Answer
(a) It takes 14.5 days for the rotor to come to rest.
(b) The rotor spins through $4.0\times 10^{10}$ revolutions.
Work Step by Step
(a) We can find the time it takes for the rotor to come to rest:
$\omega_f = \omega_0+\alpha~t$
$t = \frac{\omega_f - \omega_0}{\alpha}$
$t = \frac{0 - 5.0\times 10^5~rad/s}{-0.40~rad/s^2}$
$t = 1.25\times 10^6~s$
$t = 347.2~hours$
$t = 14.5~days$
It takes 14.5 days for the rotor to come to rest.
(b) We can find the magnitude of the angular displacement:
$\Delta \theta = \frac{1}{2}\alpha~t^2$
$\Delta \theta = \frac{1}{2}(0.40~rad/s^2)(1.25\times 10^6~s)^2$
$\Delta \theta = 2.5\times 10^{11}~rad$
We can express the angular displacement in units of revolutions:
$\Delta \theta = 2.5\times 10^{11}~rad\times \frac{1~rev}{2\pi~rad} = 4.0\times 10^{10}~rev$
The rotor spins through $4.0\times 10^{10}$ revolutions.
