From "Classical Mechanics" - Taylor, problem 3.30
Consider a rigid body rotating with angular velocity $\omega$ about a fixed axix. (You could think of a door rotating about the axis defined by its hinges.) Take the axis of rotation to be the z-axis and use cylindrical coordinates, $\rho_\alpha, \phi_\alpha, z_\alpha$ to specify the positions of the particles $\alpha = 1, \ldots, N$ that make up the body.
a. Show that the velocity of the particle $\alpha$ is $\rho_\alpha \omega$ in the $\phi$ direction.
b. Hence show that the $z$ component of the angular momentum $l_\alpha$ of particle $\alpha$ is $m_\alpha\rho_\alpha^2\omega$.
I tried solving this as follows:
- $\mathbf{v}=\boldsymbol{\omega}\times \mathbf{r}_\alpha$ with $\mathbf{r}_\alpha = (\rho_\alpha, \phi_\alpha, z_\alpha)$. And if I calculate this cross-product $(0,0,\omega)\times (\rho_\alpha, \phi_\alpha, z_\alpha)$ I get $(-\phi_\alpha \omega, \rho_\alpha \omega, 0)$. Which confirms problem part a.
- I perform the same operation $\mathbf{l}_\alpha = \mathbf{r}_\alpha \times \mathbf{p}_\alpha = m\cdot \mathbf{r}_\alpha \times \mathbf{v}_\alpha$ using the $\mathbf{v}_\alpha$ I calculated above. The result is $$\mathbf{l}_\alpha = m(-\rho_\alpha z_\alpha \omega, \phi_\alpha z_\alpha \omega, \rho_\alpha^2 \omega + \phi_\alpha^2 \omega)$$ But this does not agree with the suggested solution. Where am I wrong?
