Derive the expression for centripetal acceleration.

Explanation

In uniform circular motion the velocity vector turns continuously without changing its magnitude (speed), as shown in figure.

Note that the length of the velocity vector is not changed during the motion, implying that the speed remains constant. Even though the velocity is tangential at every point in the circle, the acceleration is acting towards the center of the circle. This is called centripetal acceleration. It always points towards the center of the circle. This is shown in the figure
The centripetal acceleration is derived from a simple geometrical relationship between position and velocity vectors

Let the directions of position and velocity vectors shift through the same angle `theta` in a small interval of time `Deltat`, as shown in figure. For uniform circular motion, `r=|vecr_(1)|=|vecr_(2)|andv|vecv_(1)|=|vecv_(2)|`. If the particle moves from position vector `vecr_(1)` to `vecv_(2)`, the displacement is given by `Deltavecv=vecr_(2)-vecr_(1)` and the change in velocity from `vecv_(1)` to `vec_(2)` is given by `Deltavecv=vecv_(2)=vecv_(1)`. The magnitudes of the displacement `Deltar` and `Deltav` satisfy the following relation
`(Deltar)/(r)=(-Deltav)/(v)=theta`

Here the negative sign implies that Av points radially inward, towards the center of the circle.
`Deltav=v((Deltar)/(r))`
Then, `a=(Deltav)/(Deltat)=(v)/(r)((Deltar)/(Deltat))=-(v^(2))/(r)`
For uniform circular motion `v=omegar`, where `omega` the angular velocity of the particle about the center. Then the centripetal acceleration can be written as `a=-omega^(2)r`