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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:

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.

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

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. [/url] 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.

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.

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 - 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.

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?

I can give solution from Russian forum. You don't need to know Russian language if you're understood physics equations. http://corum.mephist.ru/index.php?showtopic=21094&st=25

Good. Back to cylinder problem. Case 1. These objects collide away from center mass. One object loose momentum. Another is getting angular and tranlslation momentums. Case 2. These objects collide directly to center mass. One object loose momentum. Another is getting only tranlslation momentums. The translation cylinder center mass velocity is different for these cases?

You mean, if these objects have no angular momentums then they may have them after collision. Correct? So. Momentum is conserve, but it may by transfer between translation and angular parts on collision action.

Any object with mass and velocity. Good. Where is the angular momentum for object with linear velocity? Classical mechanics does not cover multidimention interaction. There is just a translation and a rotation movement. The Translation with rotation is just a sum of these movements. What about laws of momentum conservation? Could be sum too?

What do you mean? One translation momentum (bullet) transfers to one transltation and one rotation momentums. One to two. What is the conculusion for this? Is angular momentum added from nowhere? Or it substracted from translation momentum initially? If yes, what is the classical mechanics rule which covers this momentum (from translation to angular) transfer? The translation momentum is not conserve anymore?

This is not mistake. How are you going to describe the translation with rotation movement for this cylinder which comes from single linear momentum?

Wrong. I mean correct for this classical physics laws. However conceptually wrong for multidimensional interaction. The classical physics disallow convert translation momentum to angular momentum. Because the classical physic allow to convert one type of movement to ONLY one other. But it can’t be start two movements together. However Mother Nature gives us many examples where one type of movement interacts to two types of movements together. The problem with cylinder - one of them. This is not covering by classical mechanics laws. It just substitutes solutions for any of these types of movements. We should choose this type of movements first. This is the problem of classical mechanic laws. The momentum conserve, however the generic law of momentum conservation for multidimensional movement is covering linear and angular momentums.

Cool It means the translation momentum is different P=mv. V different - P different. P=Ft=mv outcomming V different. The other part of force goes to rotation part. Momentum is conserve however it's splitting to translation and rotation parts for case 2. The cylinder translation velocities will be different for case 1 and 2. If argue about translation momentum part conserving then this process is part of perpetual mobile

Please be patient. All physics laws work. Don’t worry I’m just asking about difference between velocities. Nothing else Merged post follows: Consecutive posts mergedSo. If calculate this problem using clasical mechanics laws, the linear momentum should be conserve. However, these cases takes a different energies. The cylinder with translation and rotation movement takes 2 parts(translation and rotation) of energies. The cylinder with translation takes only one(translation) part. These translation kinetic energy parts should be equal. Where the rotation knetic energy part comes from? From nowhere? What about angular momentum? How it starts rotation if all momentum transfers to translation part? Any answers? Merged post follows: Consecutive posts merged Let's solve simple problem. Given a long thin cylinder with length L, radius R and mass M. Case 1. This cylinder takes momentum P into his center mass. Case 2. This cylinder takes momentum P on a some distance from his center mass. The cylinder start rotation with translation movement. How many times the translation velocity for case 2 is lower than translation velocity for case 1? ====================================================================== If calculate this problem using classical mechanics laws, the linear momentum should be conserve. However, these cases take different energies. The cylinder with translation and rotation movements will take two parts (translation and rotation) of energies. The cylinder with translation takes only one( translation) part. These translation kinetic energy parts should be equal. Where this rotation kinetic energy come from? From nowhere? What about angular momentum? How it starts rotating if all momentum transfers to translation part? If assume for case 2 the incoming momentum vector dividing for two parts initially then it will explain everything. The first part – translation momentum gives ability the cylinder moves. The second part – angular momentum gives ability the cylinder rotates. The sum of these parts kinetic energies is equal to full translation kinetic energy from case1. The case 2 cylinder translation velocity lower than case 1 translation cylinder velocity. The equation translation velocity is: Et - kinetic energy translation part Er - kinetic energy rotation part Let’s make a simple experiment with pencil. Case 1: Let’s hit the pencil on his middle – The pencil fly away on a long distance. Case 2: Let’s hit the pencil on his edge – The pencil starts rotation and fly away not so far. The classical mechanics solution says – For both cases the pencil should fly away on a long distance with same high translation velocity. No matter how fast it is rotating. Is it really true? I don’t see this result.

Really? Let's solve simple problem. Isolated System. No external forces. A Cylinder with length L, radius R and mass M. Case 1. This cylinder takes momentum P into his center mass. Case 2. This cylinder takes momentum P on a distance L/4 from his center mass. The cylinder start rotation with translation movement. How many times the translation velocity for case 2 is lower than translation velocity for case 1?

Whould you tell me please, why are these squeres rotating slowly after collision? Is angular momentum depend on angular velocity? Fine. What is a different for this case energy transfer instead momentum transfer if this squares repulsing with higher linear velocity than linear velocity before colision starts. === You may try it. This experiment file locates on my site. http://knol.google.com/k/alex-belov/paradox-of-classical-mechanics-2/1xmqm1l0s4ys/9# Evaluation version of this simulator is free.

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.

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#

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. Merged post follows: Consecutive posts mergedA 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. Merged post follows: Consecutive posts mergedLet 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.

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#

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

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#