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Calculating the angular momentum of rotating objects

  1. Nov 6, 2009 #1
    Calculating the angular momentum of rotating objects.

    When the angular momentum is computed you add all the individual angular momenta of infinitesimally small masses tilll you get the total, right ?

    But doesn't that give a wrong answer - because each tiny piece of the object (rotating disk or whatever) is not just obiting the center, it's also spinning once on its own axis for every revolution of the disk.
    So the total angular momentum of real objects will be higher than that for a bunch of particles circling a central axis but maintaining their orientation with no spin.

    Do I have this right or am I all wrong ? ;-)
     
  2. jcsd
  3. Nov 6, 2009 #2

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    You have it wrong. The angular momentum of a real object (i.e., a non-point mass) is the angular momentum due to the linear momentum of the object's center of mass plus the rotational motion of the object about its center of mass:

    [tex]\vec L_{\text{tot}} = m \vec r_{\text{cm}}\times \vec v_{\text{cm}} + \boldsymbol I \vec{\omega}[/tex]

    This is equal to the sum of the angular momenta of the components of the object.
     
  4. Nov 6, 2009 #3
    Sorry, I can't see those figures on my computer, they just look like crazy squiggles on a black background and i can't make them out. Is there an easy way to see them in some other format or whatever ?

    Thanks for the answer btw.
     
  5. Nov 6, 2009 #4

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    You must be using a very old version of Internet Exploder.
     
  6. Nov 6, 2009 #5
    version 6
     
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