Answer
The magnitude of the angular acceleration is $~~9.30~rad/s^2$
Work Step by Step
In part (a), we found that the angular speed of the tires is $~~59.2~rad/s$
We can let $\omega_0 = 59.2~rad/s$
We can find $\theta$ as the car comes to a stop:
$\theta = (2\pi~rad)(30.0) = (60.0~\pi)~rad$
We can find the angular acceleration:
$\omega_f^2 = \omega_0^2+2\alpha \theta$
$\alpha = \frac{\omega_f^2 - \omega_0^2}{2~ \theta}$
$\alpha = \frac{0 - (59.2~rad/s)^2}{(2)(60.0~\pi~rad)}$
$\alpha = -9.30~rad/s^2$
The magnitude of the angular acceleration is $~~9.30~rad/s^2$
