Centripetal force is the force required to move a body
uniformly in circle, this force act along the radius towards the center of
circle.

If the body of mass m is moving in horizontal circle of
radius of r with constant speed v the equation for centripetal force is given
by

Fc = mv

^{2}/r
Fc= mr

^{2}ω^{2}/r
Fc= mrω

^{ 2}##
#Derivation Of Centripetal Acceleration And Force

Consider a body of mass m is moving in horizontal circle
with a constant speed v and angular velocity ω at time t.

The particle is at point p were radius vector op= r after
the time t +ΔT particle reached at point Q and angle made by the radius vector
at the center of the circle is Δ θ

Angular velocity ω = Δ θ / Δ T

Δ θ = ω Δ T --------------------(2)

Let V

_{1}and V_{2}be the velocity vector at location P and Q respectively.
In vector of the body is always directed towards the
tangent.

In circular motion the speed of the particle remains
constant.

∴
|PA| = |QB| = |V|

|V

_{1}| = |V_{2}| = |V| ----------------(3)
For the calculation we take on external point P

^{1}and we draw P^{1}A^{1}and P^{1}B^{1}parallel to PA and PB respectively.
∴ ∠ A

^{1}P^{1}B^{1}= Δθ
From triangle law of vector

P

^{1}A^{1}+ P^{1}B^{1}= P^{1}B^{1}
Or

A

^{1}B^{1}= P^{1}B^{1 }- P^{1}A^{1}
Or

A

^{1}B^{1}= V_{2}– V_{1}= ΔV --------------------------(4)
If ΔT = 0 then A

^{1}and B^{1}comes closer to each other and the point can be contact by an arc of circle.
∴ Angle Δθ = A

^{1}B^{1}/ P^{1}A^{1}= ΔV/|V|
Putting the value of Δθ form equation (2)

ω.Δt = ΔV/|V|

ω.|V| = ΔV/| Δt |

ωrw = rω

^{2}---------------(5)
If LT ΔV/Δt the instantaneous Δt = 0 acceleration at
particle at point P which called centripetal acceleration.

∴ ac = rω

^{2}----------------------(6)
According to Newton 2

^{nd}law of the magnitude
Fc = mac = mrω

^{2}------------------------(7)