Derivation of precession rate of a wheel gyroscope 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.
 A: 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 $\omega_{\text{precession}}$.
