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Re: 2nd Work-Energy Relation




Al wrote:
* * *
(1) Newton's Second Law: Mi dVi = Fei dt + Sj Fji dt
for a collection of particles, where
Fei is the external force acting on particle i,
Sj is summation over index j (usually symbol Sigma),
Fji is the force ON particle i, due to particle j
Now define: M = Si Mi , M Xcm = Si Mi Xi , Vcm = dXcm/dt
Ri = Xi - Xcm, vi = dRi/dt
Multiplying (1) by Vcm and summing over i,
we obtain: Si Mi Vcm dVcm = Si Fei dXcm + Si Sj Fji dXcm
(2) which yields: dK = Fext dXcm
where Fext = Si Fei, and Si Sj Fji = 0 by Newton's 3rd Law.

The second Work-Energy relation is obtained by multiplying (1) by vi,
and summing: Si Mi vi dvi = Si Fei dRi + Si Sj Fji dRi
(3) which can be written as: dk = Si Fei dRi - dU,
where k is the kinetic energy associated with motions relative to the CM,
and where the internal forces are associated with changes in potential
energies that are functions of the relative positions (Ri-Rj).

Several special cases are clarified by particular forms of (2) and (3).
When the external force is exerted at a single point, Fext = Fo.
dK = Fo dXcm and dk = Fo dRo - dU
A. If point o actually moves, dXo n.e. 0, then Fo does work on the system.
dK = Fo dXo - Fo dRo and dk = +Fo dRo - dU
This more clearly shows the partition of the work done by Fo between
'internal' and 'external' modes and highlights the role of deformations.

B. If point o does NOT move, dXo = 0, and Fo does no work on the system:
dK = -Fo dRo and dk = +Fo dR - dU
Now energy may still be transferred from internal to external modes
through the presence of Fo AND the deformations in the system.

C. We can also formulate these relations to eliminate Fo when particle o
remains stationary since Fo + Sj Fjo = 0 (ie the external force
balances the internal forces acting ON particle o.)
dK = Sj Fjo dRo and dk = -Sj Fjo dRo - dU = Sj Si>o Fji dRi
We read these as showing that the internal forces acting at point o can
accelerate the entire body, while the REMAINING internal forces change the
energy internal to the body, AS THE BODY IS DEFORMED.
* * *

Al,
When translated into vector notation (eg "multipling" , above, means
the dot product of two vectors), this is a very lucid exposition of
what is going on. I am grateful for it. But I still object to the
semantic penchant of making a litmus test for doing work out of the
non-zero motion of the "point of application" of a force. Except in
such "contact force" models, the term is murky and would not enter
into your equations. (It really is a nice analysis!)

Bob Sciamanda sciamanda@edinboro.edu
Dept of Physics
Edinboro Univ of PA http://www.edinboro.edu/~sciamanda/home.html