I searched on the web and all I could find is the precession rate of a spinning top. But what I want is the derivation of the precession rate of a wheel hanging from a rope, as shown below:
which is taken from this video.
Professor Walter Lewin says in his video that $$\omega_{\text{precession}} = \frac{\tau}{L_{\text{spin}}} $$
But the derivation is not given in that video. So it would be much appreciated if someone will give me the derivation.
There is a more formal/mechanical way to do it, but I think this might be simple enought while being correct. There might be also a more direct way to do it.
I'll use cylindrical coordinates.
We know the torque (about the origin is) is \begin{equation} \vec{\tau} = \tau \, \hat{\phi} \end{equation} We also know, becuase it's a flat symmetric top, that the angular velocity of the body is parallel to the angular momentum of the body. Then, \begin{equation} \vec{L} = L \, \hat{\rho} \text{.} \end{equation} The relation between angular mometum and torque is \begin{equation} \frac{d}{dt} \vec{L} = \vec{\tau} \text{.} \end{equation} Note that the magnitude of the angular momentum is constant, since both torque and angular mometum are perpendicular for all time. \begin{align} L \frac{d}{dt} \hat{\rho} &= \tau \hat{\phi} \\ L \left(\frac{d}{dt} \hat{\rho}\right) \cdot \hat{\phi} &= \tau \\ \left(\dot{\phi}\hat{\phi}\right) \cdot \hat{\phi} &= \frac{\tau}{L} \\ \dot{\phi} &= \frac{\tau}{L} \end{align} Since $\dot{\phi}$ is constant (both $\tau$ and $L$ are), we call them $\omega_{\text{precession}}$.
