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In page 296 and 297 of Kleppner and Kolenkow, The author goes over an example of a massless tilted rod rotating about the z-axis as shown in the figure having two point masses at each end:

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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

tryst with freedom
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  • You got this one dear? – Kashmiri Jun 26 '21 at 16:04
  • Dear? well yeah sort of. I thought about it a bit more and understood it like this: Imagine the mass had some size, then imagine partioning that mass into small fragments, consider a vector from the origin to one of the fragment which is off-axis , then you will see that position vector and angular velocity is not completely parallel, so the parallel component of angular velocity will also contribute angular momentum. Only on reducing the size of particle into point do we get that the angular momentum only gets contribution from perpendicular part @Kashmiri – tryst with freedom Jun 26 '21 at 16:07

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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.)

Luke
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  • So, you're saying small masses were said in statement of the cross product definition? – tryst with freedom Oct 14 '20 at 04:22
  • 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. – Luke Oct 15 '20 at 22:21