Lagrangian equations of motion for ball rolling on turntable
The equations governing the motion of a ball of mass m, radius R rolling on a table rotating at constant angular velocity $ \Omega $ which are derived using <PERSON>'s laws are: (I present these for comparison) \begin{align} (I+mR^2) \dot{\omega}_x &= (I+2mR^2)\Omega \omega_y-mR \Omega^2y \ (I+mR^2) \dot{\omega}_y &= -(I+2mR^2)\Omega \omega_x + mR \Omega^2 x \ \dot{x} &= R \omega_y \ \dot{y} &= -R \omega_x \end{align} Where $x,y,\omega_x,\omega_y$ are absolute values measured in the rotating frame ($x,y$ being positions and $\omega_x,\omega_y$ angular velocities of the ball). To express the position, velocity, etc in the inertial $XYZ$ frame we can perform a change of variables: \begin{align} X=x \cos \theta - y \sin \theta \ Y=x \sin \theta + y \cos \theta \end{align} The equations written above are correct as far as I know. Now I've tried to derive these equations using the Lagrangian approach, but my equations differ slightly from the above. 101
I'll share my work here:
We start with the standard formulation of the Lagrangian equations of motion: $$\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_i}} \right) - \frac{\partial L}{\partial q_i}=Q_i$$ For this system, there are no non-conservative forces doing work at any time so $Q_i=0$ (Assuming no-slip here). Now the kinetic energy of the system is: $$T=\frac{1}{2}m \| v \| ^2 + \frac{1}{2} I \| \omega \|^2 $$ Where $v$,$\omega$ are the absolute linear and angular velocities of the ball, $I$ is the moment of inertia of the center of mass. I'll proceed in the rotating frame basis $xyz$, where the position $r$ of the ball is given by $$\vec{r}=x \hat{i} + y \hat{j}$$ Using velocity kinematics in rotating frames, the absolute velocity of the ball is given by: $$\vec{v}=\vec{\Omega} \times \vec{r} + \vec{v}{xyz}$$ Where $\vec{v}{xyz}$ is the velocity of the ball in the rotating frame: $$\vec{v}{xyz} = \dot{x} \hat{i} + \dot{y} \hat{j}$$ So the absolute velocity $\vec{v}$ after expanding gives: $$\vec{v} = (\dot{x}-\Omega y) \hat{i} + (\dot{y} +\Omega x) \hat{j} $$ It follows from taking the magnitude of the absolute velocity: $$\| v \|^2 = (\dot{x}-\Omega y)^2 + (\dot{y}+\Omega x)^2$$ The first unknown term in the Lagrangian equations of motion. 101