Kinematic Equations for Circular Motion
Circular motion with constant acceleration:
The simplest case of circular motion is that of a particle moving with constant angular acceleration, either zero or nonzero. The goal of this section is to derive the equations giving the angular velocity and angular position of the particle as a function of time.
We consider a particle moving counterclockwise along a circular path with an initial angular velocity ${\omega }_0$ starting at location ${\theta }_0$ at $t=0$.
Case 1: angular acceleration is zero: $\alpha =0\ rad/s^2$
The angular velocity of the particle is then constant and equal to its initial angular velocity ${\omega }_0$ i.e.
\omega \left(t\right)={\omega }_0
The angular position $\theta \left(t\right)$ of the particle is then derived as follows
\begin{aligned} \omega =\frac{d\theta }{dt}\ \ \ \ &\Rightarrow \ \ \ \ \ d\theta =\omega \left(t\right)dt \\ \\ &\Rightarrow \ \ \ \ \ d\theta ={\omega }_0dt \\ \\ &\Rightarrow \ \ \ \ \ \int^{\theta \left(t\right)}_{{\theta }_0}{d\theta }=\int^t_{t=0}{{\omega }_0dt} \\ \\ &\Rightarrow \ \ \ \ \ \theta \left(t\right)-{\theta }_0={\omega }_0t \\ \\ &\Rightarrow \ \ \ \ \ \theta \left(t\right)={\omega }_0t+{\theta }_0 \end{aligned}
In this case, the angular velocity $\omega \left(t\right)$ is constant and the angular position $\theta \left(t\right)$ varies linearly as shown in the graphs below.
Case 2: angular acceleration is a non-zero constant: $\alpha \ne 0$
The angular velocity $\omega \left(t\right)$ of the particle is then derived as follows
\begin{aligned} \alpha =\frac{d\omega }{dt}\ \ \ \ \ &\Rightarrow \ \ \ \ \ d\omega =\alpha dt \\ \\ &\Rightarrow \ \ \ \ \ \int^{\omega \left(t\right)}_{{\omega }_0}{d\omega }=\int^t_{t=0}{\alpha dt} \\ \\ &\Rightarrow \ \ \ \ \ \omega \left(t\right)-{\omega }_0=\alpha t \\ \\ &\Rightarrow \ \ \ \ \ \omega \left(t\right)=\alpha t+{\omega }_0 \end{aligned}
The angular position $\theta \left(t\right)$ of the particle can then be derived from $\omega \left(t\right)$ as follows
\begin{aligned} \omega =\frac{d\theta }{dt}\ \ \ \ &\Rightarrow \ \ \ \ \ d\theta =\omega \left(t\right)dt \\ \\ &\Rightarrow \ \ \ \ \ d\theta =\left(\alpha t+{\omega }0\right)dt \\ \\ &\Rightarrow \ \ \ \ \ \int^{\theta \left(t\right)}_{{\theta }_0}{d\theta }=\int^t_{t=0}{\left(\alpha t+{\omega }_0\right)dt} \\ \\ &\Rightarrow \ \ \ \ \ \theta \left(t\right)-{\theta }_0=\frac{1}{2}\alpha t^2+{\omega }_0t \\ \\ &\Rightarrow \ \ \ \ \ \theta \left(t\right)=\frac{1}{2}\alpha t^2+{\omega }_0t+{\theta }_0 \end{aligned}
In this case, the angular velocity $\omega \left(t\right)$ varies linearly and the angular position $\theta \left(t\right)$ varies quadratically as shown in the graphs below.
Case 3: a third useful formula derived from $\theta \left(t\right)$ and $\omega \left(t\right)$
This common substitution allows for a relationship between the speed of the particle and its angular displacement $\mathrm{\Delta }\theta =\theta \left(t\right)-{\theta }_0$ that is independent of time.
We consider a particle that starts at time $t_i$ at an angular position ${\theta }_0$ with an initial angular velocity ${\omega }_0$ and reaches at time $t_f$ the angular position ${\theta }_f$ with a final angular velocity ${\omega }_f$. Therefore, we can express $t_f$ as follows
{\omega }_f=\alpha t_f+{\omega }_0\ \ \ \ \Rightarrow \ \ \ \ \ t_f=\frac{{\omega }_f-{\omega }_0}{\alpha }
Substituting into $\theta \left(t\right)$ then yields
\begin{aligned} \theta \left(t_f\right)=\frac{1}{2}\alpha t^2_f+{\omega }_0t_f+{\theta }_0\ \ \ \ &\Rightarrow \ \ \ \ {\theta }_f-{\theta }_0=\frac{1}{2}\alpha t^2_f+{\omega }_0t_f \\ \\ &\Rightarrow \ \ \ \ {\theta }_f-{\theta }_0=\frac{1}{2}\alpha {\left(\frac{{\omega }_f-{\omega }_0}{\alpha }\right)}^2+{\omega }_0\left(\frac{{\omega }_f-{\omega }_0}{\alpha }\right) \\ \\ &\Rightarrow \ \ \ \ \ {\theta }_f-{\theta }_0=\frac{{\omega }^2_f+{\omega }^2_0-2{\omega }_0{\omega }_f}{2\alpha }+\frac{{\omega }_0{\omega }_f-{\omega }^2_0}{a_x} \\ \\ &\Rightarrow \ \ \ \ \ {\theta }_f-{\theta }_0=\frac{{\omega }^2_f-{\omega }^2_0}{2\alpha } \\ \\ &\Rightarrow \ \ \ \ \ {\omega }^2_f={\omega }^2_0+2\alpha \left({\theta }_f-{\theta }_0\right) \\ \\ &\Rightarrow \ \ \ \ \ {\omega }^2_f={\omega }^2_0+2\alpha \mathrm{\Delta }\theta \end{aligned}
For an object covering an angular displacement $\mathrm{\Delta }\theta $ with a constant angular acceleration $\alpha $, you may write
{\omega }^2_f={\omega }^2_0+2\alpha \mathrm{\Delta }\theta
Summary of formulas:
The formulas derived above are summarized below as they are fundamental formulas for circular motion kinematics with constant angular acceleration.
\boxed{ \begin{aligned} \theta \left(t\right)&=\frac{1}{2}\alpha t^2+{\omega }_0t+{\theta }_0 \\ \\ \omega \left(t\right)&=\alpha t+{\omega }_0 \\ \\ {\omega }^2_f&={\omega }^2_0+2\alpha \mathrm{\Delta }\theta \end{aligned} }