Timothy Golden BandTechnology.com wrote:
Ditto wrote:
Timothy Golden BandTechnology.com wrote:
Ditto wrote:
Fallingeagle wrote:
Can someone explain Why the total angular momentum of a system wi=
th a
pully, and two masses joined by a massless string overt the pully
equals the angular momentum of the pully + the angular momentum =
of the
two masses?
why: L(sys) =3D L(pully) + L(masses)
I omega + m1 * v * r + m2 * v * r
This is what I do not understand:
the spin (AM) of the pully is caused by the the two masses, why i=
s the
total spin not just the spin of the pully?
also: what is spin (angular momentum)....really??
For a point particle, angular momentum is defined as the cross prod=
uct
of it's radial position vector r and it's linear momentum. i.e L =
=3D r x
p.
For a system of particles, total angular momentum =3D sum(k =3D 1 t=
o n)
[L_k]
So for your pully system, Ls =3D Lp + Lm1 + Lm2
Thing to note is that angular momentum of a system depends on choic=
e of
origin of coordinate system, and that a particle moving in a straig=
ht
line has a non-zero angular momentum.
If you are right then my explanation must be wrong.
OK. Let's suppose that the system starts out without any velocity in
any part of the system and that the first mass is more massive than t=
he
second. So the first mass starts to move downward due to
gravity(assuming that is part of the probelm) As the masses accelerate
the pulley is accellerating. Therefore if angular momentum is conserv=
ed
and the pulley is exhibiting angular momentum then the masses must be
as well.
Yes. But angular momentum isnt conserved here because there are
external torques acting on the system due to the weight of the two
masses and the pulley.
Until one of these weights collides with the pulley no rotational
components will be present, unless you have them swinging about like
pendulums.
Rotational components components will still be present even if the
masses are travelling in straight lines. The rotational component is
normal to the radius vector, and the linear part parallele to the
radius vector.
I think this problem neglects these effects. Am I wrong in
thinking that no angular velocity means zero angular momentum?
I dont think you are wrong here. Angular velocity =3D r x v/ (|r|^2), L=
=3D
r x (mv)
From
http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html
I see
L =3D mvr sin =CE=B8
and the definition
L =3D r x p
The 2D version should be adequate for this problem.
I dont think there is a 2D version of the cross product because the
direction is normal to the plane with the 2 vectors.
Here already we see
a theoretical conflict in the definition. The cross product exists for
a 3D system only.
I dont understand how there is a theoretical conflict.
The point is that the definition is not general. A theoretical 5
dimensional system will not fit into the fundamental definition of
L =3D r X p .
It cannot work because the cross product is not defined. We can readily
see that momentum can still work and since momentum is what angular
momentum is built from it will be applicable to addressing rotational
energy of rigid bodies. I don't claim to understand how such bodies
would rotate; they could have more freedoms. But the method of study
that would be most consistent would be to fall back to linear momentum
analysis and plain old acceleration.
OK, and angular momentum is still a very uselful concept in our 3D
world for solving mechnical problems where the forces between the
constituent particles are central.
It seems to me that people have gotten a bit carried away with angular
momentum.
This problem is a clear example.
It depends what you want from the problem. The OP asked :
Can someone explain Why the total angular momentum of a system with a
pully, and two masses joined by a massless string overt the pully
equals the angular momentum of the pully + the angular momentum of the
two masses?
So you have to use angular momentum to answer his question.
Sure you can argue that the weight has
angular momentum, but that form of angular momentum is just linear
momentum. So long as conservation of energy is obeyed this kid will do
just fine. For him to be convinced of the weight's not having angular
momentum is a more valid position than your claim that it does in my
opinion.
Since angular momentum is defined as r x p, then even if the weight is
moving in a straight line, it has angular as well as linear momentum.
Even if it is moving in a circle, at has linear as well as angular
momentum. It's just a definition.
Get rid of the pulley and replace it with an equivalent amount of
friction.
The situation is the same to the weight yet the same student
would not be trained to use angular momentum to solve the problem.
The moment of inertia of the pulley will affect the tension seen by the
weights.
I guess that means that I'm breaking a conservation law. Since that
conservation law comes from one below it and I am obeying that one then
I have this caveat.
If you mean the conservation of angular momentum coming from the
conservation of linear momentum, then both are violated because of the
weights of the masses.
Angular momentum is a shortcut so that we don't have to think about all
of the little bits of the pulley accelerating around in a circle.
You might as well say that linear momentum is a shortcut so that we
don't have to think about all of the little bits of the weights
accelerating. Angular momentum makes certain problems easier to solve
by taking advantage of the fact that T =3D dL/dt for systems where the
forces are central, as in the case of rigid bodies.
There
is no uniqueness between angular momentum and linear momentum. The
special part is that these objects hold together so that these systems
are well behaved.
I think what you are trying to say is that Newton's Laws of Linear
Motion aren't unique to linear systems in being able to solve any
problem there, and not to rotational ones which is true.
I am still open to being disproved and it is really beyond my abilities
to make such a claim as to invalidate conservation of angular momentum.
Only if there are no external torques acting on the system.
But if we allow room for interpretation both positions can coexist. I
am simply saying that I prefer straight momentum and defending it. I've
written out the cancellation above in this thread that demonstrates
your way and mine.
I suppose the question would be what breaks without conservation of
angular momentum?
Perhaps it is here in this problem and I'm not seeing it.
I guess the best is to do it out thoroughly in both contexts and see if
there is any difference.
Maybe tomorrow I'll work on that.
-Tim
Therefore no general dimension solution exists under
this definition.
If the mass travels a straight line then a pure radial measure would
require r at infinity.
theta would then be zero for any finite velocity. This could be a neat
calculus puzzle but since the center of motion is not strictly
specified this may be a broken study of the system.
I dont understand the point your are making here.
Angular momentum is
merely a shortcut method of handling the component momenta of a rigid
object's constituent particles and their corresponding accelerations.=
A
rotating object is an accelerated system.
The idea of angular momentum arose from Kepler's law of Equal Areas
which is just a restatement of the Conservation of Angular Momentum. It
can be applied to rotating bodies because the internal forces are
central forces meaning that it acts along the radius vector between
particles.
This complexity is at the edge of my understanding and I look forward
to learning something here. I am entirely open to being wrong. Please
find where I have gone astray.
I get the feeling that you dont realize that when talking about angular
momentum of a system, you have to take the individual momentum of the
parts of the system about the same point and cannot change it when
calculating it for another part of the system.
=20
-Tim
=20
Hope this helps.
.
