Exercise 4

Again we start from the Lagrangian

\[ L = \frac{m}{2} ( \dot{X}^2 + \dot{Y}^2 ) + \frac{m\omega^2}{2} ( X^2 + Y^2 ) + m \omega ( \dot{X}Y - \dot{Y}X) \]

Instead of using cartesian coordinates \(X\) and \(Y\) we now use polar coordinates \(R\) and \(\Theta\). The equations for transformation are as follows: \[ X = R \sin \Theta \] \[ Y = R \cos \Theta \]

Using Pythagoras \(\cos^2 \Theta + \sin^2 \Theta = 1\) we get: \[ X^2 + Y^2 = R^2 \]

The equations for transformation of the velocity are as follows: \[ \dot{X} = \dot{R} \cos \Theta - R \dot{\Theta} \sin \Theta \] \[ \dot{Y} = \dot{R} \sin \Theta + R \dot{\Theta} \cos \Theta \] \[ \Rightarrow \dot{X}^2 +\dot{Y}^2 = \dot{R}^2 + R^2 \dot{\Theta}^2 \]

In the Lagrangian there is the expression \(\dot{X}Y - \dot{Y}X\). Using the equations for transformation we get: \[ \dot{X}Y - \dot{Y}X = - R^2 \dot{\Theta} \]

Now we have all expressions in the Lagrangian written with polar coordinates. \[ L = \frac{m}{2} ( \dot{R}^2 + R^2 \dot{\Theta}^2) + \frac{m\omega^2}{2} R^2 - m \omega R^2 \dot{\Theta} \]

For clarification: in Lenny’s system the particle is moving with a constant speed on a straight line without forces acting on it. The Lagrangian we derived describes the same motion of the same particle, but viewed from George’s system, which rotates with angular velocity \(\omega\) with respect to Lenny’s system. In addition, the Lagrangian is written in polar coordinates. This Lagrangian describes the motion of a particle (in Lenny’s system) with constant speed on a straight line

For the special case \(\omega = 0\), George’s system does not rotate. In this case the Lagrangian is: \[ L = \frac{m}{2} ( \dot{R}^2 + R^2 \dot{\Theta}^2) \]

This Lagrangian describes the motion with constant speed on a straight line, using polar coordinates.

A great advantage of the Euler-Lagrange equation is their form invariance, they have the same form whatever coordinates are used. The equations of motion are as follows: \[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{R}} = \frac{\partial L}{\partial R} \] \[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{\Theta}} = \frac{\partial L}{\partial \Theta} \]

Calculation of the equation of motion for \(R\): \[ \frac{\partial L}{\partial \dot{R}} = m \dot{R} \] \[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{R}} = m \ddot{R} \] \[ \frac{\partial L}{\partial R} = mR \dot{\Theta}^2 + m \omega^2 R - 2 m \omega R \dot{\Theta} \] \[ = m R ( \dot{\Theta}^2 - 2 \omega \dot{\Theta} + \omega^2 ) \] \[ = m R ( \dot{\Theta} - \omega)^2 \] \[ \ddot{R} = R ( \dot{\Theta} - \omega)^2 \]

Calculation of the equation of motion for \(\Theta\): \[ \frac{\partial L}{\partial \dot{\Theta}} = m R^2 \dot{\Theta} - m \omega R^2 = m R^2 ( \dot{\Theta} - \omega) \] \[ \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{\Theta}} = 2 m R \dot{R} ( \dot{\Theta} - \omega) + mR^2 \ddot{\Theta} \] \[ \frac{\partial L}{\partial \Theta} = 0 \] \[ 2 m R \dot{R} ( \dot{\Theta} - \omega) + mR^2 \ddot{\Theta} = 0 \] \[ \Rightarrow \ddot{\Theta} = - \frac{2 \dot{R} (\dot{\Theta} - \omega)}{R} \]