#### Answer

(a) $\omega = 640~rpm$
(b) v = 40 m/s
(c) v = 0

#### Work Step by Step

(a) We can find the angular velocity in rad/s
$\omega = \frac{v}{r}$
$\omega = \frac{20~m/s}{0.30~m}$
$\omega = 66.7~rad/s$
We can convert the angular velocity to units of rpm.
$\omega = (66.7~rad/s)(\frac{60~s}{1~min})(\frac{1~rev}{2\pi~rad})$
$\omega = 640~rpm$
(b) At the top edge of the tire, the speed from the rotation is in the same direction as the speed $v_{cm}$ of the center of mass of the tire. We can find the speed of a point at the top of the tire.
$v = \omega~r+v_{cm}$
$v = (66.7~rad/s)(0.30~m)+20~m/s$
$v = 40~m/s$
(c) At the bottom edge of the tire, the speed from the rotation is in the opposite direction from the speed $v_{cm}$ of the center of mass of the tire. We can find the speed of a point at the top of the tire.
$v = v_{cm}-\omega~r$
$v = (20~m/s) - (66.7~rad/s)(0.30~m)$
$v = 0$