Question about odd and even sets

[ABV]

Hi.

First. I apologize if this thread breaks any rules of this forum.

 

I’m working on rolling body (thin ring) problem and found interesting thing.

Main question is: Could be even and odd sets transform to each other?

If answer is no then it means broken rolling ring to chain would loose momentum on surface by this insufficient sets transformation.

 

This is the main problem:

 

The idea is very simple.

If spit a rolling ring to small parts set of n elements (1,2,3,...,n) with mass m then each of them conduct linear and circular movement on surface.

Each piece has constant angular velocity. And each piece has variable linear velocity at surface point. Once per circle each element stop on surface. At surface point this element has linear velocity value equal to zero at this time. If break chain then all of these elements except one is continue to move. If calculate net linear momentum of these element this should be equal to this ring initial linear momentum. But one of these elements is stop already and net momentum will be for n-1 elements. In this case one element attached to the surface and mass is M(suface) + m(element). Set of elements has a mass equal to (n-1)*m now. It’s change initial condition. The surface is still keeping same momentum and increase own mass. Set of elements n-1 should hold same momentum, but lost one piece with mass m.

 

Is this net of linear momentums for set of elements n-1 with net mass (n-1)*m is equal to the ring initial linear momentum?

 

My suggestion:

If ring has set of n elements then for surface point only ring's set of n-1 elements is always move. But one element with zero value of velocity stand down and it will be always part to surface.

For this particular case for this broken thin ring or chain, the set of n-1 elements not equivalent to set of n elements. It means net momentum of set of n elements won’t be equal to net momentum of set of n-1 elements.

 

 

The original problem location:

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

 

There also review and other solution by RPenner.

[J.C.MacSwell]

Hi.

First. I apologize if this thread breaks any rules of this forum.

 

I’m working on rolling body (thin ring) problem and found interesting thing.

Main question is: Could be even and odd sets transform to each other?

If answer is no then it means broken rolling ring to chain would loose momentum on surface by this insufficient sets transformation.

 

This is the main problem:

 

The idea is very simple.

If spit a rolling ring to small parts set of n elements (1,2,3,...,n) with mass m then each of them conduct linear and circular movement on surface.

Each piece has constant angular velocity. And each piece has variable linear velocity at surface point. Once per circle each element stop on surface. At surface point this element has linear velocity value equal to zero at this time. If break chain then all of these elements except one is continue to move. If calculate net linear momentum of these element this should be equal to this ring initial linear momentum. But one of these elements is stop already and net momentum will be for n-1 elements. In this case one element attached to the surface and mass is M(suface) + m(element). Set of elements has a mass equal to (n-1)*m now. It’s change initial condition. The surface is still keeping same momentum and increase own mass. Set of elements n-1 should hold same momentum, but lost one piece with mass m.

 

Is this net of linear momentums for set of elements n-1 with net mass (n-1)*m is equal to the ring initial linear momentum?

 

My suggestion:

If ring has set of n elements then for surface point only ring's set of n-1 elements is always move. But one element with zero value of velocity stand down and it will be always part to surface.

For this particular case for this broken thin ring or chain, the set of n-1 elements not equivalent to set of n elements. It means net momentum of set of n elements won’t be equal to net momentum of set of n-1 elements.

 

 

The original problem location:

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

 

There also review and other solution by RPenner.

 

For an idealized ring rolling on an idealized surface, 1 and only 1 point is at rest. So no element is ever at rest, if n is a finite number.

[ABV]

For an idealized ring rolling on an idealized surface, 1 and only 1 point is at rest. So no element is ever at rest, if n is a finite number.

 

Correct, but even and odd sets are a big difference. These sets (in generic) can't covert to each other without extra transformation (loses).

This means set with one less element won't return whole momentum back.

Just keep in mind, mass of chain and surface changed.

 

Transformation of this chain started when ring stopped and cut the thin. This transformation move net momentum to n-1 set.

Other words, (I'm not sure about it) if a thin ring do geometry transformation during riding time then this thin ring will stop without any extra forces.

 

 

Would be some math specialists can help?

Edited June 17, 2009 by ABV

[ABV]

My suggestion:

 

If ring has set of n elements then for surface point only ring's set of n-1 elements is always move. But one element with zero value of velocity stand down and it will be always part to surface.

 

For this particular case, this set of n-1 elements (broken thin ring or chain) not equivalent to set of n elements (initial ring). Base on law of momentum conservation net momentum for set n-1 elements would have same initial momentum, but momentum density for each element of this set n-1 will change. The surface will take all elements momentums back when they’ll stop. Base on law of momentum conservation, from same momentum the body with higher mass will take lower velocity then body with lower mass.

 

P = m*V = const

 

m1*V1 = m2*V2

 

The surface with new mass will take a velocity V1 from set of elements n-1. This velocity is different from initial surface velocity V0, because the surface mass has been changed.

 

Follow the law of momentum conservation, even all elements of broken ring will stop on the surface, this surface will have linear velocity more than zero.

[ABV]

Please look on this picture for better understanding.

[J.C.MacSwell]

Please look on this picture for better understanding.

 

Not sure what you are getting at but I hope you don't think the rolling object will come to a halt without transferring any of it's momentum to the surface.

[ABV]

Not sure what you are getting at but I hope you don't think the rolling object will come to a halt without transferring any of it's momentum to the surface.

 

1. Just static friction should be there.

2. Translation with rotation motion gives different linear velocity at surface point. Red piece shows V=0. 2V(not shown) on the top of this ring.

image from upenn:

http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsubsection4_1_4_3.html

3. The piece with zelo linear velocity must be join to the surface as solid material at stop point (just for modeling).

 

The Idea is stop red piece and cut the ring at the same time.

Rest of the ring return whole momentum to the surface back. But surface mass(M(surface) + m(red piece) is different from initial mass M(surface) at this time.

The surface velocity V1<>0 will be different from initial velocity V0=0 because momentum returns back to different platform (the surface mass has been changed).

P=mV=const

m1V1=m2V2

Edited June 23, 2009 by ABV

[J.C.MacSwell]

1. Just static friction should be there.

2. Translation with rotation motion gives different linear velocity at surface point. Red piece shows V=0. 2V(not shown) on the top of this ring.

image from upenn:

http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsubsection4_1_4_3.html

3. The piece with zelo linear velocity must be join to the surface as solid material at stop point (just for modeling).

 

The Idea is stop red piece and cut the ring at the same time.

Rest of the ring return whole momentum to the surface back. But surface mass(M(surface) + m(red piece) is different from initial mass M(surface) at this time.

The surface velocity V1<>0 will be different from initial velocity V0=0 because momentum returns back to different platform (the surface mass has been changed).

P=mV=const

m1V1=m2V2

 

So...the momentum does get transferred to the surface and the surface (which has increased in mass) is now moving...at the same velocity it had prior to the little human on the left giving the ring the initial push.

 

No problem.

[ABV]

So...the momentum does get transferred to the surface and the surface (which has increased in mass) is now moving...at the same velocity it had prior to the little human on the left giving the ring the initial push.

 

No problem.

 

Yeah, but little human push rolling body with mass n*m.

And surface accept momentum back from chain with mass (n-1)*m.

In this case chain return back momentum to surface with different mass M(surface)+m(red element).

No body turn red element to linear velocity zero. No extra momentum for this action. It means the surface won't return to initial velocity after all.

 

----

If red element takes mass zero then the surface return back to initial velocity. But on real world each element has a mass.

[J.C.MacSwell]

Yeah, but little human push rolling body with mass n*m.

And surface accept momentum back from chain with mass (n-1)*m.

In this case chain return back momentum to surface with different mass M(surface)+m(red element).

No body turn red element to linear velocity zero. No extra momentum for this action. It means the surface won't return to initial velocity after all.

 

----

If red element takes mass zero then the surface return back to initial velocity. But on real world each element has a mass.

 

The red element has no linear momentum.

 

Opposite the red element, at the highest point on the ring, there is an element with double the average linear momentum of the ring.

 

Removing the red element from consideration increases the average linear momentum of the remaining elements.

 

Which means the surface will return to initial velocity after all.

[ABV]

The red element has no linear momentum.

 

Opposite the red element, at the highest point on the ring, there is an element with double the average linear momentum of the ring.

 

Removing the red element from consideration increases the average linear momentum of the remaining elements.

 

Which means the surface will return to initial velocity after all.

 

The Chain n-1 will have a ring net momentum.

 

But the surface won't return to initial velocity because the surface mass has been changed.

The red element already on surface.

[J.C.MacSwell]

The Chain n-1 will have a ring net momentum.

 

But the surface won't return to initial velocity because the surface mass has been changed.

 

Yes it will because the additional mass, the ring, was originally at rest with respect to the surface.

[ABV]

Yes it will because the additional mass, the ring, was originally at rest with respect to the surface.

The ring with mass n*m has a movement with some linear momentum.

After conversion without extra linear momentum.

The chain with mass (n-1)*m has a movement with some linear momentum.

 

Both of them part othe isolated system. But this is not means if little human pushh tje ring system will stay. System will start move to oppsite direction. If little human catch the ring system will stop. Push and catch this ring to opposite direction will return system to initial point.

 

Here is complete different behavior with ring and chain. The little human push the ring with one mass, but chain is stopping with other mass. Even if momentums are same, the surface mass will change at conversion moment. The red element stop without extra momentum.

[J.C.MacSwell]

The ring with mass n*m has a movement with some linear momentum.

After conversion without extra linear momentum.

The chain with mass (n-1)*m has a movement with some linear momentum.

 

Both of them part othe isolated system. But this is not means if little human pushh tje ring system will stay. System will start move to oppsite direction. If little human catch the ring system will stop. Push and catch this ring to opposite direction will return system to initial point.

 

Here is complete different behavior with ring and chain. The little human push the ring with one mass, but chain is stopping with other mass. Even if momentums are same, the surface mass will change at conversion moment. The red element stop without extra momentum.

 

I'm sorry ABV but if I guessed what the above meant, I think I would be doing you a disservice to answer. You are probably doing an excellent attempt at translation, much better than I could do, but I cannot tell exactly what you mean. Most of it I think I understand.

 

I will add this. If the isolated system, in summation, has no net linear momentum, the center of mass will never move, it will constantly remain in the same place.

[ABV]

I'm sorry ABV but if I guessed what the above meant, I think I would be doing you a disservice to answer. You are probably doing an excellent attempt at translation, much better than I could do, but I cannot tell exactly what you mean. Most of it I think I understand.

 

I think it's pretty simple.

 

The rolling ring has n*m mass and momentum P and surface with mass M and momentum -P

After stop(catch by surface) red element and cut the ring.

The chain has a mass (n-1)*m and surface with mass M+m

 

The surface would stop if it keep same initial mass M, but the surface mass with given velocity has been changed to M(surface)+m(red element)

 

P.S.

Thank you for your posts anyway.

I think it's a good problem for brainwashing

Edited June 24, 2009 by ABV

[ABV]

The ground links(red bit) momentum transfers to the links on the top of ring. The momentum is always constant for ring. Even it cut (and no push between links) rest of the chain (n-1), the chain will have initial ring's momentum. This means links on the chain will increase velocity after cut to keeping same net momentum. But even if apply this momentum to the surface with new mass, it on't

return surface to initial velocity. Because the surface velocity wasn't change during the surface apply ground link(red one) of chain.

 

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The surface will return back to initial velocity V=0 if the rest of chain could increase momentum. But it's nonsense. Base on law of momentum concervation chain should keep same initial ring's momentum.

[ABV]

If someone looks closely to this problem, it's easy to see difference between how to accelerate ring and how this ring stop.

During acceleration time whole ring moves on surface and rich const rotation and translation velocities.

During collision time the pseudo chain return momentum to the surface through one chains element (red bit on picture).

Is it makes any sense? Yes. Because this red bit caught with zero value of velocity V=0 and it didn’t transfer any momentum to the ground. Otherwise rest of the chain will return whole initial ring’s momentum. But those elements of chain return momentums to the surface with new mass. The red bit element was already stopped at begin of collision time. It means the surface join the red bit before the rest of chain returned the momentum and ground didn’t change own velocity.

If apply momentum to the surface with new mass then the surface will conduct movement after chain stop.

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Rotation with translation movement without slipping on the surface - each piece of the ring has velocity zeto V=0 on the ground.

 

3 actions.

 

1. Ring and surface are getting momentums.

 

2. Cut and hold red bit on surface (red bit has velocity zero V=0 on surface. Red bit don't transfer momentum to the ground)

 

3. The pseudo chain transfer momentum to the ground.

 

These 3 actions happen in different time.

[D H]

But it's nonsense.

That is a correct synopsis of this entire thread.

 

Moved to pseudoscience.

[ABV]

That is a correct synopsis of this entire thread.

 

Moved to pseudoscience.

 

Question to you. Why?

 

May be I wrong with something, but I'm talking about physic. Beside you, who just keep silence and move thread without any reason.

 

I need explanation first.

 

Otherwise, I would mention you virtual name as a person who mislead others without any reason.

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Originally Posted by ABV

But it's nonsense.

 

That is a correct synopsis of this entire thread.

 

Moved to pseudoscience.

 

You mean, law of momentum conservation doesn't work? Or you didn't understand the post?

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Ok.

This is what other says about this model.

 

"If the platform is completely free to move (say floating in outer space) momentum conservation requires that it will end up with a positive forward velocity V=(nm/(M+nm) )v. Kinetic energy is not conserved because as each link slaps down on the surface some energy is converted to heat.

 

For a full ring rolling at constant velocity there's no horizontal force between the bottom of the ring and the surface but that requires the ring to be balanced (rotationally symmetric). As links become missing from the circle that's no longer true so the succeeding links that hit the surface do have a forward pull on them accelerating the platform forward."

 

P.S.

Please check this out for details on knol site again.

I did a lot of changes there.

 

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

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Just keep in mind. This model has 3 phases. Bodies at phase1 is different from bodies at phase3. The law of momentum conservation works, but it gives different result for bodies with different mass.

Edited July 1, 2009 by ABV
Consecutive posts merged.

[ABV]

Hi.

I think this is a clue for understanding.

 

The element with zero values of linear velocity on the surface is too small. Base on math the geometrical size of this element strives to zero limit. However, this is the physical element, and it has its own geometrical size.

 

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

[ABV]

These thoughts can be used for a rolling body with a rolling friction.

 

 

The diagram has shown a rolling disk which contains n sectors with mass m. This (centre mass) CM disk has a translation velocity V. The red sector has linear velocity zero. Base on previous explanations, only n-1 sector on the rolling disk has velocity more then zero. These moving sectors transfer linear momentum dP to the surface with mass M+m per time frame dt.

If use classical model: The disk with linear velocity V transfer momentum to the surface. The total velocity on the end of action is:

 

V1=((n*m)/(n*m+M))*V

 

If discount the red sector with zero linear velocity: The total velocity on the end of action is:

 

V1’=(((n-1)*m)/((n-1)*m+M+m))*V.

 

The velocity difference between these two models is:

 

V1-V1’=m/(n*m+M)*V

 

If sectors geometrical size strives to zero, then sectors mass strive to zero also. For velocity difference equation, it gives a zero result.

 

V1-V1’=0.

 

At a math point of view with a very small sector with zero linear velocity the surface takes equation

 

V1=((n*m)/(n*m+M))*V.

 

However, the physical elements have its own geometrical size and end up velocity for this action is:

 

V1’=(((n-1)*m)/((n-1)*m+M+m))*V

 

Velocity difference is:

 

V1-V1’=m/(n*m+M)*V

[ABV]

For this model, а repulsion and а collision comes on different planes.

 

The vertical plane doesn’t have any contacts with the rolling body.

Otherwise the horizontal plane always has a physical contact with rolling body.

 

From vertical plane point of view, all elements of rolling body have a movement.

If calculate a rolling body translation momentum with respect to vertical plane then all elements of rolling body must be included.

 

Otherwise if look on horizontal plane, one element of rolling body stays on the surface all the time.If calculate a rolling body translation momentum with respect to horizontal plane then one element of rolling body must be included to the mass of horizontal plane.

 

 

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

[ABV]

Quote from forum.

“Follow by law of momentum conservation

V0x(mx(n-1))=V1x(all mass)

If discount one element from the ring then and average speed for (n-1) elements is changing.

The rest or ring elements have average velocity is V'0=(n/(n-1))xV0!”

 

This is absolutely correct. The momentum is conserve. The average velocity is changing.

P=const=mV=mVo=((n-1)m)x(n/(n-1))Vo. Just one little obstacle this model has. One ring’s element stays on the ground with velocity zero.

To solve this problem let reverse frame of reference for this model. The platform moves horizontally with mass M and velocity V. The ring with mass n*m stays and does rotation movement.

For vertical plane (a wall) the platform has a translation momentum P=MV. However, the translation momentum for horizontal plane (the platform surface) counts ring’s element, which is joining to the surface with velocity V. In this case the translation momentum is P=(M+m)V. The average velocity is equal to V.

How the momentum can be different for these planes? Answer is horizontal and vertical planes have a different frame of reference. To reach same momentum P=MV. The frame reference center should move with velocity V/n.

WAIT A SECOND!

Are these vertical and horizontal frames of references move relatively to each other?

 

THIS IS THE ONE.

If action starts from zero velocity on one frame of reference and this action finish with zero velocity for another frame of reference then relativity to the first frame of reference the system will continue to move on the end of action!

 

The momentum is conserve. The frame of reference is changing. This frame of reference exchange gives the system ability to move.

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A few philosophy thinks.

Collisions may be classified in two groups. Explicit and implicit.

 

1. Explicit collisions – happens between objects which conducts a simple movements relativity to each other.

For example. A collision between rolling body and wall on the surface. Law of momentum conservation is working.

 

2. Implicit collisions – happens between objects which conducts a complicate movements relativity to each other.

For example. A collision between a rolling body and a surface (It’s kind of weird thing)

This it may happen in 2 cases.

a. Between these objects distortions. Rolling friction.

b. These objects may have shared points to each other during movement action. Coupling through construction elements.

For ideal model the rolling body is conducting rotation with translation movement without any collisions on straight line surface. The rolling body with complicated movement has a parallel tangential line to the surface.

If use same frame of reference for implicit and explicit collisions the momentum will have a different value.

 

To avoid this law of momentum conservation problem, the model can implement two possibilities.

a. The model should transform own frame of reference.

If a platform (the surface) is a center of frame of reference then the calculated average velocity of rolling ring is higher than rolling ring’s center mass velocity.

If reverse this frame of reference and put rolling ring into center then the platform velocity should be reduced for calculations. From the other word this frame of reference should base on law of momentum conservation. Base on this frame of reference momentum measurement other parameters (like velocity) should be readjusted.

b. The model should include a dark matter.

This dark matter with own mass gives ability to law of momentum conservation works without frame of reference replacement.

For rolling ring model can happening two things. The dark matter may reduce the platform mass or be between objects to compensate extra momentum.

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Let imagine observer and two objects (A and B). One object A with simple movement can move only into one plane. The observer in own frame of reference sees only this plane also.

Another object B with complicated movement can move in two planes and observer can see only one of these planes.

The object A has position without movement for observer. The object B has a movement. The object B has component velocity for each plane. However, observer can see only one component velocity of objects B . The object A takes momentum from object B after collision and objects A velocity may be higher than inspected.

a. The observer can assume the object A has a huge mass. Or if observer knows the object B mass then he can assume include dark matter for collision process.

b. If frame of reference base on law of momentum conservation then using all know parameters easy to calculate another plane velocity component of object B which is invisible for observer. This velocity component for observer has imaginary character. From the other words, the frame of reference use law of momentum conservation to readjust another object parameters.

 

 

A few words about movement without external forces. The object can rotate on own center mass without external forces. However, if alternate simple and complicated movement on object collisions then for observer frame of references it may possible exchange rotation to translation movement.

Everything Is Relative.

[ABV]

This is a good example is shown transfer between angular and linear momentums.

Unfortunately the simulator does not allow giving different angular velocities for squares.

Anyway, for this model easy to understand these squares with different angular velocities will take a different linear velocities on the end of action.

Base on my theory about frame of reference conversion for complicated movement, the law of momentum conservation works, however for correct calculation need use imaginary part of velocity.

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

[ABV]

Another good example:

 

Let’s repulse a long cylinder inside Isolated System. The repulse should hit the cylinder away from object center mass. This hit energy splits for two object movements. This cylinder starts rotation with translation movement. The Isolated System starts move to opposite cylinder translation movement direction. During this object movement let transform this cylinder to a sphere. This sphere is rotating faster than cylinder because this object moment of inertia less than cylinder moment of inertia. However this rotation does not make any sense for translation momentum transfer. The sphere translation momentum is equal to cylinder translation momentum. However from cylinder repulse action the system takes higher translation momentum than sphere has. The sphere translation momentum won’t be enough to stop the Isolated System.