Re: first law of thermo

["Carl E. Mungan" <mungan@USNA.EDU>]

>  > Free-body diagram. The net force on the block is -umg. This equals
> >  ma. Thus a = -ug which is a constant. Thus we can use the equation of
> >  kinematics, v^2 = v0^2 + 2ax. But v = 0.
>
> I recognize that as an equation from point-particle
> kinematics.  If you claim it applies to objects with
> nontrivial internal structure, please provide a
> derivation or a pointer to a derivation.

Which equation are you referring to when you say "that"? Everything
above holds for a block with internal structure.

>  >  We insist there is a logically sound theorem
> >  applicable to the block as a whole.

Consider a complex object of total mass m. Let a = dv/dt =d^2 r/dt^2
be the acceleration (v = velocity, r = position) of its center of
mass (measured in an inertial frame). Let F be the sum of all forces
acting on the object. (The internal forces will of course cancel.)
Then

F = m dv/dt => F dot dr = m v dot dv

Integrate both sides to get desired result W = delta(K). Add
subscript "com" to W and subscript "TR" to K if you prefer (but you
said this wasn't an argument over nomenclature).

> Hint:  I've got a pocketful of counterexamples, so
> any such derivation is going to be quite a rare bird.

Fire away.

> Since when is the definition of kinetic energy a
> matter of personal preference?

K = mv^2/2 using the above symbols. I believe this follows
conventional notation.

> The attempted
> calculation in the previous note miscalculated
> the kinetic energy by roughly a factor of two.

Show me.

> Step 1:  Impart energy to the object while it is disconnected
> from the table.  In particular, set it spinning, so that
> there is
>  -- rotational kinetic energy in the blocks, plus
>  -- potential energy in the spring.
>
> Note:  The energy is not imparted to the center-of-mass
> motion of the object.  This emphasizes the pedagogical point
> that there is nothing special about the center-of-mass mode.
> I can choose any mode that has lots of energy but low
> entropy.

Fine so far.

> Note:  I have imparted energy while the object is disconnected
> from the table, so that the blocks can be treated as pointlike
> (no relevant internal structure) to an adequate approximation
> _during this step_ of the operation.

Okay with me, but maybe not for the same reasons as you. Read on.

> Step 2:  Let the spinning object touch the table.  It will
> spin around for a while and dissipate its energy as heat.

I understand what you mean, but I personally would look at it
differently. First of all, I wouldn't use the word "heat" at all.
Here are two views I would take to learn different things about the
energy accounting:

VIEWPOINT 1: VIEW THE ENTIRE SYSTEM (two blocks, spring, and table)
AS ONE BULK WHOLE.
What I'd say is that the interaction with the table has converted
*one form* of internal energy (namely macroscopic rotational and
vibrational) to *another form* of internal energy (namely microscopic
kinetic and potential). Bulk energy refers purely to translational KE
of the system's com BECAUSE OF OUR VIEWPOINT. This is zero both
before and after the interaction, so W-K correctly predicts 0=0.
Also, there are no external forces on the system, so FLT correctly
predicts 0=0. This view is thus not very helpful. Let's go on.

VIEWPOINT 2: BREAK THE SYSTEM INTO MACROSCOPIC PARTS.
Suppose we're not happy with calling the bulk rotational and
vibrational "internal energy". Fine, I can break the system into
three parts: the two masses and the table. If you will permit me, I
will take the spring to be massless (ideal) so it doesn't count as a
part. In that case, we have mechanical energy = sum of K and U for
the parts = mv^2 + kx^2/2 where I assumed for simplicity that each
block had the same mass m and speed v and the "centrifugal" stretch
of the spring is x. And we have initial internal energy = E1 (which
reflects the temperatures and phases of the blocks and table). W_ext
= delta (E_mech) + delta (E_int) now predicts 0 = -mv^2 - kx^2/2 + E2
- E1 where E2 is the final internal energy and where I assumed spring
relaxes fully. (It's not actually obvious that this will happen, as
friction may hold the spring slightly stretched.) This matches your
solution.

> Note:  I have not said anything about the details of the
> frictional process.

But we can learn more. Friction due to the table does com work on
each block and W-K predicts umgd = mv^2/2 (ignoring the relaxation of
the spring) where d will tell me the number of revolutions the blocks
make before coming to rest. You didn't ask about this, but it is an
interesting aspect of the problem that I would certainly want to
treat in a mechanics course.

> I have not calculated the transfer of
> anything across the block/table boundary, not Q, not W.

But that's because you only considered the system as a whole. Suppose
I wanted to know how much the blocks and table separately warm up
*and* suppose I insert a thermal barrier (somehow without altering
the frictional force!) between the blocks and table. (One approximate
way to realize this would be to assume the blocks have very high
thermal conductivities, large masses, and that I lift the object off
the table at the instant it stops. I'm taking advantage of the small
temperature rise of the blocks and slow timescale for subsequent
thermal conduction to the table compared to the time to come to
rest.) Carl
--
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/