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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:

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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.

Vincent Thacker
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Sophile
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1 Answers1

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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 $\omega_{\text{precession}}$.

Gilgamesh
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  • It seems that this is incorrect. Because direction of precessional motion is in direction of torque, but direction of precessional angular speed is perpendicular to torque and angular momentum, $\vec v=\vec\omega \times\vec r$. – Neil Libertine Nov 10 '22 at 06:34
  • Precessional angular speed $\mathbf{\omega}{\text{press.}} \propto \hat{z}$ is perpendicular both to torque ($\hat{\phi}$) and angular momentum ($\hat{\rho}$). I understand that by ''precessional motion'' you mean $\mathbf{\omega}{\text{press.}} \times \mathbf{r} \propto \hat{\phi}$, in which case I agree it has the same direction than torque. I fail to see what's your point, thought. Could you point out which step you find incorrect? – Gilgamesh Nov 10 '22 at 07:25
  • You wrote above, $\tau$ is in $\hat\phi$ direction. In last two steps you derived precrssional angular speed in same direction as torque, while it is perpendicular to both torque and angular momentum. – Neil Libertine Nov 10 '22 at 07:47
  • $\tau$ is indeed in $\hat{\phi}$ direction. Last three equations are scalar equations about their magnitudes, not vectorial. I didn't wrote $\vec{\omega}_{\text{press.}} \propto \vec{\tau}$, if that's what you mean. – Gilgamesh Nov 10 '22 at 19:19
  • $\tau$ is in $\phi$'s direction. $L$ is in $\rho$. In direction of torque is $v$ so, $\vec\omega=\frac{v}{\rho}(\hat\rho\times \hat\phi)$. So precessional angular speed is in direction perpendicular to $\phi$ and $\rho$, that is vertical direction or $\hat z$. – Neil Libertine Nov 11 '22 at 03:59
  • Yes, I agree and I have never said otherwise. Again, last three equations are not vectorial but scalar (I took the inner product). That is, they are relating the magnitude of the vectorial quantity precessional angular speed with the magnitudes of the vectorial quantities angular momentum and torque. – Gilgamesh Nov 12 '22 at 03:07