\subsection{Kinematics}

Idea: Analogs
Rotational kinematics variables are analogous to translational ones: \begin{align*} \Delta \theta &\leftrightarrow \Delta x
\vec \omega &\leftrightarrow \Delta \vec v
\vec \alpha &\leftrightarrow \vec a. \end{align*} This gives us familiar looking kinematics equations: \begin{align*} \Delta \theta &= \omega_0t + \frac 12 \alpha t^2
\omega_f^2 &= \omega_0^2 + 2\alpha \Delta \theta. \end{align*}
The $F=ma$ will not really test the use of such equations, however. The focus is more on tricky setups and rolling without slipping.

Theorem: Instantaneous Center of Rotation
We know from the definition of angular velocity, that if we have a rigid body rotating around a fixed point with angular velocity $\omega$, then the velocity of a given point $\mathbf r$ from the center has velocity $$\mathbf v = \mathbf \omega \times \mathbf r.$$ However, it turns out the general motion of any rigid body (not necessarily fixed to a pivot) has an instantaneous center of rotation. A general rigid body has a center of mass with some velocity which is also spinning. So, the velocity of a point $\mathbf r$ from the com has velocity: $$\mathbf v = \mathbf v_{\text{cm}} + \omega \times \mathbf r = \omega \times (\mathbf r - \mathbf r_{\text{cor}}),$$ where $\mathbf r_{\text{cor}}$ is the center of rotation. Finding this center of rotation is often helpful for intuition.
A particular case where this is helpful is when an object is rolling without slipping. This means the objects contact point with the ground has 0 instantaneous velocity, which makes it the center of rotation!
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This let's us write down $$v = \omega r \implies a = \alpha r.$$ Further, because the bottom is the instantaneous center, this let's us write $v_{\text{too}} = \omega \cdot (2r) = 2\omega r = 2v.$

Tip: Reference Frames
You can use reference frames just as you did in regular kinematics! Moving into a rotational reference frame with speed $\omega$ would just mean subtracting off that angular velocity on all the other objects, as usual.
\subsection{Intro to Dynamics}

Definition: Torque \& Moment of Inertia
The rotational analog of $F=ma$ is torque, $$\tau = I\alpha = \mathbf r \times \mathbf F,$$ where $I$ is the moment of inertia, which is defined as $$I = \sum m_i r_i^2.$$
The $F=ma$ will often have you calculate the moment of inertias of funky objects. In those cases, the following theorems are useful.
Theorem: MOI theorems
The parallel axis theorem tells us $$I = I_{\text{cm}} + Md^2,$$where $d$ is the distance of the new axis from the COM. The perpendicular axis theorem tells us that for a flat 2D object, the moment of inertia about the perpendicular axis is equal to the sum of two in-plane axes. That is, if the object lies in the $x$-$y$ plane in any orientation, $$I_z = I_x + I_y.$$
Tip
Don't be afraid to write down lots of equations for rotational dynamics problems. You will usually need an $F=ma$, a $\tau = I\alpha$, and also a constraint such as $a = \alpha r$.