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

  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,

--- On Sun, 1/10/10, Bernard Cleyet <> wrote
 (Sunday, January 10, 2010, 12:30 AM):
bc still wants somene to write the diff. eq. to include the buoyancy