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Re: Problem: acceleration perpendicular to velocity



At 05:58 PM 9/18/01 -0400, Hugh Haskell wrote:
And if I calculate the acceleration from the
velocity, it is always -g, irrespective of the launch angle.

Right. That's the velocity.

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.

Right. Speed is not the same as velocity.

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.

Ho, ho, ho. We are in rather deep water here. Paddle vigorously, but
don't rock the boat.

The norm of a 2-dimensional vector is
|v| = sqrt(vx^2 + vy^2)

Now in physics we are accustomed to making first-order approximations to
everything in sight. We are accustomed to taking derivatives. Unless it
is an unphysical situation, the derivative has to exist, right? The Taylor
series has to exist, and to converge, right?

Wrong!!!!!!!!

The sqrt() function does not have a Taylor series if we expand it around
the place where its argument goes to zero. The first derivative doesn't
exist. Your intuition about first-order expansions is going to get you
nowhere.

The situation is no better in D=1. The absolute value function doesn't
have a derivative at the origin, either.

In the vertical launch scenario, the rate-of-change of speed is -g on the
way up and +g on the way down. Right at the turn-around point, it is
UNDEFINED. The derivative doesn't exist. You can define it by fiat if you
like; any value in the interval [-g, +g] is plausible, and there are
several reasons for preferring the zero value.

One time-honored trick for dealing with nasty situations of this sort is to
redefine
|V| = sqrt(vx^2 + vy^2 + epsilon^2)
figure out what's going on, and then see what happens if you let epsilon go
to zero. In this case the trick does the right thing; it coughs up the value
(d/dt)speed = 0
at the turnaround point, as desired.

We need this trick because the definition of speed is a bit hokey, a bit
unphysical. It doesn't usually cause trouble except near speed=0. The
other reason it doesn't usually cause trouble is that most important
physical laws are stated in terms of velocity, not speed.

But be warned: there are situations in physics where the derivatives do
not exist. There are situations where this reflects the real physics, and
has nothing to do with hokey definitions. Critical phenomena (critical
exponents) are a prime example. Do not assume everything of interest is
differentiable!

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.

That's an incisive analysis. The epsilon-trick passes the test. All
observers will agree that (d/dt)speed=0 at the turnaround point.

If the original problem had asked us to calculate the rate-of-change of the
ENERGY (as opposed to the speed), there would have been no nasty sqrt(0)
problems deal with. The designers of the question probably thought they
were safe from sqrt(0) problems because they specified a nonzero horizontal
velocity-component at the turnaround point. They didn't anticipate that
the customers (us) would be adventurous enough to shift into an observer
frame where the horizontal velocity-component (and therefore the total
speed) went to zero.