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Re: Problem



At 12:47 -0700 9/18/01, John Mallinckrodt wrote:

Yes, the speed is "a function of time" and is, therefore, "not
constant", but there can still be *instants* of time *at* which
its time derivative is zero and, therefore, *at* which the speed
is "not changing."

OK. I went back and did the arithmetic very carefully and I see that
if the launch angle is anything but 90 degrees that the rate of
change of the speed is 0 when t=v(0)/sin(theta). But when either of
the follwoing conditions is met, theta=90 degrees, or theta=0 and
v(0)=0 (i.e., a free fall drop from rest), the rate of change of
speed is -g, always. And if I calculate the acceleration from the
velocity, it is always -g, irrespective of the launch angle. In other
words, for a vertical trajectory the rate of schange of speed is
always -g, but once the trajectory has any angle at all different
from 90 degrees, the rate of change of speed is 0 instantaneously at
the top of the trajectory.

That's what the math says, unless I have set the problem up wrong.
But that doesn't square with the physics as I know it. Why should the
presence of a rate of change of speed at the top of the trajectory be
dependent on whether the trajectory was vertical or not? This says
that different inertial observers would see different rate of change
of speed at the top depending on whether or not their speed relative
to the projectile was v(0)cos(theta) or not. That makes no sense to
me at all.

For the record, in UCM, this situation does not arise at all. The
math gives a constant speed and from the velocity equation we find
the acceleration is just v^2/r, or the standard centripetal
acceleration. The results of doing the problem by speed and by
vectors are entirely consistent.

As I think more about this, perhaps there is something to it. The
acceleration is constant, independent of launch angle, as can be seen
by differentiating the velocity expression. But rate of change of
speed is not necessarily acceleration (although in one dimensional
problems it is the same thing). My first clue to this led me to go
back and change all the references to acceleration associated with
speed to "rate of change of speed," to remind me that they are not
necessarily the same thing. Perhaps a simple-minded way to think
about it is this: acceleration changes either speed or direction or,
most generally, both. In a trajectory problem, most of the time it is
changing both, in different proportions. On the way up, it is slowing
the object down and changing it's direction, at the top is is
changing only the direction (this being true in the vertical case, as
well, where the direction changes from "up" to "down"), and on the
way down the process is reversed. At the time that the speed change
has gone to zero, the angular rate of change is greatest (the largest
curvature of a parabola is at its apex). Where the rate of change of
speed is greatest, the rate of change of angle is least, and vice
versa, leading to, in the trajectory case, a constant overall
acceleration.

OK, I think I am now convinced that in the problem originally posed,
the answer at all times is that the rate of change of speed is zero.
And my confusion is grounded in mixing the rate of change of speed
with acceleration.

Maybe we should be teaching all our students quantum mechanics,
instead of basic physics. Some of the knottiest problems I have run
into are at the very basic levels, and after 45 years in the game, I
still manage to surprise myself occasionally with what I thought I
understood, but really didn't. I've done the math for tangential
acceleration and centripetal acceleration in generalized curvalinear
motion many times, but this reminds me that I sure didn't understand
it nearly as well as I thought I did. And I doubt that the students I
taught this stuff to over those years went out of my class with any
better understanding, either, although I hope they got their
misperceptions corrected before I did. I suspect a worthwhile problem
for students at the intro. calculus physics level, a really
interesting problem would be to calculate the tangential and
centripetal accelerations for a parabolic trajectory in a
graviational field and show that the total acceleration is always
vertical and equal to "g."

And we worry about the students not being able to figure out what
questions to ask.

Hugh
--

Hugh Haskell
<mailto://haskell@ncssm.edu>
<mailto://hhaskell@mindspring.com>

(919) 467-7610

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