## Essential University Physics: Volume 1 (3rd Edition)

We know that $(\alpha)=18rp\frac{m}{s^2}=(2)(\pi)(\frac{18}{60})\frac{rad}{s^2}=1.885\frac{rad}{s^2}$ The angle is $\theta=2(2\pi)=4\pi=12.566rad$ We can find the angular speed as $\omega_f^2=2\alpha \theta$ This simplifies to: $\omega _f=\sqrt{2\alpha \theta}$ We plug in the known values to obtain: $\omega_f=\sqrt{2(1.885)(12.566)}=6.9\frac{rad}{s}$ (b) We can find the required time as $t=\frac{\omega_f-\omega_i}{\alpha}$ We plug in the known values to obtain: $t=\frac{6.9-0.0}{1.885}=3.7s$