I Component of angular momentum perpendicular to rotation axis

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1. Apr 2, 2016

Soren4

Consider the rotation of a rigid body about a fixed axis $z$, not passing through a principal axis of inertia of the body. The angular momentum $\vec{L}$ has a parallel component to the $z$ axis (called $\vec{L_z}$) and a component perpendicular to it (called $\vec{L_n}$). I have some doubts on $\vec{L_n}$.

From the picture we have that, taking a singular point of the rigid body $P_i$,

$\mid \vec{L_{n,i}}\mid=m_i r_i R_i \omega cos\theta_i\implies \mid \vec{L_n}\mid=\omega \sum m_i r_i R_i cos\theta_i$

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

So is it possible to say that $\mid \vec{L_n}\mid \propto \mid \vec{\omega} \mid$ (1)?

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

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

If (1) holds then $\frac{d\mid \vec{L_n} \mid}{dt} \propto \mid \vec{\alpha} \mid$

But what about the direction of $\vec{L_n}$ and of its derivative?

2. Apr 4, 2016

BiGyElLoWhAt

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.