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Re: ENERGY WITH Q

On Sat, 1 Dec 2001, John S. Denker wrote:

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?

Gene is referring to the simple relationship obtained by
integrating Newton's second law which, when applied to a complex
system, says

net force on system = system mass * acceleration of system CM

The result is a simple and highly useful relationship between what
some of us like to call "pseudowork" or "center of mass work" and
the change in the bulk kinetic energy of the system. Because so
many people seem to get *so* exercised by the name "pseudowork"
I'll give the result first avoiding any jargon:

The integral of the net force acting on a system dotted with the
displacement of the system center of mass is equal to 1/2 the system
mass times the change in the square of its translational speed.

When I (and I dare say when Gene M., Carl M., Joel R., and Bob S.)
say something like "pseudowork = delta(K_trans)" or "center of
mass work equals the change in the bulk translational kinetic
energy" *this* is the relationship we are referring to.

Perhaps _I_ am missing something.
Did somebody prove a CM-work / translational-KE theorem
when I wasn't looking?

No, it has been proven in *lots* of places. You can prove it
yourself in about one line. It is by far the simplest of the many
possible "work-energy type" relationships.

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".

Great! This is a completely acceptable alternate specification of
the problem which will yield the same fundamental physics.

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

Fine, as long as it is possible at some later instant of time for
the mass on the spring to reach a position that is L/4 from the
"back end" of the cylinder and to be moving at the same speed as
the cylinder as was the case in my problem.

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).

Correct. This is the same as saying

F = (2m)X'' [3_alt]

It is Newton's Second Law for this "complex" system.

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

Correct to here.

v = F/m t [7]

No. I get

v = F/2m t [7_corr]

(Note, it's far easier to use the impulse-momentum theorem
which says Ft = 2mv.)

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

No. I get

X = x1 - L/8 [8_corr]

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

O.K.

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

No. You seem to have made a cancelling error in plugging in the
result of your equation [8] and then discarding the factor of 1/4.
At any rate, using the corrected equations [7_corr] and [8_corr],
I get

X = -L/8
+ m/F v^2
+ 1/2 integral (x1' - x2') dt

Now the last term is !!not!! guaranteed to be zero.

Indeed, we can easily perform the integral and find that

1/2 integral (x1' - x2') dt = 1/2 (x1 - x2)
= 1/2 (L/4)
= L/8

Thus,

X = mv^2/F

a result that is FAR more simply arrived at using the highly
useful concept of "pseudowork" and its easily proven relationship
to the change in bulk translational kinetic energy. We get

FX = mv^2

in agreement with the corrected result of the prior lengthy
analysis.

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.

I hope my explanation has been satisfactory.

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.

On the contrary, I think this exercise has now served to fully
demonstrate the usefulness of the concept. By allowing us to get
to the desired result so quickly, it radically reduces the chance
of algebraic error.

John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm