I How to introduce dissipation to a spinning top

AI Thread Summary
The discussion focuses on modifying the Lagrangian of an axisymmetric spinning top to account for dissipation at the pivot point. It emphasizes that the Lagrangian cannot be altered directly for dissipative systems, as they do not conform to Hamiltonian mechanics. Instead, general equations are suggested to describe the system's dynamics, incorporating a dissipation torque. The relationship between the generalized forces and the velocities is highlighted, indicating that dissipation results in non-positive work. Overall, the conversation centers on the complexities of incorporating dissipation into the Lagrangian framework.
etotheipi
A axisymmetric spinning top is pivoted at O. The components of the inertia tensor ##I_O## at the point ##O##, with respect to the principal axes, are denoted ##A##, ##A## and ##C##. It's Lagrangian is$$\mathcal{L}(\mathbf{q}, \dot{\mathbf{q}}) = \frac{1}{2} A\dot{\theta}^2 + \frac{1}{2}A(\dot{\phi} \sin{\theta})^2 + \frac{1}{2}C(\dot{\psi} + \dot{\phi} \cos{\theta})^2 - mgh\cos{\theta}$$How can the Lagrangian be modified to account for dissipation at the pivot? We can't use the Rayleigh function here, because that is for velocity-dependent dissipation. Are there some references?
 
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etotheipi said:
How can the Lagrangian be modified to account for dissipation at the pivot?
You can not modify the Lagrangian in such a way because the system with dissipation is not a Hamiltonian system. Use general equations:
$$\frac{d}{dt}\frac{\partial L}{\partial \dot x^i}-\frac{\partial L}{\partial x^i}=Q_i(t,x,\dot x)$$
Dissipation means that ##Q_i\dot x^i\le 0##

If say you apply a dissipation torque ##\boldsymbol \tau## then
$$Q_i=\Big(\boldsymbol\tau,\frac{\partial\boldsymbol\omega}{\partial\dot x^i}\Big)$$
 
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