The Partitioning of Kinetic Energy

In summary, the conversation discusses the partitioning of kinetic energy between rotational and translational modes, specifically in regards to bullets and their spinning motion. The question arises about whether the energy put into spinning the bullet takes away from the energy used for translation and therefore affects the distance traveled by the bullet. The answer is that while some energy is lost through spinning, it is compensated for by the increased stability and accuracy of the bullet. The equation for rotational energy is also mentioned.
  • #1
In_Development
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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 traveled 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.
 
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  • #2
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 traveled by the bullet is less than it could be?


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 let's 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 traveled 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.

Don't worry, I won't be pestering everybody too often as most of my research will be carried out in the laboratory,

You're allowed to use a gun in you lab?
 
  • #3
Wouldn't this mean that the distance traveled by the bullet is less than it could be?
But only in a vacuum is the translational energy the sole factor that controls how far it would travel!
 
  • #4
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!

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

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
In_Development said:
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 traveled 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.
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:

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

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

AM
 
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1. What is the partitioning of kinetic energy?

The partitioning of kinetic energy is a process in which the total kinetic energy of a system is divided into different forms or types of energy. These forms can include translational, rotational, vibrational, and electronic energy.

2. How is kinetic energy partitioned in a gas?

In a gas, kinetic energy is partitioned between translational and rotational energy. This is because the molecules in a gas are free to move and rotate in all directions.

3. What is an example of kinetic energy partitioning?

An example of kinetic energy partitioning is the energy distribution of molecules in a gas. At room temperature, most of the kinetic energy is in the form of translational energy, while at higher temperatures, rotational energy becomes more significant.

4. Why is the partitioning of kinetic energy important?

The partitioning of kinetic energy is important because it allows us to understand the different types of energy present in a system and how they contribute to the overall energy of the system. This information can be used in various scientific and technological applications.

5. How does the partitioning of kinetic energy affect energy transfer?

The partitioning of kinetic energy affects energy transfer by determining the pathways through which energy can be transferred. For example, in a gas, energy can be transferred through collisions between molecules, which is dependent on the distribution of kinetic energy between translational and rotational motion.

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