We will do a short derivation for the kinematics equations for rotations when the instantaneous angular acceleration, $\alpha$ is constant.
$$\begin{aligned} \alpha \, &= \frac{d \omega}{dt} = \text{const} \\ d \omega \, &= \alpha \, dt \\ \int\limits_{\omega_{0}}^{\omega} \, d \omega \, &= \int\limits_{0}^{t} \alpha \, dt’ \\ \omega \, &= \omega_{0} + \alpha t \end{aligned}$$
That is the first equation. Now for the second:
$$\begin{aligned} \omega \, &= \frac{d \theta}{dt} \\ d \omega \, &= \omega \, dt \\ d \omega \, &= (\omega_{0} + \alpha t) \, dt \\ \int\limits_{\theta_{0}}^{\theta} \, d \theta \, &= \int\limits_{0}^{t} (\omega_{0} + \alpha t’) \, dt’ \\ \theta \, &= \theta_{0} + \omega_{0}t + \frac{1}{2} \alpha t^{2} \end{aligned}$$
Just use the two equations above to eliminate the t to obtain the third equation:
$$\omega^{2} = \omega_{0}^{2} + 2 \alpha ( \theta – \theta_{0} )$$
You will notice that the kinematics equations for rotations are extremely similar to that for translational motion.
Specifically, there is a one-to-one correspondence between:
$$\begin{aligned}\theta \, &\Leftrightarrow x \\ \omega \, &\Leftrightarrow v \\ \alpha \, &\Leftrightarrow a \end{aligned}$$
For comparison:
$$\begin{aligned} v = v_{0} + at \, &\Leftrightarrow \, \omega = \omega_{0} + at \\ x = x_{0} + v_{0}t + \frac{1}{2} a t^{2} \, &\Leftrightarrow \, \theta = \theta_{0} + \omega_{0} t + \frac{1}{2} \alpha t^{2} \\ v^{2} = v_{0}^{2} + 2a (x-x_{0}) \, &\Leftrightarrow \, \omega^{2} = \omega_{0}^{2} + 2 \alpha (\theta – \theta_{0}) \end{aligned}$$
Small note about $\omega$ and $\alpha$:
$\omega$ and $\alpha$ are vectors.
The direction of $\vec{\omega}$ is determined by right hand rule.
The direction of $\vec{\alpha}$ follows from its definition $\frac{d \vec{\omega}}{d t}$. If the angular motion is speeding up, $\vec{\alpha}$ will be in the same direction as $\vec{\omega}$. Different direction if the angular motion is slowing down.
Rotational Energy
Consider a rigid body as a collection of small particles rotating about a fixed z-axis at angular speed $\omega$:
The kinetic energy of the $i \text{th}$ particle is given by:
$$\begin{aligned} K_{i} &= \frac{1}{2} m_{i}v_{i}^{2} \\ &= \frac{1}{2} m_{i} r_{i}^{2} \omega^{2} \end{aligned}$$
In order to find the total kinetic energy, we just have to sum up all the particles:
$$\begin{aligned} K_{\text{R}} &= \sum\limits_{i} K_{i} \\ &= \frac{1}{2} \sum\limits_{i} m_{i}r_{i}^{2} \omega^{2} \end{aligned}$$
