Rod / Disk Combo Dynamics

Let's solve this using Lagrangian mechanics. This is possible because there are no losses (since the disk rolls without slipping). To solve, we will first need some expressions for the kinetic energy of the rolling disk, the kinetic energy of the rod, and the potential energy of the rod.

Start with the kinetic energy of the disk. We must account for rotation and translation:

$E_D = \frac{1}{2} M \, \dot{x_0}^2 + \frac{1}{2} I \omega^2 \\ = \frac{1}{2} M \, \dot{x_0}^2 + \frac{1}{2} (\frac{1}{2} M R^2) (\frac{\dot{x_0}}{R})^2 \\ = \frac{3}{4} M \dot{x_0}^2$

Next, find the kinetic energy of the rod. Begin by writing expressions for the position and velocity of an infinitesimal portion of the rod of length $d \alpha$ , located a distance $\alpha$ from the center. Recall that $\theta$ is the angle of the rod with the vertical. We are also at liberty to choose a vertical reference position for the potential energy, so take the center of the disk to be $y = 0$ .

$x = x_0 + \alpha \, sin \theta \\ y = \alpha \, cos \theta \\ \dot{x} = \dot{x_0} + \alpha \, cos \theta \, \dot{\theta} \\ \dot{y} = -\alpha \, sin \theta \, \dot{\theta}$

The velocity of the infinitesimal rod portion is:

$v^2 = \dot{x}^2 + \dot{y}^2 = \dot{x_0}^2 + 2 \alpha \, cos\theta \, \dot{x_0} \dot{\theta} + \alpha^2 \dot{\theta}^2$

The mass of the infinitesimal rod portion is:

$dm = \frac{d \alpha}{R} M$

The kinetic energy of the infinitesimal rod portion is:

$dE_R = \frac{1}{2} \, dm \, v^2 = \frac{M}{2 R} [\dot{x_0}^2 + 2 \alpha \, cos\theta \, \dot{x_0} \dot{\theta} + \alpha^2 \dot{\theta}^2] \, d \alpha$

The total rod kinetic energy is:

$E_R = \frac{M}{2 R} \int_0^R [\dot{x_0}^2 + 2 \alpha \, cos\theta \, \dot{x_0} \dot{\theta} + \alpha^2 \dot{\theta}^2] \, d\alpha \\ = \frac{M}{2 R} [\dot{x_0}^2 \, R + R^2 \, cos\theta \, \dot{x_0} \dot{\theta} + \frac{1}{3} R^3 \, \dot{\theta}^2 ] \\ = \frac{1}{2} M \dot{x_0}^2 + \frac{1}{2} M R \, cos\theta \, \dot{x_0} \dot{\theta} + \frac{1}{6} M R^2 \, \dot{\theta}^2$

The total system kinetic energy is:

$E = E_D + E_R = \frac{5}{4} M \dot{x_0}^2 + \frac{1}{2} M R \, cos\theta \, \dot{x_0} \dot{\theta} + \frac{1}{6} M R^2 \, \dot{\theta}^2$

The gravitational potential energy of the disk is irrelevant, because it does not change (per the Euler-Lagrange equations). The gravitational potential energy of the rod is:

$U = M g \, \frac{R}{2} \, cos \theta$

System Lagrangian:

$L = E - U = \frac{5}{4} M \dot{x_0}^2 + \frac{1}{2} M R \, cos\theta \, \dot{x_0} \dot{\theta} + \frac{1}{6} M R^2 \, \dot{\theta}^2 - \frac{1}{2} M g \, R \, cos \theta$

The Euler-Lagrange equations describe the system dynamics. They are:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x_0}}} = \frac{\partial{L}}{\partial{x_0}} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{\partial{L}}{\partial{\theta}}$

Computing this in pieces for the $x_0$ EL equations:

$\frac{\partial{L}}{\partial{\dot{x_0}}} = \frac{5}{2} M \dot{x_0} + \frac{1}{2} M R \, cos\theta \, \dot{\theta} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x_0}}} = \frac{5}{2} M \ddot{x_0} + \frac{1}{2} M R \,[ cos\theta \, \ddot{\theta} - sin\theta \, \dot{\theta}^2] \\ \frac{\partial{L}}{\partial{x_0}} = 0 \\ \frac{5}{2} M \ddot{x_0} + \frac{1}{2} M R \,[ cos\theta \, \ddot{\theta} - sin\theta \, \dot{\theta}^2] = 0 \\ \frac{5}{2} \ddot{x_0} + \frac{1}{2} R \ cos\theta \, \ddot{\theta} = \frac{1}{2} R sin\theta \, \dot{\theta}^2$

Computing this in pieces for the $\theta$ EL equations:

$\frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{1}{2} M R \, cos\theta \, \dot{x_0} + \frac{1}{3} M R^2 \, \dot{\theta}\\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{1}{2} M R \, [ cos\theta \, \ddot{x_0} - sin\theta \, \dot{x_0} \dot{\theta} ]\ + \frac{1}{3} M R^2 \, \ddot{\theta} \\ \frac{\partial{L}}{\partial{\theta}} = -\frac{1}{2} M R \, sin\theta \, \dot{x_0} \, \dot{\theta} + \frac{1}{2} M g R \, sin \theta\\ \frac{1}{2} M R cos\theta \, \ddot{x_0} + \frac{1}{3} M R^2 \, \ddot{\theta} = \frac{1}{2} M g R \, sin \theta \\ \frac{1}{2} cos\theta \, \ddot{x_0} + \frac{1}{3} R \, \ddot{\theta} = \frac{1}{2} g \, sin \theta$

Summarizing the results:

$\frac{5}{2} \ddot{x_0} + \frac{1}{2} R \ cos\theta \, \ddot{\theta} = \frac{1}{2} R sin\theta \, \dot{\theta}^2 \\ \frac{1}{2} cos\theta \, \ddot{x_0} + \frac{1}{3} R \, \ddot{\theta} = \frac{1}{2} g \, sin \theta$