Component of angular momentum perpendicular to the rotation axis in rigid body rotation

Question

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).

Answer 1

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 m_i lies on the xz plane.

Then linear momentum of m_i is

$$ \boldsymbol{p}_i = -m_i \boldsymbol{r}_i \times \boldsymbol{\omega} = (0,m_i\, \Omega\, x_i,0)$$

The angular momentum of m_i 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 r_i) 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 m_i 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.