Acceleration of a particle moving along a circular path

Consider A and B, the two positions of a particle displaced through an angle δθ in time δt as shown in the figure:

Let, r = radius of curvature of the circular path
v = velocity of the particle at A, and
v+δv = Velocity of the particle at B.

The change of velocity, as the particle moves from A to B may be obtained by drawing in the vector triangle oab, as shown in Fig.

In this triangle, oa represents the velocity v and ob represents the velocity v+δv. The change of velocity in time δt is represented by ab.

Now, resolving ab into two components, i.e, parallel and perpendicular to oa. Let ac and cb be the components parallel and perpendicular to oa respectively.

∴ ac = oc – oa = obcosδθ – oa = (v+δv)cosδθ – v
cb = obsinδθ = (v+δv)sinδθ

Since, the change of velocity of a particle ha two mutually perpendicular components, therefore the acceleration of a particle moving along a circular path has the following two components of the acceleration which are perpendicular to each other.