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Angular momentum at a distance

  1. Dec 29, 2011 #1
    Say I have a situation like the image.

    I have a potentially complex object with moment of inertia I about its center of mass, mass M, velocity v and position r with respect to P. It is rotating with angular velocity ω about its center of mass

    What is the angular momentum about point P? I would say [tex]\vec{L} = \vec{r} \times M\vec{v}[/tex] but would that be accurate or only an approximation?
  2. jcsd
  3. Dec 30, 2011 #2
    I think L=r*m*v is an approx because what you did is that you Used L=w*I where I was substuted as I=MR^2 which is the Inertia for a point mass .If the object is a complex its Inertia would be I=cMR^2 where c is a constant (example for Rod rotating about its tip I =1/3*MR^2 , c= 1/3 ) therefore exact L for complex won't equal r*Mv :)
  4. Dec 30, 2011 #3

    Filip Larsen

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    Gold Member

    You should remember to include the objects "own" angular momentum. If the angular momentum around the CM of an object is LCM then the angular momentum of the body for any other reference point P is

    LP = LCM + rP x mv,

    where rP is the displacement vector from P to CM. This equation is exact in the sense that follows from the definitions of angular momentum and center of mass. If you know the object to be rigid you can further use LCM = Iω.
  5. Dec 30, 2011 #4
    Thanks a ton! This is exactly what I was looking for.

    Do you know of a proof of this?
  6. Dec 30, 2011 #5
    Never mind, I managed to prove it. Thanks!
  7. Dec 30, 2011 #6

    D H

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    It's pretty simple for a system of particles. Let
    [itex]\mathbf r_n[/itex] be the vector from the center of mass to nth particle,
    [itex]m_n[/itex] be the mass of the nth particle,
    [itex]\mathbf R[/itex] be the vector from the origin to the center of mass, and
    [itex]M[/itex] be the total mass of the system, [itex]M=\sum_n m_n[/itex].

    Note that [itex]\sum_n m_n \mathbf r_n = 0[/itex] by definition of the center of mass.

    The angular momentum of the system of particles with respect to the center of mass is
    [tex]L_{CoM} = \sum_n m_n \mathbf r_n \times \dot{\mathbf r}_n[/tex]
    The angular momentum of the system of particles with respect to the origin is
    L &= \sum_n m_n (\mathbf R+\mathbf r_n) \times (\dot{\mathbf R}+\dot{\mathbf r}_n) \\
    &= \sum_n m_n\mathbf R\times\dot{\mathbf R} +
    \sum_n m_n\mathbf R\times\dot{\mathbf r}_n +
    \sum_n m_n\mathbf r_n\times\dot{\mathbf R} +
    \sum_n m_n\mathbf r_n\times\dot{\mathbf r}_n \\
    &= \mathbf R\times M\dot{\mathbf R} +
    \mathbf R\times\frac{d}{dt}\left(\sum_n m_n\mathbf r_n\right) +
    \left(\sum_n m_n\mathbf r_n\right)\times \dot{\mathbf R} +
    The sum [itex]\sum_n m_n\mathbf r_n[/itex] vanishes by definition of center of mass and thus
    [tex]L = \mathbf R\times M\dot{\mathbf R} + L_{CoM}[/tex]
  8. Dec 30, 2011 #7
    Thanks. That was the same as mine but much neater.
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