Question:
A spherical rubber balloon inflated with air is held stationary, with its opening, on the west side, pinched shut.
(a) Describe the forces exerted by the air inside and outside the balloon on sections of the rubber.
(b) After the balloon is released, it takes off toward the east, gaining speed rapidly. Explain this motion in terms of the forces now acting on the rubber.
(c) Account for the motion of a skyrocket taking off from its launch pad.
Answer:
(a) The air inside pushes outward on each patch of rubber, exerting a force perpendicular to that section of area. The air outside pushes perpendicularly inward, but not quite so strongly.
(b) As the balloon takes off, all of the sections of rubber feel essentially the same outward forces as before, but the now-open hole at the opening on the west side feels no force – except for a small amount of drag to the west from the escaping air. The vector sum of the forces on the rubber is to the east.The small-mass balloon moves east with a large acceleration.
(c) Hot combustion products in the combustion chamber push outward on all the walls of the chamber, but there is nothing for them to push on at the open rocket nozzle. The net force exerted by the gases on the chamber is up if the nozzle is pointing down. This force is larger than the gravitational force on the rocket body, and makes it accelerate upward.
So I have a few questions.
(a) Why is the force outside less strong? Does the author mean it's less strong temporarily until the balloon is inflated and is in equilibrium? What provides the force from the inside — compressed air molecules? What about the outside force? Where does the tension of the ball come into play, if at all?
(b) What pushes against what here? (Drag hasn't really been explained by the author). Is it the air inside the balloon pushing westward on the air outside and the air outside pushing back on the air inside the balloon which transfers all the way to the front of the balloon, pushing it eastwards?
(c) The chamber walls are able to provide equal force in the opposite direction of the force of combustion products keeping the walls in equilibrium. But what happens at the hole? Combustion products push against the air below and the air pushes on the combustion products which transfers to the top?
Best Answer
Let's start by looking at a)
so I find the explanation quite bad that is given. The first key concept is (surface) tension. What happens when you try to stretch a balloon or rubber band, you need to exert force, the further you stretch it the more you need. Basically a balloon or rubber band wants to keep or go back to it's unstressed/untensioned form (deflated). Once you start inflating it you stretch it out. All forces need to be in balance once the balloon is stable and unchanging.
Now we know that each patch of stretched balloon is pulling on it's neighbours trying to shrink in, this act as a total inward force. We also know that there's an air pressure inside and outside the balloon as there's gas/air inside and outside the balloon. Now since there's already an inward pressure then that means that the inward acting pressure of the air outside needs to be smaller than the outward acting pressure inside the balloon, sum of inward+outward need to be zero.
As you might know pressure is provided the kinetic(thermal) energy of the molecules, put more air inside and you get a higher pressure.
b) yes basically your theory is correct. Drag is nothing more than friction in a fluid (liquid or gas) when moving.
c)Yes basically your theory is correct again. What happens at the hole, well to be extremely correct you have a transfer of momentum. But if you look at it in terms of force, then just think about the 3rd law of newton: