Component of angular momentum perpendicular to rotation axis

[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?

 

[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.