What is uniform circular motion ? By using proper figure, obtain equation of acceleration ${a_c}\, = \,\frac{{{v^2}}}{r}$ for uniform circular motion. Show that its direction is towards centre.

When an object follows a circular path at a constant speed, the motion of the object is called uniform circular motion.

Suppose an object is moving with uniform speed $v$ in a circle of radius $\mathrm{R}$ as shown in figure. Since the velocity of the object is changing continuously in direction, the object undergoes acceleration.

Let $\vec{r}$ and $\overrightarrow{r^{\prime}}$ be the position vectors and $\vec{v}$ and $\overrightarrow{v^{\prime}}$ the velocities of the object when it is at point $\mathrm{P}$ and $\mathrm{P}^{\prime}$ as shown in figure.

Velocity at a point is along the tangent at that point in the direction of motion.

The velocity vectors $\vec{v}$ and $\overrightarrow{v^{\prime}}$ are as shown $\overrightarrow{\Delta v}$ is obtained in figure (a2) using the triangle law of vector addition.

Since the path is circular, $\vec{v}$ is perpendicular to $\vec{r}$ and so is $\vec{v}^{\prime}$ to $\overrightarrow{r^{\prime}} .$ Therefore, $\overrightarrow{\Delta v}$ is perpendicular to $\overrightarrow{\Delta r}$. Since average acceleration is along the $\overrightarrow{\Delta v}\left(\vec{a}=\frac{\overrightarrow{\Delta v}}{\Delta t}\right)$, the average

acceleration $\vec{a}$ is perpendicular to $\overrightarrow{\Delta r}$.

Figure (b) shows the same quantities for smaller time interval.

$\overrightarrow{\Delta v}$ and hence $\vec{a}$ is again directed towards the centre.

In figure (c), $\Delta t \rightarrow 0$ and the average acceleration becomes the instantaneous acceleration. It is directed towards the centre.

Thus, we find that the acceleration of an object in uniform circular motion is always directed towards the centre of the circle.

The magnitude of $\vec{a}$ is $|\vec{a}|=\lim _{\Delta t \rightarrow 0} \frac{|\Delta \vec{v}|}{\Delta t}$

Let the angle between position vectors $\vec{r}$ and $\overrightarrow{r^{\prime}}$ be $\Delta \theta$.

The angle between is $\vec{v}$ and $\overrightarrow{v^{\prime}}$ is also $\Delta \theta$.