# Magnitude of the time derivative of angular momentum

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

• What do you think the result should be? Jul 31, 2022 at 7:56
• I dont know to be honest, could you give me a tip?
– IBI
Jul 31, 2022 at 8:12
• Your question is unclear. What causes trouble for you? Is it $\omega$? Is it $\cos(\theta)$? Is it the projection? Is it the fact that you're taking the derivative of a moving vector? If you don't give more details about the exact point of the computation that leaves you stuck, it's difficult to pinpoint the problem. Jul 31, 2022 at 8:20
• its why we mulitply $w$ 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 $Lcos(\theta)$ 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?
– IBI
Jul 31, 2022 at 8:31
• You should add this to your question (by editing). Comments can be deleted, so important details mustn't be there. Jul 31, 2022 at 8:35

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.