Mandlbaur

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 ?

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studiot    1131

 

 

If r = zero, angular momentum = zero.

 

The radius of a point particle is zero, by definition of a point particle.

Point particles may possess angular momentum.

 

 

You didn't address my second point.

What is the radius of a twisted space curve?

Edited by studiot
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Bender    132

 

 

If r = zero, angular momentum = zero.

If a sphere rotates around its own axis, which r do you use in your formula? (by "your", I mean the way you seem to interpret it and the fact that you give the impression that it is the only formula related to angular momentum you know about, but here you have yet another chance to prove that impression wrong)

 

I'm still looking forward to your demonstrations of how you would actually calculate anything based on your ideas about angular momentum. Have you made any progress?

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Mandlbaur    21

 

The radius of a point particle is zero, by definition of a point particle.

Point particles may possess angular momentum.

 

 

 

If the radius is zero then the angular momentum is also zero by definition.

 

 

If a sphere rotates around its own axis, which r do you use in your formula? (by "your", I mean the way you seem to interpret it and the fact that you give the impression that it is the only formula related to angular momentum you know about, but here you have yet another chance to prove that impression wrong)

 

 

If the radius is zero, the moment of inertia is zero.

Therefore the angular momentum is also zero.

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studiot    1131

 

 

If the radius is zero then the angular momentum is also zero by definition.

 

 

 

If the radius is zero, the moment of inertia is zero.

Therefore the angular momentum is also zero.

 

In my first post and several times subsequently I asked you very politely to define radius.

 

I expect you have seen the type of fairground ride where a seating capsule or capsules is rotated on the end of an metal arm or arms of fixed length L about a central hub.

 

So here is my question to you about your claim:

 

What is the radius of rotation of this mechanism ?

I am not seeking numbers, symbols in a formula will do.

Please note this forum requires you to answer this question fully in a way that I can use to calculate the radius.

 

 

 

 

"If the radius is zero then the angular momentum is also zero by definition."

 

There is where you are showing lack of understanding.

 

Consider a rotating sphere with non zero radius.

 

Every point in that sphere has zero radius but is a mass point.

The overall sphere has an angular momentum which is the sum of all the individual angular momenta of all the mass points about the central axis of that sphere.

The overall mass of that sphere is the sum of the masses of all the mass points.

 

If, as you say, all the mass points on the axis of that sphere have zero angular momentum they contribute nothing to the angular momentum of the sphere, but they do contribute mass to the mass of the sphere.

 

How can this paradox be resolved?

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Mandlbaur    21

 

In my first post and several times subsequently I asked you very politely to define radius.

 

I expect you have seen the type of fairground ride where a seating capsule or capsules is rotated on the end of an metal arm or arms of fixed length L about a central hub.

 

So here is my question to you about your claim:

 

What is the radius of rotation of this mechanism ?

I am not seeking numbers, symbols in a formula will do.

Please note this forum requires you to answer this question fully in a way that I can use to calculate the radius.

 

 

 

 

"If the radius is zero then the angular momentum is also zero by definition."

 

There is where you are showing lack of understanding.

 

Consider a rotating sphere with non zero radius.

 

Every point in that sphere has zero radius but is a mass point.

The overall sphere has an angular momentum which is the sum of all the individual angular momenta of all the mass points about the central axis of that sphere.

The overall mass of that sphere is the sum of the masses of all the mass points.

 

If, as you say, all the mass points on the axis of that sphere have zero angular momentum they contribute nothing to the angular momentum of the sphere, but they do contribute mass to the mass of the sphere.

 

How can this paradox be resolved?

 

 

Allow me to draw your attention to your own words: "Consider a rotating sphere with non zero radius."

 

So which would you like me to consider? A zero radius or a non- zero radius?

 

 

Also, since this is your example, surely it is yourself who is required to define radius.

 

You are talking complete nonsense. Please apply some reason to what you are saying.

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Mordred    833

Let me know when you plan on identifying the linear momentum vector I asked about. After all your logic argument requires that there is a linear momentum vector in the system so identify it.

 

This is now the third time I've asked you to identify this component.

Edited by Mordred

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Phi for All    4741

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Moderator Note

The OP needs to pick from a few of the many questions being asked and answer them. I see a LOT of defensiveness, but little actual defense of the idea. Support your idea by addressing the concerns of those you're discussing this idea with. Didn't you come here instead of anywhere else to be challenged with great questions? If not, perhaps you shouldn't be here, or this thread should be closed.

 

Don't bother responding to this note. Respond instead to Mordred's thrice-asked question above.

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Bender    132

If the radius is zero, the moment of inertia is zero.

Therefore the angular momentum is also zero.

The sphere in my example has a radius (let's call it A). I'm just wondering what r you would use in your formula.

I'm assuming here you merely misunderstood my question and you are not suggesting a rotating sphere has no angular momentum. If that assumption is wrong, please let me know.

 

(studiot is making a different, but also very valid point about how to apply your idea to point masses)

 

 

I'm still looking forward to your demonstrations of how you would actually calculate anything based on your ideas about angular momentum. Have you made any progress?

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Mandlbaur    21

Let me know when you plan on identifying the linear momentum vector I asked about. After all your logic argument requires that there is a linear momentum vector in the system so identify it.

 

This is now the third time I've asked you to identify this component.

 

 

The OP mentions a variable p. That variable is a linear momentum vector.

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Mordred    833

Ive already told you that p in a closed angular momentum system is angular momentum not linear momentum. P can be either linear or angular but not both at the same time in a closed system.

 

Your logic argument tries to use two simultaneous values for momentum p, simultaneous from one equation l=r×p.

 

Why can you not understand this basic concept?

 

Take a spinning wheel on a fixed axis, now locate the linear momentum? all motion in this case is angular not linear. So where is linear momentum p in this case?

 

A momentum vector describes the actual motion not the forces involved. Those are force vectors.

Edited by Mordred

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Mandlbaur    21

Ive already told you that p in a closed angular momentum system is angular momentum not linear momentum. P can be either linear or angular but not both at the same time in a closed system.

 

Your logic argument tries to use two simultaneous values for momentum p, simultaneous from one equation l=r×p.

 

Why can you not understand this basic concept?

 

Take a spinning wheel on a fixed axis, now locate the linear momentum? all motion in this case is angular not linear. So where is linear momentum p in this case?

 

A momentum vector describes the actual motion not the forces involved. Those are force vectors.

 

 

You may have said that, but what you are saying does not make any sense.

 

The p in that equation is the linear momentum vector. The angular momentum is represented by L.

 

A spinning wheel falls outside the scope of this discussion because we would use a different equation for it: L = I x W.

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Mordred    833

There you go, you just identified that p=mv does not apply in this wheel example.

 

as it uses w and not p

 

So in this system it cannot be conserved as it does not apply.

 

Your logic argument states both angular and linear is conserved. Might not have been your intention but that is how it reads.

 

Yes in a linear system, linear monentum is conserved. Yes in an angular momentum system it is conserved. However in a closed system the conservation laws only apply to systems where the two are.

 

Isolated from each other. in other words in regards to conservation of angular momentum there is no linear momentum as it uses torque not linear momentum.

 

So [latex]\tau=f*L[/latex] where L is the lever arm. [latex] L=r sin\theta[/latex] so [latex]\tau=Frsin\theta[/latex]

 

In terms of velocity conservation of angular momentum becomes

 

[latex]I_1w_1=I_2w_2[/latex]

 

in linear systems f=ma but in rotational systems [latex]f=m\alpha[/latex]

 

now the reason you can use L=r×p is due to the numerous parallels between linear and angular momentum.

 

Here is a list of the parallels

 

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin

 

Lets make this simple.

 

In physics a conserved quantity is a quantity that is constant over time.

 

the mathematical expression is with Q being the quantity that is conserved.

 

 

[latex]\frac{dQ}{dt}=0[/latex]

 

This is an expression that states q is constant over time hence it is conserved.

 

The conservation law in question is specifically the Isolated quantity that does not change over time

 

Linear systems are conserved in displacement translations. Meaning the laws of physics for this system is unchanged regardless of location. The quantity of q in the above becomes.

 

[latex]\frac{d\vec{p}}{dt}=0[/latex]

 

This is the mathematical expression for a conservation law. YOU MUST WRITE A NEW EXPRESSION for any other quantities that are constant over time

 

Hence the expression for conservation of angular momentum is

 

[latex]\frac{d\vec{L}}{dt}=0[/latex]

 

which means the laws of physics for rotations do not depend on orientation.

 

ONLY 1 quantity can be described by a conservation law. As a conservation law is a descriptive of a conserved QUANTITY

 

You cannot have two conserved quantities under the same law. (Unless that law is describing a complex conjugate) aka [latex]\vec{L}[/latex] a cross product term is a complex vector. However it is still a conserved quantity.

 

1 quantity per law =GOLDEN RULE

 

[latex]\frac{dq_1+dq_2}{dt}=0[/latex] is not allowed. As quantity is singular not plural....👹 under a law describing a conserved quantity.

 

so you cannot have two vectors (linear momentum and angular momentum under the same law. Not to mentioned the differences under Noether translations and rotations.

 

systems can have multiple conserved quantities but each is a seperate law.

 

ps not yelling just highlighting key points under bold.

 

You state angular momentum and momentum is conserved it does not equal the mathematical expression for a law describing a single conserved quantity (conservation law of quantity or value).

 

By the way this is precisely why Studiot wanted to show you how to properly define a closed isolated state or system.

 

Every conservation law examples other than in this thread.

 

lepton number

spin

parity

energy/momentum

charge

 

etc etc all satisfy

 

[latex]\frac{dQ}{dt}=0[/latex]

Edited by Mordred

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Mandlbaur    21

I have omitted irrelevant text. Providing alternative theory does not say anything about my work. I am well aware that there are alternative theories. Please address my OP.

 

There you go, you just identified that p=mv does not apply in this wheel example.


Do not put words in my mouth. I identified that the example you gave does not apply to this discussion.

 

 

 

as it uses w and not p

So in this system it cannot be conserved as it does not apply.

Your logic argument states both angular and linear is conserved. Might not have been your intention but that is how it reads.

I state that angular momentum and momentum are both accepted to be conserved. Once again putting words in my mouth.

 

 

Yes in a linear system, linear monentum is conserved. Yes in an angular momentum system it is conserved. However in a closed system the conservation laws only apply to systems where the two are.

 

If a value is conserved, it will always be conserved. It is not going to choose a system within which to be conserved.

 

 

Isolated from each other. in other words in regards to conservation of angular momentum there is no linear momentum as it uses torque not linear momentum

 

The equation which I have specified contains a linear momentum vector - are you denying this ?

 

 

 

o [latex]\tau=f*L[/latex] where L is the lever arm. [latex] L=r sin\theta[/latex] so [latex]\tau=Frsin\theta[/latex]

In terms of velocity conservation of angular momentum becomes

[latex]I_1w_1=I_2w_2[/latex]

in linear systems f=ma but in rotational systems [latex]f=m\alpha[/latex]

now the reason you can use L=r×p is due to the numerous parallels between linear and angular momentum.

 

l=rxp is the classical definition. The parallels were drawn after that. The mistaken assumption of angular momentum being conserved actually comes from those parallels being drawn.

Edited by Mandlbaur

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Mordred    833

Did you not understand a single word about how a conservation law is defined?

 

Blooming bugger I provided both verbal and the math definition.

 

No I did not put words in your mouth you posted an equation whose every term is a rotation.

 

 

You may have said that, but what you are saying does not make any sense.

 

The p in that equation is the linear momentum vector. The angular momentum is represented by L.

 

A spinning wheel falls outside the scope of this discussion because we would use a different equation for it: L = I x W.

why would a spinning wheel fall outside the scope of discussion on angular momentum?

 

This is a thread dealing with angular momentums conservation laws.

 

This quote was in responds to my query of where is the linear momentum term on tbe spinning wheel? where you stated in this quote you don't use p=mv. You use L=Iw.

 

Did I put words in your mouth?

If a value is conserved, it will always be conserved. It is not going to choose a system within which to be conserved.

 

Wrong the conserved quantities only apply when certain criteria are met. Closed systems. You can have more than one conserved quantity in the same system.

 

However a conservation law only applies to 1 quantity being conserved. definition my previous post.

 

All Other arguments are useless and moot the conservation law must satisfy

 

[latex]\frac{dQ}{dt}=0[/latex]

 

which your logic argument does not

l=rxp is the classical definition. The parallels were drawn after that. The mistaken assumption of angular momentum being conserved actually comes from those parallels being drawn.

and yet your logic argument includes the parallels.

 

Which just happens to be torque.

 

"In order to affect the component of momentum perpendicular to the radius, we have to apply a parallel component of force (Newtons first law)."

That's contradicting yourself from your own logic argument. You first state you must use them then state it is incorrect to do so.... Edited by Mordred

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imatfaal    2477

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Moderator Note

 

OK Mandlbaur. Last chance to get this thread back on the tracks. Your next post MUST include an equation (with proper naming of symbols) which you believe defines the conservation of angular momentum.

 

FYG [latex] \textbf{L} = \textbf{r} \times \boldsymbol{\rho}[/latex] says nothing about conservation - it merely relates the linear momentum of a particle to the angular momentum of that particle around a specific axis at a perpendicular distance with the magnitude the same as r

 

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studiot    1131

I looked in the back of my cupboard and found one remaining olive branch I had overlooked.

 

As I understand the situation

 

You think you have discovered some circumstances where conservation of angular momentum is not true.

 

I agreed with you there are such circumstances.

 

So I tried to explore these circumstances with you.

 

In fact I put substantial effort into this.

 

But as far as I can tell from your minimalist responses you were not interested in such a project.

 

I will make one comment on your assertion that the equation L = r x p is a definition of angular momentum.

 

It is not a definition, it is is a very simplified formula for calculating the angular momentum in some circumstances.

 

So I will leave you to consider my morning cup of tea that I have just poured.

 

In order to stir the tea into the milk I pour the stream of tea into the cup along the inside of the cup.

Thus it has linear momentum as it leaves the spout and enters the cup.

But it has zero angular momentum.

As the tea enters the cup it swirls round and ceases translation (the cup does not go anywhere)

But in swirling round it gains angular momentum but looses linear momentum.

 

So we have an everyday system that starts with all linear momentum and ends with all angular momentum.

 

So neither are conserved since linear momentum is destroyed and angular momentum is created.

 

 

 

Do you wish to discuss the analysis of this real world situation or do you wish to close the thread?

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Mandlbaur    21

I will make one comment on your assertion that the equation L = r x p is a definition of angular momentum.

 

It is not a definition, it is is a very simplified formula for calculating the angular momentum in some circumstances.

 

 

 

From Wikipedia:

In physics, angular momentum (rarely, moment of momentum or rotational momentum) is the rotational analog of linear momentum. It is an important quantity in physics because it is a conserved quantity – the angular momentum of a system remains constant unless acted on by an external torque.

The definition of angular momentum for a point particle is a pseudovector r×p,...

!

Moderator Note

 

OK Mandlbaur. Last chance to get this thread back on the tracks. Your next post MUST include an equation (with proper naming of symbols) which you believe defines the conservation of angular momentum.

 

FYG [latex] \textbf{L} = \textbf{r} \times \boldsymbol{\rho}[/latex] says nothing about conservation - it merely relates the linear momentum of a particle to the angular momentum of that particle around a specific axis at a perpendicular distance with the magnitude the same as r

 

 

 

I am of the opinion that your moderation of this thread is biased.

 

What you should be doing is picking out Mordred for posting off topic, accusing me of saying things I have not, posting nonsense and refusing to respond to the OP as requested.

 

Instead you are asking me to define something that my OP has proven to be non-existent and making an ultimatum of it in order to have an excuse to censor.

 

I feel that your actions are despicable.

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studiot    1131

 

From Wikipedia:

In physics, angular momentum (rarely, moment of momentum or rotational momentum) is the rotational analog of linear momentum. It is an important quantity in physics because it is a conserved quantity – the angular momentum of a system remains constant unless acted on by an external torque.

The definition of angular momentum for a point particle is a pseudovector r×p,...

 

What is clear from this quote is that r x p is not a fundamental definition of angular momentum in the eyes of the author, since it is specified as being only applicable to a point particle, and nothing else.

 

So there must be a more comprehensive definition which includes the angular momentum of thigns which are not point particles.

 

I also note the linked Wikipedia of point particles which clearly indicates that r = 0 for a point particle and you so rudely denied.

 

 

Wikipedia

its defining feature is that it lacks spatial extension: being zero-dimensional, it does not take up space.[

 

You were offered, and declined, convivial rational discussion at this more fundamental level.

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Mordred    833

 

 

From Wikipedia:

In physics, angular momentum (rarely, moment of momentum or rotational momentum) is the rotational analog of linear momentum. It is an important quantity in physics because it is a conserved quantity the angular momentum of a system remains constant unless acted on by an external torque.

The definition of angular momentum for a point particle is a pseudovector r×p,...

 

 

 

I am of the opinion that your moderation of this thread is biased.

 

What you should be doing is picking out Mordred for posting off topic, accusing me of saying things I have not, posting nonsense and refusing to respond to the OP as requested.

 

Instead you are asking me to define something that my OP has proven to be non-existent and making an ultimatum of it in order to have an excuse to censor.

 

I feel that your actions are despicable.

I showed your OP post does not match the definition of a law of conservation.

 

Can you argue the definitions I provided? I prefer a good argument on physics. Let me know when your ready to provide one instead of attacking other posters character.

Edited by Mordred

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Klaynos    717
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Moderator Note

Closed for failing to engage with criticism as the speculation rules dictate even after multiple chances.

Do not reintroduce this topic.

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