#### Answer

(a) $\alpha = 420~rad/s^2$
(b) The drill turns through 8.4 revolutions in the first 0.50 seconds.

#### Work Step by Step

(a) We first convert the final angular velocity to units of rad/s as:
$\omega_f = (2000~rpm)(\frac{1~min}{60~s})(\frac{2\pi~rad}{rev})$
$\omega_f = 209.4~rad/s$
We then find the angular acceleration.:
$\alpha = \frac{\omega_f-\omega_0}{t}$
$\alpha = \frac{209.4~rad/s-0}{0.50~s}$
$\alpha = 420~rad/s^2$
(b) We first find $\theta$ during the first 0.50 seconds;
$\theta = \frac{1}{2}\alpha~t^2$
$\theta = \frac{1}{2}(420~rad/s^2)(0.50~s)^2$
$\theta = 52.5~rad$
We then use $\theta$ to find the number of revolutions $N$:
$N = (52.5~rad)(\frac{1~rev}{2\pi~rad})$
$N = 8.4~rev$
The drill turns through 8.4 revolutions in the first 0.50 seconds.