On Apr 13, 12:29 pm, "PD" <TheDraperFam...@gmail.com> wrote:

On Apr 13, 10:57 am, "Peter" <Poakfi...@msn.com> wrote:

On Apr 13, 10:32 am, "PD" <TheDraperFam...@gmail.com> wrote:

On Apr 13, 8:57 am, "Peter" <Poakfi...@msn.com> wrote:

On Apr 13, 7:51 am, "PD" <TheDraperFam...@gmail.com> wrote:

On Apr 13, 2:00 am, "Greg Neill" <gneill...@VEsympatico.ca> wrote:

<mme...@cars3.uchicago.edu> wrote in messagenews:7ADTh.90$25.220@news.uchicago.edu...

In article <461ee083$0$11811$9a6e1...@news.newshosting.com>, "Greg Neill"

<gneill...@VEsympatico.ca> writes:

"Peter" <Poakfi...@msn.com> wrote in message
news:1176421490.201417.260000@o5g2000hsb.googlegroups.com...

For a point mass, r x p and mr^2w are the same thing.

Only for a circular orbit, where r and p are
perpendicular.

Nah, where did you get it from. In the above, w is the angular
velocity, defined as w = (r x v)/|r|^2. So,

r x p = m*(r x v) = m*|r|^2*w. For any motion, not just circular.
This doesn't change the fact, of course, that angular momentum is
*defined* by r x p (not the most general definition, but it'll suffice
for here) while in the special case of point mass it *evaluates* to
mr^2w. "Defined as" and "evaluates to" is not the same thing.

You're right of course.

I thought that Peter was considering mr^2w to consist of
only scalar values (based upon the apparent level of his
arguments), and w in particular to be a scalar constant

But any idiot can look at animations illustratingKepler'ssecondlaw
and see that w is not a scalar constant. I believe that was preciselyKepler'spoint.

http://home.cvc.org/science/kepler.gifhttp://www.glenbrook.k12.il.us/...

as in the usual high school circular motion analysis.
This, I figured, was his bid to show that mr^2w is not
constant for an elliptical orbit since r varies. Perhaps
I jumped to that conclusion prematurely.- Hide quoted text -

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- Show quoted text -- Hide quoted text -

- Show quoted text -

PD, I'll tell you why I say mr^2w is incorrect. I just saw a device,
which will soon be mass-produced, that shows the correct
expression for the angular momentum of a point mass is mrw, not mr^2w.
It is an elongated circular track, like a racetrack;
its length is twice its width. But it is a virtually frictionless air
track, with air walls at its semicircular ends. A relatively
large puck is released with a certain velocity on the middle of the
rectilinear part of the track. It moves practically with the
same velocity until it reaches the curved part of the track, where the
airflow from the wall changes the direction of its
velocity, without changing its magnitude; thus, the puck goes around
the track and keeps going with the same speed.
Clearly, the angular momentum of the puck, with respect to the center
of the track, when it is released on the middle
of the straight part of the track, is mrw, and when it reaches the
center of the curved part of the track is m2r(w/2), because
its speed has not changed. Thus, its angular momentum has been
conserved. If mr^2w were correct, angular momentum
would appear not to be conserved, but it is, because no forces act in
the direction of its motion. I hope this convinces you.

Peter

While this is amusing, Peter, you are off the mark on several fronts.
- The forces acting on the puck in this device are not *central*
forces. Angular momentum is conserved under a *central* force. There
is no guarantee that angular momentum is conserved if the forces
acting on a body are not central. There is no single point in this
device that the net force acting on the puck points toward. It simply
is not an example of a central force situation. In the case that you
are talking about, if you choose the center of the track to be your
axis, then when the puck hits the circular arc, the force that acts on
the puck does not point toward the center of the track -- it points at
the center of the circular arc, which is not in the same place.
- You say the angular momentum of the puck on the straight portion of
the track is mrw. This is simply wrong, and I have no idea why you
think it's correct. The *linear* momentum of the puck on the straight
portion of the track is p=mv, which (depending on how you are defining
w) is numerically equal to mrw. The *angular* momentum of the puck on
the straight portion of the track is L = r x p. Please do not confuse
the two or imagine that only one can apply at a time or that somehow
angular momentum *turns into* linear momentum. It doesn't.

PD- Hide quoted text -

- Show quoted text -

The forces acting on the puck when it is going around the circular
part of the track, are central to its motion at that time. The linear
(tangential) velocity of any object going around any reference point
is rw = v; so its linear (tangential) momentum is mrw = mv. I think
there is only one kind of momentum mv,

And this *precisely* where you are wrong. There are two kinds of
momentum. Each of them follows an independent conservation law. Your
attempts to force one to become "really" the other are unwarranted.

which is a vector, because
velocity is a vector, regardless of how the object moves, and is
conserved if no forces in the in direction of its motion act on it. I
think it is funny that apparently nobody knows who defined angular
momentum as mr^2w. I would like to see someone deriving it from
scratch.

It isn't a matter of derivation. It's a matter of *observation*. That
is true for linear motion as well. No one *derived* the fact that
momentum is conserved. It is simply observed that it happens to *be*
conserved. Newton made the first observation that there is a quantity,
which we now call momentum (he used a different Latin word), that
stays the same unless there is a net force acting on it, in which case
the relationship is F = dp/dt. (Note this is what Newton said. Current
textbooks present it differently to connnect it better with kinematic
variables that are studied first, which is why you see it as F=ma. But
Newton described his law as F=dp/dt.) Likewise, when Newton's laws for
linear motion were studied in detail and found to match observation
exceedingly well, some of those people noticed, "Hey, there's another
quantity that seems to be conserved in rotational cases we've
measureed. Let's call it angular momentum."

The one interesting thing that has come out of this thread is that
I can't seem to find who introduced the angular quantities
into mechanics. Who is responsible for the concepts of
angular momentum and torque, for conservation of angular
momentum, etc? Lagrange perhaps?

On Apr 13, 12:19 pm, "Randy Poe" <poespam-t...@yahoo.com> wrote:

On Apr 13, 11:57 am, "Peter" <Poakfi...@msn.com> wrote:

On Apr 13, 10:32 am, "PD" <TheDraperFam...@gmail.com> wrote:

On Apr 13, 8:57 am, "Peter" <Poakfi...@msn.com> wrote:

On Apr 13, 7:51 am, "PD" <TheDraperFam...@gmail.com> wrote:

On Apr 13, 2:00 am, "Greg Neill" <gneill...@VEsympatico.ca> wrote:

<mme...@cars3.uchicago.edu> wrote in messagenews:7ADTh.90$25.220@news.uchicago.edu...

In article <461ee083$0$11811$9a6e1...@news.newshosting.com>, "Greg Neill"

<gneill...@VEsympatico.ca> writes:

"Peter" <Poakfi...@msn.com> wrote in message
news:1176421490.201417.260000@o5g2000hsb.googlegroups.com...

For a point mass, r x p and mr^2w are the same thing.

Only for a circular orbit, where r and p are
perpendicular.

Nah, where did you get it from. In the above, w is the angular
velocity, defined as w = (r x v)/|r|^2. So,

r x p = m*(r x v) = m*|r|^2*w. For any motion, not just circular.
This doesn't change the fact, of course, that angular momentum is
*defined* by r x p (not the most general definition, but it'll suffice
for here) while in the special case of point mass it *evaluates* to
mr^2w. "Defined as" and "evaluates to" is not the same thing.

You're right of course.

I thought that Peter was considering mr^2w to consist of
only scalar values (based upon the apparent level of his
arguments), and w in particular to be a scalar constant

But any idiot can look at animations illustratingKepler'ssecondlaw
and see that w is not a scalar constant. I believe that was preciselyKepler'spoint.

http://home.cvc.org/science/kepler.gifhttp://www.glenbrook.k12.il.us/...

as in the usual high school circular motion analysis.
This, I figured, was his bid to show that mr^2w is not
constant for an elliptical orbit since r varies. Perhaps
I jumped to that conclusion prematurely.- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

PD, I'll tell you why I say mr^2w is incorrect. I just saw a device,
which will soon be mass-produced, that shows the correct
expression for the angular momentum of a point mass is mrw, not mr^2w.
It is an elongated circular track, like a racetrack;
its length is twice its width. But it is a virtually frictionless air
track, with air walls at its semicircular ends. A relatively
large puck is released with a certain velocity on the middle of the
rectilinear part of the track. It moves practically with the
same velocity until it reaches the curved part of the track, where the
airflow from the wall changes the direction of its
velocity, without changing its magnitude; thus, the puck goes around
the track and keeps going with the same speed.
Clearly, the angular momentum of the puck, with respect to the center
of the track, when it is released on the middle
of the straight part of the track, is mrw, and when it reaches the
center of the curved part of the track is m2r(w/2), because
its speed has not changed. Thus, its angular momentum has been
conserved. If mr^2w were correct, angular momentum
would appear not to be conserved, but it is, because no forces act in
the direction of its motion. I hope this convinces you.

Peter

While this is amusing, Peter, you are off the mark on several fronts.
- The forces acting on the puck in this device are not *central*
forces. Angular momentum is conserved under a *central* force. There
is no guarantee that angular momentum is conserved if the forces
acting on a body are not central. There is no single point in this
device that the net force acting on the puck points toward. It simply
is not an example of a central force situation. In the case that you
are talking about, if you choose the center of the track to be your
axis, then when the puck hits the circular arc, the force that acts on
the puck does not point toward the center of the track -- it points at
the center of the circular arc, which is not in the same place.
- You say the angular momentum of the puck on the straight portion of
the track is mrw. This is simply wrong, and I have no idea why you
think it's correct. The *linear* momentum of the puck on the straight
portion of the track is p=mv, which (depending on how you are defining
w) is numerically equal to mrw. The *angular* momentum of the puck on
the straight portion of the track is L = r x p. Please do not confuse
the two or imagine that only one can apply at a time or that somehow
angular momentum *turns into* linear momentum. It doesn't.

PD- Hide quoted text -

- Show quoted text -

The forces acting on the puck when it is going around the circular
part of the track, are central to its motion at that time. The linear
(tangential) velocity of any object going around any reference point
is rw = v; so its linear (tangential) momentum is mrw = mv. I think
there is only one kind of momentum mv, which is a vector, because
velocity is a vector, regardless of how the object moves, and is
conserved if no forces in the in direction of its motion act on it. I
think it is funny that apparently nobody knows who defined angular
momentum as mr^2w. I would like to see someone deriving it from
scratch.

Regardless of what you think about the universe, in the actual
universe there is another quantity with the units of momentum called
angular momentum, defined as r x p, and which is conserved
in the absence of torque, e.g. under central forces. This is a simple
consequence of Newton's laws, and it is empirically verified.

You have seen the simple derivation of mr^w2 from r x p. Several
times.

Vector velocity is not "conserved" in the absence of forces in
the same direction. If there is no force in the direction of motion,
then no work is done and energy is conserved, speed is
conserved. But vector velocity is altered.

Every time you write you introduce more misconceptions, and
you are ignoring what every one else is writing. It is clear
you aren't here to get questions answered. I agree with others
that you are trolling.

- Randy- Hide quoted text -

- Show quoted text -

I understand only Newton's laws govern all motion, and torque is alien
to them, and unnecessary.

Velocity and momentum are true vector, but
torque and angular momentum are not.

Could you please explain what you
mean by
a central force?

Could you please clarify the following, I don't get
it:

Vector velocity is not "conserved" in the absence of forces in

the same direction.

If there is no force in the direction of motion,

then no work is done and energy is conserved, speed is
conserved. But vector velocity is altered.

Peter- Hide quoted text -

- Show quoted text -

I understand only Newton's laws govern all motion,

and torque is alien to them, and unnecessary.