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I saw this question in a test. I would have answered kinectic energy due to rotation and translation. It that correct. Else what is the answer?

Oh no, i forgot to mention it was objective type question.There were three other options.I just looked them up

1)kinetic energy due to rotation 2)kinetic energy due to translation 3)none of the above

the question looked confusing and the choices looked misleading at that time. Thanks for all your resopnses

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as you can see from the range of answers below, it rather depends on what the physics level of this test was. If it was an atomic physics test then look at Marek's answer, if it was basic mechanics I would prefer Georg's. You might need to refine your question based on the physics level in future. –  Roy Simpson Mar 21 '11 at 12:06
Roy is right. In any case, even if you don't want to talk about internal energy (which is fine for many purposes) you might be still interested in stuff such as friction, adhesion of wheels to road, shock absorptions, etc. Unless of course this is a question about idealized wheel from basic mechanics class disregarding every interaction besides the ones you mentioned. –  Marek Mar 21 '11 at 13:04
@Georg If you think the question is too basic, downvote it rather than arguing with answerers. –  mbq Mar 21 '11 at 15:20

4 Answers 4

They also posses internal energy consisting of chemical energy (from interaction between molecules), kinetic energy of individual molecules, mass energy of individual elementary particles, strong interaction energy between quarks and gluons et cetera et cetera. There will also be some interaction energy coming from the interface of wheels with materials it is in contact with (be it air or road). And of course also interaction energy between wheel and its axis (relevant for shock absorption, etc.)

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@Marek, this is the question of a undergraduate obviously? What is Your aim with this "I know better" overkill? –  Georg Mar 21 '11 at 11:42
@Georg: huh? This is what you've taken from this? This is a correct answer. Sweeping everything under rug and staying at the level of elementary school isn't the way we do things around here ;) –  Marek Mar 21 '11 at 11:44
""chemical energy between molecules," ah, is this the well known phlogiston? –  Georg Mar 21 '11 at 12:32
What about the kinetic energy of the earth itself? The solar system? The galaxy? The cluster of galaxies? As relevant as the internal energies to an undergraduate test question. Friction and axle losses are fine. –  anna v Mar 21 '11 at 12:54
@Marek of course the velocity of the car trapped in the field of the earth is part of the energy of the wheels, and the rotational velocity too, which I forgot. We are all hurtling through space at enormous velocities and each of us has the corresponding energy to that velocity, let alone the wheels. It is just that it is irrelevant, as the internal energies are irrelevant to a mechanics question. –  anna v Mar 21 '11 at 13:19

I think there's an important lesson for the OP lurking in this question.

Anything as large as a car wheel is almost certainly extremely complicated as well. The wheel has more atoms than a beach has grains of sand. It's made of long, convoluted polymer molecules that were formed in complicated chemical processes, as well as from air and metals, themselves interesting materials to understand. The wheel deforms and bounces as it rotates. It dissipates energy into the road and its surroundings, and takes energy in from the motor. If you wanted to understand everything about the wheel, by the time you are done, you'll have a learned a great deal from all over the natural sciences. I once read an entire book called "The Bicycle Wheel" that really only covered a few select topics, like the tension in the spokes and the alignment. Everything that can be said would take many books.

On the other hand, physicists do not want to know everything about the wheel. Part of the art of physics is figuring out the bare minimum amount of information about the wheel to take into consideration. What to consider depends on what sort of question you're answering.

Suppose you are testing the brakes, and you want the car to stop very suddenly. Then the brakes must dissipate all the kinetic energy of the car. In that case, you might wonder, "Can I calculate this energy using simply $KE = 1/2 mv^2$, or do I need to consider the extra energy of the rotating wheels?" Here we would ignore most sorts of energy, and focus only on the translational and rotational kinetic energies.

In another situation, we might ask, "How will my gas mileage improve if I inflate my tires another 5 psi?" (In the US, we use absurd units for everything.) In that case, we would be concerned with how the energy of the wheel is being turned into heat, and we'd have to consider the energy involved in deforming the wheel repeatedly as it goes around - stretching the rubber and slipping against the road, etc.

It's true that there are many forms of energy in the wheel, and so the full picture is complicated. However, the goal is to ignore as many complications as is reasonable at any given time. Figuring out which things to consider and which to ignore is a difficult skill that physicists gain only by dint of long and diligent practice.

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They are in a gravitational field, so they have some gravitational potential energy.
The atoms etc in the wheels have nuclear binding energy.
There is some stretched rubber which carry elastic potential energy.
They have temperature which amounts to kinetic energy. They have electric charge, so there is some electric potential energy. As you mentioned, they have translational and rotational kinetic energy

If you take the union (some are partly overlapping) of all these energies you find their total energy E=mc^2.

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Sme overkill as Mareks! –  Georg Mar 21 '11 at 11:43
@Georg: it's not called overkill. It's called a correct answer. You don't make your answer better by forgetting about everything non-trivial. –  Marek Mar 21 '11 at 11:45
On top of that "cookie monster" rechurned, that is what I hate most, when people are too lazy to read a thread in toto. –  Georg Mar 21 '11 at 12:43
This proves that the question is not a good one. –  delete Mar 21 '11 at 15:42

In the idealized case of "the wheels of a moving car" and assuming this is an introductory mechanics course, the author of the question is probably looking for you to notice that the whole of the mechanical energy can be written as a rotational term ($\frac12 I_{tangent} \omega^2$) about the point of contact with the road.

Now, by the parallel axis theorem and the rolling-without-slipping condition, this will be the same value as a translational terms ($\frac12 m v^2$) plus a rotational term about the center ($\frac12 I_{center} \omega^2$), but it can arguably be expressed entirely as option (2) on the list.

Not a good test question, in my opinion, but a useful "make them think" question because it illuminates the internal consistency that arises from the parallel axis theorem.

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