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Brian W wrote:
>>Not much air resistance in an oil bath, I would have thought, Jack.
>>Could it be that it makes the sums work a little better?
In fact the drag force on a "slowly moving" sphere is proportional to
velocity and not the square of velocity. This applies for motions in which
the Reynolds number (i.e. 2xradiusxvelocity/kinematic viscosity)is small.
For a sphere of one mm radius, moving in water, the velocity must be less
than 0.2 cm/s. See, for example, L.M. Milne-Thomson, "Theoretical
Hydrodynamics", fifth edition, 1968, pp 681,682.
Don Polvani
-----Original Message-----
From: Brian Whatcott [mailto:inet@INTELLISYS.NET]
Sent: Tuesday, November 27, 2001 1:46 PM
To: PHYS-L@lists.nau.edu
Subject: Re: MATH PHYSICS
Brian W
At 09:15 AM 11/27/01, you wrote:
> You're missing low velocity viscous drag, as used in analyzing the
>Millikan oil-drop experiment. See chapter 12 of "The Mechanical Universe"
>(standard edition).
> Regards,
> Jack
>
>On Mon, 26 Nov 2001, Brian Whatcott wrote:
>
> > >David Abineri wrote:
> > > >
> > > > If one assumes that a projectile encounters an air resistance
> > > > proportional to velocity, one can write a differential equation like
> > > > mr''=-mgj - kr' which can be solved for r using an integrating
factor
> > > > e^(kt/m).
> > > >
> > > > The final solution for r, however, does not admit an interpretation
for
> > > > k=0. Why is it that one does not get the ideal case to come from
this
> > > > more general case when k=0?
> > > >
> > > > I hope that the question makes sense.
> >
> >
> > Not to me. The assumption is unphysical, in my view.
> > Cd is proportional to V^2 ??
> > What am I missing, would you say?