Lagrangian of a precessing wheel: gyroscope in Lagrangian mechanics?

Let's consider this classical example:

I would write the Lagrangian as the one of the spherical pendulum plus the rotational energy of the wheel, but I fail to see how the precession motion would arise from that. I feel like the two components should be somehow coupled. I would greatly appreciate some guidance.

Here is my attempt which I believe is wrong, as I fail to see how that would create the precession motion. Taking the wheel as an ideal ring with radius $$R$$ and mass $$m$$ and indicating with $$r$$ the arm length:

$$\displaystyle L=L_\text{pend} + L_\text{rot}$$

where:

$$\displaystyle L_\text{pend}=\frac{1}{2} mr^2\left( \dot{\theta}^2+\sin^2\theta\ \dot{\phi}^2 \right) + mgr\cos\theta$$

and

$$\displaystyle L_\text{rot} = \frac{1}{2}I\dot\alpha^2 = \frac{1}{2}m R^2\dot\alpha^2$$

• Can you show your attempt of the Lagrangian? I guess it might miss something in the kinetic energy part. Apr 12 '20 at 13:00
• You can find a great solution in Understanding Precession (chapter 3, Solving the Bicycle Wheel Gyroscope with Lagrangian Mechanics). Apr 12 '20 at 15:08
• @ThomasFritsch Please see the edit Apr 12 '20 at 15:25
• @ThomasFritsch Wow, supercool link, thank you! Apr 12 '20 at 15:29

I) choose coordinate system $$(x,y,z)$$

II) choose the generalized coordinates $$\varphi\,,\psi$$

Gyroscope

III) create the rotation matrix $$R$$ via the generalized coordinates and in your case also time depended $$\Omega\,\tau$$ . where $$\Omega$$ is the wheel rotation about the y axis and $$\tau$$ is the time

$$R=S_x(\varphi)\,S_z(\psi)\,Sy(\Omega\tau)$$

(I choose this sequence of the rotation matrix to avoid singularity ,this is important for numerical simulation )

Where: $$S_x=\left[ \begin {array}{ccc} 1&0&0\\0&\cos \left( \varphi \right) &-\sin \left( \varphi \right) \\ 0 &\sin \left( \varphi \right) &\cos \left( \varphi \right) \end {array} \right]$$

$$S_z=\left[ \begin {array}{ccc} \cos \left( \psi \right) &-\sin \left( \psi \right) &0\\\sin \left( \psi \right) &\cos \left( \psi \right) &0\\ 0&0&1\end {array} \right]$$

$$S_y=\left[ \begin {array}{ccc} \cos \left( \Omega\,\tau \right) &0&\sin \left( \Omega\,\tau \right) \\ 0&1&0 \\ -\sin \left( \Omega\,\tau \right) &0&\cos \left( \Omega\,\tau \right) \end {array} \right]$$

IV) obtain the angular velocity vector $$\vec{\omega}$$ out of the rotation matrix $$R$$

$$\vec{\omega}=J_R\,\vec{{\dot{q}}}+\vec{\omega}_\tau$$

where

$$J_R= \left[ \begin {array}{cc} \cos \left( \psi \right) \cos \left( \Omega \,\tau \right) &-\sin \left( \Omega\,\tau \right) \\ -\sin \left( \psi \right) &0\\ \cos \left( \psi \right) \sin \left( \Omega\,\tau \right) &\cos \left( \Omega\,\tau \right) \end {array} \right]$$

$$\vec{{\dot{q}}}= \left[ \begin {array}{c} \dot{\varphi} \\ {\it \dot{\psi}} \end {array} \right]$$

$$\vec{\omega}_\tau=\left[ \begin {array}{c} 0\\ \Omega \\ 0\end {array} \right]$$

V) Kinetic Energy

$$T_g=\frac{1}{2}\vec{\omega}^T\,J\,\vec{\omega}$$

where J is the inertia tensor of the wheel with $$J_z=J_x$$

$$J=\left[ \begin {array}{ccc} J_{{x}}&0&0\\ 0&J_{{y}}&0 \\0&0&J_{{x}}\end {array} \right]$$

VI) generalized torques

the torque about the x axis is $$\tau_\varphi=-m\,g\,a$$

thus the generalized torques are:

$$\begin{bmatrix} \tau_x \\ \tau_z \\ \end{bmatrix}=J_R^T\,\begin{bmatrix} \tau_\varphi \\ 0 \\ 0\\ \end{bmatrix}= \left[ \begin {array}{c} -\cos \left( \psi \right) \cos \left( \Omega \,\tau \right) a\,m\,g\\ \sin \left( \Omega\,\tau \right) a\,m\,g\end {array} \right]$$

To get the generalized torques in the equations of motion,you have to extend the kinetic energy $$T_g$$ to $$T_g\mapsto T_g+\tau_x\,\varphi+\tau_z\psi$$

Thus the Lagrange $$L$$ is:

$$L=T_g$$

I admit that this is quite elaborate calculations,so use symbolic manipulated program like Maple to obtain the results.