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Re: The drag force -- a correction squared



Dear Justin-
Well, yes and no.
The original problem, as stated in your previous post, gives
the equation of motion that is applicable to the Millikan oil-drop
experiment. If you want to talk about forces and explicitly write
the coeficient of v as proportional to the mass, then the mass
cancels out of the acceleration equation: dv/dt = -g - kv. The
solution is v(t)=-V(1 - exp(-kt)), where V is the terminal velocity,
and is equal to g/k. The quantity V is measurable for any given
oil drop (the time to reach terminal velocity is negligible), so
in this sense it was reasonable to write the force as proportional
to m because you don't need to know m to do the experiment. The quantity
m cancels out of the kinematic equation.
If, on the other hand, you try to measure the quantity k for a
variety of oil drops, you will find that k depends on the ratio of
the drop's radius to its mass. That is, k is inversely proportional
to the product of the oil density times the surface area of the drop.
You will find all this discussed in <The Mechanical Universe> by
Olenick, et al., Chapter 12 on the oil-drop experiment.
Moral: please provide the full context when quoting others.
Regards,
Jack

Adam was by constitution and proclivity a scientist; I was the same, and
we loved to call ourselves by that great name...Our first memorable
scientific discovery was the law that water and like fluids run downhill,
not up.
Mark Twain, <Extract from Eve's Autobiography>

On Sun, 8 Oct 2000, Justin Parke wrote:

It seems to me that the drag force is in fact not dependent on the mass,
rather it is the terminal velocity which is mass dependent. Consider an
object with small mass provided with some means of propulsion down, a small
rocket fired from a height toward the earth, for instance. In this case we
can make the drag force as large as we wish by simply increasing the speed of
the rocket. Mass has nothing to do with it. It just so happens that in the
case of free-fall the drag force ceases to increase when terminal velocity is
reached since the particle is now in equilibrium. We have reason to expect
this because of Newton's laws. There is no reason to expect, however, that
in general the drag force should be mass dependent.

For that reason it seems incorrect to say that the drag force is mass
dependent.

As far as examples which include a mass dependent resistive force to simplify
the math with the understanding that the student "knows better," if the
student knows better then they should not need the simplification in the
first place! (IMHO)

Under the circumstances, I would be interested to hear
from Justin as to whether or not he has yet had his question
answered with sufficient directness and, if so, by whom.

Unfortunately (or probably for the better, to be honest) I do not remember
what was said by whom. I have learned not to expect a simple answer though!

Thanks for all the input (and I will still continue to post even though I am
"only" a high school teacher)!

Justin Parke
Oakland Mills High School