Answer
(a) $v = 1.70~m/s$
(b) $\omega = 94.4~rad/s$
Work Step by Step
(a) We can find the tangential speed of a point on the edge of the shaft. Note that this will be the same speed as the belt.
$v = \omega ~r = (60.0~rev/s)(2\pi~rad/rev)(0.45~cm)$
$v = 170~cm/s = 1.70~m/s$
(b) We can use the tangential speed to find the angular velocity of the wheel.
$\omega = \frac{v}{r} = \frac{1.70~m/s}{0.0180~m}$
$\omega = 94.4~rad/s$
