Definition:
When a particle moves in a circular path then acceleration act on the particle which has a direction toward the center of the circle. This acceleration is called centripetal acceleration.
Derivation of Centripetal Acceleration: Let us consider, A particle that has mass $m$ moving with velocity $v$ in a circular path of radius $r$.
If a particle is moving from point $P_{1}$ to point $P_{2}$ by covering distance $\Delta s$ on the circumference of the circle by making an angular displacement of $\theta$ at the center $O$ of the circle.
The direction of velocity of the particle at point $P_{1}$ and $P_{2}$ is $\overrightarrow{v_{1}}$ and $v_{2}$.
Now take the change in velocity from point $P_{1}$ to $P_{2}$ by vector subtraction method as shown in figure below:
To find the expression for the centripetal acceleration, Now take two similar triangles $\Delta OP_{1}P_{2}$ and $\Delta ABC$ from the figure:
$\frac{OP_{1}}{AB}=\frac{P_{1}P_{2}}{BC}$
Now substitute the values from the figure in the above equation i.e.
$\frac{r}{v}=\frac{\Delta s}{\Delta v}$
$\Delta v = \frac{v}{r} \Delta s $
Now divide by $\Delta t$ into both sides the above equation can be written as
$\frac{\Delta v}{\Delta t}=\frac{v}{r} \frac{\Delta s}{\Delta t}$
If $\Delta t$ is tends to zero i.e. $\Delta t \rightarrow 0$ then
$\underset{\Delta t \rightarrow 0}{Lim} \: \frac{\Delta v}{\Delta t}=\frac{v}{r} \: \underset{\Delta t \rightarrow 0}{Lim} \: \frac{\Delta s}{\Delta t}$
Where
$\underset{\Delta t \rightarrow 0}{Lim} \: \frac{\Delta v}{\Delta t} \rightarrow$ Instantaneous Acceleration. It is also known as Centripetal Acceleration $(a)$
$\underset{\Delta t \rightarrow 0}{Lim} \: \frac{\Delta s}{\Delta t} \rightarrow$ Instantaneous Velocity $(v)$
Now the above equation can be written as
$a = \frac{(r\omega)^{2}}{r} \quad \left( \because v=r\omega \right)$
To find the expression for the centripetal acceleration, Now take two similar triangles $\Delta OP_{1}P_{2}$ and $\Delta ABC$ from the figure:
