Centripetal Acceleration Formula
We can derive a formula for the centripetal acceleration. Consider the case of a stone tied to a rope of length r. It is swung round and round in a horizontal circle, at a constant speed v.
Imagine looking down from the top. At one instant, the stone moves in one direction, perpendicular to the rope. After a short time t, the rope moves through an angle θ, and the stone moves over a short distance s. Since d is quite short, lets take it to be approximately straight.
Draw a picture of the initial direction of the stone, the initial and final positions the rope, and the line for d. Using geometry, it is possible to show that the angle between the new path d, and the initial direction, is half the angular displacement of the rope, i.e. θ/2.
Exercise. Draw this picture.
So instead of going straight, the stone moves, say, to the left by a small angle θ/2.
To find the sideways acceleration, we need to know the distance of the stone from the original direction. From the initial position, think of a distance d along the original direction, and the line d along the new direction. The sideways displacement is roughly the arc formed by these two lines. This arc length is d x θ/2.
Summarising, after a short time t, the stone is displaced sideways by dθ/2. So what is the acceleration?
Recall the acceleration formula: s = ut + at2/2.
The displacement s is dθ/2 for the stone. Initial sideways velocity u is 0. Substituting, dθ/2 = at2/2. Solving, a = dθ/t2.
Writing in this form: a = (d/t)(θ/t), we find that a = vω. This is the formula for the centripetal acceleration.
It is also commonly written in two other forms, using either v = rω or ω = v/r. Substituting one or the other, we find:
a = ω2r, or a = v2/r.
Copyright 2010 by Kai Hock. All rights reserved.
Last updated: 22 September 2010.