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Rotation about the center of mass and spin angular momentum

  1. Jun 17, 2009 #1
    I needed to refresh my classical physics knowledge and I was going through Prof. Walter Lewin's physics videos at ocw.mit.edu and at some point he proposed the following theorem without proof:

    "The angular momentum due to rotation about the center of mass is universal and does not depend on the relative choice of coordinate axis, unlike, say orbital angular momentum"

    It might be really trivial, but I need to see a mathematical proof of this statement, can anyone help?
     
  2. jcsd
  3. Jun 17, 2009 #2
    You can find proof in Kleppner & Kolenkow (pp. 262-4) or David Morin's book (pp. 380-1).
     
  4. Jun 17, 2009 #3
    I took the time to find both books with great effort, and I checked the pages you refer me to.

    None of them has the proof to the theorem I asked.

    I wish you had taken the time to more carefully read what I asked.
     
  5. Jun 18, 2009 #4
    Regarding Kleppner & Kolenkow, the correct pages are 260-2, not 262-4, so it seems it's my mistake. However, on p. 263 they state and reason quite clearly that "rotational motion about the center of mass depends only on the torque about the center of mass, independent of the translational motion ... "

    Morin proves on page 381 that [itex]\mathbf{L}=M(\mathbf{R} \times \mathbf{V}) + \mathbf{L}_{CM}[/itex], where [itex]\mathbf{L}_{CM}=\int \mathbf{r'} \times (\mathbf{\omega} \times \mathbf{r'}) dm[/itex].

    He states clearly that r' and omega' are both measured relative to the CM. How can something defined relative to the CM be dependent on the origin? Just take a moment to visualize this - you can move your frame of reference around, yet your CM won't budge, right? How can anything defined relative to the CM change when some other arbitrary coordinate system is moved around?

    If you need to see this formally, then note he r' = R - r, where R is the CM coordinate and r the particle's position. Moving your frame of reference moves both R and r by the same amount and this cancels out: r' = (R+a) - (r+a) = R-a, independent of a. The particle's velocity in the CM frame (v' = omega' cross r' = V - v) is also independent of shifting the origin of your frame of reference for the same reason.

    -----
    Assaf
    http://www.physicallyincorrect.com" [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Jun 18, 2009 #5
    This was extremely helpful. Thank you very much
     
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