Conservation of energy - olympiad problem

AI Thread Summary
The discussion centers on the elastic energy of a rubber balloon when inflated, specifically addressing the relationship between work done (dW) and change in potential energy (dU). Participants clarify that the correct relationship is dW = dU, rather than dW = -dU, emphasizing the directionality of forces involved. The elastic force, derived from the potential energy change, is equated to the pressure difference acting on the balloon's surface. There is a consensus that using the energy conservation equation simplifies understanding the forces at play. The conversation highlights the importance of accurately defining "elastic force" in the context of balloon inflation.
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When a rubber balloon of spherical shape with un-stretched radius 0 r is inflated to a sphere of radius r ( ≥ r0 ), the balloon surface contains extra elastic energy due to the stretching. In a simplistic theory, the elastic energy at constant temperature T can be expressed byU= 4\Pi {r_0}^2kRT(2 {\lambda}^2+\frac{1}{\lambda^4}-3)

(c) Express ΔP in terms of parameters given in Eq. (2.2), and sketch ΔP as a function of λ = \frac {r}{r_0}In the solution the problem is solved calculating the work, dW, and then making it equal to

\frac {dU}{dr}.dr

But this isn't true. The true equation is

dW=-dU.

Are the solutions wrong?
 
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Why do you think dW=-dU? :smile:
 
jpas said:
But this isn't true. The true equation is

dW=-dU.
Careful. While the force associated with the potential--in this case an elastic force--is given by F = - dU/dr, the change in potential energy is the work done against this force.
 
Hi,

I think I got it. I'm calculating the force associated with the pressure difference, not the force associated with the potential. These have different directions. Then, because

F_{elastic}=-\frac{dU}{dr} (hooke's law)

\Delta P A = \frac{dU}{dr}

Is that it?
 
That's right. The force needed to expand the balloon is equal and opposite to the elastic force.
 
Doc Al said:
That's right. The force needed to expand the balloon is equal and opposite to the elastic force.

I think we need to clarify "elastic force" here a bit. The "elastic force" we derive from F=-dU/dr is a mathematical quantity equal to -delta(P)*A, while the actual elastic force is tangent to the surface. For that reason, I think we should avoid using F=-dU/dr and instead, use the energy conservation equation dW=dU.
 
hikaru1221 said:
I think we need to clarify "elastic force" here a bit. The "elastic force" we derive from F=-dU/dr is a mathematical quantity equal to -delta(P)*A, while the actual elastic force is tangent to the surface.
Good point. My point was that the force required to expand the balloon will equal dU/dr, not -dU/dr.
For that reason, I think we should avoid using F=-dU/dr and instead, use the energy conservation equation dW=dU.
My point was that dW equals (dU/dr)dr , not (-dU/dr)dr. (Equivalent to what you're saying, I think.)
 
Doc Al said:
My point was that dW equals (dU/dr)dr , not (-dU/dr)dr. (Equivalent to what you're saying, I think.)

Yes, they're equivalent; both leads to the same result. I just wanted to advise the OP not choose the harder path as it's more complicated to explain the term "elastic force", though your explanation goes to the root.
 
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