## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$\omega = 0.5~rad/s$
The area under the torque versus time graph is equal to the change in angular momentum. We can find the area under the graph. $area = \frac{1}{2}(2~N~m)(2~s)$ $area = 2~kg~m^2/s$ The total change in angular momentum from t= 0 to t = 3 s is $2~kg~m^2/s$. Since the object started from rest, the object's angular momentum at t = 3.0 s is $2~kg~m^2/s$. We can find the angular velocity at t = 3.0 s. $L = I~\omega$ $\omega = \frac{L}{I}$ $\omega = \frac{2~kg~m^2/s}{4.0~kg~m^2}$ $\omega = 0.5~rad/s$