Answer
a) $\alpha=15\frac{rad}{s^2}$
b) $\theta-\theta_{\circ}=4.0\times 10^3 rad$
Work Step by Step
(a) We can determine the angular acceleration as
$\alpha=\frac{\omega-\omega_{\circ}}{t}$
We plug in the known values to obtain:
$\alpha=\frac{550-430}{8.2}$
$\alpha=15\frac{rad}{s^2}$
(b) We know that
$\theta-\theta_{\circ}=\frac{1}{2}(\omega_{\circ}+\omega)t$
We plug in the known values to obtain:
$\theta-\theta_{\circ}=\frac{1}{2}(430+550)(8.2)$
$\theta-\theta_{\circ}=4.0\times 10^3 rad$
