J.C.MacSwell Posted July 22, 2009 Share Posted July 22, 2009 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# There is no transfer of angular to linear momentums. Both are constant for the system. (or would be with accurate simulator representation) You are showing (attempting but with the limitations of the simulator) a transfer of transfer of rotational kinetic energy and linear kinetic energy. Link to comment Share on other sites More sharing options...
ABV Posted July 22, 2009 Author Share Posted July 22, 2009 There is no transfer of angular to linear momentums. Both are constant for the system. (or would be with accurate simulator representation) You are showing (attempting but with the limitations of the simulator) a transfer of transfer of rotational kinetic energy and linear kinetic energy. 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. Link to comment Share on other sites More sharing options...
insane_alien Posted July 22, 2009 Share Posted July 22, 2009 the total angular (and linear) momentum of the entire system in that situation before and after the collision is zero. this is because they are rotating and moving in opposite directions. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 22, 2009 Share Posted July 22, 2009 the total angular (and linear) momentum of the entire system in that situation before and after the collision is zero. this is because they are rotating and moving in opposite directions. Exactly. The angular momentum of one block can be transferred to the other and vice versa but the sum total does not change. The linear momentum of one block can be transferred to the other and vice versa but the sum total does not change. But the angular momentum cannot be transferred to linear. It is the rotational kinetic energy that is transferred to linear kinetic energy. Link to comment Share on other sites More sharing options...
ABV Posted July 22, 2009 Author Share Posted July 22, 2009 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? Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 22, 2009 Share Posted July 22, 2009 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? Obviously P=0 if it's an isolated system with no external forces, so nothing happens in either case. Link to comment Share on other sites More sharing options...
ABV Posted July 23, 2009 Author Share Posted July 23, 2009 Obviously P=0 if it's an isolated system with no external forces, so nothing happens in either case. 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. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 24, 2009 Share Posted July 24, 2009 (edited) 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? Good questions. Just keep in mind that these are not isolated systems and external force or forces are involved, as inputs to the original conditions. and angular and linear momentum are vectors and different concepts (You cannot add them together or transfer them, like you can with rotational and translational energies, which are scalars) 1. Where does the rotational energy come from? An external force. Same place the translational energy comes from. Can you not see that there are different inputs in each case? They are not isolated systems. In each case an external force is involved. If you assume the same force, it must act through more distance to transfer the same linear momentum in the second case, where the force does not act through the center of mass. This requires more energy. 2. What about angular momentum? How it starts rotation if all momentum transfers to translation part? Any answers? Again keep in mind these are separate concepts, angular and linear momentum are different. You cannot transfer only linear momentum in the second case, with a force not acting through the center of mass. As stated above, the same force (same vector quantity in the same direction) will act through more distance in order to transfer the same linear momentum. This takes more energy which accounts for the additional rotational kinetic energy. There is now angular moment in the system (second case) where there was none originally. Where did it come from? Now I will ask you, if you understand all of the above, how is angular momentum conserved? Look at the reaction to the force, the equal but opposite reaction. It had to have something to react against, something to accelerate in the opposite direction. Perhaps the simplest to understand would be a reaction against an identical, but mirror image system. Can you see how linear and angular momentum are each conserved in this case? The net sum of each for the combined system (which would be considered an isolated system with no external forces)would be zero, though twice the energy would be required. Merged post follows: Consecutive posts merged 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. If you hit it with the same energy, the same force for the same distance, the classical mechanics solution does not say it will fly away with same high translation velocity. It does however say that if you impart the same linear momentum, the same force for the same amount of time, it will in fact fly away with same high translation velocity. Edited July 24, 2009 by J.C.MacSwell Consecutive posts merged. Link to comment Share on other sites More sharing options...
ABV Posted July 24, 2009 Author Share Posted July 24, 2009 If you hit it with the same energy, the same force for the same distance, the classical mechanics solution does not say it will fly away with same high translation velocity. Cool It means the translation momentum is different P=mv. V different - P different. It does however say that if you impart the same linear momentum, the same force for the same amount of time, it will in fact fly away with same high translation velocity. 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 Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 24, 2009 Share Posted July 24, 2009 Cool It means the translation momentum is different P=mv. V different - P different. This is correct. 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. This is wrong. Force over time will affect translational momentum regardless of whether it is affecting angular momentum as well. They do not share a balance. They each must balance independently. 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 If you input the same linear momentum, a force for a given time, the translational velocities will be the same. If you input the same energy, force over distance, they will be different. Link to comment Share on other sites More sharing options...
ABV Posted July 25, 2009 Author Share Posted July 25, 2009 This is correct.If you input the same linear momentum, a force for a given time, the translational velocities will be the same. 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. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 25, 2009 Share Posted July 25, 2009 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. Momentum, both angular and linear, have precise definitions in classical physics. You are going to continue making the same mistakes until you understand them. Link to comment Share on other sites More sharing options...
ABV Posted July 25, 2009 Author Share Posted July 25, 2009 Momentum, both angular and linear, have precise definitions in classical physics. You are going to continue making the same mistakes until you understand them. This is not mistake. How are you going to describe the translation with rotation movement for this cylinder which comes from single linear momentum? Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 25, 2009 Share Posted July 25, 2009 (edited) This is not mistake. How are you going to describe the translation with rotation movement for this cylinder which comes from single linear momentum? Look at everything involved. The sum total of translational momentum will remain unchanged. The sum total of rotational momentum will remain unchanged. If you see rotational momentum where there was none before, then look carefully at the whole system, everything that was involved, and you will find something that will account for it, so that you are looking at the whole system, and you will see that angular momentum is conserved. A "single linear momentum" input, added other than in a direction intersecting the centre of mass of a system, always adds or subtracts angular momentum to/from the system, so you have to look at the bigger picture to see where it came from. Edited July 25, 2009 by J.C.MacSwell Link to comment Share on other sites More sharing options...
ABV Posted July 27, 2009 Author Share Posted July 27, 2009 Look at everything involved. The sum total of translational momentum will remain unchanged. The sum total of rotational momentum will remain unchanged. What do you mean? One translation momentum (bullet) transfers to one transltation and one rotation momentums. One to two. A "single linear momentum" input, added other than in a direction intersecting the centre of mass of a system, always adds or subtracts angular momentum to/from the system, so you have to look at the bigger picture to see where it came from. 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? Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 27, 2009 Share Posted July 27, 2009 What do you mean?One translation momentum (bullet) transfers to one transltation and one rotation momentums. One to two. bullet? 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? Every action has an equal but opposite reaction. You cannot create angular momentum without creating an equal but opposite angular momentum. Link to comment Share on other sites More sharing options...
ABV Posted July 27, 2009 Author Share Posted July 27, 2009 bullet? Any object with mass and velocity. bullet?Every action has an equal but opposite reaction. You cannot create angular momentum without creating an equal but opposite angular momentum. 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? Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 27, 2009 Share Posted July 27, 2009 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? If you have two objects that, in isolation, have no angular momentum, they may still in combination have angular momentum. Merged post follows: Consecutive posts merged 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? If you have two objects that, in isolation, have no angular momentum, they may still in combination have angular momentum. Link to comment Share on other sites More sharing options...
ABV Posted July 27, 2009 Author Share Posted July 27, 2009 If you have two objects that, in isolation, have no angular momentum, they may still in combination have angular momentum. 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. Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 27, 2009 Share Posted July 27, 2009 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. No. If they are moving with respect to one another, but not directly toward (collision course for respective centers of mass) or directly away, then they have angular momentum about their combined center of mass. The total angular momentum about their combined center of mass will not change in the event they did collide (of course it would be an off center collision in this case) nor would there be any change in their combined linear momentum. Link to comment Share on other sites More sharing options...
ABV Posted July 27, 2009 Author Share Posted July 27, 2009 No. If they are moving with respect to one another, but not directly toward (collision course for respective centers of mass) or directly away, then they have angular momentum about their combined center of mass. The total angular momentum about their combined center of mass will not change in the event they did collide (of course it would be an off center collision in this case) nor would there be any change in their combined linear momentum. 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? Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 28, 2009 Share Posted July 28, 2009 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? This cylinder problem? (below is your post currently numbered 30) 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? Link to comment Share on other sites More sharing options...
ABV Posted July 28, 2009 Author Share Posted July 28, 2009 This cylinder problem? (below is your post currently numbered 30) 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 Link to comment Share on other sites More sharing options...
J.C.MacSwell Posted July 28, 2009 Share Posted July 28, 2009 (edited) 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 What is the question? (your remarks for case 1 and 2 of post # 47 do not make sense for the way you presented them in post 30) Edited July 28, 2009 by J.C.MacSwell Link to comment Share on other sites More sharing options...
ABV Posted July 28, 2009 Author Share Posted July 28, 2009 What is the question? 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? Link to comment Share on other sites More sharing options...
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