Euler Lagrange equations with viscous dissipation

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SUMMARY

The discussion focuses on deriving the Lagrangian for a system with viscous dissipation, specifically using the Euler-Lagrange equation with dissipation. The Lagrangian is defined as ##\mathcal{L} = T - U##, where the kinetic energy is expressed as ##\frac{1}{2}J\dot{\theta}^2## and the potential energy includes gravitational and spring components. The potential energy is calculated as ##mg(1-\cos(\theta)) = \frac{mg\theta^2}{2}##, while the dissipative energy is represented as ##D = c\frac{\dot{x}^2}{2}##. The discussion emphasizes the importance of correctly identifying potential energy signs in the context of this system.

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  • Knowledge of viscous dissipation in mechanical systems
  • Basic concepts of kinetic and potential energy
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Homework Statement


The system can pivot at point O and I am taking small angle approximations.
I am trying to determine the Lagrangian, ##\mathcal{L} = T - U## for the following system:
2nh6RzK.png


Homework Equations


E-L equation with dissipation: ##\frac{\partial\mathcal{L}}{\partial q_i} - \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{q}_i} + \frac{\partial D}{\partial\dot{q}_j} = 0##

The Attempt at a Solution


I am going to use the generalized coordinate theta.

For the circular mass, I have the potential energy to be ##mg(1-\cos(\theta)) = \frac{mg\theta^2}{2}## and the kinetic energy is ##\frac{1}{2}J\dot{\theta}^2## where J is the mass moment of inertia. The potential energy of the spring is ##\frac{1}{2}kx^2##, and the dissipative energy is ##D = c\frac{\dot{x}^2}{2}##.

Before I convert the xs to thetas am I missing a kinetic or potential energy?
 
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Don't think you are missing anything. I do worry about the potential energy signs.
 

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