When a particle speeds up or slows down in a circle both magnitude and direction of its velocity change. If the particle moves with constant speed in a circle, only the direction of its velocity changes not the magnitude.

The acceleration in circular motion has two components. One is parallel (${{a}_{\parallel }}$) or tangential and another is perpendicular (${{a}_{\bot }}$) or centripetal (radial) component. In Figure 1 a car (not shown in figure) moves in a circular path and has both perpendicular and parallel components of acceleration.

The parallel component means the magnitude of velocity changes and the perpendicular component means the direction of velocity changes. If there is no parallel component of acceleration, the speed of the car remains the same and there will be only the perpendicular component which leads to uniform circular motion. When a body moves with constant speed in a circle of radius $r$, the motion is uniform circular motion.

Note that in Figure 1 the car is speeding up and the acceleration vector lies ahead the normal and if the car had slowed down, the acceleration vector would have been behind the normal. And in Figure 2 the car moves with constant speed so there is no parallel component of acceleration and but there is a perpendicular component along the normal which is the same as the major acceleration vector. In case of uniform circular motion the instantaneous acceleration (not average acceleration) has direction towards the centre of the circle which is called centripetal acceleration or radial acceleration. What about the instantaneous acceleration for non-uniform circular motion?

The average and instantaneous accelerations are not the same in uniform circular motion. In Figure 3 the car (not shown in figure) is moving with constant speed but the direction changes. Let $\Delta t$ is the time during the motion along the curve between the points ${{p}_1}$ and ${{p}_2}$. So the average acceleration vector due to the change in velocity ${{\vec{a}}_{\text{av}}}$ is,

\[{{\vec{a}}_{\text{av}}}=\frac{\Delta \vec v}{\Delta t}\]

Note that in Figure 3 that the direction of average acceleration vector is the same as the direction of $\Delta \vec v$ which lies ahead the normal. But in the limit that ${\Delta t}$ approaches zero the point ${{p}_2}$ moves closer and closer to the point ${{p}_1}$ and the angle between ${{\vec{v}}_{1}}$ and ${{\vec{v}}_{2}}$ is nearly zero, so the difference $\Delta \vec v$ becomes $d\vec{v}$ whose direction is towards the centre of the circle. Therefore, in uniform circular motion the instantaneous acceleration at any point on the circle points towards the centre of the circle. Do not forget that if we are only saying the word *acceleration* we are talking about instantaneous acceleration.

One misunderstanding may arise here lies within the figures Figure 1 and Figure 3 as the acceleration vector lies ahead the normal in both uniform and non-uniform circular motion. In the circular motion shown in Figure 1 the car is speeding up but the car is moving at constant speed in Figure 3. The difference is that the acceleration vector shown in Figure 1 is instantaneous acceleration vector but the acceleration vector shown in Figure 3 is average acceleration vector. Non-uniform circular motion has both parallel and perpendicular components of acceleration but uniform circular motion does not have the parallel component of acceleration.

The next approach is to get the acceleration of the car or any object that moves in a circle in uniform circular motion. Suppose a particle moves in a circle of radius $r$ with constant speed as in Figure 4. In Figure 4(a) let the chord joining the points $A$ and $B$ is $\Delta s$, $\angle AOB=\alpha $ and $r$ is the radius of the circle. The exaggerated view of the subtraction $\vec{\Delta v}={{\vec{v}}_{2}}-{{\vec{v}}_{1}}$ is shown in Figure 4(b).

The angle $\angle AOB$ is equal to $\angle MLN$ which is in turn equal to $\angle LNO$. Notice that $\triangle AOB$ and $\triangle LNO$ are similar triangles and the ratios of corresponding sides in similar triangles are equal. So you can write,

\[\frac{\Delta v}{\Delta s}=\frac{{{v}_{1}}}{r}=\frac{{{v}_{2}}}{r} \tag{1} \label{1}\]

You know that the magnitude of $-{{\vec{v}}_{1}}$ is ${{v}_{1}}$ and the magnitude of ${{\vec{v}}_{2}}$ is ${{v}_{2}}$. If the particle moves from point $A$ to point $B$ along the circle in time interval $\Delta t$, the average acceleration is,

\[{{a}_{av}}=\frac{\Delta v}{\Delta t} \tag{2} \label{2}\]

From Eq. \eqref{1}, put the value of $\Delta v=\frac{{{v}_{1}}}{r}\Delta s$ to Eq. \eqref{2}, and you get,

\[{{a}_{av}}=\frac{{{v}_{1}}}{r}\frac{\Delta s}{\Delta t} \tag{3} \label{3}\]

In the limit that $\Delta t$ approaches zero ${{a}_{av}}$ in Eq. \eqref{3} becomes instantaneous acceleration. When $\Delta t$ approaches zero the point $B$ moves closer and closer to the point $A$ and the two points are nearly identical so,

\[{{a}_{\text{ins}}}=\underset{\Delta t\to 0}{\mathop{\lim }}\,\frac{{{v}_{1}}}{r}\frac{\Delta s}{\Delta t}=\frac{{{v}_{1}}^{2}}{r} \tag{4} \label{4}\]

Note that in Eq. \eqref{4} ${{v}_{1}}$ represents the magnitude of velocity at any point on the circle (in uniform circular motin the speed is constant), so we can write, ${{v}_{1}}=v$. The instantaneous acceleration in the uniform circular motion always points towards the centre of the circle called centripetal acceleration. Eq. \eqref{4} can be rewritten as,

$${a_{{\text{cen}}}} = \frac{{{v^2}}}{r} \tag{5} \label{5}$$

The relationship in Eq. \eqref{5} shows that the acceleration in uniform circular motion is depends on the speed and the radius of the circle.