Component of angular momentum perpendicular to the rotation axis in rigid body rotation I have difficulties in understanding, in the rotation of a rigid body, the properties of the component of the angular momentum vector $ \vec {L} $ which is perpendicular to the fixed axis of rotation $ z $. I will call this component $ \vec {L_n } $.
Suppose that the angular velocity  $ \Omega $ is constant in direction but can vary in magnitude.

$ | \vec {L_ {n, i}} | = m_i r_i R_i \Omega cos \theta_i \implies | \vec {L_n} | = \Omega \sum m_i r_i R_i cos \theta_i$
Can I therefore say that $ | \vec {L_n} | \propto | \vec {\Omega} | $ (1)?
If so, suppose to apply a torque perpendicular to the $z$ axis and parallel to $ \vec {L_n} $, so that the magnitude of this vector increases. Follows from (1) that there should be an angular acceleration $ \vec {\alpha} $, although we are in the absence of a torque with an axial component.
This would go against the fact that $ I_z \vec {\alpha} = \vec { M_z}$ (Where $I_z $ is the moment of inertia with respect to the $z$ axis and $ M_z $ is the axial component of the exerted torque).
 A: Place a coordinate system at O oriented such that $\boldsymbol{r}_i = (x_i,0,z_i)$. The body is rotating with $\boldsymbol{\omega} = (0,0,\Omega)$ and accelerating with $\boldsymbol{\alpha} = (0,0,\dot\Omega)$ about the z axis, and the mass mi lies on the xz plane. 
Then linear momentum of mi is
$$ \boldsymbol{p}_i = -m_i \boldsymbol{r}_i \times \boldsymbol{\omega} = (0,m_i\, \Omega\, x_i,0)$$
The angular momentum of mi about the origin O is
$$ {\boldsymbol{L}_i}_O = \boldsymbol{r}_i \times \boldsymbol{p}_i = (-m_i \Omega x_i z_i,0,m_i \Omega x_i^2) $$
The component you are talking about is along the y axis (perpendicular to rotation and ri) and it is zero. But that does not mean the torque about this axis is zero (actually it guarantees it is not zero). In addition, the component towards P is $\propto \Omega$ and the rotation radius $x_i$.
The total force applied on mi is found from the derivative on a rotating frame with $\Omega$ non-constant.
$$ \boldsymbol{F}_i = \frac{{\rm d}}{{\rm d}t} \boldsymbol{p}_i = \left(\frac{\partial \boldsymbol{p}_i}{\partial \Omega}\right) \dot{\Omega} + \boldsymbol{\omega} \times \boldsymbol{p}_i = (-m_i \Omega^2 x_i , m_i \dot{\Omega} x_i ,0) $$
The above is basically centrifugal force along x and tangential force along y.
The torque about the origin is similarly
$$ {\boldsymbol{M}_i}_O = \frac{{\rm d}}{{\rm d}t} {\boldsymbol{L}_i}_O = \left(\frac{\partial {\boldsymbol{L}_i}_O}{\partial \Omega}\right) \dot{\Omega} + \boldsymbol{\omega} \times {\boldsymbol{L}_i}_O = ( -m_i \dot{\Omega} x_i z_i, -m_i \Omega^2 x_i z_i, m_i \dot{\Omega} x_i ) $$
If the body is not rotationally accelerating the the torque is only about the y axis ${\boldsymbol{M}_i}_O = (0,-m_i \Omega^2 x_i z_i,0)$. So here is a situation with a moment applied perpendicularly to the rotation axis, and the magnitude of the rotation does not change.
I think you have a math error somewhere if you are arriving at a different conclusion. You can check the above by deriving the 3×3 inertia matrix about O as $${\mathtt{I}_i}_O = m_i \begin{vmatrix} y_i^2+z_i^2 & -x_i y_i & -x_i z_i\\
-x_i y_i & x_i^2+z_i^2 & -y_i z_i \\ -x_i z_i & -y_i z_i & x_i^2+y_i^2 \end{vmatrix} $$
You can verify that $${\boldsymbol{M}_i}_O = {\mathtt{I}_i}_O \boldsymbol{\alpha} + \boldsymbol{\omega} \times {\mathtt{I}_i}_O \boldsymbol{\omega}$$
which is Euler's equations of rotational motion.
PS. Trying to do 3D dynamics component by component is tedious and error prone.  
