Answer
$\omega = 98~rpm$
Work Step by Step
The change in angular velocity is equal to the area under the angular acceleration versus time graph from t = 0 to t = 3.0 s.
$\Delta \omega = \frac{1}{2}(4~rad/s^2)(2~s)$
$\Delta \omega = 4~rad/s$
We can convert this to units of rpm.
$\Delta \omega = (4~rad/s)(\frac{1~rev}{2\pi~rad})(\frac{60~s}{1~min})$
$\Delta \omega = 38~rpm$
We can find the angular velocity at t = 3.0 s.
$\omega = \omega_0+\Delta \omega$
$\omega = 60~rpm+38~rpm$
$\omega = 98~rpm$
