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.

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