0
$\begingroup$

What will be the work done in radially stretching a rubberband it can't be zero as there is potential energy being stored in it All I came up with it that there would be increase in overall length so assuming that initial length before expansion was $$2πR$$ And after its expansion it's$$ 2πx $$ I came up with the net energy or work done by external agent to be $$(K.2π(x^2-R^2))/2$$ Is it correct or there is something else to be done that I am missing. And if it were the case of an elastic wire have some Young's modulus what would be the energy stored in that case?

$\endgroup$
  • $\begingroup$ Even if you could treat the rubber as a linearly spring, the $2\pi$'s should also be squared. $\endgroup$ – Chet Miller Dec 25 '18 at 19:08
0
$\begingroup$

In the initial state (radius $R$) the rubber band is at rest. It's from that configuration that energy is required for extending it. Then elastic energy should increase quadratically from there. I would take as potential elastic energy $${\textstyle{1 \over 2}} K\,[2\pi\,(x - R)]^2.$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Is it correct or there is something else to be done that I am missing.

What you're missing is that the stress-strain curves of rubbers tend not to be linear. Here are two examples:

Stress strain

Source.

That means that these materials don not have a constant Young's modulus, but rather:

$$E=f(\gamma)$$

Similarly, the spring 'constant' of a real rubber band will not be constant but:

$$K=g(x)$$

The potential energy stored upon extension is found from:

$$\Delta U=\int_0^xg(x)dx$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But if hypothetically assuming that it's linear then will the expression I derived is correct? $\endgroup$ – aditya prakash Dec 25 '18 at 17:10
  • $\begingroup$ @adityaprakash Yes. $\endgroup$ – alephzero Dec 25 '18 at 18:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.