1.2 Kinematic equation for circular motion

1.1 Angular Variables
1.2 Kinematic equation for circular motion
1.3 Relation between angular and linear variables
1.4 Unit vectors along the radius and tangent
1.5 Velocity and acceleration of particle in circular motion

2.1 Centripetal force
2.2 Application of circular motion

3.1 Conditions for a particle in a circular motion

4.1 Conditions for a rod in a vertical circular motion

5.1 By friction
5.2 By banking of roads
5.3 By both friction and banking of road

We can write following kinematic equation for circular motion when the angular acceleration is constant,

Let us derive the above equations for better understanding.

Assume a particle moving in a circular path of radius $r$ with its center at $O$. The particle accelerates with constant angular acceleration $\alpha $ and ${\omega _0}$ is the angular velocity at time $t=0$.

For constant angular acceleration we know, $$\frac{{d\overrightarrow \omega }}{{dt}} = \overrightarrow \alpha $$

Integrating the above equation with proper limits we get,

$$\begin{equation} \begin{aligned} \left[ {\overrightarrow \omega } \right]_{{\omega _0}}^\omega = \overrightarrow \alpha \left[ t \right]_0^t \\ \overrightarrow \omega - {\overrightarrow \omega _0} = \overrightarrow \alpha \left( {t - 0} \right) \\ \overrightarrow \omega = {\overrightarrow \omega _0} + \overrightarrow \alpha t \\\end{aligned} \end{equation} $$

As we know, $$\begin{equation} \begin{aligned} \overrightarrow \omega = {\overrightarrow \omega _0} + \overrightarrow \alpha t\quad ...(i) \\ \overrightarrow \omega = \frac{{d\overrightarrow \theta }}{{dt}}\quad ...(ii) \\\end{aligned} \end{equation} $$

From equation $(i)$ and $(ii)$ we get, $$\frac{{d\overrightarrow \theta }}{{dt}} = {\overrightarrow \omega _0} + \overrightarrow \alpha t\quad {\text{or}}\quad d\overrightarrow \theta = \left( {{{\overrightarrow \omega }_0} + \overrightarrow \alpha t} \right)dt$$

Integrating the above equation with proper limits we get, $$\begin{equation} \begin{aligned} \int\limits_0^\theta {d\overrightarrow \theta = \int\limits_0^t {\left( {{\omega _0} + \alpha t} \right)} dt} \\ \overrightarrow \theta = {\overrightarrow \omega _0}t + \frac{1}{2}\overrightarrow \alpha {t^2} \\\end{aligned} \end{equation} $$

As we know, $$\begin{equation} \begin{aligned} \frac{{d\overrightarrow \omega }}{{dt}} = \overrightarrow \alpha \quad {\text{or}}\quad d\overrightarrow \omega = \overrightarrow \alpha dt\quad ...(i) \\ \frac{{d\overrightarrow \theta }}{{dt}} = \overrightarrow \omega \quad {\text{or}}\quad dt = \frac{{d\overrightarrow \theta }}{{\overrightarrow \omega }}\quad ...(ii) \\\end{aligned} \end{equation} $$

From equation $(i)$ and $(ii)$ we get,

$$d\overrightarrow \omega = \frac{{\overrightarrow \alpha d\overrightarrow \theta }}{{\overrightarrow \omega }}\quad {\text{or}}\quad \overrightarrow \omega d\overrightarrow \omega = \overrightarrow \alpha d\overrightarrow \theta $$

Integrating the above equation with proper limits we get,

$$\begin{equation} \begin{aligned} \int\limits_{{\omega _0}}^\omega {\overrightarrow \omega d\overrightarrow \omega } = \overrightarrow \alpha \int\limits_0^\theta {d\overrightarrow \theta } \\ {\omega ^2} = \omega _0^2 + 2\overrightarrow \alpha .\;\overrightarrow \theta \\\end{aligned} \end{equation} $$

Question 1. A particle moving in a circular path with an initial velocity of $2\ rad/s$ and the angular position at any time $t$ can be written as $\theta = {t^3} - 2{t^2} + ct$. Find the angular velocity and angular acceleration at time $t=2\ s$ (where $c$ is a constant).

Solution: From the question,

$\begin{equation} \begin{aligned} {\omega _0} = 2\;rad/s \\ \theta = {t^3} - 2{t^2} + ct \\\end{aligned} \end{equation} $

Therefore, our first target is to find the constant $c$. So, differentiating the above equation w.r.t time $t$ we get,

$$\frac{{d\theta }}{{dt}} = \omega = 3{t^2} - 4t + c$$

At $t=0$, $\omega = {\omega _0}$ So,

$$\begin{equation} \begin{aligned} 2 = 0 - 0 + c \\ c = 2\;rad/s \\\end{aligned} \end{equation} $$

Therefore $\omega $ at time $t=2\ s$ we get, $$\begin{equation} \begin{aligned} \omega = 3{t^2} - 4t + 2 \\ \omega = 12 - 8 + 2 \\ \omega = 6\;rad/s \\\end{aligned} \end{equation} $$

Also, $\alpha $ at time $t=2\ s$ we get, $$\begin{equation} \begin{aligned} \frac{{d\omega }}{{dt}} = 6t - 4 \\ \alpha = 12 - 4 \\ \alpha = 8\;rad/{s^2} \\\end{aligned} \end{equation} $$