Question

Derive the expression for centripetal acceleration.

Answer 1

In a uniform circular motion, the velocity vector turns continuously without changing its magnitude (speed), as shown in the figure.

 

Velocity in a uniform circular motion

Note that the length of the velocity vector is not changed during the motion, implying that the speed remains constant. Even though the velocity is tangential at every point in the circle, the acceleration is acting towards the center of the circle. This is called centripetal acceleration. It always points towards the center of the circle. This is shown in the figure.

`(DeltavecV)/(Deltat) = veca`

Centripetal acceleration

The centripetal acceleration is derived from a simple geometrical relationship between position and velocity vectors.

The geometrical relationship between the position and velocity vectors

Let the directions of position and velocity vectors shift through the same angle θ in a small interval of time ∆t, as shown in the figure. For uniform circular motion, r = `|vecr_1|` = `|vecr_2|` and v = `|vecv_1|` = `|vecv_2|`. If the particle moves from position vector `vecr_1` to `vecr_2`, the displacement is given by `Deltavecr = vecr_2 - vecr_1`,. and the change in velocity from `vecv_1` to `vecv_2` is given by `∆vecv = vecv_2 – vecv_1`,. The magnitudes of the displacement ∆r and of ∆v satisfy the following relation. `(Δr)/r = (−Δv)/v = θ` Here the negative sign implies that ∆v points radially inward, towards the center of the circle.

`Deltav = -v((Deltar)/r)`

a = `(Deltav)/(Deltat) = v/r((Deltar)/(Deltat)) = -v^2/r`

For uniform circular motion v = cor, where co is the angular velocity of the particle about center. Then the centripetal acceleration can be written as.
a = -ω²r