#### Answer

(a) $v = 12~m/s$
(b) The disk rotates through 36 revolutions.

#### Work Step by Step

(a) We can find the angular speed after the $\frac{1}{2}~second$ acceleration period.
$\omega = \alpha ~t$
$\omega = (600~rad/s^2)(\frac{1}{2}~s)$
$\omega = 300~rad/s$
We can find the speed of the dot on the edge of the disk.
$v = \omega ~r$
$v = (300~rad/s)(0.040~m)$
$v = 12~m/s$
(b) We can find the angle $\theta_1$ through which the disk rotates in the first $\frac{1}{2}~second.$
$\theta_1 = \frac{1}{2}\alpha~t^2$
$\theta_1 = \frac{1}{2}(600~rad/s^2)(0.5~s)^2$
$\theta_1 = 75~rad$
We can find the angle $\theta_2$ through which the disk rotates in the next $\frac{1}{2}~second.$
$\theta_2 = \omega_f~t$
$\theta_2 = (300~rad/s)(0.5~s)$
$\theta_2 = 150~rad$
The disk rotates through a total angle of $75~rad+150~rad$ which is $225~rad$.
We can express this in revolutions.
$\theta = (225~rad)(\frac{1~rev}{2\pi~rad})$
$\theta = 36~rev$
The disk rotates through 36 revolutions.