Re: [Phys-l] basic laws of motion +- vectors +- angular momentum

["LaMontagne, Bob" <RLAMONT@providence.edu>]

I agree with others who have expressed an interest in revisiting this topic. 
 
I agree with your "don't do that" quip. In fact, that is the nut of the unease I have with trying to go from conservation of linear momentum to conservation of angular momentum for an extended object. I see how all the position vectors for the attachment points for the forces forming ( r /\ Fdt)  terms can be added to form a simple sum, however, I see no way to use those points to form a sum of r /\ dp because the rigid object has many dp associated with parts of the object not at attachment points. So sum( r /\ Fdt) does not seem to lead to a writeable sum(r /\ dp). It seems a leap of faith to associate the first sum with an angular momentum regardless of how angular momentum is defined. In fact, it doesn't appear that the right hand sum can be written in an explicit form except for the very simplest cases where each element of mass has an attached force associated with it. Even appealing to internal forces does not seem to me to lead to an explicit expression.
 
Bob at PC
________________________________

From: phys-l-bounces@carnot.physics.buffalo.edu on behalf of John Denker
Sent: Tue 12/9/2008 5:22 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] basic laws of motion +- vectors +- angular momentum




> However, the other side of Fdt=dp does not have an equivalent set of
> attachment points and lever arms if we are dealing with a rigid
> continuous object.

Let's be careful.  If we're summing over force vectors
strictly speaking, force vectors don't have attachment
points.  So, as Henny Youngman said, don't do that.
Instead let's sum over "physical interactions" or some
such, where each physical interaction has two vectors,
i.e. a force vector and a point-of-attachment vector.

> It appears that the operations on the two sides of N2
> are not equivalent in any obvious way.

Lost me there.  The LHS is _equal_ to the RHS.  Is that
trying to draw a distinction between equal and "equivalent"?
I'm not following that.

Or is that trying to say that the LHS is not not _compatible_
with the RHS?  I'm not following that, either.  I hope nobody
thought I was suggesting using angular momentum as a direct
plug-in replacement for linear momentum.  You can derive
conservation of linear momentum starting from conservation
of angular momentum, but they are not equivalent.  They don't
even have the same dimensions.