Physics of Balloon?

Question

Trying to udwrstand the physics involved in balloon. Consider a small blob of radius r1. You blow it to radius r2. Now at r2, the balloon is trying to get to r1. What keeps it at r2?

Also, say the ballon suddenly develops a tendency to shrink (i.e. its surface area begins to shrink linearly in time) then what would be required to keep it at r2? (You can assume a blower inside)

Would appreciate equations. Thanks.

To clarify: you are putting in energy on the surface area so that it keeps reducing linerly in time--trying to get to radius r1 --its original state.

There is no diffusion. Assume the surface area of ballon to be a mass M trying to collapse to the center.

Answer 1

the elasticity of the stretching balloon exerts increasing pressure as the balloon stretches

the air in the balloon then has to be at higher pressure to overcome the increased force of the stretched balloon

the elasticity equations of a real balloon are very complex, but a perfectly spherical balloon with perfectly symetrical stretch will have increase that can be modeled with a constant like a spring constant, only using area instead of length

the pressure of the air exerts force equal to the balloon and is calculated using the ideal gas law pv=nrt

in this case we can assume that temperature (t) doesn't change

n, the amount of air (usually in moles) increases as you inflate the balloon

if you know the amount of air and the volume of air in the balloon, you can then calculate the pressure

the r in the equation is the gas constant, you can look it up in the units you want to work with

Answer 2

Pressure is the force per unit area applied on a surface in a direction perpendicular to that surface.

P = F/A

where,
p is the pressure
F is the normal force
A is the area.

When the balloon is not filled with no stress on the balloon material, the pressure inside and outside the balloon is equal.

When the balloon goes to radius 1, the pressure inside the balloon is slightly greater than the typical 14 pounds per square inch at sea level outside air (keep in mind you need a greater force than the outside force to keep the balloon filled because of the force needed to stretch the balloon material).

When you either heat the air or add more air, the pressure inside the balloon increases again allowing the size of the balloon to go to radius 2. Again the reason the pressure inside the balloon must have increased in order to forceably move the balloon to a larger shape, creating more structural stress on the balloon.

The mathematics is slightly more complex than the simple pressure formula above due to the fact you are placing engineering stresses on materials which is studied under fluid mechanics and strength of materials.

Answer 3

the amount of pressure inside pressing out as the material of the balloon is offering resistance to the matter of the gas exerting the inside and the pressure or the outs side of 14.7 lbs per sq in..thus the outside pressure is contributing to the size of the balloon, but taking up to a high altitude will allow the radius to increase in length. and submitting to the depths of the ocean will cause a decrease in the radius thus also limiting the diameter of said balloon. in transient space the balloon would cause it to POP. also remember that the effect of of gravity on the balloon is measured by the formula one over the distance squared gives the relative mass[weight] of the balloon which can also affect both the radius and diameter of the balloon plus the circumference of the balloon by the way what is the color of the balloon?

Answer 4

Unfortunately I cannot give you equations but I can explain a few things.
At r1 the balloon is in a relaxed state but at r2 the balloon is in a state of tension (e.g. outward pressure). the outward pressure is what keeps it at r2. Now if the balloon develops a tendency to "shrink", this is called diffusion (or osmosis in liquids) there are two ways to change that tendency.
1. Create another outward force equal to the rate of diffusion.
2. Change the "outside" pressure to one equal that of the "inside" pressure, but in this relaxed state (inside=outside) the balloon is again at the state of r1.
without actual figures I cannot give you equations.
-M-

Answer 5

The balloons should be similar (very similar weight mutually with string) for a honest attempt even if they are inflated to different diameters. The gasoline interior the balloons should be similar (air?). There should be no major draft interior the room (followers off? lit candle is calm?). The balloons might want to benefit static electrical energy and should be released via a moist or moist hand, etc.

Answer 6

what are you talking about???