Bouyant force accelerates released balloon

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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.
 
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.