Re: Car acceleration

["Carl E. Mungan" <mungan@USNA.EDU>]

John D, thanks for your reply, always appreciated:

> The key concept hear is _momentum_.  Momentum is
> transferred across the aforementioned system
> boundary.  Students can easily enough calculate
> the effect this momentum-change has on the
> energy budget...

Yes they can. Or they can do a Newton's second law analysis. Or they
can do a pseudowork analysis. All amount to *exactly* the same thing.
But sometimes it's helpful to have different viewpoints. I can
calculate the electric field of a sphere by integrating Coulomb's law
or by applying Gauss' law. Again, they amount to the same thing (in
this context!) but just as I hate to be told I have to choose my
system one way, so I would hate to be told I have to analyze it via
momentum transfer. (I realize you're *not* saying this. But again, I
view it as another tool in my box, ideally useful for what it's
designed for.)

> In the case of the car, there
> will be real genuine KE associated with the rotation
> of the tires, engine machinery et cetera, which
> cannot be even guessed at based on pseudowork
> arguments.

Nor via momentum transfer. But the beauty of pseudowork is it *picks*
out for us exactly the part of the KE we *are* interested in: namely
the bulk translational KE of the car. When it works (and I keep
trying to shoot problems at the list to show that that is quite
often), it is usually the simplest (ie. most intuitive to students)
method because it zeroes in on the essentials of the *bulk mechanics*.

>  > If you additionally assume F is constant, this agrees with an
> >  analysis using Newton's second law plus the equations of kinematics
> >  for constant acceleration.
>
> But what if it isn't?  What about my punter who
> only intermittently pushes on the pole?
> (This is a minor gripe, not much more than a
> nitpick, so don't feel obliged to respond.  The
> rotational KE issue is incomparably more profound.)

I can tell from your parenthetical remark that you know the answer:
integrate Fdx = d(mv^2/2) assuming you know or can figure out F(x) or
F(t).

>  > (just tooting the "pseudowork" horn again)
>
> Not music to my ears :-)

I appreciate the smiley, because I take it (wishfully?) as evidence
that while you still don't want to use pseudowork, you're at least
less hostile to it. I hope John M or someone will find the "killer
app" to convince you yet :-)) Carl
--
Carl E. Mungan, Asst. Prof. of Physics  410-293-6680 (O) -3729 (F)
      U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5026
mungan@usna.edu    http://physics.usna.edu/physics/faculty/mungan/