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?

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...

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).

Quote:Original post by arithma
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.

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).

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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.

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.

Quote:Original post by arithma
Is 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 Physics
Momentum 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.

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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]