The linear acceleration $a$ for an object rolling down an incline without slipping is: $a=\frac{g\sin \theta }{1+\beta }$ where $\beta =\frac{I}{m{r}^2}$ For a solid cylinder, $I=\frac{1}{2}m{r}^2$, so $\beta =\frac{1}{2}$. $a=\frac{g\sin \theta }{1+1/2}=\frac{g\sin \theta }{3/2}=\frac{2}{3}g\sin \theta$. Note: This acceleration is less than $g\sin \theta$ (acceleration of sliding) because some of the potential energy is converted into rotational kinetic energy.