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Angular Momentum of a particle

  1. Jan 17, 2008 #1
    Angular Momentum....

    Since magnitude of the angular momentum of a particle ( or body) depends upon the linear momentum of a particle( r x p)[ p- linear momentum, r - position vector]

    does it mean that a body rotating about a fixed axis( only rotational motion ) does not posses any angular momentum?

    i am really confused since angular momentum is the rate of change of torque...

    how is this possible... please help me...
     
  2. jcsd
  3. Jan 17, 2008 #2
    Any rotating mass has angular momentum. It's magnitude can also be given by,

    [tex] L = I \omega [/tex]

    where I is moment of inertia. The direction of the vector is given by the right hand rule.


    Try taking the derivative of the angular momentum with respect to time

    [tex] L = r \times p [/tex]
     
    Last edited: Jan 17, 2008
  4. Jan 17, 2008 #3
    If mass is not rotating, it has angular momentum as well.

    L = r x p
     
  5. Jan 17, 2008 #4
    [tex] \frac {d}{dt} \L = r \times \frac {dp}{dt} + \frac {dr}{dt} \times p [/tex]

    Usually we assume that

    [tex] \frac {dr}{dt} = 0 [/tex]


    Hence

    [tex] \tau = r \times \frac {dp}{dt} = r \times F [/tex]
     
    Last edited: Jan 17, 2008
  6. Jan 17, 2008 #5
    Yes, but it's a zero vector.
     
    Last edited: Jan 17, 2008
  7. Jan 18, 2008 #6
    I do not think so...
    If a mass does not rotate around its own center and does not rotate around other center ("O" in the picture), just moving straight, it still may have non-zero angular momentum.
     

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    Last edited: Jan 18, 2008
  8. Jan 18, 2008 #7
    Yes you are right jdg812, I kept r constant. You can zip by O in a straight light and have angular momentum with respect to O. In this case, r is a variable.

    [tex] L = mvr \sin(\phi) [/tex]

    As long as

    [tex] \sin(\phi), p, r [/tex]

    is not zero you have non-zero angular momentum.
     
    Last edited: Jan 18, 2008
  9. Jan 18, 2008 #8

    Tom Mattson

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    OK

    No, it doesn't mean that at all. You can view a rotating body as a collection of particles orbiting a common axis. What you do is compute the differential angular momentum for a particle in the body whose mass is dm. This gives you the angular momentum for just that one piece of the body. Then you integrate over the entire body to get the total angular momentum. Since all the particles are orbiting the axis of rotation in the same direction, the result can't possibly be zero.

    Other way around. Torque is the time derivative of angular momentum.
     
  10. Jan 18, 2008 #9

    Tom Mattson

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    It doesn't matter if you assume [itex]d\vec{r}/dt=0[/itex]. [itex]d\vec{r}/dt[/itex] is parallel to [itex]\vec{p}[/itex]. The cross product of any two parallel vectors is zero.
     
  11. Jan 18, 2008 #10
    thanks guys...
     
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