Derivation of angular momentum in cylindrical coordinates

Question

I tried to derive the formula for angular momentum ($\vec{l} = m\rho \phi^2 \vec{e_z}$ in the case of motion restricted to the x-y plane) in cylindrical coordinates directly from the vector cross product $m (\vec{r} \times \dot{\vec{r}})$.

Taking $\vec{r} = (\rho, \phi, z)$ and $\dot{\vec{r}} = (\dot \rho - \phi \dot\phi, \rho\dot\phi + \dot\phi, \dot z)$ as $d(\rho\vec{e_\rho})/dt = \dot\rho \vec{e_\rho} + \rho \dot\phi \vec{e_\phi}$ and $d(\phi\vec{e_\phi})/dt = \dot\phi \vec{e_\phi} - \phi \dot\phi \vec{e_\rho}$

However when taking the cross product (and taking $z = 0$ as we are moving in strictly the x-y plane), I get $\vec{l} = m(\rho \phi^2 + \rho\dot\phi - \dot\rho\phi + \phi^2\dot\phi)\vec{e_z}$ and I am unsure how to deal with the other three terms.

Is this actually the formula for angular momentum and is there just some intial assumption I am missing that gets rid of the last three terms, or have I misrepresented my vectors somehow? In the literature given to us, the derivation of the formula isn't given at all although intuitively it clearly makes sense.

Answer 1

Your expression for the velocity is just wrong. $$\tag1\dot{\vec r}=\frac{\mathrm d\vec r}{\mathrm dt}=\frac{\mathrm d\ }{\mathrm dt}\left(\varrho\hat{\vec\varrho}+z\hat{\vec z}\right)=\dot\varrho\hat{\vec\varrho}+\varrho\dot\varphi\hat{\vec\varphi}+\dot z\hat{\vec z}$$ so that the angular momentum is obtained by $$\tag2\vec\ell=m\left(\varrho\hat{\vec\varrho}+z\hat{\vec z}\right)\times\left(\dot\varrho\hat{\vec\varrho}+\varrho\dot\varphi\hat{\vec\varphi}+\dot z\hat{\vec z}\right)=m\left[-z\varrho\dot\varphi\hat{\vec\varrho}+\left(z\dot\varrho-\varrho\dot z\right)\hat{\vec\varphi}+\varrho^2\dot\varphi\hat{\vec z}\right]$$