Understanding the Forces and Effects of Motion: Balloon Experiment Explained"

[Johnny Leong]

First question:
A balloon filled with 2 m3 of Helium is attached to a rope with mass per unit length 0.14 kg / m. The mass of the balloon fabric is 0.24 kg. The rope is first attached to the ground. The balloon rises when the rope is uncoiled as shown in the figure (Please visit here)
(a) The forces acting on the balloon are named A, B and C as shown at the right hand side of the figure. What are the forces A, B and C?
(b) Calculate the length of rope which has been uncoiled when the balloon has stopped rising.
(Density of Helium = 0.13 kg / m3, Density of air = 1.30 kg / m3)
(c) If the rope is now allowed to fall away from the balloon,the balloon again rises and eventually bursts. Explain why the balloon bursts.

Answers to
(a). A is upthrust, B is total weight of the balloon, C is tension of the rope.
(b). Don't know how to do, please help.
(c). The pressure inside is greater than the pressure outside.

And another question, A big balloon, the balloon can carry people to the sky. When it is filled with hot air, it will rise and go into the sky. Why it will stop rising when the air inside it is cooled down?
Is it true that when the air is hot, hot air rises and makes the balloon rise, and when the air is cooled down, cool air descends and makes the balloon stop rising.

 

[Doc Al]

Originally posted by Johnny Leong
(b) Calculate the length of rope which has been uncoiled when the balloon has stopped rising.

The balloon will stop rising when the weight of the uncoiled rope (the tension exerted on the balloon) is enough to make the net force on the balloon zero. So the three forces you identified in part A must sum to zero. Calculate the "upthrust" (buoyant force) using Archimede's principle: it equals the weight of the displaced air.

(c). The pressure inside is greater than the pressure outside.

As the balloon rises, the outside air pressure is less causing the balloon to stretch more to contain the helium. Eventually the tension in the balloon material is too much and it bursts.

Is it true that when the air is hot, hot air rises and makes the balloon rise, and when the air is cooled down, cool air descends and makes the balloon stop rising.

The hot air is less dense than the surrounding (outside) cool air. The buoyant force (the outside air pushing on the balloon) is enough to lift the balloon plus people. When the air inside is cooled, its density increases: the buoyant force can no longer support the heavier balloon.

 

[Johnny Leong]

Thank you for your help, Doc Al!

Now, I have another question. Refer to the second question above.
If I already know that the upthrust to make the big balloon remain in the sky is 2760 N and the total weight of the air in the big balloon is 2500 N. Because there is a rope to attach to a big basket and the balloon, then the tension of the rope should be 260 N to keep the whole balloon remain at rest in the sky. And now the rope to the big basket is cut, to calculate the initial acceleration of the balloon, the answer should be calculated like this:
F = ma, F = 260 N, m = 2500 / 10 = 250 kg
then a = 1.04 m / s2.
Is my answer right?

 

[Doc Al]

Originally posted by Johnny Leong
... the answer should be calculated like this:
F = ma, F = 260 N, m = 2500 / 10 = 250 kg
then a = 1.04 m / s2.

Assuming that what you call "total weight of the air in the big balloon" also includes the weight of the balloon itself, then your answer is exactly right.