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Angular momentum problem

  1. Nov 10, 2014 #1
    1. The problem statement, all variables and given/known data
    At time t, particle with mass m has displacement ##\vec r (t)## relative to origin O. Write a formula for its angular momentum about O and discuss whether this depends on choice of origin.

    The second part is what I'm more unsure of.

    2. Relevant equations

    3. The attempt at a solution
    ##\vec l=\vec r\times\vec p## so at time t, ##\vec l=\vec r(t)\times m\vec v##

    For the second part, I wasn't sure what to do. I tried and got as far as this:
    ##\vec l_2=\vec r_2\times m\vec v_2##
    Then ##\vec l-\vec l_2=(\vec r\times m\vec v)-(\vec r_2\times m\vec v_2)##
    What now?
  2. jcsd
  3. Nov 10, 2014 #2


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    The point you're missing is that when you change the origin, the velocity doesn't change(of course if the translation is independent of time):
    \vec v '=\frac{d}{dt}\vec r'=\frac{d}{dt}(\vec r+\vec a)=\frac{d}{dt}\vec r=\vec v
  4. Nov 10, 2014 #3
    What happens if you try to describe the displacement of the particle r(t) from other point as a combination of motions (using the relative movement).

    Something like this:
    From origin (point O) you know the displacemente is r(t)
    From other origin (point A) you know the displacement is rA(t)
    Point A does not moves relative to Oringin O

    So r(t) can be described also as OA + rA(t)
    What does this say to you?
  5. Nov 10, 2014 #4


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    I don't understand what you're asking!
  6. Nov 10, 2014 #5
    Hm, well the magnitude doesn't change but wouldn't the direction with respect to the second origin be different? But, I gave it a shot (below).

    Okay, so let ##\vec r_2=\vec r+\vec a## and ##\vec v_2=\vec v## then ##\vec l_2=(\vec r+\vec a)\times m\vec v## and so ##\vec l_2-\vec l=(\vec r+\vec a)\times m\vec v- \vec r\times m\vec v=\vec a\times m\vec v## so depends on where the origin is with respect to the first origin?
  7. Nov 10, 2014 #6


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    No, both the direction and the magnitude of the velocity vector remain unchanged. My proof was vectorial!
    That's true but because in general, [itex] \vec v [/itex] changes from point to point in both direction and magnitude, you can say what happens so in general Angular momentum does change in a time independent translation of origin.
  8. Nov 10, 2014 #7
    Ok, thank you very much for the help!
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