Re: Atwood's machine problem

[John Denker <jsd@AV8N.COM>]

Ken Caviness wrote:

> Isn't the angular momentum about the pulley equal to zero?  L-vector
> = r-vector cross p-vector, so the |L| = mrv sin(angle between r and
> v).  Looks like zero to me, since r & v are parallel (or
> anti-parallel).

It sure looks like zero to me.

I agree that the angular momentum approach to the problem doesn't
work.

The way I prefer to look at it is as follows:  Any downward force
on one rope is "mirrored" as an upward force on the other rope.
Therefore the system is isomorphic to two frictionless ice skaters
with a rope between them, so whenever either of them pulls, the
two of them move towards each other.  For the ice skaters, the CM
is stationary.  For the pulley version, you can't say "the CM"
is stationary by the strict definition of CM ... but you can
instead invoke the principle that the same equations have the
same solutions ... so it comes to the same thing.  That is, the
weighted average of the two positions (weighted according to
mass) -- with due regard to minus signs -- will be constant.

This is a super-easy problem for me, because when my brother and
I were little, when we were first old enough to have an allowance,
we pooled our allowance for weeks and weeks, and the first thing
we bought was a nice big rope.  So we got lots of first-hand data
monkeying with rope.