# What variable is the conjugate momentum for angular momentum?

From the definition of conjugate momentum for a generalized coordinate we get that the conjugate for angular momentum should be proportonal to its integral with respect to time. According to my calculations it yields zero starting from $\vec{L} = \vec{r}\times\vec{p}$, but from $\vec{L} = I\vec{\omega}$ it yields $\vec{\theta}$ (angular displacement).

Since it is possible to define an angular momentum and angular velocity for any curvilinear motion, does it make sense to define angular displacement for motions other than circular?

On the other hand, $\dot{\vec{L}} = \vec{r}\times\nabla\cal{H}$. So, does this mean that $\frac{\partial}{\partial\vec{\theta}} = \vec{r}\times\nabla$?

Furthermore, translational invariance of the Lagrangian implies conservation of linear momentum, and conversely setting $\frac{\partial\cal{L}}{\partial\vec{p}} = 0$ implies conservation of the baricenter. So if rotational invariance implies conservation of angular momentum, shouldn't $\frac{\partial\cal{L}}{\partial\vec{L}} = 0$ imply conservation of "orientation"? If so, why is there little to none discussion about this conservation law? Doesn't it set new integrals of motion?