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Orbital Angular Momenta - Conversion to a Different Origin

  1. Nov 7, 2013 #1
    Suppose we have some object with angular momenta [itex]\mathbf{L}_1[/itex] about rotation center [itex]O_1[/itex] and we have another object with angular momentum [itex]\mathbf{L}^\prime_2[/itex] about rotation center [itex]O_2[/itex]. [itex]O_2[/itex] has some velocity [itex]\mathbf{v}_2[/itex] relative to [itex]O_1[/itex]. Then we wonder what is the angular momentum of the second object with respect to [itex]O_1[/itex] so that we may express the total angular momentum of the system. Is there anything to say in general about this?

    What about the case where the second object is a sphere, and it is rotating on its axis parallel to [itex]\mathbf{r} \times \mathbf{v}_2[/itex], where [itex]\mathbf{r}[/itex] is a vector from [itex]O_1[/itex] to [itex]O_2[/itex] ? For example consider a rotating sphere in a circular orbit, where the sphere's axis of rotation is parallel to the plane of the orbit. What is the total angular momentum about the center of the circular orbit?
  2. jcsd
  3. Nov 8, 2013 #2
    I don't know why this didn't occur to me until I took a shower.

    [itex]\mathbf{x} = \mathbf{x}^\prime + \mathbf{O}_2[/itex]


    [itex]\mathbf{L}_2 = \sum m_i \mathbf{x}_i \times \dot{\mathbf{x}}_i = \sum m_i (\mathbf{x}_i^\prime + \mathbf{O}_2)\times \dot{\mathbf{x}}_i [/itex]

    [itex]= \sum m_i (\mathbf{x}_i^\prime + \mathbf{O}_2)\times \left(\dot{\mathbf{x}}^\prime_i + \dot{\mathbf{O}}_2\right)[/itex]

    [itex]=\mathbf{L}_2^\prime + \sum m_i \mathbf{O}_2 \times \dot{\mathbf{O}}_2 + \sum m_i \mathbf{x}_i^\prime \times \dot{\mathbf{O}}_2 + \sum m_i \mathbf{O}_2 \times \dot{\mathbf{x}}^\prime_i [/itex]

    So at least the first two terms have a nice form.

    [itex]=\mathbf{L}_2^\prime + M_{total} \mathbf{O}_2 \times \dot{\mathbf{O}}_2 + \sum m_i \mathbf{x}_i^\prime \times \dot{\mathbf{O}}_2 + \sum m_i \mathbf{O}_2 \times \dot{\mathbf{x}}^\prime_i [/itex]
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