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Conservation of energy - olympiad problem

  1. Jul 10, 2010 #1
    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 by


    [tex] U= 4\Pi {r_0}^2kRT(2 {\lambda}^2+\frac{1}{\lambda^4}-3) [/tex]

    (c) Express ΔP in terms of parameters given in Eq. (2.2), and sketch ΔP as a function of λ = [tex]\frac {r}{r_0}[/tex]


    In the solution the problem is solved calculating the work, dW, and then making it equal to

    [tex]\frac {dU}{dr}.dr[/tex]

    But this isn't true. The true equation is

    dW=-dU.

    Are the solutions wrong?
     
  2. jcsd
  3. Jul 11, 2010 #2
    Why do you think dW=-dU? :smile:
     
  4. Jul 11, 2010 #3

    Doc Al

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    Staff: Mentor

    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.
     
  5. Jul 11, 2010 #4
    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

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

    [tex] \Delta P A = \frac{dU}{dr} [/tex]

    Is that it?
     
  6. Jul 11, 2010 #5

    Doc Al

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    Staff: Mentor

    That's right. The force needed to expand the balloon is equal and opposite to the elastic force.
     
  7. Jul 11, 2010 #6
    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.
     
  8. Jul 11, 2010 #7

    Doc Al

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    Staff: Mentor

    Good point. My point was that the force required to expand the balloon will equal dU/dr, not -dU/dr.
    My point was that dW equals (dU/dr)dr , not (-dU/dr)dr. (Equivalent to what you're saying, I think.)
     
  9. Jul 11, 2010 #8
    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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