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Where does the energy go?

  1. Apr 29, 2005 #1
    Say I am out in space rotating with my arms extended. If I slowly pull my arms in, my angular velocity speeds up to conserve angular momentum, and my rotational kinetic energy has increased. That doesn't bother me, cuz I know I had to expend some chemical energy to bring my arms in. I did work against the centrifugal force. Now suppose that instead of starting out with my arms extended, i start out rotating with them tucked against me. If I slowly extend my arms, say at a small constant speed, my angular velocity slows down and I lose kinetic energy. But i still had to do work against the centrifugal force to keep my arms from accelerating outward. So I expended chemical energy, and now I have less kinetic energy. Where did the energy go?
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  3. Apr 29, 2005 #2


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    I'm pretty sure that you're dealing with apples and oranges here. The energy used to alter the configuration of the system isn't related to the conservation laws that you're dealing with. It went within your own body, as heat and mechanical work within muscle fibres, the same way that it would through walking or just breathing. Cellular chemistry doesn't really equate with mechanical physics at this level.
  4. Apr 29, 2005 #3

    James R

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    The centrifugal force in the second case helped you to extend your arms, while in the second case you had to fight it all the way.
  5. Apr 29, 2005 #4
    When you just 'let go of your arms' their increase in kinetic energy accounts for the decrease in rotational energy. And when your arms are totally spread this energy will be dissipated in vibrational motion, sound, etc. I guess when you spread your arms slowly there will be dissipation in your muscles that accounts for the loss of kinetic energy.
  6. Apr 29, 2005 #5
    Hey, thanks for the quick replies! So I can assume that the missing energy went into heating up my body? For simplicity I'm assuming that there is no sound--that's one reason for doing it in space.

    add: what I'm getting at is this: In the first case, some chemical energy will be converted into heat. I will expend slightly more chemical energy than I gained in kinetic energy. But in the second case, it seems like more energy is going into heat. So will I be hotter in the second case than in the first? BTW, I understand that my arms are moving in the direction opposing the force in the first case, and in the direction of the force in the second case. So technically it is negative work done by me in the second case. But how could my muscles know the difference? I'm still having to expend energy to keep my arms from accelerating outwards.
    Last edited: Apr 29, 2005
  7. Apr 29, 2005 #6


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    As I stress in most of my posts, I'm not an expert in anything, including this, so consider my responses as semi-educated opinions rather than facts. Your muscles don't know anything other than what the motor neurons tell them. This doesn't even have to involve your brain. A lot of skeletal muscles and the sensory nerves in the same area have an almost instantaneous feedback connection. That's why you automatically jerk your hand off of a hot stove before the pain signal can even reach your brain. If you had to stop and think about it, you'd be damaged. Be that as it may, the heat is a function of the inefficiency of the muscular action. (Think of it as a frictional loss for simplicity.) So the energy that you assume to be missing is expended both as physical movement and as heat.
    Mainly, I don't see where you're gaining or losing kinetic energy. Your momentum is conserved. What you lose in rotational velocity as your arms extend is gained by increased speed of your arms through space, since they have a much larger 'orbit' to travel through.
  8. Apr 29, 2005 #7
    I thought this also Danger, wouldnt what he loses in angular velocity, he'd make up in MOI? I still dont see where hes losing kinetic energy.
  9. Apr 29, 2005 #8

    Thanks for the info on muscles. I didn't know about the instantaneous nerve response. That is neat. I think u are right, the missing energy is being converted into heat. The reason that kinetic energy is lost is that it depends on the square of the angular velocity, whereas the momentum depends linearly on the angular velocity. So if we denote the initial and final angular velocites by w1 and w2, and the moments of inertia by I1 and I2, we find

    L = I1*w1 = I2*w2 = constant angular momentum

    But the final kinetic energy is

    KE2 = (1/2)*I2*w2^2 = (1/2)*I1*w1*w2 < (1/2)*I1*w1^2 (which is KE1)

    because w2 < w1.
  10. Apr 29, 2005 #9
    Good question. You are making the mistake of looking at forces as being responsible for the conservation of angular momentum. That is wrong since forces can only alter the L-operator (both in magnitude and direction via an exerted torque). The actual conservation really has to do with the way mass has been distributed inside a body. This is expressed by the rotational inertia "I", which is nothing more then a tensor that expresses the directional dependence of some physical quantity (like the pressure tensor). So when you close down your arms, your total weight stays the same but the mass-distribution alters. This implies that the I-tensor will alter and thus also your angular velocity (omega) since L =I * omega.

  11. Apr 29, 2005 #10
    I don't think I said that anywhere. Angular momentum, as u pointed out, will be conserved since there is no external torque. But I still have to have forces to explain the work that went into changing the kinetic energy. I think James R and da_willem have explained it correctly, i'm just still thinking it over.

    add: I see now that the lost kinetic energy is easily dissipated as heat. I didn't realize the spec. heat of water was so high. Assuming an average size man is spinning with arms tucked in at 5 rps, and extends his arms slowly, I got that he loses about 350 J of kinetic energy. This would only raise his body temp by about .001 degrees Celsius. That seems reasonable.
    Last edited: Apr 29, 2005
  12. Apr 29, 2005 #11
    You missed my point. Indeed the will be forces at hand if you change the position of your arms but that is NOT the crucial factor which explains the increase or decrease in orbital velocity. The crucial ingredient is the changing distribution of mass of the rotating object. That's all. How that change is accomplished (through forces if you will) really is irrelevant in the physical description of this phenomenon.

  13. Apr 29, 2005 #12
    No, I didn't miss your point. My question was about energy conservation, not why angular velocity changes. I understand that much. Did you see post #8?
  14. Apr 29, 2005 #13
    Well, in this case you are constantly doing that work, otherwise your entire body would be ripped into pieces thanks to the centrifugal force. However, when you are rotating, you are in equilibrium because this centrifugal forces is equal to the centripetal forces that also acts on every part of your body. Otherwise, you would not rotate. There are no tidal effects here (i mean places where these forces are stronger then at other places on your body, otherwise you would be stretched). Again, the influence of these forces is equally big throughout your entire body because you are a rigid rotator. This is also the very reason why L = I * omega. The omega is equally big everywhere on your body, otherwise your legs would rotate faster then your head :surprised

  15. Apr 29, 2005 #14
    Actually, what I said in that statement you just quoted isn't true. As an earlier poster said, the centrifugal force helps my arms move outward. So even though I'm using up energy to keep them from accelerating outward, it is no more energy than I would dissipate by just standing still and flexing my muscles. So I'm not doing any work. I'm just dissipating the energy that would go into my arms flopping about wildly if I just let them go.
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