Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

[Phys-L] Re: conservation of angular momentum question

On 03/14/05 18:09, Daniel S. Price wrote:

A classic application of conservation of angular momentum, often cited
in physics texts, is the joining of two cylinders rotating about a
common axis. When the cylinders are allowed to meet "face-on", the
angular momentum of the system is conserved. When cylinders rotating
about parallel axes meet "edge-on", however, the angular momentum of the
system may not be conserved.

Says who?

Angular momentum always obeys a local conservation law.
Period. No exceptions. So figure out what you mean by
"the system" and account for any angular momentum that
flows across the boundary of the system.

In the latter case, imagine two cylinders, one initially rotating and
the other stationary. If they are gently brought into contact,
frictional force between the cylinders acts to slow the original
cylinder's rotation and induce rotation in the other cylinder. HRW
(fifth edition), chapter 12, question 49, is an example of this
situation, and in the problem the authors claim that angular momentum is
not conserved.

That's a totally idiotic thing for them to say.

I say angular momentum always obeys a local conservation
law. No exceptions.

If they mean that the angular momentum of the system is
not conserved, I do not see the source of the external torque (unless
the frictional force is responsible, though it would seem to be internal
to the system as it acts only between the cylinders).

I don't have the cited text in front of me. But I can
guess what the setup is. We have not just two cylinders,
but two cylinders on *axles* with *bearings*. To accelerate
one of the cylinders, it is necessary and sufficient to
have a _moment_, i.e. a pair of forces separated by a
lever-arm. It is sufficient and conventional to consider
one force applied to the rim and the other applied to
the axis. If the axle is anchored to something outside
your "system" boundary, as I suspect it is, then this is
a major pathway whereby angular momentum is transferred
across the boundary.

Another version of the question makes an interesting
exercise also: consider two cylinders *without* axles
and bearings, just floating around in interplanetary
space. When they come into contact, some angular
momentum will flow from the rotating one to the initially
nonrotating one. The angular momentum of the two-cylinder
system will be strictly conserved, since there is no
pathway for transferring it across the boundary. The
wrinkle is that some angular momentum will be lost from
the "spin" coordinates and transferred to the "orbit"
coordinates. That is, the two cylinders will pick up
some nontrivial center-of-mass velocity.

The former case is also discussed in HRW5, chapter 12, question 53; the
coupling of the cylinders "face-on" is said to maintain angular momentum
(though friction between the cylinders seems to be responsible for their
eventual rotation as a unit).

In that case, the rims are colocated and the axles are
colocated, so the moments will be equal-and-opposite.

There is nothing special about the "face on" geometry,
except that it implies that the axles are colocated.

Why are the two situations different in how conservation of angular
momentum is applied?

The statement that angular momentum obeys a local
conservation law is *not* different between the three
setups considered above. It is *always* true.

It is also independent of what you choose as your
origin of coordinates ... provided you choose one
and stick to it consistently. BTW if you happen to
be working in the CoM frame of an isolated system,
you don't even have to choose one; all the laws
are manifestly independent of the choice ... while
more generally, moving your origin from one place
to another shifts the total angular momentum by
MV /\ X, where X is the vector from the old origin
to the new one, and MV is the linear momentum of
the CoM, and of course /\ means wedge product.

As Joel R. pointed out, you can't choose your origin
to be centered on cylinder #1 and centered on cylinder
#2 simultaneously; you have to choose ... except in
the less-than-general coaxial case.

It would be a good exercise to work the non-coaxial
case choosing an origin centered on neither cylinder,
just to convince yourself that the formalism works,
independent of the choice of origin.
Phys-L mailing list