What do you understandby centripetal acceleration ? Derive formula for the centripetal acceleration of a particle moving on a circular path.

Explanation

Centripetal Acceleration : When a particle performs a uniform circular motion, its direction changes continuously though its speed remains constant. Thus its velocity changes continuously. That is there is an acceleration in circular motion. The direction of this circle. Hence it is called .centripetal acceleration..
Derivation : Suppose a particle is moving with a uniform speed v on a circular path whose radius is r and centre is O. At every point of the path the direction of particle.s motion would be tangential. Suppose the particle covers a distance `Deltas` from `P_(1)` to `P_(2)` in a small time - interval `Delta t`. Let `vec(v)_(1)` and `vec(v)_(2)` be the velocities of the particle at `P_(1)` and `P_(2)` respectively. The magnitude of both `vec(v)_(1)` and `vec(v)_(2)` is v. The velocity change from `P_(1)` to `P_(2)` is :

`vec(v)_(2)-vec(v)_(1)=vec(Deltav)`
If `vec(v)_(1)` and `vec(v)_(2)` are drawn from the same point Q and a third vector is drawn from the arrow-head of `vec(v)_(1)` to the arrow-head of `vec(v)_(2)` then this third vector will represent the velocity-change.
`vec(v)_(2)-vec(v)_(1)=vec(Deltav)`
The triangle `OP_(1)P_(2)` and the vector triangle QAB are similar.
`(P_(1)P_(2))/(P_(1)O)=(AB)/(AQ)`
`(Deltas)/(r)=(Deltav)/(v)`
`:.Deltav=(v)/(r)Deltas`
Dividing both sides by `Deltat` we have
`(Deltav)/(Delta t)=(v)/(r)xx(Deltas)/(Deltat)`
`a=(v)/(r)xxv" "[(Deltav)/(Deltat)=a,(Deltas)/(Deltat)=v]`
`a=(v^(2))/(r)`
we have, `v=omega.r`
So, `a=omega^(2).r`