Derive the expression for centripetal acceleration.

Explanation

Consider the position vectors and velocity vectors shift through the some angle `theta` in a small invertal of time `Deltat` as shown in figure

(ii) In uniform circular motion.
`r=|vec(r_(1))|=|vec(r_(2))| and v=|vec(v_(1))|=|vec(v_(2))|`
(iii) From figure, the geometrical relationship between the magntitude of position and velocity vectors is given by.
`(Deltar)/(r)=-(Deltav)/(v)=theta`
(iv) Here the negative sign implies that `thetav` points radially inward, towards the center of the circle.
`Deltav=-v((Deltar)/(r))`
(v) Dividing both sides by `Deltat,` we get,
`a=(Deltav)/(Deltat)=-(v)/(r)((Deltar)/(Deltat))`
(vi) Applying the limit `Deltat rarr0,` We get,
`(dv)/(dt)=-(v)/(r) ((dr)/(dt))`
(vii) Since `a_(c)=(dv)/(dt)` and `v=(dr)/(dt)`, we can write,
`a_(c)=-(v^(2))/(r)=(v^(2))/(r)`
where `a_(c)` is the centripetal acceleration.