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*From*: John Denker <jsd@av8n.com>*Date*: Mon, 21 Jul 2014 12:12:41 -0700

On 07/21/2014 11:38 AM, Paul Lulai wrote:

Then even though p is constant, the position vector r is constantly

changing in both magnitude and direction.

OK.

In this case, L is not constant.

OK.

No torques, no interactions with something outside of my system (the

ball and my oddly chosen origin), and L is not conserved.

Not OK.

I think the key here is the distinction between /conservation/

and /constancy/.

For all X, local conservation of X means that the amount

of X in a given region is constant /except/ insofar as

it flows across the boundary into adjacent regions.

Ordinary straight-line momentum is *always* conserved.

For a ball, it is not constant, because momentum can

flow across the boundary. Momentum flow can involve

short-range forces such as air resistance, and long-range

forces such as gravitation.

Angular momentum is *always* conserved. For a ball,

it is not constant, because angular momentum can flow

across the boundary. Angular momentum flow can involve

short-range torques such as air resistance, and long-range

torques such as gravitation.

For more about the relationship between constancy, conservation,

and continuity, see

http://www.av8n.com/physics/conservation-continuity.htm

is angular momentum only conserved if the ball is orbiting about the

center of a circular path or a foci of an ellipse?

That is neither a necessary nor sufficient condition for

conservation (or constancy) of angular momentum.

Any /central force/ automatically and obviously leaves the

angular momentum both conserved *and* constant. This includes

the Kepler problem as a special case.

However, the fact remains that *any* interaction leaves the

angular momentum conserved, whether or not it is constant,

whether or not the mechanism of conservation is obvious.

**References**:**Re: [Phys-L] angular momentum***From:*Paul Lulai <plulai@stanthony.k12.mn.us>

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