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?
 
Physics news on Phys.org
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
$$Q_i=\Big(\boldsymbol\tau,\frac{\partial\boldsymbol\omega}{\partial\dot x^i}\Big)$$
 
Last edited:
  • Informative
  • Like
Likes vanhees71 and etotheipi
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...
Back
Top