A cylindrical rod of mass M, length L and radius R has two cords wound around it whose ends are attached to the ceiling. The rod is held horizontally with the two cords vertical. When the rod is released, the cords unwind and the rod rotates the linear acceleration of the cylinder as it
falls, is:

The cylinder rotating under gravity has both translational and rotational motions. Let v be the linear velocity of its centre of mass and $\mathrm{\omega }$ its angular
velocity about the axis of rotation; then in descending a distance h it will lose PE = Mgh while gain in

$\mathrm{KE} = (\frac{1}{2}{\mathrm{I\omega }}^2+\frac{1}{2}{\mathrm{Mv}}^2)$

Now Mgh = $\frac{1}{2}{\mathrm{I\omega }}^2+\frac{1}{2}{\mathrm{Mv}}^2$

Here $\mathrm{I} = \frac{1}{2}{\mathrm{MR}}^2 \mathrm{and} \mathrm{v} = \mathrm{R\omega }$

So $\mathrm{Mgh} = \frac{1}{2}{\mathrm{Mv}}^2+\frac{1}{2}\left(\frac{1}{2}{\mathrm{MR}}^2\right)\left(\frac{{\mathrm{v}}^2}{{\mathrm{R}}^2}\right)$

        ${\mathrm{v}}^2 = \frac{4}{3}\mathrm{gh}$

Differentiating

$2\mathrm{v}\frac{\mathrm{dv}}{\mathrm{dt}} = \frac{4}{3}\mathrm{g}\frac{\mathrm{dh}}{\mathrm{dt}} \mathrm{or} \mathrm{a} = \frac{2\mathrm{g}}{3}$