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Re: [Phys-l] Linear Air Drag



A triumphant example of a Reductio argument, if I may say....


Brian W

curtis osterhoudt wrote:
Probably not. Given a ball bearing with a density very close to the shampoo (or, indeed, a spherical bubble), it won't descend at all.
/************************************
Down with categorical imperative!
flutzpah@yahoo.com
************************************/




________________________________
From: Brian Whatcott <betwys1@sbcglobal.net>
To: Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu>
Sent: Wed, November 4, 2009 4:07:53 AM
Subject: Re: [Phys-l] Linear Air Drag

Just so we are all singing from the same hymn sheet, I believe that Bob is asserting that given two steel ball bearings of the same diameter, one weighing half the other,
they both descend a shampoo column at the same rate. Is that it, Bob?

Brian

LaMontagne, Bob wrote:
Yes, the buoyant force is about 1/7 to 1/8 of the weight of the steel balls and scales with r^3 as well. This still gives a terminal speed proportional to r^2 which is what my students verify by plotting. Even though the shampoo is very viscous it has a density close to that of water.

Bob at PC

________________________________________
From: phys-l-bounces@carnot.physics.buffalo.edu [phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Ken Fox [physicsfox23@gmail.com]
Sent: Tuesday, November 03, 2009 7:47 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Linear Air Drag

I hesitiate to enter here, but I wonder whether the upward force should have
a buoyancy term as well. I am probably missing something. It would be
constant for each ball but not a general constant.

Ken Fox

On Tue, Nov 3, 2009 at 1:51 PM, LaMontagne, Bob <RLAMONT@providence.edu>wrote:

Brian,

I'm not sure what "generally" means in this case. I can email you the data.
The terminal velocity is directly proportional to r^2 for the series of
steel balls. The data form an amazingly straight line of slope 2 on a
log-log plot. If you doubt the results I suggest that you try it - it's a
fun experiment. The balls can be bought from a couple of different ball
bearing suppliers online - and all have the same density.

The resistive force can be expanded into two physically significant terms:
viscous and dynamic. The form becomes (for a sphere):

R = 6 Pi Eta r v + 0.5 Cd Rho Pi r^2 v^2

Where Eta is the viscosity of the fluid
Cd is the drag coefficient of the sphere (usually 0.5 but varies with
smoothness)
Rho is the density of the fluid

For the mini lab I do with the Pantene shampoo, the first term is many
orders of magnitude more important than the second term (as you suggest). I
also do a fun experiment where the students drop an air filled weather
balloon off the roof of our building. The second term now dominates by many
orders of magnitude. We predict the time of ball based on the weight of the
ball and the resistive force (terminal speed) and measure it. The results
are usually quite good.

These are cheap experiments/demos that produce good results and are very
repeatable. They also can be easily modeled with spreadsheet programs.


Bob at PC

-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:
phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Brian Whatcott
Sent: Tuesday, November 03, 2009 1:02 PM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Linear Air Drag

The salt-thickened synthetic detergent sold as hair shampoo is quite
viscous no doubt,
so that the usual aero formula for drag loses applicability: Force
(drag) = Cd. A. rho, V^2
wherein the Cd coefficient of drag subsumes the variation in Reynolds
number
for a customary range of fluids. (which reminds me that the
comparable equation for lift
is a source of confusion when the representative length may be taken as
an aerofoil
thickness, rather than chord...)

I don't think it's generally true however, that large balls drop faster
in viscous liquids,
though denser or heavier balls would, wouldn't they?

Brian W

LaMontagne, Bob wrote:
One of the in-class mini-labs that I do with my General Physics students
here at Providence College is to have them drop steel balls with diameters
ranging from 1/16" to 1/2" in steps of 1/16" into 2 liters of Pantene Pro V
clear shampoo.
BC is correct that drag is not a simple function of speed. The relevant
drag formula in this case is Resistance = C r v, where C is a constant and r
is the radius of the sphere. The terminal speed becomes proportional to r^2.
In a single one hour lab the students can plot log(v) versus log(r) and
verify the relationship. It's a nice break from the usual class routine. The
smallest sphere takes about 8 minutes to traverse the fluid, the largest
about 12 seconds.
Bob at PC

________________________________________
From: phys-l-bounces@carnot.physics.buffalo.edu [
phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Bernard Cleyet [
bernardcleyet@redshift.com]
Sent: Tuesday, November 03, 2009 1:11 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] Linear Air Drag

The Reynolds number for water droplets (e.g. in clouds) is within the
limit for viscous (Stokes') drag.

However, BW is quite correct (I pray!) for the usual projectiles.
Intermittent updraft in a cloud is the only case ICTO where Peter's
work would be applicable. Well, one could w/ a submerged spring gun
fire a bb up a cylinder of oil. Remember to include buoyancy!

bc must write to the "The Physics Teacher".

p.s. v^3 if extreme. Drag is not a simple function of speed.

On 2009, Nov 02, , at 19:19, Brian Whatcott wrote:


Hmmm. I'm puzzled by the concept of linear air resistance.
Air drag for projectiles is a function of v squared.