I was studying the topic -"Fixed Axis Rotation " from the book Kleppner Kolenkow -The Introduction to Mechanics. Please clarify the meaning of the $v_j$ variable which I have indicated in the photo below. If I interpret $v_j$ as the speed of particle j, even for $\vec v_j$ restricted to the $xy$ plane that seems wrong in general.
Any component of $\boldsymbol{r}_j$ that is parallel to $\boldsymbol{v}_j$ is ignored due to the cross product, and only the perpendicular distance remains.
That perpendicular distance is designated as $\rho_j$ in the book.
Updated, clearer answer
I suppress the $j$ subscript indicating the $j^{th}$ particle in the following answer, but I am implicitly addressing particle j.
Using cylindrical polar coordinates, the position $\vec r = \rho\hat h + z\hat k$ where $\rho$ is the distance from the origin in the $xy$ plane, $z$ the distance in the increasing vertical direction, $\hat h$ a unit vector in the increasing $\rho$ direction, and $\hat k$ a unit vector in the increasing $z$ direction. The velocity vector $\vec v = \dot \rho \hat h + \rho \dot \phi\hat l + \dot z\hat k$ where $\phi$ is the azimuthal angle in the $xy$ plane and $\hat l$ a unit vector in the increasing $\phi$ direction. Or we can say $\vec v = v_h \hat h + v_l\hat l + v_k\hat k$ where $v_h = \dot \rho, v_l = \rho \dot \phi, v_k = \dot z$.
The $\hat k$ component of $\vec r \times m\vec v$ is $m\rho^2\dot \phi$, or $m \rho v_l$. so, $L_z = m\rho^2\dot \phi$.
I think I see your concern. To me, $v$ indicates the magnitude of $\vec v$ which is $ \sqrt{v_h^2 + v_l^2 + v_k^2} $ in general. Even for $\vec v$ restricted to the $xy$ plane where $v_k$ is zero, $v$ is not $v_l$ unless $\vec v$ has no component in the $\hat h$ direction.
In general, $\vec v = \vec \omega \times \vec r$. However, the book says "consider a body rotating about the z axis". With this constraint, $\vec \omega = \omega \hat k$ and $\omega = \dot \phi$; $v_h$ is zero, $v = v_l = \rho \omega$, and $L_z = m \rho^2 \omega$. Note, even with this constraint, $\vec L = \vec r \times m \vec v = -mz\rho\omega\hat h + m\rho^2\omega\hat k$ so $\vec L$ is not totally in the z ($\hat k$) direction.
Since we are dealing with a rigid body, $\vec \omega$ is the same for every particle in the body.
