Topic: Rigid Bodies
Concepts: 5 kinematics equations, Angular kinematics

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Solution:

Let $t_1$ be the time it takes the object to accelerate with $\alpha$ to $\omega_0$ from rest. Let $t_2$ be the time the object stays rotating at $\omega_0$. Then the time it takes the object to go back to rest is also $t_1$ since it is decelerating at $-\alpha$ (same magnitude as before). To find $t_1$, we have

To find $t_2$, we calculate the total angular displacement $\theta_{\text{tot}}=3(2\pi)=6\pi$ in terms of $t_2$. Recall one of our kinematics equations for constant acceleration,

where $\omega_i$ is the initial angular velocity. During the first $t_1$ of time, we have

In the next $t_2$ of time,

In the last $t_1$ of time,

Thus,

The total time $T$ taken is

so the answer is E.