# Where does the lost energy go in a rubber band powering a rotating shaft?

Okay, I'm no physics whiz, and this has me stumped. You know those toy airplanes you can get with the rubber-band driven propellers? You twist the propeller a bunch of times, and this stores potential energy in the rubber band. Then when you let go of the rubber band, it drives the propeller.

Well, suppose you have two of these airplanes. They are identical except for the fact that airplane A has an extremely efficient bearing which holds the rotating propeller shaft. The other airplane (B) has a "normal" bearing, which is less efficient. Thus, the shaft on B generates more heat when it spins the shaft.

You take both systems and wind the propeller 100 times, then let go. Shaft A spins with almost no friction. It therefore uses up it's energy faster, and it spins through it's 100 counter-rotations in maybe 5 seconds. Shaft B takes 10 seconds to do the same thing. BOTH shafts started with 100 "shaft-spin units" stored in their rubber bands at the start. 10 seconds later, each has done 100 shaft spins. Shaft B is hotter, though, due to the friction.

My question: Since both systems started with the same potential energy, where did the "extra" energy go in shaft A? Remember, they both did 100 spins. Shaft A did it faster, but it still did the same amount of spins. Maybe it's leftover in shaft A's angular momentum at the end? (remember, I'm not a physics guy - be nice if it turns out "angular momentum" isn't the proper term!) :-)

Maybe shaft B only does 99.9 counter-spins, losing 0.01 spin-units to friction? Am I making some incorrect assumption?

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You're assuming that the propellor speed is dominated by the bearing friction, while in practice it's dominated by the drag on the propellor as it moves through the air. Typically the propellor will rapidly accelerate to a steady speed at which the force from the rubber band matches the aerodynamic drag, then the propellor speed will slow as the band unwinds and the force from the band decreases. The energy stored in the band ultimately goes into moving the air.

In the case of plane B the force at the propellor is slightly reduced due to friction in the bearing, so less of the rubber band energy goes into moving the air and the remainder goes into heating the bearing.

Response to comment

Suppose we remove the propellor or put the plane in a vacuum so there's no energy loss to air. In that case plane A forms an oscillating system. The rubber band exerts a torque $\tau$ on the shaft/propellor and it's angular velocity increases according to:

$$\tau = I \space \dot{\omega}$$

where $I$ is the moment of inertia of the shaft and $\omega$ is the angular velocity. Note that the torque $\tau$ will be a function of time because the torque reduces as the rubber band unwinds. When the rubber band has fully unwound all the energy has been tranferred into the shaft and it will now be rotating at a speed given by:

$$E = \frac{1}{2} I \space \omega^2$$

where $E$ is the energy you originally put into the shaft. Once the band has fully unwound the shaft will carry on turning so it will wind the rubber band in the opposite direction. If the bearing is frictionless and there are no losses in the rubber band, all the energy in the shaft will be transferred back into the rubber band, and it will end up wound up just as tightly as it started but in the opposite direction. The band will now start accelerating the shaft in the other direction, and the cycle of unwinding and rewinding will continue indefinitely.

Now consider plane B. There will be a torque $\tau_f$ due to friction that opposes the torque from the rubber band, so the equation of motion is:

$$\tau - \tau_f= I \space \dot{\omega}$$

This means the shaft in plane B will accelerate more slowly than in plane A, and when the band has fully unwound it's speed will be given by:

$$E - E_f= \frac{1}{2} I \space \omega^2$$

where $E_f$ is the energy lost to friction i.e. the energy that has gone into heating the bearing. So at this point the shaft in plane B is rotating more slowly than the shaft in plane A, and it will have taken longer to get to the point of being fully unwound.

As with plane A, the shaft will now start to rewind the rubber band in the opposite direction, but because energy has been lost it won't be able to rewind the band as tightly as it was originally wound. There will still be a cycle of unwinding and rewinding, but each step of the cycle will wind the band to a lower tension until the shaft stops rotating. At that point all the energy originally in the rubber band will have gone into heating the bearing.

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Hmmm, that's very interesting. Thanks for taking the time to answer! How about if we remove the propellers, then, and just have the shafts? I think the rest still stands (shaft A would finish quicker; shaft B would take longer and end up hotter), and the surface friction of the air on the shafts would be negligible, I think. Thanks again! –  loneboat Sep 13 '12 at 16:36
The potential energy in 1st case is fully converted only into kinetic energy that is $\frac{1}{2} I \omega^2$ and in other case, it is converted to both kinetic and heat energy, In the first case, $\omega$ is more and it completes revolutions faster but in the second $\omega$ is less so it takes more time to do it.