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Angular Momentum of a Particle Confusion

  1. Dec 12, 2014 #1
    1. The problem statement, all variables and given/known data

    I was reading the textbook section on angular momentum, and I'm having some difficulty grasping angular momentum.

    Here is a question:

    upload_2014-12-12_22-9-24.png

    In the book, it says that the angular momentum L is equal to vector r cross vector p for a particle. But, for a rigid body, the equation is L = Iw.

    In this case, why isn't the pulley treated as a rigid object. Does this mean that you can take the cross product to find the angular momentum in any case?

    If someone could please clarify this. Thank you.
     

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  3. Dec 13, 2014 #2
    L=r×p ::this equation means rotational
    Angular momentum is the. Moment of linear momentum.
    L=Iw:: this is analogous to p=MV.but
    Thing to be remembered is I can be found for rigid body only. If body is non rigid we have to find L of each particle at any
    Instance and add. So this equation is used generally for rigid bodies only. That doesn't imply u can't use the first equation for rigid bodies. only you should be carefull about taking correct value of r .
     
  4. Dec 13, 2014 #3

    ehild

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    The pulley is treated as a rigid body. What is the angular momentum for a ring? How is the angular velocity of the pulley related to the linear velocity of its rim?
     
  5. Dec 13, 2014 #4

    haruspex

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    For an assembly of particles ##\vec L = \Sigma {\vec r_i \times \vec p_i} = \Sigma {m_i \vec r_i \times \vec v_i}##. For a rigid body rotating with vector ##\vec \omega##, ##\vec v_i= \vec \omega \times \vec r_i ##, so ##\vec L = \Sigma {m_i \vec r_i \times (\vec \omega \times \vec r_i)}##. If ##\vec s_i## is the component of ##\vec r_i## orthogonal to ##\omega##, this reduces to ##\vec L = \Sigma {m_i (\vec s_i^2) \vec \omega} = \vec \omega\Sigma {m_i (\vec s_i^2)} = \vec \omega I##
     
  6. Dec 13, 2014 #5
    When pulley is treated as rigid body and if you know linear velocity if any point ,then divide that velocity with perpendicular distance from axis of rotation and you'll get angular velocity.
     
  7. Dec 13, 2014 #6

    ehild

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    @vishnu: I asked the OP, as I wanted him to answer his own question. Do not answer instead of him.
     
  8. Dec 13, 2014 #7
    Ooops
     
  9. Dec 13, 2014 #8
    The angular momentum for a ring is MR2.

    Since a ring is a rigid body, it's angular momentum is I = lw = (MR2)(w) = (MR2)(v/r) = mvr

    Ok, I see.

    So from what I can understand from the posts above, you can use r x p for a particle, or for a sum of particles, but for a rigid body, you must use I = Iw.

    And, when the velocity used in the calculations must be the tangential velocity.

    Is this correct?
     
  10. Dec 13, 2014 #9

    ehild

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    Yes, L=Iw. And you get it by summing (integrating) the contribution rxp of all particles of that rigid body, using the fact that the angular frequency is the same for all of them.
     
  11. Dec 13, 2014 #10
    Thank you.
     
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