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I Component of angular momentum perpendicular to rotation axis

  1. Apr 2, 2016 #1
    Consider the rotation of a rigid body about a fixed axis [itex]z[/itex], not passing through a principal axis of inertia of the body. The angular momentum [itex]\vec{L}[/itex] has a parallel component to the [itex]z[/itex] axis (called [itex]\vec{L_z}[/itex]) and a component perpendicular to it (called [itex]\vec{L_n}[/itex]). I have some doubts on [itex]\vec{L_n}[/itex].
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    From the picture we have that, taking a singular point of the rigid body [itex] P_i [/itex],

    [itex]\mid \vec{L_{n,i}}\mid=m_i r_i R_i \omega cos\theta_i\implies \mid \vec{L_n}\mid=\omega \sum m_i r_i R_i cos\theta_i [/itex]

    Is this correct? Now the quantity [itex]\sum m_i r_i R_i cos\theta_i [/itex] depends on the point [itex] O[/itex] used for the calculation of momenta, nevertheless it is a costant quantity, once calculated, right?

    So is it possible to say that [itex]\mid \vec{L_n}\mid \propto \mid \vec{\omega} \mid[/itex] (1)?

    If [itex]\vec{\omega}[/itex] is constant then there is an external torque [itex]\vec{\tau}=\frac{d\vec{L}}{dt}=\frac{d\vec{L_n}}{dt}=\vec{\omega}\times\vec{L_n}[/itex], perpendicular both to [itex]\vec{\omega}[/itex] and [itex]\vec{L_n}[/itex].

    Now suppose that the direction of [itex]\vec{\omega}[/itex] is still the same (the axis is fixed) but [itex]\mid\vec{\omega}\mid[/itex] changes. Then the external torque has two components. For the component parallel to the axis we can write [itex]\vec{\tau_z}=\frac{d\vec{L_z}}{dt}=I_z \vec{\alpha}[/itex], once calculated the moment of inertia [itex]I_z[/itex] with respect to the axis of rotation, while for the orthogonal component we have [itex]\vec{\tau_n}=\frac{d\vec{L_n}}{dt}[/itex], but what are the characteristics of the component [itex]\vec{L_n}[/itex] in that case?

    If (1) holds then [itex]\frac{d\mid \vec{L_n} \mid}{dt} \propto \mid \vec{\alpha} \mid[/itex]

    But what about the direction of [itex]\vec{L_n}[/itex] and of its derivative?
     
  2. jcsd
  3. Apr 4, 2016 #2

    BiGyElLoWhAt

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    Gold Member

    So, just to clear this up, we're calculating the angular momentum about an axis other than the axis it's rotating about? If that's the case, I think you're missing a trig function. One for the ##r_i cos(\theta_I) ## and another for ##R_i ## chunk. I might be misunderstanding what you're trying to do here, however.
    Regardless, if you start from ##I = \int_B r^2dm##, and you calculate that about an axis, that should give you the proper results for any scenario, it's just that some are easier to calculate than others.
     
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