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Re: Newton's second law



Justin wrote:

F = dp/dt= mdv/dt + vdm/dt

In the third term, what is v?

This is a more interesting question than some might think.

Newton's second law says that the net external force on "a system" is
equal to the rate of change of momentum of "the system." (Actually,
Newton's form of "Newton's second law" says that the net external
impulse--in his words, "action"--on a system is equal to the change
of momentum--in his words, "motion"--of the system. But that's
another story!)

In Newtonian physics, the only way for the momentum of "a system" to
change, is for the velocities of its components to change. If the
mass of a system changes, then the system itself has changed and you
have violated the conditions for applying Newton's second law. As a
result it makes no sense to treat "mv" by the product rule and to
write d/dt(mv) = m(dv/dt) + v(dm/dt) except insofar as you require
that dm/dt = 0.

We get around this formal limitation and solve problems involving
"variable mass systems" (e.g., rocket problems) by applying Newton's
second law in differential form to an infinitesimal period of time
during which an infinitesimal part of the system with mass dM
"discontinuously" changes its velocity by an amount u while the bulk
of the system with mass m - dM changes its velocity by an
infinitesimal amount dv.

F_ext dt = dp = dp1 + dp2 = dM(u) + (m-dM)dv

where all F's, p's, v's, and u's are to be taken as vectors.

Ignoring the doubly infinitesimal quantity we find

F_ext dt = dM u + m dv

or

F_ext - u(dM/dt) = m(dv/dt)

We then sometimes *call* the second term on the left "the thrust" and
note that in cases where the system of interest is shedding mass,
dM/dt = -dm/dt. This allows us to write the last equation in terms
of m and dm/dt, but we should be under no illusions that the dm/dt
came from taking a time derivative of mv.

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