More on Mechanics? Nov 21, 2016

See below.

Explanation:

We will be using the so called Euler Lagrange formulation

$\frac{d}{\mathrm{dt}} \left(\frac{\partial L}{\partial {\dot{q}}_{i}}\right) - \frac{\partial L}{\partial {q}_{i}} = {Q}_{i}$

where $L = T - V$. In this exercise we have $V = 0$ so $L = T$

Calling ${x}_{a}$ the center of left cylinder coordinate and ${x}_{b}$ the rigth one, we have

${x}_{b} = {x}_{a} + R \cos \theta + L \cos \alpha$

Here $\sin \alpha = \frac{R}{L} \sin \theta$ so substituting for $\alpha$

${x}_{b} = {x}_{a} - R \cos \theta + \sqrt{{L}^{2} - {R}^{2} {\sin}^{2} \theta}$

now deriving

${\dot{x}}_{b} = {\dot{x}}_{a} + R \sin \left(\theta\right) \dot{\theta} - \left(\frac{{R}^{2} \cos \left(\theta\right) \sin \left(\theta\right)}{\sqrt{{L}^{2} - {R}^{2} {\sin}^{2} \left(\theta\right)}}\right) \dot{\theta}$

but

$T = \frac{1}{2} J \left({\omega}_{a}^{2} + {\omega}_{b}^{2}\right) + \frac{1}{2} m \left({v}_{a}^{2} + {v}_{b}^{2}\right)$

Here $J$ is the inertia momentum regarding the mass center. Also,

${v}_{a} = {\dot{x}}_{a} = R \dot{\theta}$
${\omega}_{a} = \dot{\theta}$

so, after substitutions and calling $\xi \left(\theta\right) = 1 - \frac{R \cos \left(\theta\right)}{\sqrt{{L}^{2} - {R}^{2} {\sin}^{2} \left(\theta\right)}}$ we have

$T = \frac{1}{2} \left(J + m {R}^{2}\right) \left(1 + {\left(1 + \sin \left(\theta\right) \xi \left(\theta\right)\right)}^{2}\right) {\dot{\theta}}^{2}$

We choosed $\theta$ as the generalized coordinate. So we will reduce $F$ actuating in the coordinate $x$ to an equivalent force in $\theta$. This coordinate acts rolling wise so we need a generalized momentum regarding the contact point in the floor, which is

${Q}_{\theta} = F R \left(1 + \sin \theta\right)$

The movement equations are obtained after

$\left(J + m {R}^{2}\right) \left(\left(1 + \sin \left(\theta\right) \xi \left(\theta\right)\right) \left(\cos \left(\theta\right) \xi \left(\theta\right) + \sin \left(\theta\right) \xi ' \left(\theta\right)\right) {\dot{\theta}}^{2} + \left(1 + {\left(1 + \sin \left(\theta\right) \xi \left(\theta\right)\right)}^{2}\right) \ddot{\theta}\right) = F R \left(1 + \sin \left(\theta\right)\right)$ now solving for $\ddot{\theta}$

$\ddot{\theta} = \frac{F R \left(1 + \sin \left(\theta\right)\right) - \left(J + m {R}^{2}\right) \left(1 + \sin \left(\theta\right) \xi \left(\theta\right)\right) \left(\cos \left(\theta\right) \xi \left(\theta\right) + \sin \left(\theta\right) \xi ' \left(\theta\right)\right) {\dot{\theta}}^{2}}{\left(J + m {R}^{2}\right) \left(1 + {\left(1 + \sin \left(\theta\right) \xi \left(\theta\right)\right)}^{2}\right)}$

Attached two plots. The first shows $\theta$ evolution and the second is for $\dot{\theta}$

Value of parameters:

$R = 0.5 , J = 1 , m = 1 , L = 2$ The applied force is shown in dased red.  