Bouyant force accelerates released balloon

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Homework Help Overview

The discussion revolves around the initial upward acceleration of a helium-filled balloon when released in air. Participants are examining the buoyant force and its relationship to the weights of the air displaced and the helium inside the balloon, using given densities of air and helium.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of buoyant force and its impact on acceleration, questioning the inclusion of the weight of the helium in the balloon. There are discussions about the effects of temperature on air density and attempts to reconcile calculated values with a reference answer from a textbook.

Discussion Status

Several participants have offered different formulations for calculating acceleration, leading to a range of values. There is recognition of the complexity of the problem, with some participants suggesting that it involves multiple forces and considerations. The discussion is ongoing, with no explicit consensus reached.

Contextual Notes

Participants note variations in air density with temperature and the implications of gravitational forces on buoyancy, particularly in different environments such as underwater scenarios. There is an acknowledgment of the need to consider the weight of the balloon itself in calculations.

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Homework Statement


A small balloon filled with helium is released. What is it's initial upward acceleration? The only data given is ρ air = 1.29 kg/m^3 and ρ helium is .18 kg/m^3

I assume the buoyant force ( FB ) = the weight of the air displaced by the balloon

so BF = W(air) = ρ(air) * g * VolB = MassB * Accel

and VolB = MassB / ρ(helium)


Homework Equations



so Accel = [ ρ(air) / ρ(he) ] * g = ( 1.29 / .18 ) * 9.81 = 70.3 m/s^2



The Attempt at a Solution



the answer in the book is 61 m/s^2, and I found that ρ(air) and ρ(he) change with temperature, but I have not been able to get the answer in the book. At ρ(air) 20degrees C, I get Acc = 65.4 m/s^2 which is still too high.
 
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johns123 said:
I assume the buoyant force ( FB ) = the weight of the air displaced by the balloon
Don't forget the weight of the balloon (well, the helium inside) itself.
 
Maybe something like BF = W(air) - W(he) = g * VolB( p(air) - p(he) ) = MassB * Acc

then Acc = [ p(air) - p(he) / p(he) ] * g = [ (1.29 - .18) / .18 ] * 9.81 = 60.5 !

You sir, are a freakin' genius :-)
 
Finally, I see the concept: It's 2 problems in one. Net Force vertical .. and .. buoyant force vertical.

Net Force = BF(air) - W(he) = W(air) - W(he) = p(air)*g*VolB - p(he)*g*VolB = MassB*Acc = p(he)*VolB*Acc

then Acc = g * [ p(air) - p(he) ] / p(he) = 60.5 m/s^2 which is the correct answer.

I wonder if this is also true of an object under water ? And we just ignore the gravitational attraction since it is only a very small part of the answer ? Like I've been doing.
 
For something filled with gas, under water, you can usually neglect the gravitational force on the gas, right.
 
Yes. But when I needed to not ignore it, I didn't have a clue. It still seems a bit strange, but I finally realized that the gravitational force on the object was independent of it's volume and had to be considered separately. I might see this problem again with a heavy submarine .. and I'll be watching for it. Thanks MFB.
 

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