Deriving centripetal acceleration

When I was an A-level physics student (many, many years ago, when the world was young LOL) I found the derivation of the centripetal acceleration formula really hard to understand. What follows is a method that I have developed over the years that seems to work well. The PowerPoint is included at the end.

Step 1: consider an object moving on a circular path

Let’s consider an object moving in circular path of radius r at a constant angular speed of ω (omega) radians per second.

The object is moving anticlockwise on the diagram and we show it at two instants which are time t seconds apart. This means that the object has moved an angular distance of ωt radians.

Step 2: consider the linear velocities of the object at these times

The linear velocity is the speed in metres per second and acts at a tangent to the circle, making a right angle with the radius of the circle. We have called the first velocity v1 and the second velocity at the later time v2.

Since the object is moving at a constant angular speed ω and is a fixed radius r from the centre of the circle, the magnitudes of both velocities will be constant and will be given by v = ωr.

Although the magnitude of the linear velocity has not changed, its direction most certainly has. Since acceleration is defined as the change in velocity divided by time, this means that the object has undergone acceleration since velocity is a vector quantity and a change in direction counts as a change, even without a change in magnitude.

Step 3a: Draw a vector diagram of the velocities

We have simply extracted v1 and v2 from the original diagram and placed them nose-to-tail. We have kept their magnitude and direction unchanged during this process.

Step 3b: close the vector diagram to find the resultant

The dark blue arrow is the result of adding v1 and v2. It is not a useful operation in this case because we are interested in the change in velocity not the sum of the velocities, so we will stop there and go back to the drawing board.

Step 3c switch the direction of velocity v1

Since we are interested in the change in velocity, let’s flip the direction of v1 so that it going in the opposite direction. Since it is opposite to v1, we can now call this -v1.

It is preferable to flip v1 rather than v2 since for a change in velocity we typically subtract the initial velocity from the final velocity; that is to say, change in velocity = v2 – v1.

Step 3d: Put the vectors v2 and (-v1) nose-to-tail

Step 3e: close the vector diagram to find the result of adding v2 and (-v1)

The purple arrow shows the result of adding v2 + (-v1); in other words, the purple arrow shows the change in velocity between v1 and v2 due to the change in direction (notwithstanding the fact that the magnitude of both velocities is unchanged).

It is also worth mentioning that that the direction of the purple (v2v1) arrow is in the opposite direction to the radius of the circle: in other words, the change in velocity is directed towards the centre of the circle.

Step 4: Find the angle between v2 and (-v1)

The angle between v2 and (-v1) will be ωt radians.

Step 5: Use the small angle approximation to represent v2-v1 as the arc of a circle

If we assume that ωt is a small angle, then the line representing v2-v1 can be replaced by the arc c of a circle of radius v (where v is the magnitude of the vectors v1 and v2 and v=ωr).

We can then use the familiar relationship that the angle θ (in radians) subtended at the centre of a circle θ = arc length / radius. This lets express the arc length c in terms of ω, t and r.

And finally, we can use the acceleration = change in velocity / time relationship to derive the formula for centripetal acceleration we a = ω2r.


Well, that’s how I would do it. If you would like to use this method or adapt it for your students, then the PowerPoint is attached.

Please Like or leave a comment if you find this useful 🙂

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s