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Re: [Phys-l] Separating inertial mass and g mass. Was: Re: adifferent kind of math background quiz

Also, my apprehensions regarding the missing friction term now seem to be groundless, since the the velocity field in a viscous fluid must be continuous function, and the drag force must take into account all resistances to motion.

Moses Fayngold,

--- On Wed, 1/13/10, Bernard Cleyet <> wrote:

From: Bernard Cleyet <>
Subject: Re: [Phys-l] Separating inertial mass and g mass. Was: Re: adifferent kind of math background quiz
To: "Forum for Physics Educators" <>
Date: Wednesday, January 13, 2010, 2:16 AM


I assumed a simple p.  So:

L^2 * m * theta double dot = -l(M(p)g - M(h2o) * sine (theta) - 2 * gamma * l^2 * theta dot   ignored all dissipation except the linear.

M(p) is the mass of the bob and M(h20) the mass of water displaced.    I have data for a bob displacing 10 gm.  and not.

My solution is omega zero squared is g/l [M(p) - M(h2o)]/M(p)   and omega one squared = omega zero squared - gamma^2 / M(p)^2

I haven't checked this frequency w/ my data or fitted it to:  A * exp(-c1 * t) * sine (c2 * t + c3)   Where c2 should equal omega1.

I think the solution of Coulomb + linear dissipation is:   [exp (- c1*t)] * [1- c2 * t]     quadratic dissipation is

1 / [1 + c * t]      Fits (to other pendula and spring oscillators) using various combinations of these result in better RMSE or R^2  than just the linear alone.   Whether this means they are valid is another question.

I've dropped this to explore the change in frequency w/ various amounts of the dissipation constant (gamma) using flat airfoils attached to the rod. (compound, above and below the suspension)  I've managed to get constants ~ 0.007 and 0.07  (the exponent constant, zero and 90 deg attack angle)

bc  coffee table physicist

p.s.  Coulomb dominates at the end of the free decay, and linear at the beginning.   I'm using a Vernier rotary motion detector (optical encoder)  With the best condition I obtain about a 1k average Q.

On 2010, Jan 11, , at 08:06, Moses Fayngold wrote:

Then you just replace m -----> m -- rho V, where rho is the fluid density and V - the pendulum's volume. The additional term rho V g will represent the buoyant force (assuming we can neglect the pressure gradient in view of the relatively small l). With the buoyancy term, the equation (1) can be written as

   J  d^2 theta / dt^2  +  k(S) l^2 d theta /dt --- (m -- rho V) gl sin theta  =  0           (2)

There is still at least one more term missing - that of friction force, but I could not so far figure out the expression for it; it might be considered as absorbed by k(S) were it not for the fact that it does not depend on v in this approximation.

Moses Fayngold,

Forum for Physics Educators