear momentum.

 

You have to be careful because I think you believe these linear momentum inputs can be brought about in the same way. The second case requires more energy, yet I think you are picturing an equal energy input.

If that was the case there would be less linear momentum transferred in the second case, but you have explicitly stated otherwise as a condition of each example.

[ABV]

This is due to the angular moment input due to the "off center" linear momentum input. It has no affect on the final linear momentum.

correct

You have to be careful because I think you believe these linear momentum inputs can be brought about in the same way. The second case requires more energy, yet I think you are picturing an equal energy input.

If that was the case there would be less linear momentum transferred in the second case, but you have explicitly stated otherwise as a condition of each example.

 

Let's reverse time for asymmetric hit action.

Rotation with translation movement stops in one hit and transfer momentum to another object. Is any law which cover this situation? No

The law of (any)momentum conservation is reversible for linear and rotation movement. If translation with rotation movement is combination for asymmetric hit action then why this movement combination can't transfer in one time?

Actually It can in specific condition. However it won't be for any of these movement axis. Not for linear coordinates. Not for radius vector. I should be own coordinates for this movement. Other words, this object after asymmetric hit moves on own frame of reference. If we want to describe law of momentum conservation for this object it should be used his frame of reference.

This could be unify law of momentum conservation for any movements. Momentum is conserve for any object on his frame of reference.

[J.C.MacSwell]

correct

 

 

Let's reverse time for asymmetric hit action.

Rotation with translation movement stops in one hit and transfer momentum to another object. Is any law which cover this situation? No

The law of (any)momentum conservation is reversible for linear and rotation movement. If translation with rotation movement is combination for asymmetric hit action then why this movement combination can't transfer in one time?

Actually It can in specific condition. However it won't be for any of these movement axis. Not for linear coordinates. Not for radius vector. I should be own coordinates for this movement. Other words, this object after asymmetric hit moves on own frame of reference. If we want to describe law of momentum conservation for this object it should be used his frame of reference.

This could be unify law of momentum conservation for any movements. Momentum is conserve for any object on his frame of reference.

 

Yes. No matter what inertial frame of reference you use the conservation of linear momentum applies and also, separately, the conservation of angular momentum applies.

 

You cannot present an example where it does not. All you are doing is misunderstanding the exact meaning.

[ABV]

Yes. No matter what inertial frame of reference you use the conservation of linear momentum applies and also, separately, the conservation of angular momentum applies.

 

You cannot present an example where it does not. All you are doing is misunderstanding the exact meaning.

 

The center mass rotated body moves slowest than body without rotation at repulsion action.

The statement of proof.

 

 

Let’s take two identical thin cylinders which stay initially relatively to an observer. These cylinders will repulse to each other.

Let’s make two experiments.

 

The first experiment.

The cylinders repulse to each other from their center of mass. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation, their linear velocities are identical relatively to the observer. Let’s measure their kinetic energies. The full kinetic energy is equal to:

As it shown on equation their energies are equal.

Let’s take the derivatives from their parts of energy.

As it shown on equation their momentums are equal.

This experiment action is symmetric relatively to observer.

 

The second experiment.

The cylinders repulse to each other from their different parts. One cylinder is repulsing form his center mass. Another cylinder is repulsing from his edge. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation let’s assume their linear velocities are identical relatively to the observer. Only one of these cylinders rotates. Let’s measure their kinetic energies. The full kinetic energy is equal to:

As it shown on equation their energies are not equal.

Let’s take the derivatives from their parts of energy.

 

As it shown on equation their linear momentums are equal. However the angular momentums are not equal. Only one of these cylinders has angular momentum.

This experiment action is not symmetric relatively to observer.

This assumption broke law of angular momentum conservation. The body can’t start own rotating without symmetrical action.

This mean the assumption about identical linear velocity was wrong.

These experiments is explaining the translation with rotation movement is standalone natural phenomenon. And law of momentum conservation should cover this movement.

 

My assumption this movement have a linear and angular momentum together.

This is the law of momentum conservation for translation with rotation movement.

 

Edited July 29, 2009 by ABV

[J.C.MacSwell]

 

As it shown on equation their linear momentums are equal. However the angular momentums are not equal. Only one of these cylinders has angular momentum.

This experiment action is not symmetric relatively to observer.

This assumption broke law of angular momentum conservation. The body can’t start own rotating without symmetrical action.

This mean the assumption about identical linear velocity was wrong.

These experiments is explaining the translation with rotation movement is standalone natural phenomenon. And law of momentum conservation should cover this movement.

 

My assumption this movement have a linear and angular momentum together.

This is the law of momentum conservation for translation with rotation movement.

 

[/left]

 

You are wrong. Angular momentum is conserved.

 

You note that only one half of the combined system has angular momentum.

 

This is only true if you consider each half in isolation.

 

The centers of mass of each half have angular momentum wrt the center of mass of the combined system (which is still at rest). This is exactly equal but opposite that of the angular momentum of the rotating half.

 

The net angular momentum of the combined system is zero.

 

Angular momentum is conserved.

[ABV]

You are wrong. Angular momentum is conserved.

 

You note that only one half of the combined system has angular momentum.

 

This is only true if you consider each half in isolation.

 

The centers of mass of each half have angular momentum wrt the center of mass of the combined system (which is still at rest). This is exactly equal but opposite that of the angular momentum of the rotating half.

 

The net angular momentum of the combined system is zero.

 

Angular momentum is conserved.

 

The andular and translation movements just a simplifyed cases from translation with rotation movement. They movement located on own isolated frames of references from each other. The translation with rotation is main movements. this is the prodd version 2

 

The statement of proof.

 

 

Let’s take two identical thin cylinders which stay initially relatively to an observer. These cylinders will repulse to each other.

Let’s make two experiments.

 

The first experiment.

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The cylinders repulse to each other from their center of mass. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation, their linear velocities are identical relatively to the observer. Let’s measure their kinetic energies. The full kinetic energy is equal to:

As it shown on equation their energies are equal.

Let’s take the derivatives from their parts of energy.

As it shown on equation their momentums are equal.

This experiment action is symmetric relatively to initial event (repulsing) and observer.

 

The second experiment.

The cylinders repulse to each other from their different parts. One cylinder is repulsing form his center mass. Another cylinder is repulsing from his edge. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation let’s assume their linear velocities are identical relatively to the observer. Only one of these cylinders rotates. Let’s measure their kinetic energies. The full kinetic energy is equal to:

As it shown on equation their energies are not equal.

Let’s take the derivatives from their parts of energy.

 

As it shown on equation their linear momentums are equal. However the angular momentums are not equal. Only one of these cylinders has angular momentum.

This experiment action is not symmetric relatively to initial event (repulsing) and observer.

This assumption broke the law of angular momentum conservation. The body can’t start own rotating without symmetrical action.

 

This mean the assumption about identical linear velocity was wrong.

 

These experiments is explaining the translation with rotation movement is standalone natural phenomenon. And law of momentum conservation should cover this movement.

 

My assumption this movement have a linear and angular momentum together.

This is the law of momentum conservation for the translation with rotation movement.

 

[J.C.MacSwell]

As it shown on equation their linear momentums are equal. However the angular momentums are not equal. Only one of these cylinders has angular momentum.

This experiment action is not symmetric relatively to initial event (repulsing) and observer.

This assumption broke the law of angular momentum conservation. The body can’t start own rotating without symmetrical action.

 

This mean the assumption about identical linear velocity was wrong.

 

 

No. Nothing need be symmetric. How many times have I said that the energy inputs to each cylinder must be different? The body can start rotating without symmetric action.

 

Angular momentum in the complete system, everything involved, is maintained.

 

Angular momentum is not just the sum of the individual angular momentums of each individual cylinder.

 

Look at the angular momentum of the two cylinders centers of mass with respect to each other. Their linear momentum vectors are exactly opposite but displaced.

 

Can you not see this component? Is it not exactly opposite that of the rotating cylinder?

[ABV]

No. Nothing need be symmetric. How many times have I said that the energy inputs to each cylinder must be different? The body can start rotating without symmetric action.

Would you please give me example of assymetric action relativy to event for only rotation and only translation movement please.

 

Angular momentum in the complete system, everything involved, is maintained.

 

Angular momentum is not just the sum of the individual angular momentums of each individual cylinder.

 

Look at the angular momentum of the two cylinders centers of mass with respect to each other. Their linear momentum vectors are exactly opposite but displaced.

 

Can you not see this component? Is it not exactly opposite that of the rotating cylinder?

 

For symmetric action the angular momentums opposite for each other too. Is it nessesary keep center mass respectly to each other. What about complicated action. The center mass may move after linear repulsion.

 

 

The symmetric action where translation with rotation movement is the sum of two.

The cylinders start rotation during angular repulsing action. After this action another linear repulsing action gives these cylinders to move with linear velocities. In this case you see all symmetric relativelly to initial event. Linear and angular velocities are present for both cylinders.

 

If compare this symmetric action with previous expereiment you'll see the difference. One of these cylinder has no rotation.

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Merged post follows:

Consecutive posts merged

My assumption this movement have a linear and angular momentums together.

This is the law of momentum conservation for the translation with rotation movement:

 

 

This law has a complex number and it has linear and angular momentums.

One of these parts linear or angular should be imaginary number. It depends on observer which frame of reference he uses.

Full momentum transfer will be by absolute value:

 

 

======

Base on previous statement it possible to say the translation with rotation movement is covering translation and rotation movements together. One of these movements is the simple version of main complicated movement.

[ABV]

To keep eqaution clear for units, I added Ru- unit radius.

http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9#

A Rotation with Translation Movement is standalone natural phenomenon.

 

Statement of proof.

 

Let’s take two identical thin cylinders which stay initially relatively to an observer. These cylinders will repulse to each other.

Let’s make two experiments.

 

The first experiment.

 

The cylinders repulse to each other from their center of mass. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation, their linear velocities are identical relatively to the observer. Let’s measure their kinetic energies. The full kinetic energy is equal to:

 

 

As it shown on equation their energies are equal.

Let’s take the derivatives from their parts of energy.

 

 

 

As it shown on equation their momentums are equal.

For this experiment these cylinders have symmetric action as translation movement relatively to initial event (repulsing) and observer.

 

The second experiment.

 

The cylinders repulse to each other from their different parts. One cylinder is repulsing form his center mass. Another cylinder is repulsing from his edge. These cylinders start moving relatively to observer and they move opposite relatively to each other. Base on law of momentum conservation let’s assume their linear velocities are identical relatively to the observer. Only one of these cylinders rotates. Let’s measure their kinetic energies. The full kinetic energy is equal to:

 

 

 

As it shown on equation their energies are not equal.

Let’s take the derivatives from their parts of energy.

 

 

 

As it shown on equation their linear momentums are equal. However the angular momentums are not equal. Only one of these cylinders has angular momentum.

This experiment action is not symmetric relatively to initial event (repulsing) and observer.

This assumption broke the law of angular momentum conservation. The body can’t start own rotating without symmetrical action.

 

This mean the assumption about identical linear velocity was wrong.

 

One even can't reproduce two type of movements together for one object. One event can reproduce only one type of movement per object.

On experiment 1 both objects have only one translation movement.

The experiment 2 shows only one type of movements for one object and two types of movements for other object. However these objects taked only one event(repulsing). It means it should be one type of movements per object and translation with rotation movements for this experiment should be described as new type of moments.

The experiment 2 shows situation where the translation with rotation movement as a standalone natural phenomenon. This movement should have own momentum and law of momentum conservation.

 

My assumption this movement have a linear and angular momentums together.

The full momentum of rotation with translation movement is:

 

note: Pj - linear momentum Lk - angular momentum Ru - unit radius

 

This is the law of momentum conservation for the translation with rotation movement:

 

 

This law has a complex number and it has linear and angular momentums.

One of these parts linear or angular should be imaginary number. It depends on observer which frame of reference he uses.

 

Full momentum transfer for this movement has calculation by absolute value:

 

 

Full kinetic energy of rotation with translation movement is:

 

 

 

Follow law of momentum conservation and using complex number for translation with rotation momentum, the cylinders translation velocities have a different value on experiment 2.

 

======

Base on previous statement it possible to say the translation with rotation movement is covering translation and rotation movements together. One of these movements is the simple version of main complicated movement.

[ABV]

For example: if reverse time back for experiment 2 then one cylinders angular L2 and translation P2 momentums compensated by another cylinders only translation momentum P1.

For this case equation between momentums is:

[ABV]

Another version:

 

 

My assumption this movement have a linear and angular momentums together.

The full momentum of rotation with translation movement is:

 

 

note: Pj - linear momentum Lk - angular momentum Ru - unit radius

 

This is the law of momentum conservation for the translation with rotation movement:

 

 

This movement has 2 movement components. A rotation and a translation parts.

How are correlating translation and angular momentum parts to full momentum?

This picture shows a simple rotating body diagram.

All mass of rotation body concentrate on unit radius Ru. An initial momentum hit body on distance h from their center mass.

Knowing moment of inertia equal to

 

 

then

 

 

 

Let's assume this part of momentum goes to angular part:

 

 

Then angualr momentum equal to:

 

 

Then translation momentum equal to:

 

Lets sum these parts and check full momentum:

 

 

 

 

Follow law of momentum conservation for translation with rotation momentum, the cylinders translation velocities have a different value on experiment 2.

 

The velocities equal to:

 

 

 

 

======

Base on previous statement it possible to say the translation with rotation movement is covering translation and rotation movements together. One of these movements is the simple version of main complicated movement.

Edited August 11, 2009 by ABV