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Canonical definition of Angular Momentum, need help

  1. Aug 14, 2015 #1

    guv

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    Let's start with an arbitrary solid body rotating around a fixed axis of rotation with angular velocity ##\vec \omega## in the ## \hat z## direction. For simplicity, let's say the origin O is on the axis of rotation. Take a look at the picture I sketched in the next post. Tried my best to be clear about the notations I used in my derivation.

    By definition, $$\vec L = \int_{\Omega} \vec r \times (dm \; \vec v) = \int_{\Omega} \vec r \times (\vec \omega \times \vec r) dm$$

    Where ##\vec r## is is position position vector of dm from origin O and ##\Omega## is the domain of integration (the entire solid body).

    If the angle between axis of rotation and position vector is ##\theta## (imagine ##\hat z## points upward from O, ##\vec r## points to the upper right from O), then
    $$\vec L = \int r^2 \omega sin(\theta) dm \hat L$$

    where ##\hat L## is a unit vector from dm perpendicular from ##\vec r## (imagine this unit vector from dm going upper left from dm)

    We can decompose this ##\hat L## into its z direction component and a radial component.
    $$\vec L = \int r^2 \omega sin^2\theta \hat z dm + \int r^2 \omega sin\theta cos\theta (-\hat p) dm$$

    Here is ##\hat p## is a unit vector pointing outward from dm perpendicular to axis of rotation.

    If ##\omega## is a constant, then
    $$\int r^2 \omega sin^2\theta \hat z dm = \omega \int r^2 sin^2\theta dm \hat z = I_{zz} \omega \hat z$$

    This is the z component of the angular moment since the ##\vec \omega = (0, 0, \omega)##.

    My problem is I couldn't really simply ##\int r^2 \omega sin\theta cos\theta (-\hat p) dm## into 0 when center of mass is on the axis of rotation. Can someone help? Thanks,

    guv
     
    Last edited: Aug 14, 2015
  2. jcsd
  3. Aug 14, 2015 #2

    guv

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  4. Aug 15, 2015 #3

    guv

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    The Center of Mass ##\vec R_{cm}## is defined by ##\int_{\Omega} (\vec r - \vec R_{cm}) \; dm \equiv 0##.

    My question is inspired by the derivation in Louis Brand's "Vector Analysis" book, p.176-177 when the author jumps from eq. (2) to eq. (3). I couldn't complete this derivation.
     
  5. Aug 15, 2015 #4
    Is the body rotating?? In that case doesn't sin[tex]\theta [/tex] and cos[tex]\theta [/tex] becomes sin pi/2 and cos pi/2??? so one of the trigonometric ratio becomes 0. Not sure though, gotta think. Better wait for someone else's reply.
     
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