Mandlbaur Posted June 3, 2017 Share Posted June 3, 2017 Abstract:Both angular momentum and momentum are accepted to be conserved values and both of these are contained within the equation L = r x p. Assuming the implied rotation around a central point, they cannot both be conserved when the magnitude of the radius changes. The generally accepted principle is that momentum must change in order to conserve angular momentum. However it is logically proven that it is the component of momentum perpendicular to the radius which must be conserved.Introduction:Whilst working on a project which did not achieve the results predicted by physics, I stumbled upon this.Proof:For the equation L = r x p1. Assuming the implied rotation around a central point.Premise 1: There is a force at all times directed from the point mass along the radius toward the centre of rotation (centripetal force).Premise 2: A change in the magnitude of radius is conducted by altering the magnitude of this force.Premise 3:There can be no component of this force perpendicular to the radius. Premise 4: In order to affect the component of momentum perpendicular to the radius, we have to apply a parallel component of force (Newton’s first law).Deduction: A change in the magnitude of the radius cannot affect the component of momentum perpendicular to the radius.Conclusion:In the equation L = r x p, assuming the implied rotation around a central point, it is the component of momentum perpendicular to the radius which must be conserved when the magnitude of the radius changes. References:1) D.Halliday & R.Resnick, Fundamentals of Physics, second edition, extended version (John Wiley & Sons, Inc , New York, 1981) p. 181. Link to comment Share on other sites More sharing options...
KipIngram Posted June 3, 2017 Share Posted June 3, 2017 In a central force system p changes to compensate for changes in r and L is conserved. And you are right - that force cannot change the components of p perpendicular to the radius, so the overall change to p necessary to conserve L is achieved by affecting the p component parallel to radius. Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 3, 2017 Share Posted June 3, 2017 In a central force system p changes to compensate for changes in r and L is conserved. And you are right - that force cannot change the components of p perpendicular to the radius, so the overall change to p necessary to conserve L is achieved by affecting the p component parallel to radius.just a question Kipingram isn't that dynamic acceleration? Link to comment Share on other sites More sharing options...
KipIngram Posted June 3, 2017 Share Posted June 3, 2017 Well, it's acceleration. If an orbit is circular (eccentricity = 0) then the planetary speed never changes, but there's still an acceleration arising from changing the direction of the momentum vector. I'm not sure what you mean by "dynamic" acceleration. Link to comment Share on other sites More sharing options...
studiot Posted June 3, 2017 Share Posted June 3, 2017 I think there must be some misunderstanding here (Nver suprising with Resnick and Halliday). Moment of momentum and Momentum may well refer to different things. A spinning top or ball has moment of momentum but zero momentum. A ball travelling in straight line, without rotating on its axis, has momentum but zero moment of momentum. A ball travelling around on a whirling string has both momentum and moment of momentum. Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 3, 2017 Share Posted June 3, 2017 (edited) I think there must be some misunderstanding here (Nver suprising with Resnick and Halliday). Moment of momentum and Momentum may well refer to different things. A spinning top or ball has moment of momentum but zero momentum. A ball travelling in straight line, without rotating on its axis, has momentum but zero moment of momentum. A ball travelling around on a whirling string has both momentum and moment of momentum. Your quote] A spinning top or ball has moment of momentum but zero momentum. My quote But yes it has momentum when the top start warbling . Edited June 3, 2017 by Roger Dynamic Motion Link to comment Share on other sites More sharing options...
KipIngram Posted June 3, 2017 Share Posted June 3, 2017 Oh, yeah, I tend to call what you're calling "moment of momentum" "angular momentum." Link to comment Share on other sites More sharing options...
Mandlbaur Posted June 3, 2017 Author Share Posted June 3, 2017 In a central force system p changes to compensate for changes in r and L is conserved. And you are right - that force cannot change the components of p perpendicular to the radius, so the overall change to p necessary to conserve L is achieved by affecting the p component parallel to radius. If the p component perpendicular to radius is conserved and L and p cannot both be conserved, how is it possible to conserve L? I think there must be some misunderstanding here (Nver suprising with Resnick and Halliday). Moment of momentum and Momentum may well refer to different things. A spinning top or ball has moment of momentum but zero momentum. A ball travelling in straight line, without rotating on its axis, has momentum but zero moment of momentum. A ball travelling around on a whirling string has both momentum and moment of momentum. Your quote] A spinning top or ball has moment of momentum but zero momentum. My quote But yes it has momentum when the top start warbling . Does the specific equation I am using not clarify which of your examples applies? Link to comment Share on other sites More sharing options...
KipIngram Posted June 3, 2017 Share Posted June 3, 2017 The p component parallel to the radius is not conserved. The perpendicular components are, but p as a full entity is not. Also bear in mind that the p components that are perpendicular to the radius at time t no longer are time step later, since the radius is constantly changing direction. L is conserved by definition, since it takes a torque to change angular momentum and a central force can't make a torque. Link to comment Share on other sites More sharing options...
Mandlbaur Posted June 3, 2017 Author Share Posted June 3, 2017 (edited) The p component parallel to the radius is not conserved. The perpendicular components are, but p as a full entity is not. Also bear in mind that the p components that are perpendicular to the radius at time t no longer are time step later, since the radius is constantly changing direction. L is conserved by definition, since it takes a torque to change angular momentum and a central force can't make a torque. If the p component perpendicular to the radius is conserved, then when the radius stops changing, the magnitude of p will be the same as it was initially. i.e.: p is conserved and L therefore cannot be. L is in fact defined by the radius and therefore will change when the radius changes. Edited June 3, 2017 by Mandlbaur Link to comment Share on other sites More sharing options...
KipIngram Posted June 3, 2017 Share Posted June 3, 2017 L is in fact defined by the radius and therefore will change when the radius changes. That is patently incorrect. The conservation of L is essentially Kepler's second law. From fundamental classical mechanics, dL/dt = torque, and torque = Fxr (the cross product of the force vector and the radius vector). But in a central force system F and r are collinear, and their cross product vanishes. L is "defined" not only by the radius but also by the component of momentum perpendicular to the radius. When the radius becomes smaller, the perpendicular momentum component becomes larger - their product is unchanged. Link to comment Share on other sites More sharing options...
studiot Posted June 3, 2017 Share Posted June 3, 2017 If the p component perpendicular to radius is conserved and L and p cannot both be conserved, how is it possible to conserve L? Does the specific equation I am using not clarify which of your examples applies? No, it should however have prompted you to provide the full information about your scenario. I should not have to guess at it. RogerDynamicMotion My quote But yes it has momentum when the top start warbling . A musical top huh? Consider this instead. A mass point P is moving with constant velocity v along line AB, as shown. At the same time the line AB is rotated with constant agular velocity [math]\Omega [/math] about centre O in the direction shown. P will be subject to an acceleration [math]2v\Omega [/math], again in the direction shown. Link to comment Share on other sites More sharing options...
Roger Dynamic Motion Posted June 3, 2017 Share Posted June 3, 2017 (edited) If the p component perpendicular to radius is conserved and L and p cannot both be conserved, how is it possible to conserve L? Does the specific equation I am using not clarify which of your examples applies? Uff.. what is that supposed to mean. A spinning top or ball has moment of momentum but zero momentum. ok I see it now Edited June 3, 2017 by Roger Dynamic Motion Link to comment Share on other sites More sharing options...
Mandlbaur Posted June 3, 2017 Author Share Posted June 3, 2017 That is patently incorrect. The conservation of L is essentially Kepler's second law. From fundamental classical mechanics, dL/dt = torque, and torque = Fxr (the cross product of the force vector and the radius vector). But in a central force system F and r are collinear, and their cross product vanishes. L is "defined" not only by the radius but also by the component of momentum perpendicular to the radius. When the radius becomes smaller, the perpendicular momentum component becomes larger - their product is unchanged. Does providing an alternative theory as an argument prove my work wrong ? No, it should however have prompted you to provide the full information about your scenario. I should not have to guess at it. If you have to guess at it then you do not know your angular momentum equations. The equation L = r x p refers specifically to rotation of a point mass. Should you wish to refer to a rotating object, you would use L = I x w. -1 Link to comment Share on other sites More sharing options...
studiot Posted June 3, 2017 Share Posted June 3, 2017 Actually it doesn't refer to anything unless you define your symbols. Stating an equation without doing this is unacceptable. Notice I defined all my symbols in my brief post. Actually it doesn't refer to anything unless you define your symbols. Stating an equation without doing this is unacceptable. Notice I defined all my symbols in my brief post. Actually it doesn't refer to anything unless you define your symbols. Stating an equation without doing this is unacceptable. Notice I defined all my symbols. Further I expected some kind of explanation/description of the physical system your equation models. Or do I have do guess at that as well? Link to comment Share on other sites More sharing options...
Mandlbaur Posted June 4, 2017 Author Share Posted June 4, 2017 Actually it doesn't refer to anything unless you define your symbols. Stating an equation without doing this is unacceptable. Notice I defined all my symbols. Further I expected some kind of explanation/description of the physical system your equation models. Or do I have do guess at that as well? Thank you for your advice and opinion about the lack of definition within my work. I was under the impression that since the first object within the title is "angular momentum equation" and since this is a 300 year old, well known equation, I had provided sufficient definition of the symbols. Am I seriously out of line here ? If I were to make the title read: In the angular momentum equation, L = radius® x momentum(p), when the magnitude of the radius changes, which one of the remaining variables is correctly conserved ? Would that be acceptable? or do I need to go as far as: In the angular momentum equation, angular momentum(L) = radius® x momentum(p), when the magnitude of the radius changes, which one of the remaining variables is correctly conserved ? Please advise? My argument is general and applies to all physical systems in which this equation might be applied. The only requirement being a rotation around a central point as stipulated. Link to comment Share on other sites More sharing options...
Mordred Posted June 4, 2017 Share Posted June 4, 2017 (edited) Well if your trying to go for a mathematical proof that conservation of angular momentum is not conserved you best provide thorough details on the mathematical proofs of the angular momentum equation. Then detail a mathematical proof of where it in error. Simply naming premises etc and only referring to the angular momentum equation doesn't particularly count as a proof unless fully shown. After all the equation has been around 300 years as you say. It will take far greater attention to mathematical detail than you have in the above. Edited June 4, 2017 by Mordred Link to comment Share on other sites More sharing options...
Mandlbaur Posted June 4, 2017 Author Share Posted June 4, 2017 (edited) Well if your trying to go for a mathematical proof that conservation of angular momentum is not conserved you best provide thorough details on the mathematical proofs of the angular momentum equation. Then detail a mathematical proof of where it in error. Simply naming premises etc and only referring to the angular momentum equation doesn't particularly count as a proof unless fully shown. After all the equation has been around 300 years as you say. It will take far greater attention to mathematical detail than you have in the above. 1) My work is a logical proof, not a mathematical one. 2) Is it necessary for me to disprove every other derivation ever created on an individual basis before you will consider my work? 3) The definition of logical proof is a deduction based on valid premises. One would have to find a premiss to be invalid or the logic to be flawed in order to dismiss a conclusion arrived at by this method. 4) Am I correct when I say that logic is the cornerstone of science? Edited June 4, 2017 by Mandlbaur Link to comment Share on other sites More sharing options...
Mordred Posted June 4, 2017 Share Posted June 4, 2017 (edited) However it is logically proven that it is the component of momentum perpendicular to the radius which must be conserved. I am specifically asking you to demonstrate a working knowledge of why this is the case... Under Noethers theorem if you assign the referred to axis as N then the action L is conserved along the N axis. see rotational invariance https://en.m.wikipedia.org/wiki/Noether%27s_theorem In particular "In other words, the component of the angular momentum L along the n axis is conserved." Edited June 4, 2017 by Mordred Link to comment Share on other sites More sharing options...
Mandlbaur Posted June 4, 2017 Author Share Posted June 4, 2017 Your trying to compete with 300 years of research on the topic. Do you honestly believe a logical non mathematical proof is sufficient by itself? Funny think about physics, sometimes it surpise you on logic. Particularly if you restrict yourself to one equation without looking at the torque and moments of momentum and inertia aspects. Lets see the conservation law states " The conservation of angular momentum is conserved if and only if no external torque is applied" Now apply that specifically (torque) to the line Is there any mention or suggestion of a torque being applied within my work ? Link to comment Share on other sites More sharing options...
Mordred Posted June 4, 2017 Share Posted June 4, 2017 (edited) that is precisely my point. The conservation law specifies torque. Your post does not. If your not including torque your not discussing the conservation of angular momentum. Edited June 4, 2017 by Mordred Link to comment Share on other sites More sharing options...
Mandlbaur Posted June 4, 2017 Author Share Posted June 4, 2017 that is precisely my point. The conservation law specifies torque. Your post does not. If your not including torque your not discussing the conservation of angular momentum. The conservation law specifies that there should be no torque. Since I do not mention torque, would that not suggest that there is no torque within my argument? Unless you can point out anything within my work which suggests the slightest hint of any torque being applied, your line of argument here is nonsense. Link to comment Share on other sites More sharing options...
studiot Posted June 4, 2017 Share Posted June 4, 2017 (edited) In the angular momentum equation, L = r x p, when the magnitude of the radius changes, which one of the remaining variables is correctly conserved ? Neither is conserved. The conservation laws apply to a system, not an equation. This is why I have been trying to get you to describe your system. Consider the following system. A light, stiff, circular hoop is rolling along a flat horizontal ground surface with forward velocity v. Attached to the circumference is a mass point P. What are the angular and linear momenta of P 1) At the point of contact with the gorund 2) At the top of its travel around the hoop. Why are they different? How this fit with your equation and conservation? Edited June 4, 2017 by studiot Link to comment Share on other sites More sharing options...
Mandlbaur Posted June 4, 2017 Author Share Posted June 4, 2017 Neither is conserved. The conservation laws apply to a system, not an equation. This is why I have been trying to get you to describe your system. Consider the following system. A light, stiff, circular hoop is rolling along a flat horizontal ground surface with forward velocity v. Attached to the circumference is a mass point P. What are the angular and linear momenta of P 1) At the point of contact with the gorund 2) At the top of its travel around the hoop. Why are they different? How this fit with your equation and conservation? Does this example exhibit a rotation around a central point? Link to comment Share on other sites More sharing options...
studiot Posted June 4, 2017 Share Posted June 4, 2017 Does this example exhibit a rotation around a central point? What do you think the phrase rolling along a flat horizontal ground surface means? Link to comment Share on other sites More sharing options...
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