Angular momentum and torque

In my most recent post I presented Newton's second law in polar coordinates, \begin{align} \vec F &= m\cdot\left(\left(\ddot r - r\dot \theta^2\right) \hat r + \left(2\dot r\dot \theta + r\ddot\theta\right)\hat\theta\right)\,, \end{align} and used this to show the angular momentum $\vec L= \vec r\times\vec p = m r^2 \dot\theta\hat z$ is conserved for a particle acted on by radial forces.

I neglected to include another immediate and very important result of this, namely the relation between torque and angular momentum. The aim of this post is to correct this omission.

Any force can be cast as $\vec F = F_r\hat r + F_{\theta}\hat\theta$; the tangential term $F_{\theta}$ vanishes for a purely radial force. The tangential term enters the cross product defining the torque $\vec \tau$, \begin{align} \vec \tau &= \vec r\times\vec F = mr\left(2\dot r\dot \theta + r\ddot\theta\right) \hat z\,. \end{align}

This is seen to be the time derivative of angular momentum, \begin{align} \frac{\mathrm d \vec L}{\mathrm d t} &= m\left(2 r \dot r \dot \theta + r^2 \ddot\theta\right)\hat z = \vec\tau\,, \end{align} and we have thus shown that torque $\vec \tau$ is the rate of change of angular momentum $\vec L$: \begin{align} \vec \tau = \frac{\mathrm d \vec L}{\mathrm d t}\,. \end{align}