# The Lagrangian approach

Let’s call the distance to the central axis $d$, the radius of the wheel $r$, its moment of inertia $I$ and its mass $m$. If the wheel rotates with angular speed $\omega = \dot{\theta}$, the linear speed of its center of mass will have to be $\omega r$ to avoid slipping. The potential energy will be constant and can be set to 0, while the kinetic energy $T$ can be split in two parts:

$\displaystyle T = \frac{1}{2}m(\dot{\theta} r)^2 + \frac{1}{2}I\dot{\theta}^2 = \left(m r^2 + I \right)\frac{\dot{\theta}^2}{2}$

The Lagrangian will be

$\displaystyle L = T - U = \left(m r^2 + I \right)\frac{\dot{\theta}^2}{2}$,

so $\theta$ is a cyclic coordinate and $\dot{\theta} = 0$.