0
$\begingroup$

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?

"From the book Classical Mechanics by David Morin"

Figure from the book "Classical Mechanics" by David Morin

$\endgroup$
6
  • $\begingroup$ What do you think the result should be? $\endgroup$
    – Miyase
    Jul 31, 2022 at 7:56
  • $\begingroup$ I dont know to be honest, could you give me a tip? $\endgroup$
    – IBI
    Jul 31, 2022 at 8:12
  • $\begingroup$ 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. $\endgroup$
    – Miyase
    Jul 31, 2022 at 8:20
  • $\begingroup$ 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? $\endgroup$
    – IBI
    Jul 31, 2022 at 8:31
  • $\begingroup$ You should add this to your question (by editing). Comments can be deleted, so important details mustn't be there. $\endgroup$
    – Miyase
    Jul 31, 2022 at 8:35

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Thank you, that general formula was very helpful, i have tried it again my way and the result was exaclty the same. Thank you very much! Im sorry that i dont have enough reputation to upvote your post but ill accept this answer. $\endgroup$
    – IBI
    Jul 31, 2022 at 9:04
  • $\begingroup$ Keep in mind that the vector derivative formula has an extra term in the most general case. See this page. $\endgroup$
    – Miyase
    Jul 31, 2022 at 9:07
  • $\begingroup$ ahh yes, ill check that page, thank you again! $\endgroup$
    – IBI
    Jul 31, 2022 at 9:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.