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 ?

[studiot]

In which case can we please follow the rules of this forum in which the promoter either completely sets out his stall in the first post or defines his terms in subsequent answers to questions about that which is missing from the opening proposal.

 

I have asked repeatedly adn without success for proper definitions of all terms employed, the scope of all equations used and a full and accurate description of at least one system conforming to the proposed analysis.

[swansont]

Allow me to bring your attention back to the fact that my paper as presented is a logical proof.

 

In order to dismiss the conclusion drawn one would have to invalidate my premises or fault my logic.

 

While Mordred is wuite correct that showing your conclusion to be wrong is sufficient, I already pointed out the flaw in your OP. Linear momentum is not claimed to be conserved. Angular momentum is conserved when there is no net torque.

 

Your flawed response was that since you did not mention torque, one can't exist. But you can write anything you want. It doesn't become correct because it's written down. It's trivially easy to make conflicting claims. You have to actually show a system behaves as you insist.

[Mordred]

On a much more technical note, also an fyi for other readers.

 

Conservation laws are identifying what is referred to as a conserved quantity.

https://en.m.wikipedia.org/wiki/Conserved_quantity

 

"For any continuous symmetry of a physical system, there exists a corresponding conserved quantity.

 

"In physics, a conservation law states that a particular measurable property of an isolated physical system does not change as the system evolves over time. "

 

Key Note "An isolated physical system"

 

So one must keep linear momentum isolated from angular momentum by the above. Mixed system is a combination of both.

 

Now the above all apply to Noether's Theorem. The first quote being one of its primary axioms.

 

Under symmetry relations Noethers theorem shows that linear momentum has translational transformations, while angular momentum undergoes rotational transformations.

Edited June 8, 2017 by Mordred

[Mandlbaur]

I repeat: In order to dismiss my conclusion, one would have to invalidate my premisses or fault my logic.

 

The only other option is to accept my conclusion.

 

Any other response is nonsense.

[swansont]

I repeat: In order to dismiss my conclusion, one would have to invalidate my premisses or fault my logic.

 

The only other option is to accept my conclusion.

 

Any other response is nonsense.

 

 

And I have done that. Your responses have been less than adequate.

[Mandlbaur]

 

 

And I have done that. Your responses have been less than adequate.

 

Please point out exactly which premiss you have defeated?

[swansont]

 

Please point out exactly which premiss you have defeated?

 

 

For the third time?

 

Your abstract is flawed

http://www.scienceforums.net/topic/106547-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/page-2?p=994379#entry994379

 

 

(1) Linear momentum is conserved when there is no net external force.

(2) Angular momentum is conserved when there is no net external torque.

Under the presence of a force, which must be present for rotational motion, condition (1) is violated.

When the magnitude of p changes as r changes, they are inversely related such that L is conserved.

Your individual premises need to be addressed on a case-by-case basis, because violations depend on the system, and there is not a general statement one can make about them. IOW, an orbit behaves differently from a rock on a rope, swinging in a circle, where the rope is being pulled in or let out.

———

 

Further, your 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."

 

This is something you know to be false, since you had a previous thread where you acknowledged that when you have an object swinging on a rope, and you pull the rope in, the object speeds up. It has to, since pulling the rope in represents work done on the system. Since the conclusion is false, the premises or the logic must be mistaken. (Probably premise 3)

[Mandlbaur]

If my abstract is flawed (which I disagree with), then it needs to be re-written. That makes no argument against my proof.

 

Premise 3 states that there can be no component of centripetal force perpendicular to it.

 

You are talking nonsense.

 

You have not defeated premise 3.

 

Which premise do you believe that you have you defeated?

[Klaynos]

!

Moderator Note

Putting your fingers in your ears and claiming you're correct is not how we do things here. Stop it.

[KipIngram]

You still haven't told us what the point of this is - you're completely ignoring that question.

[Mordred]

Premise 4:

In order to affect the component of momentum perpendicular to the r

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.

 

It should have been obvious that premise 4 is incorrect.

 

There is no linear momentum. There you go not only is your conclusions incorrect but now we can also show your premises themselves is incorrect.

 

You also have no premise to define the closed system nor the conditions that the conservation law applies to. So they are also imcomplete.

Edited June 8, 2017 by Mordred

[Mandlbaur]

"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."

 

This is something you know to be false, since you had a previous thread where you acknowledged that when you have an object swinging on a rope, and you pull the rope in, the object speeds up. It has to, since pulling the rope in represents work done on the system. Since the conclusion is false, the premises or the logic must be mistaken. (Probably premise 3)

 

 

Clearly I need to explain again: since the radius is reduced, the circumference is reduced which means that the object will complete more revolutions in the same time period even if it does not speed up. You read the previous thread incorrectly.

It should have been obvious that premise 4 is incorrect.

 

There is no linear momentum. There you go not only is your conclusions incorrect but now we can also show your premises themselves is incorrect.

 

Absolute nonsense.

[Mordred]

Ir doesn't make a hoot of difference. Your comparing linear monentum in a system that has no linear monentum with angular momentum.

 

 

Its comparing apples to oranges they may have similarities but they are completely different types of fruit.

 

In your case your comparing two different conservation laws, Then making the mistake that both are in the same system.

Absolute nonsense.

No what is absolute nonsense is your refusal to listen to the 3 posters with Ph.ds and one with a Masters degree as well as an another who is an Engineer that applies these conservation laws on a daily basis. Nor even learn from the numerous links and references showing your wrong.

 

Utterly stupid if you ask me. Tell me is the best defense you can up with.?

 

Prove us wrong define linear momentum in your closed and isolated system. Show where a linear vector applies in your system or admit your wrong. (don't bother with the tangent vector for moment of inertia) that is a complex vector not linear.

 

Thus far your only defense has been strictly ignoring any response countering your misguided notions and resorting to foolish and literally dumb assertions.

 

Premises are not Facts until they are proven to properly define the system in question. Your usage of linear momentum is not part of the system. That proves your article is wrong.

 

I'm done I'm not wasting my time on stubborn and stupid false assertions.

 

I would recommend reading the modnotes if you don't want this thread gettimg locked but that is up to you.

Edited June 8, 2017 by Mordred

[Strange]

I repeat: In order to dismiss my conclusion, one would have to invalidate my premisses or fault my logic.

 

The only other option is to accept my conclusion.

 

Any other response is nonsense.

This is obviously wrong.

 

1. You can be shown to be wrong without explaining why you are wrong.

 

2. Several people have shown where your error(s) lie.

[Mandlbaur]

Ir doesn't make a hoot of difference. Your comparing linear monentum in a system that has no linear monentum with angular momentum.

 

 

Its comparing apples to oranges they may have similarities but they are completely different types of fruit.

 

In your case your comparing two different conservation laws, Then making the mistake that both are in the same system.

No what is absolute nonsense is your refusal to listen to the 3 posters with Ph.ds and one with a Masters degree as well as an another who is an Engineer that applies these conservation laws on a daily basis. Nor even learn from the numerous links and references showing your wrong.

 

Utterly stupid if you ask me. Tell me is the best defense you can up with.?

 

Prove us wrong define linear momentum in your closed and isolated system. Show where a linear vector applies in your system or admit your wrong. (don't bother with the tangent vector for moment of inertia) that is a complex vector not linear.

 

Thus far your only defense has been strictly ignoring any response countering your misguided notions and resorting to foolish and literally dumb assertions.

 

Premises are not Facts until they are proven to properly define the system in question. Your usage of linear momentum is not part of the system. That proves your article is wrong.

 

I'm done I'm not wasting my time on stubborn and stupid false assertions.

 

I would recommend reading the modnotes if you don't want this thread gettimg locked but that is up to you.

 

 

Please explain what "p" refers to in the equation?

This is obviously wrong.

 

1. You can be shown to be wrong without explaining why you are wrong.

 

2. Several people have shown where your error(s) lie.

Google Logical proof:

 

Logical proof is proof that is derived explicitly from its premises without exception. Logical proof is not the same as factual proof. In formal logic, a valid argument is an argument that is structured in such a way that if all it's premises are true, then it's conclusion then must also be true.

[Mordred]

You have been asked before if you understood the difference between moment of momentum and momentum. In the angular momentum system p stands for angular momemtum. The equation does not have both linear and angular momentum

 

In a linear momentum system p stands for linear momentum.

 

 

http://www2.eng.cam.ac.uk/~hemh1/gyroscopes/momentmomentum.html

 

 

The linear momentum vector is not a cross product. I already provided the correct definition.

 

Tell me something do you not know that equation is only an analog to linear momentum? not identical?

 

In linear momentum Newtons second law is [latex] f=ma[/latex] the rotational law is

 

[latex]\tau=I\alpha[/latex]

 

I already provided links detailing these distictions. Here this wiki covers these two equations.

https://en.m.wikipedia.org/wiki/Angular_momentum

 

Please note it involves Torque.

 

 

This is linear momentum.: By physics definition.

 

"Linear momentum is a vector whose direction is parallel to the velocity of the particle. with relations p=mv " So

 

 

so I challenge you to identify this component..in your system

Please explain what "p" refers to

In this case angular momentum not linear momentum. If you apply that formula to a linear momentum system its linear momentum not angular momentum.

 

Not both both when applying the conservation laws

Edited June 8, 2017 by Mordred

[studiot]

I am going to go right back to the beginning and start again.

 

I shall work through your proposition a line at a time trying to knock it into proper shape so you don't end up with fallacious conclusions.

 

You are welcome to come along for the ride, you never know you may learn something to your advantage.

 

 

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.

 

So the first line

 

"Both angular momentum and momentum are accepted to be conserved values"

 

This, like many of your statements are half truths. I am sure this is unintentional and based on inadequate study material.

 

Are accepted ?? by whom ??

 

Are conserved values ?? Always?? What is meant by a conserved value in this case??

 

I am glad you mentioned linear momentum as well as angular momentum as I can start with linear momentum.

This is much easier but clearly points the way to go.

 

A point to remember that will become important further down the line.

 

Both linear momentum and angular momentum are vectors. Vectors have magnitude and direction.

 

So to be conserved both magnitude and direction have to be conserved.

 

Let us consider a fairly general system of particles which comprises our system for analysis.

This may only be a single particle or it may be many, operating as a single unit.

The centre of mass moves as if the whole mass of all the particles were concentrated at this point.

The motion of this centre is completely independent of any internal forces (internal forces are those acting between the particles) since by Newton's third law every such action is countered by an equal and opposite reaction.

 

If M is the total mass of the system and F_E the external forces then by Newton's second Law

 

[math]\sum {{F_E}} = M\ddot \bar \alpha = M\frac{{d\bar v}}{{dt}}[/math]

 

I note that there is already confusion between you and Mordred on the meaning of the symbol r, which is why I asked you to define your symbols before.

By convention r is used for a position or displacement vector, not radius.

 

In either event I have used the greek letter alpha to avoid confusion.

In any case we can quickly move away from this variable.

 

The equation states Newton's second Law that the sum or resultant of the forces equals the total mass times the acceleration, where the acceleration is the second time derivative of the position vector alpha.

This is shown by the double dots over the symbol.

The symbol also has a bar over it to show it refers to the centre of mass position.

 

OK That was a big chunk, now for the cool bit.

 

In the case where the vector component of the forces in a particular direction is zero

[math]M\ddot \bar x = M\frac{{{d^2}\bar x}}{{d{t^2}}} = 0[/math]

 

I have taken the x axis to be in this direction so moved away from alpha as promised.

 

This is a very simple high school differential equation which when integrated (solved) has the solution

 

[math]M\dot \bar x = M\frac{{d\bar x}}{{dt}} = M\bar v = {\rm{a}}\;{\rm{Constant}}[/math]



and [math]\bar v[/math] is the component of the velocity of the centre of mass in the x direction.

 

This is hugely significant because Mv is the component of the linear momentum of the system in the x direction and it is constant.

 

Constant means it does not change and we have fixed the x direction.

 

So this is a form of conservation for linear momentum.

 

But it only works because a term in our equation of motion is zero or null.

 

It is not true in general.

 

This type of 'conservation' is called a null based conservation.

 

[math]{{\rm{p}}_x} = M\dot \bar x = {\rm{a}}{\kern 1pt} \;{\rm{Constant}}[/math]

Where P_x is the linear momentum in the x direction.

 

That is enough for now, but I can tell you that conservation of angular momentum is similarly based but the derivation is more complicated as there is another term in the equation of rotational motion to consider.

 

How are we doing?

Edited June 8, 2017 by studiot

[Mordred]

Thank you Studiot I was about to post a more complete detail on my last post but your post is better than what I eas going to post.

 

I will wait on the OP's response to avoid added confusion.

I am going to go right back to the beginning and start again.

 

good advise right there lol. Perhaps a new approach is needed +1 Edited June 8, 2017 by Mordred

[Strange]

Logical proof is proof that is derived explicitly from its premises without exception. Logical proof is not the same as factual proof. In formal logic, a valid argument is an argument that is structured in such a way that if all it's premises are true, then it's conclusion then must also be true.

 

 

Exactly. And as your conclusion is wrong, then either your premises or logic must be wrong. It would still be wrong even if no one had identified where the flaw was. But, in fact, they have.

[swansont]

Clearly I need to explain again: since the radius is reduced, the circumference is reduced which means that the object will complete more revolutions in the same time period even if it does not speed up. You read the previous thread incorrectly.

 

I would rather see physics than a hand-wave

[Mandlbaur]

I am going to go right back to the beginning and start again.

 

I shall work through your proposition a line at a time trying to knock it into proper shape so you don't end up with fallacious conclusions.

 

You are welcome to come along for the ride, you never know you may learn something to your advantage.

 

 

 

So the first line

 

"Both angular momentum and momentum are accepted to be conserved values"

 

This, like many of your statements are half truths. I am sure this is unintentional and based on inadequate study material.

 

Are accepted ?? by whom ??

 

Are conserved values ?? Always?? What is meant by a conserved value in this case??

 

I am glad you mentioned linear momentum as well as angular momentum as I can start with linear momentum.

This is much easier but clearly points the way to go.

 

A point to remember that will become important further down the line.

 

Both linear momentum and angular momentum are vectors. Vectors have magnitude and direction.

 

So to be conserved both magnitude and direction have to be conserved.

 

Let us consider a fairly general system of particles which comprises our system for analysis.

This may only be a single particle or it may be many, operating as a single unit.

The centre of mass moves as if the whole mass of all the particles were concentrated at this point.

The motion of this centre is completely independent of any internal forces (internal forces are those acting between the particles) since by Newton's third law every such action is countered by an equal and opposite reaction.

 

If M is the total mass of the system and F_E the external forces then by Newton's second Law

 

[math]\sum {{F_E}} = M\ddot \bar \alpha = M\frac{{d\bar v}}{{dt}}[/math]

 

I note that there is already confusion between you and Mordred on the meaning of the symbol r, which is why I asked you to define your symbols before.

By convention r is used for a position or displacement vector, not radius.

 

In either event I have used the greek letter alpha to avoid confusion.

In any case we can quickly move away from this variable.

 

The equation states Newton's second Law that the sum or resultant of the forces equals the total mass times the acceleration, where the acceleration is the second time derivative of the position vector alpha.

This is shown by the double dots over the symbol.

The symbol also has a bar over it to show it refers to the centre of mass position.

 

OK That was a big chunk, now for the cool bit.

 

In the case where the vector component of the forces in a particular direction is zero

[math]M\ddot \bar x = M\frac{{{d^2}\bar x}}{{d{t^2}}} = 0[/math]

 

 

I have taken the x axis to be in this direction so moved away from alpha as promised.

 

This is a very simple high school differential equation which when integrated (solved) has the solution

 

[math]M\dot \bar x = M\frac{{d\bar x}}{{dt}} = M\bar v = {\rm{a}}\;{\rm{Constant}}[/math]

 

 

 

and [math]\bar v[/math] is the component of the velocity of the centre of mass in the x direction.

 

 

This is hugely significant because Mv is the component of the linear momentum of the system in the x direction and it is constant.

 

Constant means it does not change and we have fixed the x direction.

 

So this is a form of conservation for linear momentum.

 

But it only works because a term in our equation of motion is zero or null.

 

It is not true in general.

 

This type of 'conservation' is called a null based conservation.

 

[math]{{\rm{p}}_x} = M\dot \bar x = {\rm{a}}{\kern 1pt} \;{\rm{Constant}}[/math]

 

Where P_x is the linear momentum in the x direction.

 

That is enough for now, but I can tell you that conservation of angular momentum is similarly based but the derivation is more complicated as there is another term in the equation of rotational motion to consider.

 

How are we doing?

 

 

Please continue

[Mordred]

good response. +1 a lot of people tend to jump ahead, glad to see you didn't. I will let Studiot move to the next lesson to avoid confusion. (its best to stick to one style) avoids potential on confusion. I'm confident Studiot will show the nitty gritty distictions between the two conservation laws. (linear, angular).

Edited June 9, 2017 by Mordred

[imatfaal]

!

Moderator Note

 

Roger's offtopic remark hidden. This thread is fraught enough without hard to interprete asides.

 

Do not respond to this moderation within the thread

 

[studiot]

To move on to angular momentum and examine the question ‘what is angular momentum?’ I am going to go back to statics and my post#64

 

 

Post#64

There is a subtle difference between angular momentum and moment of momentum.

Do you know what it is?

Hint it corresponds to the differnce between a couple and the moment of a force.

 

So consider a general force vector as in Fig 1.

Like all vectors, F acts only along straight lines, shown dashed in the figure.

F is both the force and its point of application.

 

 

 

 

Introduce some general points, S, T, U and W in Fig 2.

The moment of F about a general point is defined as the product of the perpendicular distance from the line of action to the point, as shown.

Note that there does not have to be any physical object at any of these points.

If there is a physical connection between F and the point (say S) about which the moment is taken then that moment is applied at S. Otherwise the moment is notional, but still exists mathematically.

 

Fig 2 also shows some particular points.

 

1. The magnitude of the moment is, in general, different for every point. So the inequality shows that the moment about S is numerically different from the moment about T

2. The direction of the moment is clockwise or anticlockwise depends upon which side of the line the point lies.

3. The moment at U has the same magnitude and direction as the moment at S and U lies on a line parallel to F.
So moments are only the same if they lie on a line parallel to the line of action of F

4. The moment at W is zero or does not exist and W lies on the line of action of F.

Fig 3 shows that if, instead of a force we apply a couple at F we find that

1. The couple exerts the same turning effect about every point (including S, T, W and U).

2. The couple has the same direction about every point.

3. There are no points with zero moment.

Quite different from the turning effect (moment) of a single force, as promised.

I will not prove these unless you specifically ask, although it is an easy proof to demonstrate.

 

A real world physics demonstration of this can be had by loosely nailing a plywood lamina to a post and comparing the turning effects of a single force via a spring balance with the turning effect of a large screwdriver, both applied at various points.

A further important difference is that a single force acts asymmetrically, a couple can only be realised by a pair of forces or multiple pairs of forces, which then act symmetrically. The ultimate being an infinite number of pairs or a ring force as found in a rope round a capstan or windlass.

 

I have mentioned symmetry because it connects directly with Noether’s theorem that Mordred referred to.
Noether was a very theoretical Mathematician and it is pleasing that her theorem (1918) manifests itself in physical reality but there is no reason to expect physical reality to follow modern theoretical mathematics. There is plenty of such modern mathematics that has no physical counterpart.
Earlier Mathematics was structured to follow physical reality.

 

Force is a vector and if we now replace this vector with another one, either velocity or linear momentum we find what is known as ‘moment of momentum’. This is an instantaneous quantity and the point it is taken about is called the instantaneous centre.
For a general motion this will change with the motion of the body at F. It is unsymmetrical, like the moment described above and in the same way that the moment changes, the moment of momentum changes.

 

But I mentioned two modes of turning motion and the second is called spin.
This corresponds to the couple in statics and is a symmetrical mode.

 

A spinning body possesses angular momentum S, by virtue of its spin as well as moment of momentum by virtue of its translational motion. Its total angular momentum is then the (vector) sum of these.

Hence my previous comment that there is an extra term in the equation.

We can explore a simplified version of this in the next instalment and find out the conditions that lead to conservation of total angular momentum and also the complete muddle that modern presentations have got themselves into.

 

We can also explore why adding couples and the moments of forces in the first part is simpler and easier that adding moment of momentum and spin.

 

Now that we are on better speaking terms I'm sure there must be questions, don't hesitate to ask.

[Mordred]

Excellent job thus far. +1