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Whats the equation for rotational momentum?

  1. Jun 6, 2015 #1
    1. The problem statement, all variables and given/known data
    I'm confused with when to use L=Iw (inertia times angular speed) for momentum and when to use L=rxP (r cross p) for inertia. Can someone please explain to me what each one is?

    2. Relevant equations
    Just a conceptual question.

    3. The attempt at a solution
    Just a conceptual question.
     
  2. jcsd
  3. Jun 6, 2015 #2

    stevendaryl

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    Angular momentum is always calculated relative to a point, called the origin. The angular momentum is different depending on what point you choose to consider the origin. So let's take the example of the Earth. The Earth spins on its own axis. That produces an angular momentum relative to the center of the Earth of

    [itex]\vec{L}_{spin} = I \vec{\omega}[/itex]

    But the Earth also is orbiting around the sun. So if we are computing the origin to be the sun, there is an additional term, the orbital angular momentum, given by:

    [itex]\vec{L}_{orbital} = \vec{r} \times \vec{p}[/itex]

    The total angular momentum about the sun is the sum of these two terms:

    [itex]\vec{L}_{total} = \vec{L}_{spin} + \vec{L}_{orbital}[/itex]

    By picking the center to be the center of mass of the object, you can make the orbital angular momentum zero.
     
  4. Jun 6, 2015 #3
    oh ok, so basically you use L=Iw if an object is rotating on resect to its own axis and use L=rxp if an object is rotating respect to another objects axis (and if an object does both,,,, rotate on its own axis as well as another object) then you add the two correct?
     
  5. Jun 6, 2015 #4

    haruspex

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    Sort of. It depends what you mean by I here. If you are defining I as the MoI about the object's mass centre then yes, you need to add the contribution from the linear motion. Alternatively, you can use the parallel axis theorem to find the MoI about the reference axis, then you don't need to add in the linear contribution. The two methods are equivalent.
     
  6. Jun 6, 2015 #5
    hey haruspex, you also replied to my other problem where I asked someone to check my solution to a rotations problem. So in that case, for the r component of the orbital angular momentum be the distance from the point to the center of the stick or would it be from the point to the fixed origin (which was where the center of the stick INITIALLY was).... it would be to where the fixed origin (where the center of the stick INITIALLY was) right?
     
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