Answer
(a) $\omega = 3.75~rad/s^2$
(b) $\omega = 5.0~rad/s^2$
(c) $\omega = 5.0~rad/s^2$
Work Step by Step
The change in angular velocity is equal to the area under the angular acceleration versus time graph. Note that the initial angular velocity is zero.
From t = 0 to t = 2 s:
$\alpha(t) = (\frac{-5}{2}~t+5)~rad/s^2$
$\omega(t) = \omega_0+\int_{0}^{t}\alpha(t)~dt$
$\omega(t) = 0+\int_{0}^{t}(\frac{-5}{2}~t+5)~rad/s^2~dt$
$\omega(t) = (\frac{-5}{4}~t^2+5t)~rad/s$
(a) At t = 1 s:
$\omega = [\frac{-5}{4}~(1)^2+(5)(1)]~rad/s$
$\omega = 3.75~rad/s^2$
(b) At t = 2 s:
$\omega = [\frac{-5}{4}~(2)^2+(5)(2)]~rad/s$
$\omega = 5.0~rad/s^2$
(c) From t = 2 s to t = 3 s, the area under the angular acceleration versus time graph is zero. Therefore, the angular velocity does not change from t = 2 s to t = 3 s.
At t = 3 s:
$\omega = 5.0~rad/s^2$
