Mandlbaur

The ice skater and the ball on a string and the professor on a turntable all "spin faster" when the radius is reduced, but if you measure them, they do not "spin faster" enough. Not a little discrepancy that can be explained by blurting friction. We are talking about a discrepancy so large that it is a contradiction.

Why is there no lab experiment verifying any of these demonstrations directly?

How can we call it science if we have no experiment?

http://www.baur-research.com/Physics/MPS.pdf

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.

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.

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.

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.

 

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.

 

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.

Anyone has to learn to walk before they can run.

 

I offered you the opportunity to work through the walking stage to reach your equation in considerably more detail than you will find in any single textbook or three.

 

And the only running you did was to run away from this opportunity.

 

So I will leave you with the the following thoughts.

 

Your very very elementary equation is not always true.

 

For instance it is not true in the case of point particles with mass, spinning on their own axes.

These have angular momentum, but r = zero.

 

 

If r = zero, angular momentum = zero.

Others have already done that quite adequately and patiently.

 

 

I suggest we start over and focus on to these two questions:

1) Are engineers making enormous errors when designing machines based on their understanding of angular momentum?

2) Do astronomers completely fail to predict the movement of planetary objects?

Incorrect. No-one has dismissed my argument adequately.

To make that unsubstantiated claim that is irrational, negligent, wishful thinking and typical of a person suffering from confirmation bias.

Although it is absolutely not necessary for me to respond to your out of scope questions, I will nevertheless provide responses:

1) Engineers do not use conservation of angular momentum when designing machines. If they did, their machines would not work properly. I have had this discussion on various occasions with engineers and none have been able to provide any evidence of anything which varies in radius that was designed using conservation of angular momentum. They use conservation of energy which predicts substantially different results.

2) Astronomers do fail to predict the movements in which they are using conservation of angular momentum in their calculations and the radius is variable. I have had this discussion with various astronomers and they have failed to produce any data which confirms actual measurements of planetary movement against predictions. There are also various examples of planetary motion discrepancies.

Now let's go back to the reality of the argument provided in the OP:

Logic is the cornerstone of science.

I have provided a logical argument.

To dismiss it one would have to invalidate the premisses or fault the logic.

There has been no argument levelled against this logic which has not been directly tackled and dismissed.

The only thing outstanding is for all of you to accept the conclusion drawn.

You cannot teach to someone unwilling to learn.

 

Absolutely agreed.

So since our discussion in March, it appears very little has changed. Let's summarise, so we can pick up where we left off:

 

1) You base all this on very poorly executed "experiments", most of which aren't closed systems at all and without any quantitative data.

 

2) Pulling in a spinning mass with a string makes the mass move on a spiral. This means the force is not perpendicular to the motion and linear momentum is not conserved.

 

3) All equations of angular momentum and the conservation thereof can be derived from Newton's laws of motion. Claiming that angular momentum is not conserved implies that you also claim Newton's laws of motion are incorrect.

 

4) You never considered these points (perhaps best to start with these):

This post is not the same as my previous posts.

None of your arguments presented here are valid against this OP.

I have presented a logical argument.

To dismiss this argument, it is necessary to invalidate the premisses or fault the logic.

Alternatively it is mandatory to accept the conclusion drawn.

Any other response is irrational.

Thank you for telling me that you have been wasting my time.

It is a classic symptom of confirmation bias to accuse your opponent of your own uncooperative behaviour.

 

What, no technical questions at all?

 

Well I will ask one then.

 

Under what conditions is angular momentum conserved for a particle proceeding along a straight line as shown in post#99?

 

There is enough information in post#99 to fully answer this.

 

However it is only true in 2 dimensions.

 

So a second question

 

Why is it not true then in a general 3 dimensional case?

Your questions are irrelevant because they fall outside the scope of my OP.

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"

 

 

Can we move on to the next line?

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

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.

"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.

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?

 

 

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

Please point out exactly which premiss you have defeated?

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.

What you are doing is called ad-hominem.

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.

Providing alternative theories does not do it. Accusing me of having a lack of understanding and calling me names doesn't do it.

Accusing me of being the aggressor in this discussion is simply incorrect. I am defending my position from your aggression. It is yourselves who are being combative.

This is a result of your cognitive dissonance due to your inability to accept or defeat my work.

It is unfortunate, but it is also true.

If there was any way for me to present this without triggering confirmation bias, believe me, I would try that.

The reality is that you are all making fools of yourselves and I am sorry that it has to be this way.

Veritas omnia vincit

 

You're right: a x b = - b x a. So they need a negative sign in one spot. But then they proceed to show that the affected term is zero. So the proof still works even if you correct the sign error.

 

Out of all the "combative" threads I've run across on this forum, this one takes the cake. And you refused to answer me when I asked you why you want conservation of angular momentum to be wrong so badly - I suspect you're trying to cook up a free energy device of some kind. I'm really sorry that the universe doesn't work the way you want it to; if I could edit the laws of physics there'd be some changes I'd make too. Oh well...

Flawed is flawed. You would not grant me any leeway if there were such an infantile error in my work.

Here is a mathematical proof showing the equation and definition I posted above.

 

http://www.citycollegiate.com/centre_of_mass3.htm

 

I was digging for a decent proof without applying Noether's this one will suit

This "proof" is flawed.

A vector cross product is not commutative.

You have defeated absolutely nothing. All you've done is shown that you do not understand the conditions that apply to conservation of angular momentum.

 

If you could address our questions with proper knowledge on the topic you would have.

 

 

 

 

This very line is full of mistakes already mentioned numerous times.

Please point out any mistake in that line?

The reality is that there are actually none.

I admitted a mistake (The "fixed") because it was easier to do that (and slightly restrict the application of my paper) to defeat yet another invalid argument levelled against my work than to try to explain that the example provided in the detractors argument actually uses a different equation and then go down a whole other avenue.

Just saying I should have used "fixed" solved the problem even if it slightly restricted the application of my paper.

In reality the "fixed" is not necessary and it is not actually a mistake. My paper is perfectly valid whether the "fixed" is there or not.

But you have claimed that my abstract is "full of mistakes" mentioned "numerous times".

Frankly this claim is complete nonsense. You will not be able to back it up. You should be moderated out of this discussion for making invalid claims which you are not prepared to support.

 

Well, in point of fact you do - you have to get others to choose to take you seriously in order to get your work considered. I think today hasn't been a very good day for you in that respect.

 

One last time: angular momentum is conserved in all closed systems. If you have accurately measured some apparatus exhibiting a changing angular momentum, then it is not a closed system. End of story, and I am done here.

My point is that you do need to show the required details not just hand wave them away. It is simple applications of force.

 

You should have absolutely no problem in showing how torque is involved. Or even showing how [latex] F_{21}=-F_{12}[/latex]

 

You posted one formula that doesn't even detail the conservation of angular momentum. It simply applies it.

 

Do you even know the proper conservation of angular momentum equation?

 

Your absolute refusal to even look at the mathematical proofs tells otherwise and proves to me that you do not know the details. If you did you would have had no problem posting such. There is classical proofs that don't require latex to be legible.

 

Simply looking at L=R×P isn't the full story.

 

when [latex] \vec{L}[/latex] is constant when net [latex]\tau=0[/latex] (torque) is your specific conservation of angular momentum relation. That is what you have to prove as false.

 

Or more accurately start with [latex]\vec{\tau}=\frac{d\vec{L}}{dt}[/latex] where net torque equals zero and [latex] \Delta L=constant[/latex]

 

 

 

But this is not what you originally said.

 

 

 

 

 

 

Originally you said the above applies to any rotation (around a central point but what other sort of rotation is there?)

Later you changed this to a fixed central point, without defining fixed.

 

 

So I have a list of excluded examples of systems to which your statements can be applied , but no included ones.

 

 

You have not provided a single example of a single system to which your analysis applies, despite several requests for one.

 

 

So I have explored a number of examples of genuinely rotating systems with you, and you have rejected all of them as not analysable by your method.

 

 

You are adamant that it is applicable to a system with a centre that is fixed or not going anywhere, yet you are equally adamant that this system with zero translational velocity has a momentum (note not angular momentum)

 

 

How is this possible?

 

 

You claim that the generally accepted principle is that momentum should change if the radius changes.

 

But the system, by definition, has no momentum to change.

 

It only possesses angular momentum.

 

 

 

I have no idea what books you have been reading about mechanics, but your opening statement suggests to me that you have only partly understood them.

 

In particular it is true that the most common cited example of the effect you describe incorrectly is that of the skater spinning on the spot.

 

 

The skater's angular velocity increases as she draws her arms inwards.

She has no momentum because she is spinning on the spot, but she has angular momentum, which remains constant (is conserved) so as the radius decreases the angular velocity increases, just as you say.

But her momentum does not increase, which is not as you say.

This is all possible because of the very low friction on ice so we assume no driving forces are required.

 

 

 

In the case of carousel or weight on the end of a rope, a driving force is required.

 

 

 

And this must be included in the system for analysis, which you are not doing, but everyone here is telling you that you need to do.

 

 

 

We have been trying to help you create a correct analysis of rotating systems that can be applied as widely as possible, not one that has to exclude every example that is presented.

 

 

Since you have obviously already put in a great deal of effort into this, I can only urge you to study some basic mechanics texts.

I suggest engineering science ones would be better as they are more self contained and plain speaking than applied maths or physics ones which also rely heavily on a whole raft of peripheral background material.

 

This would help you understand when momentum and /or angular momentum is destroyed and when they are conserved and how to include all the necessary items in a system.

Every argument you guys have presented, I have defeated. In my view this is the point at which you should begin to take me seriously. However you have have claimed more than once to have defeated me on multiple occasions. This can only be the result of a confirmation bias.

Since you are exhibiting confirmation bias behaviour, it is impossible for me to get you to choose to take my work seriously. In fact you will choose the opposite in order to support your position despite the facts of the matter.

You are making unsupported claims and backing them up with invalid argument and then denying to yourself that your arguments have been shown invalid. This is the classical biased behaviour that I have faced continually over the past year.

The fact is that you have never even tried to create an apparatus to test this world view of yours yet you will adamantly claim that my various apparatus were somehow flawed.

How do you suggest I tackle this?

I am prepared to put up a scientific wager that an apparatus which you design (I'll even cover the costs of it) will prove my work to be correct.