Angular momentum of a rotating skew rod

Homework Statement  The Attempt at a Solution

The angular momentum of the upper particle is given as
## \vec L_u = m \vec r \times ( \vec \omega \times \vec r )= m[ \vec \omega r^2 - \vec r ( \vec r . \vec \omega )] = m[ r^2 \vec \omega - \omega r \cos \alpha \vec r ]##
The angular momentum of the lower particle is same to that of the upper particle.
Hence, ##\vec L = 2 \vec L_u ##
Is this correct so far?

haruspex
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Homework Statement

View attachment 209494
View attachment 209495

The Attempt at a Solution

The angular momentum of the upper particle is given as
## \vec L_u = m \vec r \times ( \vec \omega \times \vec r )= m[ \vec \omega r^2 - \vec r ( \vec r . \vec \omega )] = m[ r^2 \vec \omega - \omega r \cos \alpha \vec r ]##
The angular momentum of the lower particle is same to that of the upper particle.
Hence, ##\vec L = 2 \vec L_u ##
Is this correct so far?
Check your expansion of ##\vec r . \vec \omega ##.

Check your expansion of ⃗r.⃗ωr→.ω→\vec r . \vec \omega .
After correction:
##
\vec L_u = m \vec r \times ( \vec \omega \times \vec r )= m[ \vec \omega r^2 - \vec r ( \vec r . \vec \omega )] = m[ r^2 \vec \omega - \omega r \sin \alpha \vec r ]##
Is this correct?

haruspex
Homework Helper
Gold Member
After correction:
##
\vec L_u = m \vec r \times ( \vec \omega \times \vec r )= m[ \vec \omega r^2 - \vec r ( \vec r . \vec \omega )] = m[ r^2 \vec \omega - \omega r \sin \alpha \vec r ]##
Is this correct?
Yes.

Yes.
But the angular momentum should be perpendicular to ## \vec r ##. Here, I am getting non-zero component in the direction of ##\hat r##.

haruspex