# Circular Motion (Uniform and Non-Uniform)

In circular motion, an object moves in a circular path around a fixed point (centre). The object is constantly changing its direction of motion. Its velocity is tangential to the path. The acceleration of an object in circular motion is known as centripetal acceleration, and it is directed towards the center of rotation.

The circular motion of a particle is best represented in polar coordinates. Let O be the origin and P be the position of the particle. The distance OP is $r$ and the unit vector along $\vec{\mathrm{OP}}$ is $\hat{r}$. The line OP makes an angle $\theta$ with a reference line. The unit vector $\hat{\theta}$ is perpendicular to $\hat{r}$ and it point towards increasing value of $\theta$.

The position of the particle is specified by the polar coordinates $(r,\theta)$. The position vector of the particle is given by \begin{align} \vec{r}=\vec{\mathrm{OP}}=r\,\hat{r}. \end{align} Note that the directions $\hat{r}$ and $\hat{\theta}$ varies as the particle changes its position (this is unlike Cartesian coordinates directions $\hat{\imath}$ and $\hat\jmath$). If the particle is confined to move in a circle then $r$ is constant (i.e., $\mathrm{d}r/\mathrm{d}t=0$) but $\theta$ varies with time. The time variation of $\hat{r}$ and $\hat{\theta}$ are given by \begin{align} &\frac{\mathrm{d}\hat{r}}{\mathrm{d}t}=\frac{\mathrm{d}\theta}{\mathrm{d}t}\,\hat{\theta},\nonumber\\ &\frac{\mathrm{d}\hat{\theta}}{\mathrm{d}t}=-\frac{\mathrm{d}\theta}{\mathrm{d}t}\,\hat{r}.\nonumber \end{align} The angular velocity is defined as $\omega=\mathrm{d}\theta/\mathrm{d}t$. The angular acceleration is defined as $\alpha=\mathrm{d}\omega/\mathrm{d}t$.

## Veclocity and Acceleration

The velocity of the particle is given by \begin{align} \vec{v} & =\frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \\ &=\frac{\mathrm{d}r}{\mathrm{d}t}\,\hat{r}+r\,\frac{\mathrm{d}\hat{r}}{\mathrm{d}t} \\ &=r\frac{\mathrm{d}\theta}{\mathrm{d}t}\hat{\theta} \\ &=r\omega\hat{\theta}, \end{align} where we used $\mathrm{d}r/\mathrm{d}t=0$ and $\mathrm{d}\hat{r}/\mathrm{d}t=(\mathrm{d}\theta/\mathrm{d}t)\hat\theta=\omega\hat\theta$. The magnitude of the velocity is $|\vec{v}|=\omega r$ and its direction is tangential to the circle i.e., along $\hat\theta$.

The acceleration of the particle is given by \begin{align} \vec{a}&=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t} \\ &=\omega\frac{\mathrm{d}r}{\mathrm{d}t}\hat{\theta}+r\frac{\mathrm{d}\omega}{\mathrm{d}t}\hat{\theta}+r\omega\frac{\mathrm{d}\hat{\theta}}{\mathrm{d}t}\nonumber\\ &=-r\omega^2\,\hat{r}+r({\mathrm{d}\omega}/{\mathrm{d}t})\,\hat{\theta}\nonumber\\ &=a_r\hat{r}+a_t\hat{\theta}\nonumber\\ &=-a_c\hat{r}+a_t\hat{\theta}. \end{align} where we used $\mathrm{d}r/\mathrm{d}t=0$ and $\mathrm{d}\hat{\theta}/\mathrm{d}t=-(\mathrm{d}\theta/\mathrm{d}t)\hat{r}$. The acceleration component in the direction $\hat{r}$ is called radial acceleration, $a_r=r\omega^2$. The acceleration component tangential to the circle is called tangential acceleration, $a_t=r(\mathrm{d}\omega/\mathrm{d}t)=r\alpha$. The acceleration component towards the centre of the circle is called centripetal acceleration, $a_c=\omega^2r$.

## Uniform Angular Acceleration

Let a particle starts its circular motion with initial angle zero (i.e., it is located on the reference line) and initial angular velocity $\omega_0$. If the particle rotates with a constant angular acceleration $\alpha$ then its angular velocity $\omega$ and angle $\theta$ at time $t$ are given by

\begin{align} &\omega=\omega_0+\alpha\, t, \\ &\theta=\omega_0\, t+\tfrac{1}{2}\alpha\, t^2. \end{align} From above equations, we get \begin{align} \omega^2=\omega_0^2+2\alpha\theta. \end{align}

## Problems from IIT JEE

Problem (IIT JEE 2011): A wire, which passes through the hole in a small bead, is bent in the form of a quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is,

Solution: Let $v$ be speed of the bead when it makes an angle $\theta$ with the vertical direction (see figure).
The forces acting on the bead are its weight $mg$ and normal reaction $N$. Since all the forces acting on the bead are conservative, the mechanical energy of the bead is conserved i.e., \begin{align} mgr(1-\cos\theta)=\tfrac{1}{2}mv^2. \end{align} The radially inward components of force provide centripetal acceleration to the bead i.e., \begin{align} mg\cos\theta-N={mv^2}/{r}. \end{align} Eliminate $v^2$ from above equations to get, \begin{align} N=mg(3\cos\theta-2). \end{align} From this equation, normal reaction $N$ is positive (radially outward) if $\theta < \cos^{-1}({2}/{3})$ and negative (radially inward) if $\theta > \cos^{-1}({2}/{3})$. By Newton's third law, force applied by the bead on the wire is $N^\prime=-N$, which is radially inwards initially and radially outwards later.