# Angular Momentum again

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I'm sorry that I have to ask about the same subject again, but what am I gonna do? In short, the Angular Momentum of a particle around the origin is defined as L = r cross p, where r is the vector from the reference point and p is the linear momentum of the particle. I am asking about the reasoning behind such a physical quantity. What intrigues me is that you get the same quantity of "L" if r is multiplied by a factor (f for example) and p is divided by that same factor. Interestingly, when that is done, energy is not necessarily conserved. Plz help me to end this question once and for all.

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True. So try this experiment.

Stand with your arms stretched out wide and start spinning round (you might like to do this in a relatively private space), fast enough so your arms stay flat without effort. When you've reached a steady and comfortable speed pull your arms in quickly so they are straight against your sides.

What you should find is you rotate faster as you mostly conserve angular momentum. Probably not 100% due to friction exerted as you try to maintain your balance, but it's the gain in speed that makes you unsteady. You also gain energy - your calculations are correct. This energy comes from the work done briging your arms in - if you're going fast enough it can be quite an effort.

The Winter Olympics are almost on us, so if you're sensibly wary of doing science experiements on yourself just watch the figure skating, where the same experiment is done repeatedly by well balanced atheletes on a a near frictionless surface.

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How can you expect energy to be conserved when particles can instantaneously change their velocity like that? The fact that the change conserves angular momentum is irrelevant. We could also have doubled the distance from the origin and kept the velocity constant, conserving energy but not angular momentum. Yet you're not asking for reasoning behind energy?

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The answer to why is simple: it's useful.

Angular momentum is useful for describing the motion of revolving systems. It really needs no other justification.

Is there a more specific question you want to ask?

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What makes the angular momentum impotent is that it is a conserved quantity.
Quote:
 ...We could also have doubled the distance from the origin and kept the velocity constant, conserving energy but not angular momentum....

This is not true! You forgot the sin factor in L=r coss p...

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Quote:
Original post by Kambiz
What makes the angular momentum impotent is that it is a conserved quantity.
Quote:
 ...We could also have doubled the distance from the origin and kept the velocity constant, conserving energy but not angular momentum....

This is not true! You forgot the sin factor in L=r coss p...

I'm not sure what you are referring to. Multiplying the length of r does not change the angle between r and p, so the sine of the angle between them remains the same, and r cross p will be doubled. The origin we are using to calculate angular momentum does matter. Change the origin used, and the angular momentum changes (the angular momentum for a closed system about a given point is constant, but it is different for different points).

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Quote:
 Original post by arithmaIn short, the Angular Momentum of a particle around the origin is defined as L = r cross p, where r is the vector from the reference point and p is the linear momentum of the particle.

Do you see the usefulness of angular position/velocity/acceleration? Or, more to the point I suppose, do you see the usefulness of rotational kinetic energy?

Quote:
 I am asking about the reasoning behind such a physical quantity.

Don't know if it'll help, but here's another thing to think about in addition to the ice skater examples:

Consider a definition of torque:

tau = dL/dt

So torque is the time rate change of angular momentum. This is nice if only because it's analogous to linear force:

F = dp/dt

But, if r is constant:

tau = dL/dt = d[r x p]/dt = r x dp/dt = r x F

Which is the first definition of torque you usually see (because the simplest situation is when you're trying to rotate something through a circle about a point, i.e. with constant r). So linear momentum is nice because it gives us a more general definition of torque (for non-constant r, e.g. elliptical paths).

Quote:
 What intrigues me is that you get the same quantity of "L" if r is multiplied by a factor (f for example) and p is divided by that same factor. Interestingly, when that is done, energy is not necessarily conserved.

Why should energy be conserved? It's not conserved if you magically multiply mass by a factor and divide velocity by the same factor. But I think the deeper concern is that r is arbitrary. If I'm three feet away, why can I double its angular momentum by taking one step back? Think of trying to apply a torque to that object to give it a certain angular momentum. You'll have to push harder (apply a greater torque) the further you are from the object (try it). Since:

tau = dL/dt

apply a greater torque for the same amount of time should give the object a greater angular momentum.

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Is angular momentum an encapsulation around the concept of linear momentum of particles of a rigid body, or not?

WHAT I mean is that if a rigid body is at rest (in translation sense) but rotating around its origin then we have two options to represent that rotation: either by finding a formula for linear momentum for each particle of that body or EQUIVALENTLY by finding the angular momentum of the hole body.

I am 100% sure of the linear momentum thingy. However I just can't figure out the jump from that to angular momentum (which I already use, but hate because I don't know what it represents exactly). What I can't figure out is how can I conclude the angular momentum formulation from that of linear momentum.

I'll be thankful very much if someone provides a theoretical approach since I really UNDERSTAND physics using math more than anything else.

[EDIT]
I also drop my questions about correlating the conservation of angular momentum to conservation of energy... It is obvious that it an invalid question when compared to linear momentum and Linear Kinetic Energy which are not necessarily conserved together.

[EDIT]
I also don't question the USEFULLNESS of angular momentum, never.. And I know that it is already founded science..

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Let O=(0,0,0) be the origin and r(t=0)=(1,0,0) the position of out mass point m at the time t=0 and let v=(0,1,0) the velocity of the mass point( v:constant).
The distance |O-r| is equal to 1 now(t=0).
L(0)=r coss p = m.(0,0,1)
after sqrt(3) seconds(t=sqrt(3));
r(sqrt(3))=r(0)+sqrt(3).v=(1,sqrt(3),0)
the distance |O-r| is now equal to 2 so we have doubled the distance, so let us calculate L:
L=m.(1,sqrt(3),0) corss (0,1,0) = (0,0,1)
as you see L is conserved.
(Do not forget L is conserved for the whole n body system... you have to sum over all Ls if you have more than one mass point)
You can not double the distance by a not physical way and say that L is not conserved.

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Quote:
 Original post by arithmaIs angular momentum an encapsulation around the concept of linear momentum of particles of a rigid body, or not?

No, not really. As you pointed out, there's that whole thing about linear momentum being affected by your distance to the object.

I think, if you understand the linear system (x, v, a, m, F, p) then you should understand the angular system as an analog (theta, omega, alpha, I, tau, L). m is the resistance to a change in linear motion, I is the resistance to a change in angular motion.

Let me ask this: How do you think of momentum (p)?

Now, take your answer to that, and replace the linear parts with their angular analogs.

Quote:
 World of PhysicsMomentum is a fundamental quantity in mechanics that is conserved in the absence of external forces. .... A similar quantity conserved in the absence of noncentral forces is called angular momentum.

Quote:
 WHAT I mean is that if a rigid body is at rest (in translation sense) but rotating around its origin then we have two options to represent that rotation: either by finding a formula for linear momentum for each particle of that body or EQUIVALENTLY by finding the angular momentum of the hole body.

Look at it this way, if the body is at rest but rotating, then the sum of the linear momenta will be 0 by definition, so finding a formula for the linear momentum of each particle is not equivalent to finding the angular momentum.

I forget offhand, but I think there's a relation to the area of the sector swept by the object. I've got to go, but I'll check into it (it's been a while since I've had to do a lot with angular motion).

EDIT: Cross product gives you the area of the parallelogram created by the two vectors and a direction perpendicular to that parallelogram. Thus,

r 'cross' dx = r 'cross' (v dt) = (r 'cross' v)dt

Gives twice the area of the infinitessimal sector swept by the object in the infinitessimal time step dt (see first link below).

Also, I liked these:
Conservation of Angular Momentum.
Angular Momentum.

[Edited by - Way Walker on January 27, 2006 2:02:47 PM]

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Quote:
 Original post by Way WalkerLook at it this way, if the body is at rest but rotating, then the sum of the linear momenta will be 0 by definition, so finding a formula for the linear momentum of each particle is not equivalent to finding the angular momentum.

Well am not proposing to find a relation between "sum of linear momentum" and "angular momentum". What am proposing is starting the reasoning from the (1) body being rigid and (2) the sum of linear momentum is zero. If these conditions must be satisfied, then the body MUST be rotating. Relying on this we MUST be able to find a formula for p given r (using same meanings for symbols). I believe if somebody may be able to find such a relation, it will provide instantaneous insight into angular momentum...

Moreover, you must understand that angular momentum is definetly not a concept by itself and apart from linear momentum. It is only a simplification. That emerges from the fact the Newton's Three Laws are enough to explain any phenomenon in classical mechanics; thus angular momentum is a derived concept. Yet we never see any literature concerning the "DERIVATION" of a concept. This was expectable at high school but I never imagined that it may prevail in the engineering physics courses. I don't know about physics majors: so is this really the situation?

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Quote:
 Original post by arithmaWell am not proposing to find a relation between "sum of linear momentum" and "angular momentum". What am proposing is starting the reasoning from the (1) body being rigid and (2) the sum of linear momentum is zero. If these conditions must be satisfied, then the body MUST be rotating. Relying on this we MUST be able to find a formula for p given r (using same meanings for symbols). I believe if somebody may be able to find such a relation, it will provide instantaneous insight into angular momentum...Moreover, you must understand that angular momentum is definetly not a concept by itself and apart from linear momentum. It is only a simplification. That emerges from the fact the Newton's Three Laws are enough to explain any phenomenon in classical mechanics; thus angular momentum is a derived concept. Yet we never see any literature concerning the "DERIVATION" of a concept. This was expectable at high school but I never imagined that it may prevail in the engineering physics courses. I don't know about physics majors: so is this really the situation?

Perhaps this is a case of putting the horse behind the cart... the conservation laws are more general than Newton's laws and are given higher precedence (here). In other words, Newton's laws should be derived from conservation of momentum, not the other way around.

 perhaps Noether's theorem is what you are looking for?

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Quote:
Original post by arithma
Quote:
 Original post by Way WalkerLook at it this way, if the body is at rest but rotating, then the sum of the linear momenta will be 0 by definition, so finding a formula for the linear momentum of each particle is not equivalent to finding the angular momentum.

Well am not proposing to find a relation between "sum of linear momentum" and "angular momentum". What am proposing is starting the reasoning from the (1) body being rigid and (2) the sum of linear momentum is zero. If these conditions must be satisfied, then the body MUST be rotating. Relying on this we MUST be able to find a formula for p given r (using same meanings for symbols). I believe if somebody may be able to find such a relation, it will provide instantaneous insight into angular momentum...

Given L and r, you can't find p. Cross product isn't reversible. You realized this when you pointed out that doubling r and halving p gives the same result. Doubling r and changing the angle from 90 to 30 would also give the same result. Cross product is lossy. You'll have to find some other source of insight.

Quote:
 Moreover, you must understand that angular momentum is definetly not a concept by itself and apart from linear momentum. It is only a simplification. That emerges from the fact the Newton's Three Laws are enough to explain any phenomenon in classical mechanics; thus angular momentum is a derived concept. Yet we never see any literature concerning the "DERIVATION" of a concept. This was expectable at high school but I never imagined that it may prevail in the engineering physics courses. I don't know about physics majors: so is this really the situation?

Engineering is dirty physics in the same way that physics is dirty math. Really, engineering is about results and doesn't much care how the physicists came up with the results. In fact, you probably won't get to the derivation until upper level, even graduate pure-(as-opposed-to-dirty/engineering-)physics courses. Our classical mechanics course wasn't until junior/senior year.

If you really want it, here's how I see angular momentum coming about:
Consider an object in a force field (e.g. gravity, part of a rigid object, etc.). Let's say it's in orbit. It's orbiting so obviously momentum isn't conserved. However, it won't just spontaneously stop spinning, so something's being conserved. Energy? Maybe, but if its orbit contracts (think figure skater) that won't be conserved. The quantity (r x p) is conserved. This is somewhat intuitive if you picture an orbitting object; in an elliptical orbit v (and thus p) decreases with r. You don't escape this problem with a circular orbit because the direction of v changes, but (r x p) will always be normal to the orbitting plane. Basically, r "corrects for" the changes in p.

Now, I bet you have a couple concerns about what I just said:
1) "But momentum is conserved if you take into account what's causing the force field". True, but the same complaint could be brought against the claim that momentum is conserved even when energy isn't. Energy is conserved, if you take everything into account. Like you said, these are simplifications. Even Newton's three laws can be considered simplifications of General Relativity.

2) "But angular momentum is only conserved in the absence of non-central forces". Yeah, but this requirement is basically satisfied by requiring it to be in an orbit.

If you want something more rigorous, I think a book or pen and paper are your best bet.

This is going to sound critical, but it's not, I'm honestly curious: Did you demand a rigorous definition of limits before you were satisfied with derivatives?

By the way, you didn't play my game: describe momentum to me. [grin]

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Quote:
 Original post by Way WalkerConsider an object in a force field (e.g. gravity, part of a rigid object, etc.). Let's say it's in orbit. It's orbiting so obviously momentum isn't conserved. However, it won't just spontaneously stop spinning, so something's being conserved. Energy? Maybe, but if its orbit contracts (think figure skater) that won't be conserved. The quantity (r x p) is conserved. This is somewhat intuitive if you picture an orbitting object; in an elliptical orbit v (and thus p) decreases with r. You don't escape this problem with a circular orbit because the direction of v changes, but (r x p) will always be normal to the orbitting plane. Basically, r "corrects for" the changes in p.

Quote:
 Original post by arithmaI also drop my questions about correlating the conservation of angular momentum to conservation of energy... It is obvious that it an invalid question when compared to linear momentum and Linear Kinetic Energy which are not necessarily conserved together.

-----

Quote:
 Original post by Way WalkerGiven L and r, you can't find p. Cross product isn't reversible....

If you follow me precisely as ask the question, I'm concerned with the transition from linear momentum to angular momentum. So your note over here may be precise but a little bit irrelevant.
-----

And regarding the orbits:
Orbits are also irrelevant since am considering total linear momentum is zero; except if you're talking about a wider system that spans other objects, but again it loses the "Rigid Body" qualification...

Quote:
 Original post by Way WalkerThis is going to sound critical, but it's not, I'm honestly curious: Did you demand a rigorous definition of limits before you were satisfied with derivatives?

Not really, because we learned about it in eleventh grade, so I didn't have enough backward curiosity if I may say. And besides, for most applications, an empirical understanding of small entity is achievable

Quote:
 Original post by Way WalkerBy the way, you didn't play my game: describe momentum to me.

I assume you mean linear momentum. If that's it then it is easy. An understanding of mass as "mass of particles" then as "quantity of mass of particle" and directly map "mass of particle" into "particle" so you have it as "quantity of particles". And velocity doesn't need to be explained, (the differential of position over the differential of time...). Now if you assume each particle has an equal linear momentum to each other particle (no rotation) then linear momentum can be understood as: the "quantity of particles" * "velocity" which could be manipulated to "quantity of velocity of particles". Viola; At least it makes sense to me :)

Quote:
 Original post by jjdPerhaps this is a case of putting the horse behind the cart...

According to your source, the validity is given the higher precedence, not actually the concept itself, which is related to conditions such as approaching speed of light and stuff; we are discussing a Classical Mechanics Universe

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Quote:
 Original post by arithmaAccording to your source, the validity is given the higher precedence, not actually the concept itself, which is related to conditions such as approaching speed of light and stuff; we are discussing a Classical Mechanics Universe

Um, what? Read the seciton entitle "Relationship to the conservation laws", it is about classical physics and has nothing to do with einstein's relativity. Also, you are trying to squeeze the conservation of angular momentum out of newton's law, which are derived from conservation of linear momentum... and you wonder why you're having trouble understanding it?!

I give up.

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Angular momentum as we define it is a derivation. So why do we care about it, as it's arguably not as fundamental as other quantities ?

Simply because in rotational problems it's the quantitiy most often conserved. E.g. a freely rotating non-symmetric rigid body will have an axis of rotation, a speed of rotation, and so an angular velocity. In general none of these is conserved - all can change over time, and in general do for irregular body. But angular momentum IS conserved, providing no external torque is applied.

As it can be related to other quantities it can tell us things about them which are difficult to deduce in other ways. In some ways the fact that it's conserved makes it more fundamental. E.g. fundamental particles have a fixed 'intrinsic angular momentum', also called spin.

You might also consider kinetic energy, but in general is not conserved in non-rigid bodies. E.g. a planet orbiting a star gains and loses kinetic energy as it moves closer to and away from the star. Or in a non-elastic collision between two bodies energy is lost but providing there are no external forces or toruqes total linear and angular momentum are conserved.

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Quote:
Original post by arithma
Quote:
 Original post by Way WalkerGiven L and r, you can't find p. Cross product isn't reversible....

If you follow me precisely as ask the question, I'm concerned with the transition from linear momentum to angular momentum. So your note over here may be precise but a little bit irrelevant.

I don't know what else you could mean by "Formula for p given r". I thought you were trying to attack the problem from another direction (i.e. instead of seeing how to get to angular momentum from linear, you'd try to see the relation from the other side). Please restate the question and I'll be happy to attempt another answer.

The only other interpretation I can think of is for a formula for p given a position r within the body, but surely that depends on the system you're considering.

Quote:
 And regarding the orbits:Orbits are also irrelevant since am considering total linear momentum is zero; except if you're talking about a wider system that spans other objects, but again it loses the "Rigid Body" qualification...

Orbits are not irrelevant. A "rigid body" forms a force field. If the object is spinning, its "rigidness" implies a force pulling each portion of the body toward the center of rotation (otherwise it would fly apart). Each of these portions is "orbitting" that center of rotation. (This sort of "dividing the body into pieces" is common in physics and why I mentioned being part of a rigid body as being in a sort of force field.) In other words, if you understand an object in orbit, it's an easy extension to an ensemble of objects in orbit. You've merely placed the constraint (wolog, because we can simply have a moving frame of reference) that the individual linear momenta sum to 0.

Quote:

Quote:
 Original post by Way WalkerThis is going to sound critical, but it's not, I'm honestly curious: Did you demand a rigorous definition of limits before you were satisfied with derivatives?

Not really, because we learned about it in eleventh grade, so I didn't have enough backward curiosity if I may say. And besides, for most applications, an empirical understanding of small entity is achievable

My point was, you use (and apparently still use) an intuitive understanding of limit without seeing how this concept comes from more basic concepts. A rigorous definition of limits gives the assurance that what you're doing is mathematically sound, but isn't usually of any use so long as you have the intuitive understanding.

Quote:
 I assume you mean linear momentum. If that's it then it is easy. An understanding of mass as "mass of particles" then as "quantity of mass of particle" and directly map "mass of particle" into "particle" so you have it as "quantity of particles". And velocity doesn't need to be explained, (the differential of position over the differential of time...). Now if you assume each particle has an equal linear momentum to each other particle (no rotation) then linear momentum can be understood as: the "quantity of particles" * "velocity" which could be manipulated to "quantity of velocity of particles". Viola; At least it makes sense to me :)

I'm not entirely sure "quantity of velocity of particles" makes sense, but I think I know what you mean. However, I fail to see how "quantity of angular velocity of particles" can seem so foreign if "quantity of velocity of particles" makes sense.

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To jjd:
Quote:
 From the given wiki sourceThe laws of conservation of momentum, energy, and angular momentum are of more general validity than Newton's laws, since they apply to both light and matter, and to both classical and non-classical physics. In the special case of a system of material particles interacting via instantaneously transmitted forces, Newton's second law can be viewed as a definition of force, and the third law can be derived from conservation of momentum.

And if you were right and I was wrong (in any sense you can put it) then I am sorry for any discomfort I did induce.. Thanks for your time anyway [grin]

To johnb:
I am not discussing "the usefulness of angular momentum" but asking about "the origins of its usefulness" by investigating its mathematical origin.

Quote:
 Original post by Way WalkerThe only other interpretation I can think of is for a formula for p given a position r within the body, but surely that depends on the system you're considering.

This is what I meant.

Quote:
 Original post by Way Walkerits "rigidness" implies a force pulling each portion of the body toward the center of rotation (otherwise it would fly apart).

Ok. Then I correct my statement to "non-circular orbits" are irrelevant. I think this is fair enough.

Quote:
 Original post by Way WalkerGiven L and r, you can't find p. Cross product isn't reversible.

For rigid bodies it is reversible.
vi = (m ri2)-1L cross ri

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Quote:
Original post by arithma
Quote:
 Original post by Way WalkerThe only other interpretation I can think of is for a formula for p given a position r within the body, but surely that depends on the system you're considering.

This is what I meant.

My appologies, for whatever reason the constraints placed by the sentence before the request didn't click in my head.

Quote:

Quote:
 Original post by Way Walkerits "rigidness" implies a force pulling each portion of the body toward the center of rotation (otherwise it would fly apart).

Ok. Then I correct my statement to "non-circular orbits" are irrelevant. I think this is fair enough.

Then what I said still applies (because "circular orbits" are a subset of what I described): Something is being conserved because it doesn't just spontaneously decay. It's not momentum (v can't make up its mind where to point) and it's not energy (not conserved under contraction). (r x p) gives a conserved quantity because the r compensates for changes in p.

But maybe you'll say "It's rigid, so there's no contraction and energy is conserved". All I can think to say then is that you've pretty much eliminated everything that would've lead to the development of angular momentum. It seems something like this:
Person1: Could you explain why we use 'F = dp/dt' instead of 'F = ma'?
Person2: Because the former works for changing mass. (rocket science!)
Person1: Yeah, but what if mass is constant?
Person2: Well, then 'F = ma' is just fine.
Person1: So why do we use 'F = dp/dt' instead of 'F = ma'?

If you want to use math, jjd's route is your best bet. What I said above is much less rigorous, but I think it's the way that most physicists intuitively understand the concept (although the rigorous proofs give them a warm fuzzy).

Quote:
 For rigid bodies it is reversible.vi = (m ri2)-1L cross ri

So you have what you are looking for, problem solved? I'm guessing not, but if you can tell us why this fails to answer the question we may be able to fix things.

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The formula that I posted is the reverse of what I needed [grin]

In addition, I am not asking about the validity of the formula of angular momentum. I am asking about the steps that get you from "linear momentum of particles" into "angular momentum of rigid body". This is assuming you don't use nothing but math and the concept of "linear momentum".

I drop any relevancy issues that were raised in face of using orbits. However I praise the avoidance of their usage in the discussion. After all we are trying to figure out an intuitive sense around angular momentum [smile].

Now let's get down to the dirt.
For a translationally resting rigid body the following is observed:

- Sum of pi = 0; which we get from the "translationally resting" constraint, which is only for the sake of simplification
- Every particle is connected to every other particle by a nonelastic rigid rod; to satisfy the rigidity constraint

Manipulation steps: x, y, z,...
to reach a situation where miraceously a quantity appears that simplifies things: ANGULAR MOMENTUM. This quantity must appear somewhere since the latter conditions allow rotational movement exclusively.

The outcome will allow us, instead of having to use the function pi(ri) to describe the rotation of the body, to use the constant quantity L that arrises from ??? to describe that rotation. Then the quantities pi are easily calculated given their ri and the constant alleged quantity.

There are two other things: (1) that "using system of equations of linear momentum of particles is equivalent to using the angular momentum of rigid body" to describe the rotation and (2) an answer to a similar (but simpler) question concerning the linear momentum of a rigid body.

(1)
We can describe the rotation of a rigid body (implicitly) by the following funtion:
pi(ri) = ri-2 L cross ri
This is of course in a confined situation where r and L and p are orthogonal in-pairs; but you get the idea.
------------------------------------------------------------------

(2)
How is the concept of linear momentum of a rigid body established:
Constraints: resting rotationally

resting rotationally: vi = v for any i in the body. Interestingly, that preserves the rigidness of the body.

{
sum of pi = sum of ( mi vi )
ptotal = sum of ( mi v )
ptotal = sum of ( mi ) v
ptotal = mtotal v
}
These steps are analogues to {x,y,z,...} in the answer that I seek

Please note that I allowed my self to use linear momentum of particles here because it is of low intuitive requirements. However, you shouldn't allow yourself to use angular momentum of particles in the establishment of angular momentum of a rigid body. This is because it is assumed that the whole notion of "angular momentum of anything" is not known before this step.

Please bare with me...

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Quote:
 Original post by arithmaThe formula that I posted is the reverse of what I needed [grin]

Quote:
 pi(ri) = ri-2 L cross ri

No, that is most definitely p given r under the constraints you required, which is what you asked for. What's the revese? The inverse (r given p)? But r_i(p_i) isn't well-defined. You already have L_i in terms of r_i and p_i. What do you want?

Quote:
 I drop any relevancy issues that were raised in face of using orbits. However I praise the avoidance of their usage in the discussion. After all we are trying to figure out an intuitive sense around angular momentum [smile].

It sounds like you already have an intuitive sense. The very fact that you say "Not all p_i = 0" and "Sum({p_i}) = 0" implies "Spinning" shows this.

Quote:
 Now let's get down to the dirt.

I'm not sure it can be done. Eventually, you're just going to have to pull that (r x p) out of your ass and comment that it doesn't change given the constraints. My recommendation: Try it with just two particles and note that it can be extended to more discrete particles and continuous bodies. Cylindrical coordinates can help.

If you don't want to just pull it out of your ass, you're not going to get there by your route.

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OH YEAH
I was once too intrigued by this thing and as far as i can tell:
angular momentum is NOT derived from linear momentum.

but i'll give you a clue.
imagine a long rigid body and a force that acts on a tip of it making it spin.
how does energy/linear momentum propagates between points/particles from one tip to the other? no idea? me neither. and that would be i belive the answer to your question.

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I think that I already proved that the notion of Rotation can be implied by earlier posts. Plus the relation between the single quantity called Angular Momentum and the multitudes of particles is established here and by far by earlier derivations (I don't want to take credit in any of this since nothing of it is mine... plus it is only a special case [talking about the formula]).

I have actually found a very interesting link that backs me up. It talks about Rigid Body & Linear Momentum & Angular Momentum & Torque.

It actually derives angular momentum from linear momentum and a cross product.. [grin]

The author is a professor of mathematics since 1987 (my birthdate) so I guess he know a trick or two..

I'm actually delighted to have found this website

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Quote:
 Original post by arithmaIt actually derives angular momentum from linear momentum and a cross product.. [grin]

Ahem
It does not. It derives angular momentum of a rigid body (system of points)
from angular momentum of a point.

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Quote:
 Original post by arithmaI have actually found a very interesting link that backs me up. It talks about Rigid Body & Linear Momentum & Angular Momentum & Torque.It actually derives angular momentum from linear momentum and a cross product.. [grin]

He doesn't derive it, he defines it. After getting to equation (3) he says:

Quote:
 Define:L(t) = Sum_{i = 1}^n (m_i r_i(t) x r`_i(t))T(t) = Sum_{i = 1}^n (r_i(t) x F_i(t))

And to even get to this point he says:

Quote:
 Now take the cross product of r_i(t) and equation (1_i) and sum over i.

Like I said, you have to pull it out of your ass.

I know the trouble with finding the explanation that works for you. I've often had talk to a couple professors (or fellow students) before getting an explanation that worked for me. I honestly didn't think the sum over i was the part you were struggling with. I thought you were having trouble with the conservation of angular momentum and the fact that the cross product will never just pop out of manipulations on the linear momenta. Glad you finally found the explanation that works for you.

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