# Difference between kinematic momentum and conjugated momentum in purely mechanical setup

I don't know much about physics, but I wanted to understand what was the difference between the "kinematic momentum" and the conjugated momentum. As I understand it, kinematic momentum is mass times speed and the conjugated momentum should be the one that verifies the Lagrangian equation (somehow I read it is $$\frac{\partial L}{\partial \dot{q}}$$). I was looking at this example to see when they could be different. But is there any simple purely mechanical examples, without charged particles?

The conjugate momentum corresponding to an angular coordinate is not a linear momentum but an angular momentum. For example, consider the Lagrangian of a particle of mass $$m$$ moving in a central force field described by the potential $$V(r)$$ and written in plane polar coordinates: $$L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-V(r).$$ You can easily verify that the quantity $$\frac{\partial L}{\partial\dot{r}}=m\dot{r}$$ is a linear momentum while $$\frac{\partial L}{\partial\dot{\theta}}=mr^2\dot{\theta}=mr^2\omega$$ is the angular momentum about the origin.
Addendum If $$q$$ is a generalized coordinate of linear dimension (e.g., the radial coordinate $$r$$, or cartesian coordinates $$x,y,z$$ etc), and if the potential is independent of the generalized velocities $$\dot{q}$$, the generalized velocity corresponding to $$q$$, defined as $$p_q:=\frac{\partial L}{\partial\dot{q}}$$ will always be of the form $$m\dot{q}$$, simply from dimensional analysis.
• But should it always be $m\dot{q}$, where $q$ is the generalized coordinate? – roi_saumon Mar 19 at 15:13