# Conservation of energy - olympiad problem

1. Jul 10, 2010

### jpas

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

$$U= 4\Pi {r_0}^2kRT(2 {\lambda}^2+\frac{1}{\lambda^4}-3)$$

(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?

2. Jul 11, 2010

### hikaru1221

Why do you think dW=-dU?

3. Jul 11, 2010

### 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.

4. Jul 11, 2010

### jpas

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?

5. Jul 11, 2010

### Staff: Mentor

That's right. The force needed to expand the balloon is equal and opposite to the elastic force.

6. Jul 11, 2010

### hikaru1221

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.

7. Jul 11, 2010

### 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.)

8. Jul 11, 2010

### hikaru1221

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.