Re: Newton's 2nd Law problem (resent)

[Leigh Palmer <palmer@SFU.CA>]

> To me, if an object has a mass which is increasing with time, the force to
> maintain its constant vel can be found from F=vdm/dt.
> So for the rotor blade, I don't quite see why dm/dt is the rate of mass flow
> through the rotor blades. However, in mathematical derivation, F=mdv/dt +
> v dm/dt,
> v of the object is constant in the second term (vdm/dt).
> Leigh, I can see your point in starting from first principle (i'm actually
> teaching my students to solve from first principle, i.e considering the rate
> of change of momentum (impulse-momentum form)), but I still can't reconcil
> with using vdm/dt. All the texts are using this method (find F from mdv/dt)
> without starting from first principle. Another classic example is to find
> the extra force to maintain constant vel of the conveyer belt if luggage is
> dropping on it at a constant rate (but the luggage itself increases vel from
> 0 to that of the belt).
> Leigh, could you help me to resolve this confusion, 'cause some students are
> asking me, and I think I can't say the approach of using the formulae of
> F=vdm/dt is WRONG, particularly if it still gives the correct answer?

I had really hoped someone else would answer Rom's question. Here I go:

I can only retreat to my, um, fundamentalist position. Have your student
draw a free body diagram, however trivial, identifying clearly what the
mass in question is. In Newtonian mechanics masses don't vary in time;
mass is locally conserved. A free body diagram in the case in which you
are interested may involve a small mass which is undergoing a change in
its state of motion. You may require more than one free body diagram to
specify the process if the interaction involves forces exchanged between
objects which have their states of motion changing, for example sand
poured on a conveyor belt which is not driven. Analysis and decomposition
of such systems ought to be a standard procedure. When all of this is
done one should be able to find N equations in N unknowns from the free
body diagrams. If this cannot be done either the problem was not well
posed or the analysis is incomplete. If there are more equations than
unknowns then either one or more of the equations is redundant (linearly
dependent in some simple cases) or else the problem is overdetermined in
its specification. Equations of constraint will often be needed to
complement the dynamical equations.

Once the desired balance between unknowns and equations is achieved the
"arithmetic" process starts. With luck the arithmetic will yield the
desired answer or answers. Some sets of equations are too difficult to
solve. For example I can't solve the equations of motion for the links
in a falling chain given a complete specification of its initial state
of rest, but I can construct the equations of motion by the process
described above. [It is much easier to treat this problem by a variant
of Newtonian mechanics (Lagrangian mechanics) but it is no easier to
solve the resulting equations.]

Now here's the point of all this. During the process of doing the
arithmetic on the equations it may happen that an equation which looks
like F = v dm/dt occurs. That is fine; simply continue the arithmetic
process to its conclusion. It may happen that in every conveyor belt or
rocket problem you analyze such an equation turns up at some
intermediate step. Very interesting; we may be able to make use of that
information in some way. That's fine, but what I think we must *never*
do as physics teachers is to say "Here is the rocket and conveyor belt
equation." Of course that is a useful thing to teach to, say, rocket
and conveyor belt engineers together with, of course, a careful recipe
for how it is to be applied. It is not my kind of physics, however.

A longer answer than was sought, I'm sure, but that's my opinion.