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If I've got a blimp inflated to neutral buoyancy and I hang a small mass from it, what will the acceleration of the blimp be immediately after I attach the mass?
Since we're still at zero velocity we can ignore the viscous effects of the air for just a moment. I believe that I need to know the un-inflated mass of the balloon and the payload, the volume of the helium, the mass of the helium, and the mass of the displaced air. Let's assume a very small pressure is held by the balloon so that we can think of it as simply a volume of helium at the ambient pressure. The mathematics of a simple Attwood's Machine would seem to apply.
The total mass going down:
balloon and payload
extra ballast weight (call this 'm')
Total mass going up:
mass of displaced air
Let's call the sum of all of the mass except for the ballast weight 'M'.
M = balloon + payload + helium + displaced air
The acceleration of the balloon is then:
a = g * m / (m + M)
Question 1: Of course Pascal's Principle says that air pressure will distribute itself equally on all sides. In the static case I can ignore the effects of pressure and air mass. The net force is zero (ignoring the vertical pressure gradient of the air, yes). However, for a balloon to move down, an identical volume of air needs to move up the same distance. The mass of that air cannot be ignored. Is this a valid assumption?
Question 2: Flow through a vicious fluid is typically modeled with a term which rises as a function of the square of the velocity. There is a force resisting the passage of a moving object. If it use the mass of the displaced fluid in the calculation above, am I already accounting for some of the "drag" force which is normally accounted for in this velocity squared term?
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