Mandlbaur

studiot makes a good point - in fact both of the particles in your model will move. The system will have a fixed center of mass, and both particles will move around it. Your "fixed central point" scenario is the limit as the central mass becomes >> than the other mass, but it never applies fully.

No - if your experiments seem to be showing that angular momentum isn't conserved then you are overlooking a source of external torque.

 

 

?? If your premise is that angular momentum is not conserved in a central force system, then you are simply wrong and we've pointed that out to you in a variety of ways.

 

What is your ultimate goal here? Let's say you were right, and you can construct a case where angular momentum isn't conserved. Where do you go from there? Is the next step the description of some sort of "free energy device"?

 

Answer my previous question to you. This will indicate once again that your "pointing out" is been flawed. There is nothing you have "pointed out" that I have not defeated.

 

Unfortunately I can't defeat a confirmation bias.

 

 

 

There is more than a mistake in your thinking, there is a fundamental misunderstanding.

 

Your stipulation concerning rotation about a 'fixed point' just won't wash.

 

Consider the following.

 

Somerset is famous for the largest nightime illuminated carnival in the world.

This is made from hundreds of floats which are made from enormous low loader lorries.

Mounted on these lorries are carousels, roundabouts and other rotating mechanisms, carrying carnival actors.

 

Now tell me that your analysis debars these actors from claiming that these carousels rotate about a fixed point whilst spectators on the sidewalks claim the point is moving along on the back of a lorry.

 

There is a considerable amount you clearly don't know so rudely telling a bunch of folks with Phds in engineering and/or physics that they don't know what they are talking about and only you are correct is in my view confrontational.

 

The sad thing is that your formula can be applied (if done correctly) to these carousels or my wheel, but your are refusing to allow that anyone else knows anything.

 

In more advanced mechanics this issue is resolved by means of one of two methods that of Lagrange or that of Euler, one of which takes the view of the spectator the other the view of the carousel actor.

 

Have you heard of these?

 

 

Any system to which the formula I have specified in my paper can be realistically applied is proven by my paper to not conserve angular momentum when the radius changes.

 

I am not trying to prove to anybody how clever I am.

 

I am completely dumb in your shadow.

 

Sometimes a dumb person does have a point.

 

This is one of those occasions.

Why do you think you need to disprove every single derivitative involved?

 

Have you studied the mathematical proof for conservation of angular momentum? There isn't that many formulas involved in the proof.

 

Simple denial doesn't cut it

 

Demonstrate to us a proper understanding of how conservation of angular momentum works. A logic argument isn't sufficient considering it is literally someone elses work.

 

Demonstrate you properly understand the premises involved. ie premise 1 and 3

 

Thus far all your demonstrating is a lack of understanding behind the premises you posted. (prove to me otherwise please)

I went and hit the wrong button and editted a previous post.

 

I meant this to be here lol

 

Anyways this demonstrates the premise 1 and 3

 

My point is actually that I should not be required to have to prove or disprove anything about any other theory prior to my work being considered.

 

I don't understand the rest of your post. Are you seriously claiming that I don't understand my own premisses? Surely I would not have been able to create them if I didn't understand them?

I'm not trying to be confrontational. I am trying to get you to fully examine the problem at hand. Not just handwave counter arguments.

 

Simply because you chose to not include torque does not make your logic argument correct. I quoted a particular line in your proof that would have provided insight as to why the conservation of angular momentum specifies torque in its very definition.

 

Looking at the mathematical proof would have identified the requirement on the quoted section.

 

(as already pointed out premise 1 and 3)

 

There is numerous mathematical proofs and methodologies my personal favourite being Noethers theorem under rotational invariance however there are other mathematical proofs.

 

Just a side note when trying to develop a new model or understanding its always best to provide the current understandings and mathematics then show the errors.

 

This also shows you have a decent knowledge and understanding of said topic.

 

 

I have already asked you whether it is necessary for me to prove every other derivation wrong before you will consider my work. This was a non-confrontational way of trying to point out to you that it is not necessary.

 

Seriously, others have even pointed out that your torque argument does not hold water. How does a person tackle an argument that someone refuses to give up on even when it has been defeated on multiple occasions other than by confrontation?

 

I have a knowledge of the topic because of my work which included the design, manufacture and testing of many experimental prototypes. Each one an optimised version of the previous model. All of them attempts to achieve extremely high magnitudes of angular velocity predicted by conservation of angular momentum. All of them indicating that angular momentum is not conserved in variable radii systems.

 

I have spent much time trying to understand why and have discovered the truth and pinpointed the problem. I have written three different proofs of this. This paper being the latest iteration of my second proof.

 

Clearly this is extremely difficult for people to accept. I understand that it is akin to physical pain for a person to have to change their world view. I'm sorry about that.

 

Unfortunately because I am the one who has stumbled on this, I am tasked with getting it known and therefore have no option but to continue to push this heavy stone up this endless hill until somebody listens.

 

+1

 

I was not I who chose this confrontational approach to the discussion.

 

I far prefer cooperation and pooling of knowledge.

 

You may like to know that both my questions were very very simple examples of the sort of calculations engineers do every day when considering the complicated dynamics of machinery and in fluid mechanics in general.

This would include deriving suitable equations to describe the motion of moving parts and the forces (and moments) they exert upon one another.

 

I come back to this point about conservation being a system property not an equation property.

 

The same problem bedevils another field entirely - that of Thermodynamics.

 

So many would be engineers fail simply because they do not properly identify the system.

 

 

This is why looking over the mathematical proofs for conservation of energy is important. Part of the proof is defining a closed system. The other part shows the torque aspects. The proof also identifies which vectors are involved.

 

 

The fact that I concede a mistake in my paper about referring to a fixed central point as opposed to a central point does not mean that I concede anything regarding the point that I am trying to get across.

 

I do not believe that I have been confrontational at all - that implication is a false.

 

My paper still stands, it still has not faced any argument which faults any of its premisses nor flaws any logic which is required in order to dismiss it.

 

I have faced every challenge posed and I believe that so far have defeated all of the important ones.

 

My aim is not confrontation, I am simply trying to get my point across.

 

So can I take it you are denying that a wheel rotates about its hub?

 

 

Thank you for pointing out a flaw in my work, I should have referred to a fixed central point. I will put that into my revision.

 

You did specify zero torque in your OP - premise 1 and premise 3. Each of those describes a central force. Therefore angular momentum is conserved.

 

Imagine that you lay a Cartesian x/y coordinate system down on top of your system, with the central point at the origin and such that at time t the radius vector lies along the x axis. You then stipulate that the force is directed along the x axis at that time. So the y component of momentum is conserved, but the x component is not.

 

Shortly later, the radius vector is no longer parallel with the x axis. Therefore the x component of momentum now contributes to the component of momentum perpendicular to the (new) radius vector.

 

So the force cannot affect the component of momentum perpendicular to the radius vector now, but it does affect the component of momentum that will be perpendicular to the radius vector later.

 

This is why the speed of the object can change.

 

These are really fundamental things, and we've explained them to you several ways - you're starting to come off as stubborn about this.

 

 

With this scenario that you describe, would I be correct in saying that a positive centripetal force would result in a positive contribution to the perpendicular momentum of the "radius vector later"?

 

 

We went over this in another thread, If you ignore answers you don't like , this will be an issue. You need to describe how, exactly, the radius is changing, i.e the mechanism. You will find that there is a tangential force.

 

 

Since this paper is new and had not yet been presented when we had the previous discussion, the arguments in the previous thread do not match. I believe that I have described the mechanism whereby the radius changes. However since you would like a more in depth explanation: If the centripetal force is either larger or insufficiently large enough to maintain the radius at exactly the same magnitude, then the radius will change. Since the centripetal force is acting directly along the radius, am I incorrect in believing that there can be no tangential force?

 

You can't have this condition.

 

 

Without specifying whether or not you are applying a torque, you can't tell.

 

We have covered this previously, since I do not mention a torque, I am not applying a torque.

 

 

(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.

 

 

That has already been answered. The change in p and r are inversely related.

 

This is correct and very deep, but is it not a common known fact that a perpendicular force changes the direction but not the magnitude of the velocity?

It does not exhibit rotation around a central point therefore your argument does not pertain to my paper.

 

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?

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.

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 ?

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.

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?

 

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.

 

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.

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.

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?

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.

 

 

No, the power delivered to the rocket is zero when v is zero. The power delivered depends on v. Some of the power goes into the exhaust.

 

 

If the power is zero when the velocity is zero how does accelerate in the first place?

I'll try to answer the OP as simply as I can. Therefore, I'm going to assume the rocket flies horizontal, has perfect efficiency and constant mass (in other words: it might as well be a car driving on a frictionless road in a vacuum). I have the impression that the rocket is overshadowing the essence of the question (although I could be wrong).

 

If the force is constant, energy expenditure is not. It takes more work to accelerate from 10 m/s to 20 m/s then it takes to accelerate from 0 m/s to 10 m/s.

 

Simply look at the formula for power:

P=v F

So, with constant force, the power (and thus work) increases with velocity.

 

 

Thank you for understanding the question.

If P=vF then F=P/v.

So what you are saying then is that the force/thrust that the rocket provides is dependent on the velocity ?

Also, the Force is infinitely large when the velocity is zero ?

 

 

You can't change the rules half way.

 

A rocket is basically a small payload on top of a large fuel tank.

The fuel is often 90% of the mass of the rocket.

 

At any one time the payload plus remaining fuel is being propelled forwards thus the mass of this combination diminishes rapidly with progress.

 

It is true that

 

[math]KE = \frac{1}{2}m{v^2}[/math]

But m is a function of v or the other way round.

So you need to develop a mathematical statement of this relationship to assess the KE of the part travelling forwards.

 

 

 

The equation for work is work = force x distance.

There is no way to incorporate v, m or KE into that simple equation.

Are you proposing that the equation is incorrect ?

 

Then you will realise that the mass of the rocket is not constant.

 

 

Absolutely, but how does that present any relevance to the problem I have presented?

 

 

It doesn't have constant thrust efficiency. You can't just throw that constraint into the problem, it's unphysical. A rocket where thrust=weight has 0% thrust efficiency. All of the energy is going into heat and KE of the exhaust. No work is being done, yet whatever amount of energy conversion from combustion is happening. When the rocket velocity is the same as the exhaust velocity, it has maximum efficiency. None of the energy is ending up as KE of the exhaust. You get the maximum change in KE of the rocket for a given amount of fuel combusted.

 

You are absolutely correct.

 

Let's make it an EmDrive thruster which has constant thrust and efficiency and does not use any propellant.

Your title is a good one, because I think this is a straightforward misunderstanding.

 

 

Therein lies your difficulty.

 

 

Yes the thrust (a force) is constant.

 

And if we take gravity as constant over the flight

 

The same acceleration numerically happens in each second of the flight since force is constant.

 

So each second the rocket gains say 5 m/s (or whatever) velocity.

 

So the velocity gain is additive not multiplicative as you have in your '20 times'

 

But of course if the acceleration is a

 

(a + a)² = (2a)² = 4a² and so on for higher powers.

 

It is the difference between an arithmetic progression and a geometric progression.

 

BTW

I also wonder if you actually realise what force is doing the work.

It is not the thrust alone.

 

 

Thank you for your response.

I am not quite understanding how it applies though, maybe I haven't explained my question clearly enough:

 

Work = force x distance.

In instance 1) the work is the force x 5m. In instance 2) the work is force x 100m. The difference is 100/5=20 times more work being done in the second instance as compared to the first instance.

In both instances, the engine burns for the same period (1 second) so the same energy is spent.

 

If I am doing this calculation correctly, then it doesn't make sense when you consider the work/energy principle ?

 

BTW - I have studied physics at university level.

 

 

You haven't actually shown that your scenario is correct, but in general, rocket engines are not 100% efficient at converting stored energy into the rocket's kinetic energy. There is also KE in the exhaust, and this efficiency is lowest at low speeds (and zero when the rocket is at rest)

 

https://en.wikipedia.org/wiki/Rocket#Energy_efficiency

 

 

Thank you very much for the information, but it doesn't really answer my question. Lets say for the purposes of illustration that the hypothetical rocket engine has a constant thrust and constant efficiency from the moment you light it up till it's expended in Mars orbit.

 

We have 20 times more work being done in the second section of the example for the same energy expenditure as in the first section.

 

What am I misunderstanding/miscalculating here because that does not make any sense?

If a rocket launches from earth, and lets say that it moves up 5m in the first second, then it will do work of force times distance - let's call the amount w. If we wait until it reaches a velocity of 100m/s, then we can calculate the work being done over that particular one second as 20w but the energy expenditure is the same.

 

How is this possible ?