Pulling with a Winch (Part 2)

I hacked at this like an amateur, but I got there...mostly by luck.

Here is a diagram of my initial setup:

Then from Newton's Second Law around the hinge, $\sum{T} = I\alpha$

$F\sin\beta L = \frac{1}{3}ML^2\ddot{\theta}$

Then to get $\beta$ in terms of $\theta$ use the Law of Sines:

$\displaystyle \frac{\sin\beta}{\sqrt{13}} = \frac{\sin(\pi - \arctan{\frac{3}{2}} - \theta)}{x} \Rightarrow \sin\beta = \frac{\sqrt{13}\sin(\pi - \arctan{\frac{3}{2}} - \theta)}{x}$

Then just to clean it up: $\pi - \arctan{\frac{3}{2}} = \phi$

Now to get $x$ in terms of $\theta$ using the Law of Cosines:

$x^{2} = (\sqrt{13})^{2} + L^{2} - 2L\sqrt{13}\cos(\phi-\theta) \Rightarrow x = \sqrt{ 13 + L^{2} - 2L\sqrt{13}\cos(\phi-\theta) }$

Making the proper substitutions and rearranging the second order non-linear ODE is obtained:

$\displaystyle \ddot{\theta} = \frac{3\sqrt{13}F}{ML} \frac{\sin(\phi-\theta)}{\sqrt{ 13 + L^{2} - 2L\sqrt{13}\cos(\phi-\theta) }}$

From this point I had to veer from formal technique ( I 'm sure some will greatly simplify and formalize the solution technique ). I plotted $\ddot{\theta} \,vs.\, \theta$ , broke it into 0.1 radian sections and graphically linearized each portion ( Using Excel Linear Trendlines ) The result of that process is shown below.

From here it was a matter of solving a series of 16 linear second order linear ODE's of the form:

$\ddot{\theta} = A\theta+B$ for which there are two solutions types,

$Case \,1: \underline{A>0}$

$\displaystyle \theta(t) = C_{1}e^{\sqrt{A}t}+C_{2}e^{-\sqrt{A}t} - \frac{B}{A}$

$Case \,2: \underline{A<0}$

$\displaystyle \theta(t) = C_{1}\cos(\sqrt{A}t)+C_{2}\sin(\sqrt{A}t) + \frac{B}{A}$

subject to the initial conditions $\theta(t_{o}) = \theta_{o}$ , and $\dot{\theta}(t_{o}) = \omega_{o}$ . Then I graphically determined the time $(t_{f})$ at the end of each interval using the outputs of each interval $(\theta_{f}, \, \omega_{f}, \, t_{f})$ as the inputs for the following interval.

Here are few sample calculations ( performed in Mathcad), the last of which shows my final result of 0.3649 seconds.