Angular momentum of a rotating skew rod

[Pushoam]

Homework Statement

Homework Equations

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]

Homework Statement

View attachment 209494
View attachment 209495

Homework Equations

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 $\vec r . \vec \omega$.

 

[Pushoam]

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]

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.

 

[Pushoam]

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]

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

The other term, $\vec\omega$, also has a component in the direction of $\vec r$. The two may cancel.

 

[Pushoam]

The other term$\vec\omega$, also has a component in the direction of$\vec r$. The two may cancel.

Yes, it gets canceled.
Thanks for the guidance, thanks a lot.