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Angular KE and momentum question

  1. Apr 5, 2003 #1
    Take for example the angular momentum experiment where you sit in a rotating chair and spin yerself while holding weights. If you bring the weights in closer, your angular velocity increases, but you angular momentum is conserved.

    Is angular kinetic energy also conserved with angular momentum?
     
  2. jcsd
  3. Apr 6, 2003 #2

    dg

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    No, energy is not conserved because of the work of internal forces in the system...

    Angular momentum is conserved because of the absence of external torques.
     
  4. Apr 6, 2003 #3
    The extra kinetic energy is due to the work done by the person who sit on the rotating chair to pull the weights in closer.
     
  5. Apr 6, 2003 #4
    hmm we never really talked about rotational energy in my physics class. Would it just be 1/2Iω2?

    btw. seeing linear momentum is mv. KE = 1/2mv^2. Is it just coincidence that KE is antiderivative of momentum or is it supposed to be that way. I've noted this a few times but I've never had a professor get up and say "Kinetic energy is the antiderivative of momentum" so i was just wondering.
     
  6. Apr 6, 2003 #5
    I guess I just don't understand how that work done pulling the weights closer to you would go twords increasing KE.
    Would anyone mind laying out the problem and solving?

    It just doesnt make sense that momentum can stay the same while energy changes.
     
    Last edited: Apr 6, 2003
  7. Apr 7, 2003 #6

    dg

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    Well it makes sense according to classical physics, but if you don't like it you can make your own... you must just be self-consistent and convince a few others ;)

    Besides teasing you, Let us put down a very symple system that will show you how the whole thing works.

    Something like two point-like masses hold together by a spring, rotating around an axis orthogonal to the spring with constant angular velocity. If the spring is initially extended beyong its rest lenght and kept so by a thin rod, we will have a system where we can easily describe the internal forces by a potential energy. If at some point the thin rod breaks, total energy will be conserved (no other force is involved in the system besides the conservative elastic one); since that time on the evolution of the system is dictated by conservation of angular momentum (no external momentum of force is present) and of total energy. While the two masses are coming together there is an increase of angular velocity which implies an increse of kinetic energy which is compensated (to keep total energy constant) by a reduction of potential energy. During the expansion the opposite happens.

    In the case of a human bringing two weights closer it is the chemical energy stored in some molecules that is converted to kinetic energy (and to heat), while conserving angular momentum.
     
  8. Apr 7, 2003 #7
    actually there is no such thing as Kinetic Energy or Potential Energy.
    there is only one energy which is vector product of the force vector and the so called X vector.this X vector is actually the distance of the point where the force is acting and the equilibrum point(the point where the force virtually reduce to zero).important is that the energy defined as above is always constant if the system is closed.

    i din't tought/comment of/on your wonder here but give me some finite time and i'll come up with something.
     
  9. Apr 7, 2003 #8
    Thanks dg, I think it just clicked in my mind. It's starting to make more sense now.
     
  10. Apr 8, 2003 #9
    i'll assume this discussion is not over and i'll give you my approach:
    there is only one energy in the system and it is E=FxX=const
    X is equilibrum vector starting in the position of the weight and ending in the normal projection of the same position on the axis of rotation.Now the explanation is simple as A,B,C...
    pulling the weight closer to the axis of rotation declines X but since FxX=E=const increases F.X is normal to the trajectory of the weight and F is parallel(tangential).But it also has to be FxdX=F'xdX'
    left are the old values and right are the new values.that's why the tangential displacement is larger closer to the axis of rotation.
    when you increase X then F drops and dX (the old)> dX' (the new ).

    The only thing conserved here is the energy.
    I don't know about the momentum cause at the moment like everything else in physics i'm considering time being vector.
     
  11. Apr 8, 2003 #10

    Claude Bile

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    Science Advisor

    I'm curious to know how energy can be represented as a vector
     
  12. Apr 9, 2003 #11
    here is how:
    assume you have an weight hanging on a spring in equilibrium point. then the resulting force is zero thus no movement. now pull the weight down displacing it for some X value. now the resulting force is again zero but when you release the weight the active force is same as but oposite with the one you have invested to cause the initial displacement. there are two scenarios availabe to consider:

    (1st scenario) assuming that at the moment you release the weight the action force is Fmax. in that moment the weight is in equilibrium point and the equilibrium distance is zero. the action force is causing displacement in same direction of the the force cause the job done by the action force is always positive. as you increase the equilibrium distance you decline the force.when the force reduces to zero then the equlibrium distance is maximum. the force onwards change the sing and becomes negative therefore and the equilibrium distance does the same too. when the force reaches value -Fmax the the equilibrium distance drops to zero and so on. simbolically it looks like this:
    (1 moment)F=Fmax and X=0
    (2 moment)F>0 and F->0 and X > 0 and X->Xmax
    (3 moment)F<0 and F->0 and X < 0 and X->-Xmax
    (4 moment)F=-Fmax and X=0
    (5 moment)F<0 and F->0 and X < 0 and X->-Xmax
    (6 moment)F>0 and F->0 and X > 0 and X->Xmax
    (7 moment)same as moment 1 and everything from the begining

    thusF=Fmax * cos(fi1) and X=Xmax * sin(fi1)

    (2nd scenario)the second scenario is when you assume that at the moment of release of the weight X=Xmax then simbolically it looks like this:
    (1 moment)X=Xmax and F=0
    (2 moment)X>0 and X->0 and F > 0 and F->Fmax
    (3 moment)X<0 and X->0 and F < 0 and F->-Fmax
    (4 moment)X=-Xmax and F=0
    (5 moment)X<0 and X->0 and F < 0 and F->-Fmax
    (6 moment)X>0 and X->0 and F > 0 and F->Fmax
    (7 moment)same as moment 1 and everything from the begining

    thusX=Xmax * cos(-fi1) and F=Fmax * sin(-fi1)

    what i have found is that both scenarios are corect and that they are only the same story from different aspect. the first nlue tush is for the x coordinates and the 2nd tush are the y coordinates while the z coordinates both for X and F are zeros. but therefore the x and y coordinates of energy are zeros too.the z coordinate of energy then will be

    E(x)=Fmax * sin(-fi1)*0+Xmax * sin(fi1)*0=0
    E(y)=Fmax * cos(fi1)*0+Xmax * sin(-fi1)*0=0
    E(z)=Fmax * cos(fi1)Xmax * cos(-fi1)-Fmax * sin(-fi1)Xmax * cos(-fi1)
    E(z)=Fmax*Xmax*(cos^2(fi1)+sin^2(fi1))=Fmax*Xmax

    this equation comes from vector procut of two vectors.in the end i conclude that energy is actually a vector product of force vector and the equilibrium vector.

    STICK WITH ME YOU'LL LEARN MUCH MORE!!
     
  13. Apr 9, 2003 #12
    Not every system can have the 'point of equilibrum' that you are defining.
    Take for example, a guy is sliding a box on a rough surface, where is the point of equilibrum ?

    But we can say that all the laws (as far as i know so far) of energy can be derived from the definition of Work.
     
    Last edited: Apr 9, 2003
  14. Apr 9, 2003 #13
    if you know the energy vector and the force vector then since
    E=FxX => Fx(ExF)=Fx(X) => X=ExF
    where X is the equilibrium vector.

    hey energy being vector is new to me too.

    redirecting...:
    https://www.physicsforums.com/showthread.php?s=&threadid=1000
    find out more of Kinetic Energy.
     
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