# The Partitioning of Kinetic Energy

1. Apr 15, 2006

### In_Development

The Partitioning of Kinetic Energy Between the Rotational and Translational Modes​

Alright, I'm a student studying for the International Baccalaureate and that's my title for my extended essay (a 4000 word long essay). Don't worry, I won't be pestering everybody too often as most of my research will be carried out in the laboratory, but to get me started I'd like just a little help.

First off, could someone give me any simple equations with relevance to rotational kinetic energy? It doesn't matter too much, I'll be sure to find them on my own, and I don't need them for any immediate purpose (though I will later for reference to my results). It would help though : )

Secondly, an interesting question I thought up while planning: Bullets and spinning - I know that a spin is put on bullets in order to stabilise them, but doesn't this mean that a greater quantity of the initial launch energy goes towards spinning the bullet and therefore the energy that goes into translation is less? Wouldn't this mean that the distance travelled by the bullet is less than it could be?

I'm sure the answer is something quite simple, but I can't seem to grasp it...

Also, don't worry too much if my knowledge in the area seems quite limited and the situation dire... I learn very fast.

Last edited: Apr 15, 2006
2. Apr 15, 2006

### Beam me down

Short answer: Yes, the more energy put into spinning the bullet, the less energy there is for translation movement.

Longer answer: you know that energy is conserved. When gun powder explodes in a gun the chemical enrgy is transformed into heat energy, sound energy, and mechanical energy of the bullet. It is the sum of all these new energies (for lack of a better descriptor) that must equal the chemical energy. However lets assume the amount of mechanical energy stays constant and only the proportions of RKE (rotational kinetic energy) and TKE (translational kinetic energy) vary. Therefore the more RKE the bullet has the less TKE the bullet has.

As to the distance travelled, remember that velocity is proportional to the square root of the kinetic energy (this applies for both rotational and translational velocity). Therefore the distance travelled is also proportional to the square root of the TKE.

As for stability: I'm sure that the spin adds to the stability (spinning objects are harder to move: while holding a bike wheel get someone to spin it and then try to move it) but I doubt if all of that RKE would be intentional. Though I haven't really investigated the practical design of a gun that much, I've more looked at them from an idealised physics perspective.

You're allowed to use a gun in you lab?

3. Apr 15, 2006

### Hurkyl

Staff Emeritus
But only in a vacuum is the translational energy the sole factor that controls how far it would travel!

4. Apr 15, 2006

### In_Development

Thanks for the answers! And no... I can't use guns in the lab : p (however unfortunately), enough vandalism goes on as it is!

The experiments being done in the lab will probably only go as far as a yoyo in terms of how dangerous they are... God knows you can strangle people with that rope!

Are you implying that the spin of the bullet helps to keep it aloft or that the decreased in translational energy is compensated for by the increased stability (in that the air resistance is minimised and hence the velocity is increased) or some such idea?

I'll try and look into that, unfortunately, school physics books aren't too big on guns... :(

5. Apr 15, 2006

### Andrew Mason

The rotational kinetic energy of the spinning bullet is a tiny fraction of the total kinetic energy of the bullet. So, while you are right that this does take away some bullet energy, it is more than made up for by the reduction in speed loss and improvement in accuracy. The bullet energy is 100% wasted if it misses its target (from the shooter's perspective, anyway).

Rotational energy is:

$$E = \frac{1}{2}I\omega^2$$ where $\omega = 2\pi/T$ and I is the moment of inertia.

For a cylinder of radius R, mass m, $I = mR^2/2$ for rotation about its long axis.

AM

Last edited: Apr 15, 2006