Answer
(a) The magnitude of the angular acceleration is $73.3~rad/s^2$
(b) The washer makes 23.3 revolutions.
Work Step by Step
(a) We can express the angular velocity in units of $rad/s$:
$\omega = 1400~rpm \times \frac{2\pi~rad}{1~rev} \times \frac{1~min}{60~s} = 146.6~rad/s$
We can find the angular acceleration $\alpha$:
$\omega_f = \omega_0+\alpha~t$
$\alpha = \frac{\omega_f - \omega_0}{t}$
$\alpha = \frac{146.6~rad/s - 0}{2.0~s}$
$\alpha = 73.3~rad/s^2$
The magnitude of the angular acceleration is $73.3~rad/s^2$
(b) We can find the angular displacement in this time:
$\Delta \theta = \frac{1}{2}\alpha~t^2$
$\Delta \theta = \frac{1}{2}(73.3~rad/s^2)(2.0~s)^2$
$\Delta \theta = 146.6~rad$
We can express the angular displacement in units of revolutions:
$146.6~rad \times \frac{1~rev}{2\pi~rad} = 23.3~rev$
The washer makes 23.3 revolutions.
