Answer:
(a) Approximately [tex]105\; {\rm rad \cdot s^{-1}}[/tex].
(b) Approximately [tex]15.7\; {\rm m \cdot s^{-1}}[/tex], assuming that this cylinder is rotating along the axis that goes through the center.
Explanation:
The unit [tex]{\rm rpm}[/tex] stands for "revolutions per minute", where each revolution is [tex]2\, \pi[/tex] radians.
With an angular velocity of [tex]1000\; {\rm rpm}[/tex], this cylinder would turn [tex]1000\times 2\, \pi = 2000\, \pi[/tex] radians every minute ([tex]60\; {\rm s}[/tex]). Thus, the angular velocity of this cylinder would be:
[tex]\begin{aligned} \omega &= \frac{2000\, \pi\; {\rm rad}}{60\; {\rm s}} \\ &\approx 104.720\; {\rm rad \cdot s^{-1}}\end{aligned}[/tex].
A point at the rim of this cylinder would be at a distance of [tex]r = 15.0\; {\rm cm} = 0.150\; {\rm m}[/tex] from the axis of revolution of this cylinder. If the angular velocity of this cylinder is [tex]\omega[/tex], the tangential velocity of this point would be:
[tex]\begin{aligned} v&= \omega\, r \\ &= \frac{2000\, \pi}{60\; {\rm s}} \times 0.150\; {\rm m} \\ &\approx 15.7 \; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
