Show that `a=(v^2)/r` and hence obtain an expression for centripetal force.

Explanation

Let `vecr` and `vecr`, be the position vectors and `vecv` velocities of the object when it is it point P and P. By definition, velocity at a point is along the tangent at that point in the direction of the motion. Since the path is circular, `vecv` is perpendicular to `vecr` and `vecv` is perpendicular to `vecr`
Therefore, `/_\vecv` is perpendicular to `/_\vecr`. Average acceleration `(/_\vecv)/(/_\t)` is perpendicular to `/_\vecr`
The magnitude of `veca` is, by definition, given by `|veca|=lim_(/_\t to0)(/_\vecv)/(/_\t)`
The triangle formed by the position vectors is similarly to the triangle formed by the velocity vectors.

i.e, `(|/_\vecv|)/v=(|/_\vecr|)/r " or "|/_\vecv|=v(|/_\vecr|)r`
Therefore, `|veca|=lim_(|/_\vecv)/(/_\t)=lim_(/_\t to0)(v|/_\vecr|)/(r/_\t)=(vecv)/(r)lim_(/_\t to0)(|/_\vecr|)/(/_\t)`
If `/_\t` is very small, `/_\theta` will also small. The arc PP' is approximately equal to `|/_\vecr|`
i.e., `lim_(/_\t to0)(|/_\vecr|)/(/_\t)=v` Thus, centripetal acceleration `|veca|=v/fv=(v^2)/(r)` and `veca=v/r(dvecr)/(dt)`
The centripetal acceleration is always directed towards the centre. The centripetal force=ma.