How to introduce dissipation to a spinning top

  • Context: Undergrad 
  • Thread starter Thread starter etotheipi
  • Start date Start date
  • Tags Tags
    Dissipation Spinning
Click For Summary
SUMMARY

The discussion focuses on modifying the Lagrangian of an axisymmetric spinning top to incorporate dissipation at the pivot point O. The original Lagrangian is expressed as $$\mathcal{L}(\mathbf{q}, \dot{\mathbf{q}}) = \frac{1}{2} A\dot{\theta}^2 + \frac{1}{2}A(\dot{\phi} \sin{\theta})^2 + \frac{1}{2}C(\dot{\psi} + \dot{\phi} \cos{\theta})^2 - mgh\cos{\theta}$$. It is established that the Lagrangian cannot be directly modified for dissipative forces since the system is not Hamiltonian. Instead, the general equations of motion incorporating dissipation are provided, specifically $$\frac{d}{dt}\frac{\partial L}{\partial \dot x^i}-\frac{\partial L}{\partial x^i}=Q_i(t,x,\dot x)$$, where dissipation is represented by $$Q_i\dot x^i\le 0$$.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with the inertia tensor and its components
  • Knowledge of Hamiltonian systems and their properties
  • Basic concepts of dissipation in mechanical systems
NEXT STEPS
  • Study the derivation and application of the Lagrangian equations of motion
  • Research the role of dissipation in non-Hamiltonian systems
  • Explore torque applications in rotational dynamics
  • Investigate references on dissipation in mechanical systems, particularly in spinning objects
USEFUL FOR

Physicists, mechanical engineers, and students studying dynamics and control systems, particularly those interested in the effects of dissipation in rotational motion.

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?
 
  • Like
Likes   Reactions: berkeman
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   Reactions: vanhees71 and etotheipi

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
698
  • · Replies 5 ·
Replies
5
Views
812
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
582
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 9 ·
Replies
9
Views
885
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K