Storing energy in rubber band

1. Feb 14, 2010

jeebs

Hi,
Not sure if I'm asking this in the right place, but I was wondering about the energy stored in a rubber band for a physics project I am working on. We all know that for a spring or rubber band being extended, the restoring force is given by F=-kx, and the energy stored U=0.5kx2.

However, I am considering using a rubber band to drive a propeller, in which case energy would be stored by twisting the band around many times. My question is, do you know of any way I could calculate the energy stored in the band in this way?
Alternatively, a way to calculate the torque generated by the untwisting band would be very helpful to me too.

I suspect it would depend on the constant k from the linear extension equation above, and also the length, width and thickness of the band as well as the number of turns, but I have not been able to come up with an expression myself or find one in my searches so far.
Any links to sites that you think might help me, or textbooks, or anything else would be appreciated.

Thanks.

2. Feb 15, 2010

tiny-tim

Hi jeebs!

I think it's just the length.

I suspect any extra "torsion" energy would be negligible.

So just multiply the circumference by the number of loops.

3. Feb 16, 2010

mikewofsey

I'll do the detail since I looked at this same problem many, many years ago, when I made a balsa and tissue airplane and stupidly wound the propeller too tight trying to measure the energy with a half-bowl and a balance, but I broke the plane! Anyway, the elastic potential energy stored in the band is;

U = (1/2)(kx2),

where x is the distance the band stretches from its equilibrium. Normally, that energy would be stored in the band by stretching it, but in the case of twisting the band, you're simply stretching the band in a different way. More on that in a bit.

First, you need to determine the spring constant k for the rubber band. To do that use;

F = -kx,

where you hang one end of the rubber band from a hook, and you then hang a known mass from the other end, and measure the displacement from equilibrium x. First measure the length of the unstretched band, being careful to extend it to its maximum without stretching it. Then hang the known mass on the end, and measure the new length and find the change in length, x. Then k will obviously equal F/x. (The minus sign just gets absorbed by the direction of the displacement, so you can essentially ignore it for now.) And F equals the mass of the known mass in kilograms times the acceleration of gravity, 9.81 m/s2. (There is some error here, because you're ignoring the mass of the rubber band itself, but as long as you use a reasonably heavy mass, you'll get a decent result ... use one heavy enough to stretch the band, but not so heavy that the band is maxed out.) Now you know k.

Next, go back to the energy equation,

U = (1/2)(kx2),

and now you just need to find out the change in x when you wind up the band. This is a little tricky, you should get a caliper, if you have a pair. Wind up the rubber band a set number of times, say just enough so that it doesn't start double-looping. Count up the number of complete winds. Your airplane will keep the wound rubber band of constant length, so now, the change in length x is found -- essentially -- by multiplying the number of winds times the circumference of each winding circle. Use your caliper to measure the diameter of the wound rubber band, that will be d, then apply some simple geometry, since the circumference c = 2(pi)(d/2)2.

Multiply the c times the number of winds n. This will be your new length.

Subtract the unstretched length of the rubber band from this new length and plug into your energy equation above and find the potential energy in the wound rubber band. Also notice that the caliper (inside jaws) can be good for measuring the length of the relaxed band.

Finally, remember that the additional length from the windings is from the relaxed length, so if your band is slightly stretched onto your propeller fixture, then you'll need to add that additional energy into your total energy, using the same energy equation, but using x for that one as the distance the band is stretched from equilibrium.

Best of luck, please feel free to contact me with any questions.

Last edited: Feb 16, 2010