swapping mediums

[iseason]

i was thinking about an experiment and what it would mean the other day.

If i took a balloon filled with water and held it ten metres above a pool . And held a balloon filled with air ten metres down.
shouldn't they technically break the surface together if i release them at the same time?

cheers
iseason

[PhysBrain]

There are a couple of factors which make this problem non-trivial to solve. First of all, the water-filled balloon has only one atmosphere of pressure acting on it, whereas the air-filled balloon has additional pressure on it due to being submerged in ten meters of water. Air is also compressible whereas water is not. Therefore, the volume of the air-filled balloon will expand as it rises towards the surface, which will also affect the bouyant forces acting on it.

But if we wanted to simplify the problem a bit, we could insist that the balloons are actually rigid spheres which will maintain their shapes (and hence volumes) over the range of pressures in this problem. If we hold to this assumption, we can easily compute the bouyant forces acting on the two spheres. Let F1 be the bouyant force on the water-filled sphere, and F2 be the bouyant force on the air-filled sphere.

F1 = (rho_w - rho_a)*V*g
F2 = (rho_a - rho_w)*V*g

where rho_a is the density of air, and rho_w is the density of water. It's plain to see from these formulas that the forces on the two spheres are equal in magnitude, but opposite in direction. Now, the next difficulty we encounter is the inertia of the two spheres. Because the mass of the water-filled sphere is much greater than the air-filled sphere, an equal force applied to each sphere will produce different accellerations. The less massive sphere will accellerate faster. From Newton's second law we have F=ma.

F1 = m1*a1 = (rho_w*V)*a1
F2 = m2*a2 = (rho_a*V)*a2

Solving for a1 and a2 we get

a1 = (rho_w - rho_a)*g/rho_w
a2 = (rho_a - rho_w)*g/rho_a

Or, if we set F1 = -F2, we get

a1 = -(rho_a/rho_w)*a2

which shows that for rho_w > rho_a, then |a2| > |a1|. Therefore, the air-filled sphere will accellerate faster towards the surface than will the water-filled sphere.

Of course this analysis ignores a few other factors which would influence a real experiment. For instance, we did not take into consideration the drag of the water and air on the spheres. However, the magnitudes of the drag forces are likely to be much smaller than the bouyant forces and so we can safely ignore for the purposes of answering this question.

Hope this helps.

[iseason]

The purpose was to see if I would achieve neutrality if I set up a wheel with equal balls filled with water and air. the center hub would be at the surface.The idea being that Both tide or wind can add to the action.

Because only half is exposed to wind, this prevents adverse resistence when the wheel travels towards it.

Cheers
Iseason