Re: [Phys-l] Angular momentum problem

[John Denker <jsd@av8n.com>]

On 03/24/2009 06:13 PM, pschoch@nac.net wrote:

> They had 2 rectangular blocks that would glide on an air table.  Both
> blocks have the same mass, length and width (to within our measurement
> capability).  They took Block 1 and Block 2 and oriented their long axes
> with the y axis direction, a distance apart on the air table.  They then
> gave a light push to Block 1 in the +x direction toward Block 2 which was
> still at rest (not easy on an air table, but they did quite well).  The 2
> blocks hit and stuck together at 1/4 of their length from the end; the
> bottom of Block 1 1/4 from the end hit the top of Block 2 1/4 from its
> end.  They stuck because the students used velcro.
>
> Now, the 2 blocks move off together in the +x direction, but they also
> begin to spin/rotate cw.
>
> I can explain the inelastic collision (I'm fairly sure).  However, since I
> view the 2 blocks as the system, I see no outside force and thus I can't
> explain the non-conservation of angular momentum.  I'm assuming I'm
> defining the system wrong, so that there is a torque, but I don't see it.

There's no external torque.  To say the same thing another way,
if you choose a region that contains both blocks, there is no
angular momentum crossing boundary of the region.  All the 
physics is internal to that region.

It should be qualitatively obvious that no external torque is
needed.  Consider a more extreme geometry, such as two long
skinny sticks.  Stick A is initially at rest, and stick B is
initially moving like a spear (without spin) so as to hit the 
end of stick A, forming an L-shaped composite object AB.  It
would look very weird if object AB moved away without spinning.

The physics here is that although angular momentum is conserved,
_spin angular momentum_ is not (by itself) conserved.  So the 
key point is to keep track of the angular momentum, i.e. the real 
plain old angular momentum (not just the part we call the spin 
angular momentum).

In terms of real angular momentum, initially the angular momentum
of object A is zero relative to CM(A) and the angular momentum of
object B is zero relative to CM(B), but neither of them is zero
relative to CM(AB) i.e. the CM of the combined system.

In more detail: 

The angular momentum of any given object can be separated into spin 
angular momentum and orbital angular momentum.  
 1) For any given object, the spin angular momentum is measured 
  relative to the object's center of mass.
 2) The orbital angular momentum is measured relative to some chosen 
  datum.
 1) The spin angular momentum is independent of the choice of
  datum.  
 2) The object's orbital angular momentum is independent of the 
  state of spin of the object;  it is given by r /\ p where r
  is the position of the object's CM (relative to the datum) and 
  p is the object's straight-line momentum.

In particular, there is a famous theorem that says if you switch
from one datum to another, the change in angular momentum is given
by ΔR /\ P, where ΔR is the displacement of one datum relative
to the other, and P is the overall straight-line momentum of the
system.

So the fallacy here can be seen as a mishandled change of datum:
The initial state is described in terms of two different datums
(which is already asking for trouble) and the final state is
described in terms of a third.

To say almost the same thing:  spin angular momentum is obvious,
whereas orbital angular momentum is less obvious but no less 
important.

The free choice of datum is like any other gauge freedom:  for
any _one_ object you can make its angular momentum equal to its
spin angular momentum by a suitable choice of datum ... but you
only get to do this _once_.  If you try to mix-n-match calculations
based on different datums, mayhem ensues.  (Consider the analogy 
to the choice of "ground" in electrical circuit analysis.)

Another useful concept is the projection of r(A) - r(B) onto the
direction perpendicular to the relative motion.  This is called 
the _impact parameter_.  Assuming A and B are initially spinless, 
the compound object AB will be spinless if and only if the impact 
parameter is zero.  You can show that the impact parameter is 
invariant w.r.t a change of datum and also invariant w.r.t a Galilean 
transformation of the velocities.

You don't need an air table to demonstrate the essential physics
here, and you don't need to make the objects stick together.  It
suffices to juggle one drum-major's baton by whacking it from below 
with another.  You can control the amount of spin by choosing the 
impact parameter, i.e. by choosing just where to whack it.  If you
haven't got fancy batons, broomsticks will do.  Longer is better
(within reason) in the sense that less juggling skill is required.