Figure shows two blocks, each of mass $m$, suspended from the ends of a rigid weightless rod of length ${\mathcal{l}}_1+{\mathcal{l}}_2$ with ${\mathcal{l}}_1= 20 cm$ and ${\mathcal{l}}_2=80 cm$. The rod is held in horizontal position shown in figure and then released. Calculate the accelerations of the two blocks as they starts to move.

$$\begin{gathered} Net torque, {\tau }_{net}= mg{l}_2-mg{l}_1 \\ If \alpha is the angular accelerations, then \\ \alpha =\frac{{\tau }_{net}}{l} \\ =\frac{mg\left(l_2-l_1\right)}{m{l_1}^2+m{l_2}^2} \\ =\frac{g\left(l_2-l_1\right)}{{l_1}^2+{l_2}^2} \\ =\frac{9.8 \left(0.8-0.2\right)}{0.2^2+0.8^2} \\ =8.65 rad/s^2 \\ \therefore a_1=\alpha l_1 \\ =8.65\times 0.2 \\ =1.73 m/s^2 \\ a_2=\alpha l_2 \\ =8.65\times 0.8 \\ =6.92 m/s^2 \end{gathered}$$