Is energy always conserved in a co-rotating frame?

[phantomvommand]

Is energy always conserved in a co-rotating/accelerating frame?

 

[vanhees71]

This is a bit too unspecific. How is your frame rotating? The way to check "energy conservation" is to calculate the Lagrangian of the system in terms of the generalized coordinates parametrizing the in this case non-inertial frame. If the Lagrangian is not explicitly time-dependent then the associated Hamiltonian is conserved and you could with some right of analogy call the Hamiltonian the energy of the system.

 

[Dale]

As @vanhees71 said, there is insufficient detail to answer the question. You would have to describe accurately the rotation/acceleration in question. You could do that either with a coordinate transform between an inertial frame and yours, or with the metric or Lagrangian written directly in your frame's coordinates.

 

[ergospherical]

As I think I wrote in one of your previous threads, there is a simple form of energy conservation that applies to uniformly rotating, non-translationally-accelerating frames, viz: $\dfrac{d}{dt} \left( T - \frac{1}{2}I \Omega^2 \right) - \displaystyle{\sum_a} \mathbf{F}_a \cdot \mathbf{v}_a = 0$. If all of the $\mathbf{F}_a$ are conservative then $\displaystyle{\sum_a} \mathbf{F}_a \cdot \mathbf{v}_a$ is a total time derivative and you have a conserved energy.

For other systems, whether or not you can find energy integrals depends on whether there is time dependence in the lagrangian i.e. write $H = \dot{q}^i \dfrac{\partial L}{\partial \dot{q}^i} - L$ then if $\partial L/\partial t=0$ you have$$\dfrac{dH}{dt} = \dot{q}^i \dfrac{d}{dt} \dfrac{\partial L}{\partial \dot{q}^i} + \dfrac{\partial L}{\partial \dot{q}^i} \ddot{q}^i - \dfrac{\partial L}{\partial q^i} \dot{q}^i - \dfrac{\partial L}{\partial \dot{q}^i} \ddot{q}^i$$which equals zero.