Answer
$a=19m/s^2$
Work Step by Step
The cylinder rotates at $2rev/s$ so its period $T=0.5s$
The container's radius $r=12cm=0.12m$.
We can find the speed $v$ of the cylinder by $$v=\frac{2\pi r}{T}=1.51m/s$$
Therefore, the centripetal acceleration at the outer wall is $$a=\frac{v^2}{r}=19m/s^2$$
