Answer
(a) $v = 75.1~m/s$
(b) $a_{rad} = 54200~m/s^2$
Work Step by Step
(a) The angular velocity of the motor is 3450 rev/min. Since the shaft has half the diameter, then the angular velocity is double, that is, 6900 rev/min.
$\omega = (6900~rev/min)(\frac{1~min}{60~s})(2\pi~rad/rev)$
$\omega = 722.57~rad/s$
We can find the tangential speed on the rim of the blade.
$v = \omega ~r = (722.57~rad/s)(0.104~m)$
$v = 75.1~m/s$
(b) $a_{rad} = \frac{v^2}{r} = \frac{(75.1~m/s)^2}{0.104~m}$
$a_{rad} = 54200~m/s^2$
