What is the Lagrangian of the curved shot in billiards?

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In billiards I can make a ball follow a curved trajectory if I aim at one side of the ball (left or right) at an angle at approximately 30 degrees from the ground.

I was wondering why this happens so I wanted to find the equations of motion of this system to make sense of it.

I think the kinetic energy of the system, is $$T=\frac{1}{2} m \left(\dot{x}^2 +\dot{y}^2 \right)+\frac{1}{2} I \dot{ \theta }^2 +\frac{1}{2} I \dot{ \phi }^2$$ Where $m$ is the mass of the ball, $I$ the moment of inertia, $\theta $ the angle of rotation of the ball around horizontal axis and $\phi $ the angle of rotation of the ball around the vertical axis.

The potential energy is $0$ So the Lagrangian is $$L=\frac{1}{2} m \left(\dot{x}^2 +\dot{y}^2 \right)+\frac{1}{2} I \dot{ \theta }^2 +\frac{1}{2} I \dot{ \phi }^2$$

The equations of motion are $$\frac{d}{dt} L_{\dot{x}_i} -L_{x_i}=f_i$$

Where $x_1=x$, $x_2=y$, $x_3=\theta $, $x_4=\phi $ and $f_i$ the force I am applying to each $x_i$ when I make the shot.

Now I think I must use $x=\theta r$ for $r$ the radius of the ball, to reduce the number of variables, but I am not quite sure.

My question is, is my approach correct so far?

Am I missing something?