Rotation Revision

Let the tension of each thread be $T$ and the moment of inertia of the cylinder be $I$ and its radius be $r$ . Then $I=\dfrac {mr^2}2$ and the torque of the top cylinder is given by:

$\begin{aligned} \tau & = I \alpha_1 & \small \blue{\text{where }\alpha_1 \text{ is the angular acceleration}} \\ 2Tr & = \frac {mr^2}2\alpha_1 & \small \blue{\text{of the top cylinder.}} \\ \implies \alpha_1 & = \frac {4T}{mr} \end{aligned}$

Similarly, for the bottom cylinder, $2Tr = \dfrac {mr^2}2 \alpha_2 \implies \alpha_2 = \alpha_1$ . Now consider the translation motion of the bottom cylinder which is given by the downward force:

$\begin{aligned} F_z & = m\blue a & \small \blue{\text{where }a \text{ is the resultant acceleration.}} \\ mg - 2T & = m\blue a & \small \blue{\text{As there is no slipping of the}} \\ mg - 2T & = m\blue a & \small \blue{\text{threads on the cylinders.}} \\ & = m \blue{(\alpha_1r + \alpha_2r)} \\ & = 2m \alpha_1 r \\ & = 8T \\ \implies T & = \frac {mg}{10} \end{aligned}$

Therefore $\alpha = \boxed{0.1}$ .

Solving this problem using Lagrangian mechanics. Consider the initial separation between the cylinders' COM to be $s$ . The system is set into motion, assume that the first cylinder rotates by an angle $\theta_1$ and the second cylinder rotates by an angle of $\theta_2$ . Therefore, as the string unwinds due to the system's motion, the string increases in length by $R\theta_1+R\theta_2$ as the lower cylinder falls due to its weight. Note that $s$ is treated as a constant.

Consider the axle of the first cylinder to be the origin. Then at any general time $t$ , the y coordinate of the COM of the lower cylinder is:

$y = -(s+R\theta_1+R\theta_2)$

The kinetic energy of the system is the sum of rotational kinetic energies of the two cylinders and the translational kinetic energy of the lower cylinder. This expression is:

$\mathcal{T} = \frac{1}{2}\left(\frac{mR^2}{2}\right)\dot{\theta}_1^2 + \frac{1}{2}\left(\frac{mR^2}{2}\right)\dot{\theta}_2^2 + \frac{m}{2}\dot{y}^2$ $\implies \mathcal{T} = \frac{3mR^2\dot{\theta}_1^2}{4} + \frac{3mR^2\dot{\theta}_2^2}{4} + mR^2\dot{\theta}_1\dot{\theta}_2$

The potential energy of the system is:

$\mathcal{V} = -mgy$

Applying Lagrange's equations by treating $\theta_1$ and $\theta_2$ as generalised coordinates, gives the following equations of motion. I am leaving out this computation. The equations are:

$1.5 \ddot{\theta}_1 + \ddot{\theta}_2 = \frac{g}{R}$ $\ddot{\theta}_1 + 1.5\ddot{\theta}_2 = \frac{g}{R}$

Solving for the lower cylinder's angular acceleration about it's COM gives:

$\ddot{\theta}_2 = \frac{2g}{5R}$

Now, finally, one can compute the net torque experienced by the lower cylinder about the COG as follows. Let the tension in each string is $T$ . Then:

$2TR = \frac{mR^2}{2} \ddot{\theta}_2$ $\implies T = 0.1 mg$