Magnitude of the time derivative of angular momentum

Question

This is a problem involving a stick of mass m and length l which spins with frequency ω around an axis, as shown in the figure below. The stick makes an angle θ with the axis

The goal was to find the angular momentum and the magntude of the time derivative of the angular momentum. I completely understand how to calculate the angular momentum but i am confused with the magntiude of the derivative.

I do understand that only the horizontal component of the angular momentum vector changes over time in a circular motion and the vertical component stays fixed. I do understand that the horizontal component of $L$ is $Lcos(\theta)$. I do not understand why $|dL/dt|$ is equal to $w * Lcos(\theta)$.

EDIT : Why do we multiply $w$ to the horizontal component of L to get the |dL/dt|? $w$ is the angular velocity of the horizontal component of L around that axis and $Lcos(θ)$ is the magnitude of that horizontal component, or the length of that horizontal vector. Why do wet get the magnitude $dL/dt$ if we multiply those two?

Figure from the book "Classical Mechanics" by David Morin

Answer 1

The moment of inertia of the rod for this rotation axis is: $$J=\frac{1}{12}\,ml^2\sin(\theta)$$ so its angular momentum is: $$\vec{L}=J\omega\,\vec{u}$$ with $\vec{u}$ the unit vector carrying $\vec{L}$ according to your figure.

The general formula for the derivative of a vector gives (for a vector in pure rotation): $$\frac{d\vec{L}}{dt} =\vec{\omega}\times\vec{L} =\omega L\,\vec{u}_z\times\vec{u}$$ with $\vec{\omega}=\omega\,\vec{u}_z$ the rotation vector (alongside the rotation axis). Taking the norm on both sides: $$\left\lVert\frac{d\vec{L}}{dt}\right\rVert =\omega L\left\lvert\sin\left(\frac{\pi}{2}-\theta\right)\right\rvert =\omega L\lvert\cos(\theta)\rvert$$ and since $\theta\in[0,\pi/2]$, the absolute value isn't necessary, hence the result.