Classical mechanics marion and thornton

[ehrenfest]

[SOLVED] classical mechanics marion and thornton

Homework Statement

At the beginning of section 12.4 in marion and thornton, they say that $\dot{q}_k$ and [itex]\ddot{q}_k[/tex] are both 0 at equilibrium, where these are generalized coordinates. Can someone please explain how they got that the latter? Actually and the former? Is that part of the definition of equilibrium? Where was that defined?<br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2>[/itex]

 

[Ben Niehoff]

Those are just the generalized velocity and acceleration, respectively. If an object has zero velocity, it is not moving. If it also has zero acceleration, then it is not going to start moving, either; i.e., the velocity will stay at zero. What is the definition of equilibrium? It is when a system is stationary and remaining stationary.

 

[ehrenfest]

What is the definition of equilibrium? It is when a system is stationary and remaining stationary.

I thought equilibrium was defined in terms of potential. That is, I thought an equilibrium was when the first derivative of the potential was 0.

By your logic, I would think the particle would also need to be remaining remaining stationary i.e. the third time derivative would have to be zero. But maybe they are just making an approximation... Does anyone have the book?

 

[Mindscrape]

Thornton and Marion kind of botch the chapter on coupled oscillations. You should see if your library has Taylor.

 

[nicksauce]

I thought equilibrium was defined in terms of potential. That is, I thought an equilibrium was when the first derivative of the potential was 0.

By your logic, I would think the particle would also need to be remaining remaining stationary i.e. the third time derivative would have to be zero. But maybe they are just making an approximation... Does anyone have the book?

I think that since a particle's motion is going to be defined by a second order ODE, then you just need the first two derivatives to be zero for equilibrium. Go check out uniqueness of solution theorems in an ODE book or something.

 

[D H]

I thought equilibrium was defined in terms of potential. That is, I thought an equilibrium was when the first derivative of the potential was 0.

By your logic, I would think the particle would also need to be remaining remaining stationary i.e. the third time derivative would have to be zero. But maybe they are just making an approximation... Does anyone have the book?

One does not need to look past the second time derivative if the first two time derivatives completely define the dynamics of some system. Higher order time derivatives are superfluous. If the first two time derivatives do completely define the dynamics of some system and these derivatives are both zero all higher order time derivatives are necessarily zero as well.

Where potential comes into play is in determining whether an equilibrium point is stable, quasi-stable, or unstable. The equilibrium point is stable if all sufficiently small deviations from the equilibrium point result in an increase in potential energy, unstable if some small deviation from the equilibrium point result in a decrease in potential energy, and quasi-stable if slight deviations from the equilibrium point don't affect the potential energy. Simple example: a pendulum has two equilibrium points, with the pendulum bob either directly below (stable) or directly above (unstable) the pivot point.