Re: Question about Kepler's second law

"Peter" <Poakfield@xxxxxxx> wrote in message
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On Apr 13, 10:10 pm, "Greg Neill" <gneill...@xxxxxxxxxxxxxx> wrote:
"Peter" <Poakfi...@xxxxxxx> wrote in message

news:1176504661.287112.30600@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

My incorrect conclusion is that since it is irrelevant whether there
is or not a central force for angular momentum to be conserved, then
the angular momentum of the puck in the device I described should be
conserved. What do you think should be the correct conclusion?

The problem with non-central forces is that it implies
that there are torques in the system under investigation.
In other words, the system is not isolated, so angular
momentum can come and go and conservation appears to go
out the window.

To deal with this you must expand the system under
consideration to include whatever it is that's
supplying the unbalanced forces or torques. For your
racetrack example you'd need to include the racetrack
and Earth as a platform. Then the forces will be
balanced thanks to Newton's ThirdLaw, the projectile
and the Earth will exchange momenta (linear and
angular), conservation will rule and you will be free
to choose any point of reference you want in an
inertial frame of reference to do your calculations.

Another way to deal with the problem that doesn't
involve expanding the system is to carefully choose
your origin(s) for the calculations so that the forces
observed are always central. In your racetrack
example choose an origin at the center of curvature of
one of the semi-circular ends. If you then calculate
m*r^2w for the projectile coming in on a straightaway,
going through the turn and back out along the other
straightaway, you'll find that the angular momentum
thus calculated is constant.

Change origins to the mirror image point at the other
end of the racetrack to duplicate the result for that
other end.

The force acting on the puck in the circular part of the track is
perpendicular to its motion; so, there is no torque.

There is a torque when it's viewed from anywhere but
the center of rotation; the torque is calculated
considering the chosen origin as the center of rotation.


.

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