While you are operating a Rotor (a large, vertical, rotating cylinder found in amusement parks), you spot a passenger in acute distress and decrease the angular velocity of the cylinder from $3.40 \mathrm{rad} / \mathrm{s}$ to $2.00 \mathrm{rad} / \mathrm{s}$ in $20.0 \mathrm{rev}$, at constant angular acceleration. (The passenger is obviously more of a "translation person" than a "rotation person.")
What is the constant angular acceleration during this decrease in angular speed?
Because the cylinder's angular acceleration is constant, we can relate it to the angular velocity and angular displacement via the basic equations for constant angular acceleration (Eqs. 10-12 and 10-13).

Calculations: Let's first do a quick check to see if we can solve the basic equations. The initial angular velocity is $\omega_0=3.40$
$\mathrm{rad} / \mathrm{s}$, the angular displacement is $\theta-\theta_0=20.0 \mathrm{rev}$, and the angular velocity at the end of that displacement is $\omega=2.00$ $\mathrm{rad} / \mathrm{s}$. In addition to the angular acceleration $\alpha$ that we want, both basic equations also contain time $t$, which we do not necessarily want.

To eliminate the unknown $t$, we use Eq. 10-12 to write
$$
t=\frac{\omega-\omega_0}{\alpha}
$$
which we then substitute into Eq. 10-13 to write
$$
\theta-\theta_0=\omega_0\left(\frac{\omega-\omega_0}{\alpha}\right)+\frac{1}{2} \alpha\left(\frac{\omega-\omega_0}{\alpha}\right)^2 .
$$
Solving for $\alpha$, substituting known data, and converting 20 rev to $125.7 \mathrm{rad}$, we find
$$
\begin{aligned}
\alpha & =\frac{\omega^2-\omega_0^2}{2\left(\theta-\theta_0\right)}=\frac{(2.00 \mathrm{rad} / \mathrm{s})^2-(3.40 \mathrm{rad} / \mathrm{s})^2}{2(125.7 \mathrm{rad})} \\
& =-0.0301 \mathrm{rad} / \mathrm{s}^2
\end{aligned}
$$
