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



$=\left(\frac{1}{2}\text{I}{\text{ω}}^2+\frac{1}{2}\text{M}{\text{v}}^2\right)\text{,}\text{so}$

$\text{Mgh}=\frac{1}{2}\text{I}{\text{ω}}^2+\frac{1}{2}{\text{Mv}}^2$

$\text{Here}\text{I}=\frac{1}{2}{\text{MR}}^2\text{and}\text{v}=\text{R}\text{ω}$

$\text{so}\text{Mgh}=\frac{1}{2}{\text{Mv}}^2+\frac{1}{2}\left[\frac{1}{2}{\text{MR}}^2\right]\left[\frac{{\text{v}}^2}{{\text{R}}^2}\right]$

$\text{so}{\text{v}}^2=\frac{4}{3}g\text{h}$

Now differentiating both sides with respect to time

$\text{2v}\frac{\text{dv}}{\text{dt}}=\frac{4}{3}g\frac{\text{dh}}{\text{dt}}$

$\text{or}a=\frac{2}{3}g\left[\text{as}\frac{\text{dv}}{\text{dt}}=a\text{and}\frac{\text{dh}}{\text{dt}}=\text{v}\right]$