In page 296 and 297 of Kleppner and Kolenkow, The author goes over an example of a tilted rod rotating about the z-axis as shown in the figure:

enter image description here

Details: The perpendicular to rod's length makes an angle of $\alpha $ with the z-axis

Labelled diagram:

enter image description here

The author writes the $w_{\parallel}$ doesn't contribute to angular momentum because the particles are of small size. This makes sense but he derived the angular momentum in another way by using the cross product of position vector and linear momentum. I have shown the method below:

$$ L = \sum r_i \times p_i= 2 \omega m l^2 \cos \alpha$$

The first method is said to emphasize the vector nature and the second was the regular way of following linear momentum definition. How did the regular method arrive at the same conclusion without the assumption of small particles? Or is the small particle somehow inbuilt into the cross product definition method?

The book is called an Introduction to Classical Mechanics by Kleppner and Kolenkow


The cross product equation you show is just adding up the contribution of point masses: it's just using their linear momentum $p$ and their distance from the axis of rotation $r$. There are no terms to account for anything else. As an analogy, if you model torque by taking force on a lever $\times$ length of the lever, it means you're choosing to ignore any twisting that the user might be applying to the end of the lever.

If you wanted to handle non-point masses, such that the mass's rotation about its own axis mattered, then it wouldn't be correct to represent its total contribution as $r_i \times p_i $. (You'd need to either include the angular momentum of the large mass, or split it into small point masses to analyze.)

  • $\begingroup$ So, you're saying small masses were said in statement of the cross product definition? $\endgroup$ – Buraian Oct 14 '20 at 4:22
  • $\begingroup$ yes, it's only considering their linear momentum $p$ which means it's neglecting their angular momentum. So it's assuming their angular momentum is always insignificant, which is the same as assuming they're small. $\endgroup$ – Luke Oct 15 '20 at 22:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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