Discussion
The Lagrangian of a particle of mass m moving on the xy-plane which is rotating about z-axis with angular velocity \(\omega\) is given by :
\[ \begin{aligned} L = \tfrac{1}{2} m[&(\dot{x} - \omega y)^{2} \\ &+ (\dot{y} + \omega x)^{2}] - V(x, y) \end{aligned} \]
Then :
Answer: B
The Hamiltonian H is conserved.
The Lagrangian does not contain t explicitly, so the Jacobi integral \(H = \sum p_{i}\dot{q}_{i} - L\) is a constant of the motion.
In this rotating frame H is not the total energy. Working it out,
\[ \begin{aligned} H &= \tfrac{1}{2}m(\dot{x}^{2} + \dot{y}^{2}) \\ &\quad - \tfrac{1}{2}m\omega^{2}(x^{2} + y^{2}) + V \end{aligned} \]
which contains the centrifugal term with a minus sign, so it differs from the kinetic plus potential energy, and the energy E is not separately conserved. The momenta \(p_{x}, p_{y}\) are not conserved either, because L depends on x and y through V and through the \(\omega x\), \(\omega y\) terms.
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