# Work done in circular expansion of a rubberband or an elastic wire

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?

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

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 $${\textstyle{1 \over 2}} K\,[2\pi\,(x - R)]^2.$$

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:

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:

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

• But if hypothetically assuming that it's linear then will the expression I derived is correct? – aditya prakash Dec 25 '18 at 17:10
• @adityaprakash Yes. – alephzero Dec 25 '18 at 18:12