Bouyant force accelerates released balloon

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The discussion focuses on calculating the initial upward acceleration of a helium-filled balloon when released. The buoyant force is equated to the weight of the air displaced by the balloon, leading to the formula for acceleration based on the densities of air and helium. Initial calculations yielded an acceleration of 70.3 m/s², but after considering the weight of the helium inside the balloon, a corrected value of approximately 60.5 m/s² was obtained. The conversation also touches on the relevance of gravitational forces in similar buoyancy problems, such as objects submerged in water, emphasizing the need to account for these forces separately. Ultimately, the correct approach to solving buoyancy problems involves understanding the net forces acting on the object.
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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 bouyant 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 bouyant 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 .. bouyant 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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