Re: ENERGY WITH Q

[Bob Sciamanda <trebor@VELOCITY.NET>]

To be more specific:  The line integral of the net external force over the
trajectory of the system CM is numerically equal to the increment in the
linear CM kinetic energy of the system (1/2 the system mass times the
square of the CM speed).  This follows directly from Newton's laws.  Its
usefullness (including its application to John's spring problem) was
discussed ad nauseam previously on this list.  There is no thermo here -
this is simply Newtonian mechanics.

Bob Sciamanda
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "John S. Denker" <jsd@MONMOUTH.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, October 30, 2001 10:25 AM
Subject: Re: ENERGY WITH Q


> At 09:01 AM 10/30/01 -0600, RAUBER, JOEL wrote:
> > But is it true that F dot ds_cm is Delta KE?  Which I think is the
claim
> > Carl is making.
>
> Good question.
>
> At 09:35 AM 10/30/01 -0500, Bob Sciamanda wrote:
> > The integral of the net external force over the trajectory of the
system
> > CM will ALWAYS equal the system KE change.
> > See previous (1999?) list discussions.
>
> 1) Wrong answer.
>
> Consider a flywheel, starting from rest.  I spin the thing up by
applying
> equal and opposite forces to opposite points on the rim.  The net force
is
> zero.  The net force on the CM is certainly zero.  The motion of the CM
is
> zero.  The net external force dotted on the trajectory of the CM is zero
> squared.  Yet the kinetic energy increases.
>
> 2) Even if some statement along these lines were true, it would be
useless
> except in certain unrealistically-simple situations.
>   -- It might be useful if we knew _a priori_ that nonthermal energy
> transfer were the only thing going on.
>   -- It might be useful if we knew _a priori_ that thermal conduction or
> similar purely-thermal energy transfer were the only thing going on.
>   -- But in general we don't know any such thing in typical
thermodynamic
> situations.  I apply a force to some point on the system.  I have no
idea
> what the CM is doing.  I don't know the instantaneous force on the CM,
and
> I don't know the instantaneous motion of the CM.
>
> ===============================
>
> Look, you can't do thermodynamics without entropy.  You just can't.
Give
> it up now, and save yourself a lot of bother.  Without discussing
entropy,
> you can't explain the distinction between a spinning flywheel and a warm
> flywheel no matter how carefully you keep track of the macroscopic force
> and macroscopic motion.
>    http://www.monmouth.com/~jsd/physics/thermo-laws.htm#sec-flywheel