Air Balloon Ballast Q: Find Weight to Rise 105m in 15s

  • Thread starter Thread starter KingTutATL
  • Start date Start date
  • Tags Tags
    Air
AI Thread Summary
To determine how much ballast weight must be dropped for a lighter-than-air balloon to rise 105 meters in 15 seconds, the volume of the balloon, which has a radius of 6.25 meters, needs to be calculated first. The balloon's mass must be in equilibrium with the buoyant force and the force of gravity when stationary. The required force for the ascent must account for the mass of the balloon and the mass of the ballast being jettisoned. An algebraic expression for the force in relation to the balloon's mass and the dropped mass is necessary to find the exact weight needed. Ultimately, the mass of the balloon divided by its volume should equal the density of the air, confirming the equilibrium condition.
KingTutATL
Messages
11
Reaction score
0
A lighter than air balloon and its load of passengers and ballast are floating stationary above the eart. Ballast is the wieght that can be dropped overboard to make the balloon rise. The radius of this balloon is 6.25m. Assuming a onstant value of 1.29 kg/m^3 for the density of air, determine how much weight must be dropped overboard to make the balloon rise 105m in 15.0s?

I've been working on this one for a while and just keep getting nowhere!
 
Physics news on Phys.org
KingTutATL said:
A lighter than air balloon and its load of passengers and ballast are floating stationary above the eart. Ballast is the wieght that can be dropped overboard to make the balloon rise. The radius of this balloon is 6.25m. Assuming a onstant value of 1.29 kg/m^3 for the density of air, determine how much weight must be dropped overboard to make the balloon rise 105m in 15.0s?

I've been working on this one for a while and just keep getting nowhere!
What is the volume of the balloon? What is the mass of this balloon if it is floating stationary above the Earth (not accelerating)?

What force is required for the balloon to rise 105 m in 15.0 seconds (careful: there is a bit of a trick, because the mass of the balloon accelerating depends on how much mass you drop overboard - work out an algebraic expression for the force in terms of the mass of the balloon and the mass jetisoned).

What mass must the balloon have in order to generate that buoyant force?

AM
 
How do I calculate the mass of the balloon while it is stationary. I guess it would be in equilibrium with the force of gravity and the bouyant force that oppose each other.
 
KingTutATL said:
How do I calculate the mass of the balloon while it is stationary. I guess it would be in equilibrium with the force of gravity and the bouyant force that oppose each other.
If there is no acceleration, the buoyant force and the weight of the balloon balance. What does this mean for the mass of the balloon divided by volume?

AM
 
That they should equal the density of the air!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top