Mutual Gravitational Interaction

None, except you asked for a decimal solution. Another time you could say that $\omega = \tfrac{1}{a}\sqrt{b}$ where $a,b$ are positive integers and $b$ is square-free, and ask (for example) for $a+b=38$ . I think this is a nicer way to handle exact answers.

Suppose that the normal reaction between the cylinder and the wedge is $N$ , and that the normal reaction between the wedge and the slope is $R$ . Let the friction force between the cylinder and the wedge be $F$ . Suppose that the acceleration of the wedge is $a$ , and that the angular acceleration of the cylinder is $\alpha$ . The wedge has acceleration $a\binom{-\cos\theta}{-\sin\theta}$ Since the cylinder rolls on the wedge, the acceleration of the point of contact of the cylinder with the wedge must be the same as that of the wedge. Thus the centre of mass of the cylinder has acceleration $a\binom{-\cos\theta}{-\sin\theta} + r\alpha\binom{1}{0}$ where $r$ is the radius of the cylinder. The equations of motion of the centre of mass of the cylinder and the wedge are. respectively $\begin{aligned} \binom{-F}{N - mg} & = \; ma\binom{-\cos\theta}{-\sin\theta} + mr\alpha\binom{1}{0} \\ \binom{F-R\sin\theta}{R\cos\theta - N - Mg} & = \; Ma\binom{-\cos\theta}{-\sin\theta} \end{aligned}$ and the angular equation of motion of the cylinder is $Fr \; = \; mr^2\alpha$ Thus we deduce that $F = mr\alpha$ and hence that $r\alpha = \tfrac12a\cos\theta$ , and, finally $a \; = \; \frac{2(M+m)g\sin\theta}{2(M+m) - m\cos^2\theta}$

The value of $\omega$ used here is after seeing the closed-form expression. In my iterations, I was further off the mark, so the trajectories diverged earlier. I think this also has to do with the numeric complexities. The system solution is very sensitive to the parameter $\omega$

Let $x$ be the extension of the string, so that $\ddot{x}$ is the horizontal acceleration of the centre of mass of the wheel, as well as the downward accelerations of the two masses. Let the tension in the top section of the string be $T_1$ , and the tension in the bottom section be $T_2$ . Let the angular velocity of the wheel be $\omega$ . Let the friction force acting on the wheel from the table be $F$ . The equations of motion of the centre of the wheel, $A$ and $B$ are $\begin{aligned} F + T_1 - kx & = \; m\ddot{x} \\ T_2 + mg - T_1 & = \; m\ddot{x} \\ 3mg - T_2 & = \; 3m\ddot{x} \end{aligned}$ respectively, while the angular equation of motion of the wheel is $-Fr \; = \; mr^2\dot{\omega}$ where $r$ is the radius of the wheel. Since the wheel rolls on the table, we deduce that $\dot{x} - r\omega = 0$ , so that $r\omega = \dot{x}$ , and hence $F = -m\ddot{x}$ . Adding the first three equations gives $\begin{aligned} F + 4mg - kx & = \; 5m\ddot{x} \\ 6m\ddot{x} + kx & = \; 4mg \end{aligned}$ and hence, since the system starts from rest with $x=0$ , we deduce that $x \; = \; \frac{4mg}{k}\Big[1 - \cos\left(t\sqrt{\tfrac{k}{6m}}\right)\Big]$ As might be suspected, the system oscillates. Then $T_2 \; =\; 3m(g - \ddot{x}) \; = \; mg\Big[3 - 2\cos\left(t\sqrt{\tfrac{k}{6m}}\right)\Big]$ making the maximum value of $T_2$ equal to $5mg$ .