Answer
(a) The units of $A$ are $rad/s$, and the units of $B$ are $rad/s^3$.
(b) (i) $\alpha(t) = 0$
(ii) $\alpha(t) = 15.0~rad/s^2$
(c) The flywheel turns through an angle of $9.50 ~rad$ during the first 2.00 seconds.
Work Step by Step
(a) $\omega_z(t) = A + Bt^2$
We know that the units of $\omega$ are $rad/s$.
Thus, the units of $A$ are $rad/s$ and the units of $B$ are $rad/s^3$.
(b) $\alpha(t) = \frac{d\omega}{dt} = 2Bt$
$\alpha(t) = (2)(1.50~rad/s^3)~t$
$\alpha(t) = (3.00~rad/s^3)~t$
(i) When t = 0:
$\alpha(t) = (3.00~rad/s^3)(0) = 0$
(ii) When t = 5.00 s:
$\alpha(t) = (3.00~rad/s^3)(5.00~s) = 15.0~rad/s^2$
(c) $\theta(t) = \int_{0}^{t}~\omega ~dt$
$\theta(t) = \int_{0}^{t}~A + Bt^2 ~dt$
$\theta(t) = A~t + \frac{B}{3}~t^3$
$\theta(t) = (2.75~rad/s)~t + \frac{(1.50~rad/s^3)}{3}~t^3$
We can find the angle the flywheel turns during the first 2.00 s.
$\theta = (2.75~rad/s)(2.00~s) + \frac{(1.50~rad/s^3)}{3}(2.00~s)^3$
$\theta = 9.50~rad$
The flywheel turns through an angle of $9.50 ~rad$ during the first 2.00 seconds.
