Re: [Phys-l] Coriolis effect puzzlement

[John Denker <jsd@av8n.com>]

On 12/02/2011 11:08 AM, Bob Sciamanda wrote:
> To further bring home the phenomenon at issue here, consider:
>
> A skater stands way off center on a rotating ice rink, partaking only of the 
> platform motion, and holding weights in her outstretched hands.
> If she brings in her arms, will she begin to spin about her platform 
> position?

1) That is a nice clarification of the question.

2) The answer is undoubtedly yes.

Qualitatively:  In this situation angular momentum is 
conserved.  It is conserved /separately/ from other things
like orbital angular momentum and total angular momentum.

The analysis in the inertial frame says she was spinning 
before she pulled in her arms, and she is spinning *faster* 
afterwards.

Slightly more quantitatively:

Let the position of any object (such as one of the weights)
be given by R, and decompose R as

       R = R0 + r

where R0 is the center-of-mass position of the "skater
subsystem" as a whole.

The object will have angular momentum bivector

      L = R ∧ p

where p is the ordinary linear momentum.  At this stage of
the argument, we can measure L around any axis, simply by
choosing the axis as the origin for measuring R.

Algebra tells us that we can decompose L as

    L = r ∧ p   +   R0 ∧ p          [4]

where the first term can be considered the /spin/ angular
momentum and the second term can be considered the /orbital/
angular momentum.

When the skater pulls in her arms, 
 -- The orbital angular momentum will not change.  This
  should be obvious, since R0 is the center-of-mass position,
  and is unaffected by the pull. 
 -- The spin angular momentum will also not change.  This
  should be obvious since all the other terms in eq. [4]
  are unchanging.

We conclude that spin angular momentum and orbital angular
momentum are each conserved /separately/ by the arm-pulling
process.  They are not separate in general, but they are
separate for present purposes.

We are told the skater initially partakes of the platform 
motion, which means the whole system (platform plus skater)
is undergoing rigid-body motion.

This means the spin angular momentum of the skater was 
nonzero before the pull.  It will be equally nonzero 
after the pull.  The relative radius |r| goes down, so
the angular velocity of the spin will go up.  QED.

===========

I don't know for sure, but I have a hypothesis about where 
the recent questions have been coming from ... and a theory
about why my answers were not clear.  To me it was obvious 
that spin angular momentum is /separately/ conserved in such 
a situation ... and that the orbital angular momentum could
be ignored.  This was so obvious to me that I didn't bother 
to explain it.  My bad.  I now hypothesize that it was not
obvious to everyone else.  In my previous notes I think I 
was careful to talk about the "spin".  That means what I 
said was technically correct;  it just wasn't a clear 
explanation.  Sorry.  This is an example of over-reliance 
on terminology.

The pedagogical rule is to introduce the idea /first/.
After the idea is out there, you can use terminology to
refer to the idea.  To say the same thing the other way,
throwing around terminology before the idea has been
explained is worse than useless.

===========

You can make the ideas much more quantitative if you want.
Some useful formulas are collected at:
  http://www.av8n.com/physics/clifford-intro.htm#sec-angular