Derive the expression for centripetal acceleration.

Explanation

The centripetal acceleration is derived from geometrical relation between position and velocity vectors .
Let the directions of position and velocity vectors be shifted through the same angle `theta` in a small interval of time `Delta t` , as shown in the figure. For uniform circular motion , `r = |vec(r)_(1)|= |vec(r)_(2)|and v = |vec(v)_(1)|= |vec(v)_(2)|. ` If the particle moves from position vector `vec_(v)_(1) t o vec(r)_(2)` , then the displacement is given by `Delta vec(r) = vec(r)_(2) - vec(r)_(1)` . The change in velocity from `vec(v)_(1) t o vec(v)_(2)` is given by `Delta vec(v) = vec(v)_(2) - vec(v)_(1)` . The magnitudes of the displacement `Delta r and Delta v` satisfy the following relation .
`(Deltar)/(r) = - (Delta v)/(v) = theta`
Here the nagative sign implies that `Delta v` points radially inward, towards the centre of the circle
`Delta v = - v ((Delta r)/(r))`
`:. a = (Deltav)/(Deltat) = (v)/(r) ((Deltar)/(Delta t)) = - (v^(2))/(r)`
For uniform circular motion v ` = omega r` , where `omega` is the angular velocity of the particle about the centre. Then the centripetal acceleration is given by
`a = - omega^(2) r`