Exercise 3

Our starting point is the Lagrangian given in the textbook:
\[ L = \frac{m}{2} ( \dot{X}^2 + \dot{Y}^2 ) + \frac{m\omega^2}{2} ( X^2 + Y^2 ) + m \omega ( \dot{X}Y - \dot{Y}X)\]

To get the equations of motion, we calculate the Euler-Lagrange equations: \[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{X}} - \frac{\partial L}{\partial X} = 0 \]

\[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{Y}} - \frac{\partial L}{\partial Y} = 0 \]

When calculating the partial derivative with respect to \(\dot{X}\) we consider \(X\), \(Y\) and \(\dot{Y}\) as constant values. \[ \frac{\partial L}{\partial \dot{X}} = m \dot{X} + m \omega Y \] \[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{X}} = m \ddot{X} + m \omega \dot{Y} \] \[ \frac{\partial L}{\partial X} = m \omega^2 X - m \omega \dot{Y} \]

The equation of motion for the \(X\) coordinate is: \[ m \ddot{X} + m \omega \dot{Y} = m \omega^2 X - m \omega \dot{Y} \] \[ m \ddot{X} = m \omega^2 X - 2 m \omega \dot{Y}\]

In the same way we get the equation of motion for the \(Y\) coordinate: \[ \frac{\partial L}{\partial \dot{Y}} = m \dot{Y} - m \omega X \] \[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{Y}} = m \ddot{Y} - m \omega \dot{X} \] \[ \frac{\partial L}{\partial Y} = m \omega^2 Y + m \omega \dot{X} \] \[ m \ddot{Y} - m \omega \dot{X} = m \omega^2 Y + m \omega \dot{X} \] \[m \ddot{Y} = m \omega^2 Y + 2 m \omega \dot{X}\]