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*From*: Moses Fayngold <moshfarlan@yahoo.com>*Date*: Mon, 11 Jan 2010 08:06:48 -0800 (PST)

As far as I can see it now, the buoyancy term looks simpler than the drag-force term.

You first write the equation

J d^2 theta / dt^2 = mgl sin theta -- k(S) l^2 d theta /dt, (1)

where J is rotational inertia of the pendulum about the suspension point; m is mass of the pendulum; theta is its instantaneous deviation from the vertical; l is the distance between the suspension point and the center of mass of the pendulum; k(S) is the proportionality coefficient between the drag force and velocity (v = l d theta/dt); it is a pretty complicated function of the size and shape of the pendulum. The direction of motion towards the vertical is taken as positive.

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,

NJIT

--- On Sun, 1/10/10, Bernard Cleyet <bernardcleyet@redshift.com> wrote

(Sunday, January 10, 2010, 12:30 AM):

bc still wants somene to write the diff. eq. to include the buoyancy

**Follow-Ups**:**Re: [Phys-l] Separating inertial mass and g mass. Was: Re: adifferent kind of math background quiz***From:*Bernard Cleyet <bernardcleyet@redshift.com>

**References**:**Re: [Phys-l] Separating inertial mass and g mass. Was: Re: adifferent kind of math background quiz***From:*Bernard Cleyet <bernardcleyet@redshift.com>

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