Answer
$v_A=2ft/s\rightarrow$
$v_B=10ft/s\leftarrow$
yes, the cylinder slips on the conveyor.
Work Step by Step
The required velocity can be determined as follows:
$r_{C/IC}=\frac{4}{6}=0.667ft$
We know that
$v_A=\omega(r_{A/IC})$
$\implies v_A=\omega{r-r_{C/IC}}=6(1-0.667)=2ft/s \rightarrow$
and $v_B=\omega(r_{B/IC})=\omega(r+r_{C/IC})=6(1+0.667)=10ft/s \leftarrow$
As the velocity of the contact point A is not equal to that of the conveyor, therefore, the cylinder slips on the conveyer.
