Re: centrifuge exercises

[John Denker <jsd@MONMOUTH.COM>]

At 11:30 PM 11/5/00 -0500, Ludwik Kowalski wrote:

> I calculated terminal velocity corresponding to the constant artificial
> gravity and used it to calculated the fall time.

OK.

> This is perhaps oversimplified, but it is better than ignoring
> water resistance.

That's for sure!

> I will check if the "averaging approach" is acceptable with a computer
> program (Euler integration).

Why not do the integration analytically?

Hint:  Find the time required to fall from r1 to r2
by integrating the reciprocal velocity,
(rather than finding the distance travelled between t1 and t2
by integrating the velocity).

Huge hint:  When I do the calculation, I get a result that is not only
exact, but elegant and easy to remember.  I find that the average velocity
        (r2-r1) / (t2-t1)
depends on the "average radius" but the "average radius" in question is not
(r2+r1)/2.  What sort of "average radius" am I talking about?

> But this will not tell me if Stoke's formula is
> valid for v as large as 0.014 m/s (r=1e-6) and above.

True, it won't.

But don't worry about it.  Stokes's formula should be reasonably close for
any velocity small compared to the speed of sound.  Think about the
physics!  Viscosity is due to collisions with solvent particles.

Do you read German?  There is a really cool paper on the subject.  It was
published a while ago (1905), but any university librarian should be able
to get a copy with no trouble.

=================

Also:  Based on further perusal of the references, I now believe that for a
typical bio-lab centrifuge, the sample-tube is something like 8cm in the
relevant direction, spun with its center 15cm from the spindle.  In the
common "preparatory" centrifuge the tube is held at a fixed angle to the
vertical;  this makes it utterly miserable to analyze, and inappropriate
for an intro physics course.  For intro exercises, it would be better to
think in terms of an "analytical" centrifuge which uses a _swinging bucket_
so that the axis of the tube is always aligned with the acceleration field.

Note that this 8cm differs quite a lot from the 1cm alleged in my previous
notes.  Sorry.