# Rotation about two axis and angular momentum

I've a body having initial angular velocity at $$t=0$$ as shown. The axis shown are fixed in inertial space and initially match with the principal axis. I want to find the infinitesimal change at $$t+\Delta t$$ in the angular momentum along the $$z$$ axis.

I've seen the following approach which I don't understand:

One contribution to change in $$L_z$$is due to rotation about y axis. This causes $$L_x$$ to rotate and hence a component $$-L_x \Delta{_y}$$ appears.

How do we know that $$L_x$$ will remain constant in magnitude? Also the actual motion won't be as is shown, in which the body simply goes around the y axis while maintaining it's spin $$L_x$$.

A similar method is used here by Kleppner and Kolenkow here

• I think the example probably means that the cylinder has a constant angular velocity about its axis $\omega_0$ and at time zero this is aligned with the $x$ axis. It is also being rotated about the direction normal to its axis, i.e. about the $y$ axis, at some constant angular velocity $\omega_y$. So the constant modulus of the angular velocity about the two axes is assumed. The point being made is that although the moduli of the two angular velocities are constant the direction of the angular velocity $\omega_0$ about the axis is changing and hence there is an angular acceleration. Jul 9, 2021 at 5:43
• The angular velocity are free to change, this is similar to Kleppner and Kolenkow when they try to derive Euler equations Jul 9, 2021 at 5:50

This is not an answer. I try to define clearly the problem.

The rotation fixed in the inertial coordinate (not along a principle axis): $$\vec \omega = \begin{pmatrix} \omega_x\\ \omega_y\\0 \end{pmatrix}.$$ Thus, the inertial moment is a function of time $$I = I(t)$$ with initial condition: $$I(t=0) = \begin{pmatrix} I_x&0&0\\ 0&I_y&0\\0&0&I_z\end{pmatrix}.$$

Find $$\frac{dL_z}{dt}\vert_{t}$$.

Is my description correct?

The question really confuses me. But I will add some details that may help clarify things. You may be familiar with the balance of angular momentum: $$\dot{\bf L} = {\bf T}$$, where $${\bf T}$$ is the applied torque to the body. In your example, $${\bf T} = {\bf 0}$$ and we have torque-free motion. Therefore, $${\bf L}$$ is a constant vector. It is frozen in inertial space and determined entirely by the initial angular velocities and moments of inertia. That is, if we have the inertial persepective.

If instead, we are rotating with the cylinder, and looking out at $${\bf L}$$, then $${\bf L}$$ will appear to change. How do we express this concept mathematically? Well, if I affix three vectors $${\bf e}_1$$, $${\bf e}_2$$, and $${\bf e}_3$$ which co-rotate with and are frozen in the cylinder, then I can use them as a basis to describe $${\bf L}$$: $${\bf L} = L_1 {\bf e}_1 + L_2 {\bf e}_2 + L_3 {\bf e}_3$$. It is important to note that, because we are changing our perspective to the co-rotating basis, $$L_i$$, the components of $${\bf L}$$ along $${\bf e}_i$$, appear to change, despite the fact that $${\bf L}$$ as a vector is frozen in inertial space and unchanging.

Doing the math then, we obtain $$\dot{\bf L} = \sum_i \dot{L}_i {\bf e}_i + \boldsymbol{\omega} \times {\bf L}$$. And we know this quantity must be the zero vector for torque-free motions. Now using $$L_i = I_i \omega_i$$, we obtain Euler's equations, where $$I_i$$ is unchanging along body-fixed axes, and $$\omega_i$$ are components of $$\boldsymbol{\omega}$$ along $${\bf e}_i$$.