Re: ENERGY WITH Q

["John S. Denker" <jsd@MONMOUTH.COM>]

Eugene Mosca wrote:
> It seems to me this involves an understanding of only the center-of-mass
> work-translational kinetic energy relation.  Am I missing something?

What relation is that?  Perhaps _I_ am missing something.
Did somebody prove a CM-work / translational-KE theorem
when I wasn't looking?  JM posted an excellent puzzle
  http://mailgate.nau.edu/cgi-bin/wa?A2=ind0111&L=phys-l&P=R23484
and to solve it, people seem to be applying some such
"relation" with gay abandon, without having verified
that there is any such relation.

In particular, let's approach this puzzle using standard
elementary physics methods.  In particular, since fluid
dynamics is a rather advanced subject, let us replace
the contents with a discrete mass on a spring with damping:

         ______________________________________
        |                                  |   |
        |                       ___        | m |
        |                      [   ]       |___|
        |--------spring--------[ m ]           |
        |   |             |    [___]           |
        |   |---dashpot---|                    |
        |                                      |
        |______________________________________|


Notation:
        F := applied force = given = constant
        x1 := position of container; WLoG x1 = 0 at t=0
        x2 := position of contents;  WLoG x2 = 0 at t=0
        X := position of center of mass
        L := length of container = given
        m := mass of container = given
        m also = mass of contents = given
        k := spring constant = not specified
        gamma := damping constant = not specified
        (something)' := time-derivative of (something)
        "WLoG" means "Without Loss of Generality".

Since k and gamma were not specified I reserve the right
to choose them myself, later.

The basic equations of motion are:
        F = m x1'' + k(x1 - x2) + gamma(x1' - x2')
        m x2'' = k(x1 - x2) + gamma(x1' - x2')
whence one can calculate
        F = m x1'' + m x2''     [3]
which is extremely unsurprising, and could have been
asserted independently (using conservation of momentum).

We are told that at some time t1 it is observed that
        (x1 - x2) = L/4         [4]
and
        (x1' - x2') = 0
so at this special time we can define v to be
        v = x1' = x2' = (x1' + x2')/2
and by integrating equation [3] we get
        v = F/m t               [7]

We are asked to calculate the CM position (x1+x2)/2
which by equation [4] is just
        X = x1 + L/8.           [8]

Clearly
        x1 = integral of x1' dt

We can write this in terms of the natural center
of mass coordinate (x1+x2)/2 and the internal
relative coordinate (x1-x2), to wit:
        x1 =   1/2 integral of (x1' + x2') dt
             + 1/2 integral of (x1' - x2') dt

Hence
        x1 =   1/2 integral integral F/m dt dt
             + 1/2 integral (x1' - x2') dt

        x1 =   1/4 F/m t^2
             + 1/2 integral (x1' - x2') dt

and by plugging in equation [7] and equation [8]
we see
        X =   L/8
            + m/F v^2
            + 1/2 integral (x1' - x2') dt

Now the last term is !!not!! guaranteed to be zero.
It is an oscillatory function.  In fact, it is
practically guaranteed to be nonzero (even at the
special point where the integrand happens to be zero,
as specified in this case).

Therefore I suggest that those who have "proved" that
X = L/8 + m/F V^2 have some explaining to do.


On Sat, 01 Dec 2001 06:31:25 -0700 John Mallinckrodt wrote:

> I believe it does at least point out the importance of
> the "center of mass work" (i.e., "pseudowork") concept.

I believe a correct analysis of this fine puzzle shows
the pseudowork concept to be useless or worse.