Aargh! I started out replying to a message, so styles somehow got
embedded in this!!! Try again!
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The speed of messages on this list always amazes me. Or maybe I'm
just slow. I suspect the latter. In any case....
Prompted by John M, I have re-read his paper as well as the Bernard &
Sherwood paper. In fact I did read both the last time this topic came
around on PHYS-L, but things don't really stick until you really have
to use them. Like in teaching plainly, if that is possible.
I have written up a long summary of the whole topic. So far I have
worked (excuse the pun) through levels one and two below, but I'll
include my thoughts on level three to see what you think. Before I
spring a URL for my whole article, let me start by summarizing.
We can view a system at three levels:
Level one = the whole system is one object of mass M and
(center-of-mass = cm) speed V. I don't concern myself with any
internals. We easily derive: W_cm = delta K. Here W_cm is the cm work
or pseudowork (ie. integral of F_net dot cm displacement) and K_tr is
the cm translational KE (ie. MV^2/2). This view is only useful for a
few problems, but is a helpful conceptual starting point because we
can derive this equation exactly from Newton's laws and kinematics.
Level two = I break the universe into macroscopic parts (blocks,
strings, pulleys, earth, etc) but I ignore their internals. Now I can
talk about forces between parts. If conservative, I associate this
with PE. For simplicity, let's do this even if one of the two
interacting parts is outside the system. (If this makes your skin
crawl, just choose your system to include every interacting part.)
Now sum over parts, sum delta(K_tr_i) = sum(W_cm,nc_i) +
sum(W_cm,c_i). Move the last term to the left and redefine W_nc =
sum(W_cm,nc_i), delta K = sum delta(K_tr_i) [K is NOT equal to K_tr]
and delta U = -sum(W_cm,c_i). This gives W_nc = delta E_mech, with
E_mech = K + U. Call this the work-energy theorem. This is standard
textbook stuff and works just fine for textbook problems (even when
nasties like friction are present) provided one essential point is
emphasized:
THE WORK I AM SO FAR CALCULATING IS CM WORK (PSEUDOWORK) NOT
THERMODYNAMIC WORK.
Thermodynamic work is hereby defined as the integral of a force and
the displacement of its point of application. Define W_thermo for an
object as the sum of these integrals over all forces acting on the
object. Note that we cannot relate this to the net force nor to the
displacement of the cm in general, although in some useful special
cases we can.
Level three = we break the system down to its ultimate microscopic
particles. (Whether this means small chunks, atoms, or nuclear
particles is determined by the requirement that all relevant energy
exchanges have been accounted for, ie. there's no need to include
nuclear particles if the nuclear degrees of freedom are not being
accessed.) I am going to boldly make three claims at this level:
(i) There is no heat transfer. All energy exchanges are ultimately
mediated by forces between particles. I am here trying to answer to
John D's and Jim G's desires to avoid any Q term. (Maybe I'll have
more to say after tracking down the Barrow article.) So I'll write
the first law of thermo as W_thermo = delta E_mech + delta E_int.
Here I've arbitrarily carved out some of the energy and called it
mechanical. This includes K and U of the level two parts, and at the
risk of further confusion, nothing stops me from re-defining K
*again* and including the rotational energies of the parts too. (I
can do this at level two also, by calculating the works done by the
torques.) After all, this is what we usually do in mechanics once we
hit the rotations chapter. (Vibrational kinetic energy is a bit
troublesome because I don't know whether a spring is a part, or a sum
of parts. Let me ignore that issue for now and return to it in a
later post.)
(ii) For every force at the particle level, W_cm = W_thermo, because
by definition my particles have no relevant internal structure. I
*here* make contact between the work-energy theorem and the first law
of thermo. Note that when treating say Atwood's machine, the
particles are the blocks, but when treating sliding friction of a
block, the particles are the asperities, surface layers, and the
remaining bulk (or something vaguely like that - subdivide further if
you think that's necessary, I won't quibble over details).
(iii) All inter-particle forces are conservative, so the leftover
internal energy E_int is a sum of kinetic and potential energies of
all the microscopic particles.
If you just want to do mechanics and never discuss heat, internal
energy, etc you can stop at level two. You will be able to solve all
problems just fine using the work-energy theorem. Don't listen to any
naysayers who claim you can't do work by friction, skaters pushing
off walls, etc. You can do them all fine - just remember to calculate
the cm work not the thermo work!!!
However, if you want to go on to discuss questions like what happened
to the mechanical energy dissipated by friction, then you need to
introduce the first law. You will not be able to do this in any
problem where it matters (egs. sliding friction, ball of clay falling
to the floor, adiabatic compression of an ideal gas, etc) unless you
are cognizant of *some* microscopic issues. This doesn't mean on
first encounter that you have to become an expert on quantum
mechanics or something! But you are going to need to know a few
things about internal energy of an ideal gas, asperities on the
bottom of a block, etc to be able to go beyond merely stating the
obvious and rather trite fact that "all the lost mechanical energy
ends up as internal energy."
Okay, fire away again. I'm particularly interested in level three
because I think levels one and two are clear. (Feel free to disagree
if you have additional insights about them however.)
--
Carl E. Mungan, Asst. Prof. of Physics 410-293-6680 (O) -3729 (F)
U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5026
mungan@usna.edu http://physics.usna.edu/physics/faculty/mungan/