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Angular Momentum

  1. Aug 17, 2005 #1
    I have read a statement at a physics forum:

    the net angular momentum about any point (or axis)
    is the sum of the momentum about center of mass plus the momentum that center of mass about the point (or axis).

    It sound familiar to Parallel-Axis theorem but I'm not sure it's right or wrong.

    Would someone help me to prove it by mathematical approach?

    Thanks very much!!
    (Please forgive me that my English isn't very good)
  2. jcsd
  3. Aug 17, 2005 #2


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    Here's how you can show this:
    1. Let your object consist of, say N point masses with position vectors [tex]\vec{r}_{i}, i=1,..N[/tex] from the point you are computing the angular momentum with respect to. Let the point masses be denoted by [tex]m_{i}, i=1,...N[/tex] and the velocity vector of the point mass as [tex]\vec{v}_{i}[/tex] and let C.M subscripts refer to your object's center of mass. Let [tex]\vec{R}_{i}, \vec{V}_{i}[/tex] be distances&velocities RELATIVE TO the center of mass.
    (For example: [tex]\vec{v}_{i}=\vec{v}_{C.M}+\vec{V}_{i}[/tex])
    Hence, we have for angular momentum [tex]\vec{L}[/tex]:

    Now, the first sum can be decomposed as:
    where the first sum is zero by the definition of the center of the mass, and the second sum is the angular momentum of the object as measured relative to the center of mass, within the center of mass' rest frame.

    A similar analysis can be carried out for the second sum contribution to the angular momentum.
    Last edited: Aug 17, 2005
  4. Aug 17, 2005 #3
    I got it thanks very much!!
  5. Aug 17, 2005 #4


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    Welcome to PF!
    (It seems you haven't posted here before..)
  6. Aug 17, 2005 #5
    Yes, I often surf this forum but this is my first post!

    By the way, could you please explain the second sum?
    I'm not sure that I understand it very clearly.

    thanks again!
  7. Aug 17, 2005 #6


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    I assume you mean the second sum in the expression for the angular momentum!
    We have:
    Now, the first contribution is simply:
    [tex]\sum_{i=1}^{N}\vec{r}_{C.M}\times{m}_{i}\vec{v}_{C.M}=(\sum_{i=1}^{N}m_{i})\vec{r}_{C.M}\times\vec{v}_{C.M}=\vec{r}_{C.M}\times{M}\vec{v}_{C.M}, M=\sum_{i=1}^{N}m_{i}[/tex]
    where M is the total mass of the system, and the C.M quantities are CONSTANTS in each individual term, and may be pulled out of the sum.
    The second sum is:
    and is zero, due to the definition of the center of mass.
    Thus, your final expression for the angular momentum is:
    which is the desired result..
  8. Aug 17, 2005 #7
    I did carried out the equations. But I'm little bit confusd why the two sum is zero due to the definition of the center of mass. I know this question sounds stupied, but I just can't figure it out. Maybe I'm exhausted after 3 hours of studying. Could you explain the reason for me? thanks!
  9. Aug 17, 2005 #8


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    These are quantities RELATIVE to the center of mass, right?

    By definition, we have:
    But, since [tex]\vec{R}_{i}=\vec{r}_{i}-\vec{r}_{C.M}[/tex]
    we have:
    Clearly, since [tex]V_{i}=\frac{\vec{R}_{i}}{dt}[/tex]
    we gain a similar relation here..
  10. Aug 17, 2005 #9
    Oh I see! Thank you for your patient to answer my question!

    ...can you answer me one more question?
    It's a problem my classmate ask me.

    If there are a wheel rotating at angular velocity omega
    and a smaller wheel, which is rest at first, is brought closer to the big one, and start to rotate since the friction force between the wheels.
    Is the angular momentun conserved in this case?

    And another case, if the two wheel are match together initially, but in diffirent angular velocity, will the angular momentun conserved?
  11. Aug 17, 2005 #10
    The angular momentum is conserved. It is very easy to calculate if there is no kinetic friction that causes dissipation of rotational energy as heat. If the friction is pure static friction which causes the two disks to rotate at the same rate then angular momentum of the system should rotate slower due to an increase in rotational inertia.
  12. Aug 18, 2005 #11
    Why the inertia will increase? I thought it is constant.
    If will we consider a ideal condition, I think wheel won't stop but rotate with the same linear speed at their edge, and thus no relative motion, so no friction force.
    Last edited: Aug 18, 2005
  13. Aug 18, 2005 #12
    What I meant to say was that
    Ia*wa +Ib*wb = I*w

    Where 'I' is the rotational inertia, and 'w' is the angular velocity, for disk a, disk b and the total (no subscript.)
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