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Angular momentum

  1. Mar 7, 2005 #1
    Well, this is only a theoretical question that attacked me when I'm studying the rigid body mechanics:

    When you have a rigid body rotating about an axis that passes through center of mass and the angular velocity is constant vector then the angular momentum is parallel to angular velocity. Well, but if the body rotates about other axis that don't passes through CM, what's the angular momentum? and definitely what's the angular momentum when the axis is not changing (in other words when angular velocity has a derivative non null almost)?
     
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  3. Mar 7, 2005 #2

    dextercioby

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    What's the definition of angular momentum...?For a rigid body,of course...

    Daniel.
     
  4. Mar 7, 2005 #3
    well. I have another question:

    Is there a relation between the angular momentum about an axis passing through CM and other angular momenta passing through other points of the rigid body, like the Parallel axes theorem for moments of inertia?
     
  5. Mar 7, 2005 #4

    dextercioby

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    That's exacly what it is.But how do you prove it...?You didn't give me the definition of angular momentum...

    Daniel.
     
  6. Mar 7, 2005 #5
    the angular momentum L is L=rxmr.

    for a discrete system is only the summation to all particles. but I have used then the equation of movement for rotation, that says the variation with time of angular momentum is equal to the torque.

    d/dt(sum(rxmv))=sum(rxF), d(L)/dt=T

    If I take a transformation of coordinate systems from the original inertial to the CM reference I could write:

    d/dt(sum(r(respect CM)xmv(respect CM)))=sum(r(respect CM)xF)

    the left member is the variation of L respect CM and the other is the torque respect CM.

    Then the question is if I know the L about CM, there is a relation to get L about any point with the knowledge of L about CM????
     
  7. Mar 7, 2005 #6

    dextercioby

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    Nope,i specifically said RIGID BODY.The definiton is another one...Keep searching.

    Daniel.
     
  8. Mar 7, 2005 #7
    L=Iw with w the angular velocity
     
  9. Mar 7, 2005 #8

    dextercioby

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    Yes,in simplest form...Now,have you gotten an answer to your question...?

    Daniel.
     
  10. Mar 7, 2005 #9
    yes...partially. because that ecuation states that angular momentum is parallel to angular velocity, and I know it's only valid when the rigid body is rotating around a principal axis. Then, what's the most general eq?
     
  11. Mar 7, 2005 #10

    dextercioby

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    The first assertion is very true.It is proven by the general relation
    [tex] \vec{L}:=\hat{I}:\vec{\omega} [/tex]

    ,where [tex] \hat{I} [/tex] is the moment of inertia tensor,and the ":" mean contracted tensor product.(:= is the definition symbol)


    Daniel.
     
  12. Mar 7, 2005 #11
    thanks for the help
     
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