Centripetal Acceleration Equation, Derivation & Rearrangements

Core idea

Overview

Centripetal acceleration represents the rate of change of the velocity vector's direction for an object moving in a circular path. Even at a constant speed, an object accelerates because its direction is constantly shifting toward the center of rotation.

When to use: This equation is used for any object undergoing uniform circular motion or moving along a curved trajectory with a known radius. It assumes the object's path is part of a perfect circle at the moment of measurement.

Why it matters: It is fundamental for calculating the forces needed to keep satellites in orbit, designing safe highway curves, and engineering amusement park rides. Without understanding this acceleration, we could not predict the tension in mechanical rotors or the friction required for vehicles to turn.

Symbols

Variables

a = Acceleration, v = Velocity, r = Radius

a

Acceleration

m / s^{2}

v

Velocity

m/s

r

Radius

m

Walkthrough

Derivation

Derivation of Centripetal Acceleration

Derives the acceleration of an object moving in a circle at constant speed.

Speed v is constant (uniform circular motion).

1

Relate Change in Velocity to Angle Swept:

Using similar triangles: the velocity triangle is similar to the radius/arc triangle, so $Δ$ v/v equals $Δ$ $θ$ , and $Δ$ $θ$ = $Δ$ s/r.

\frac{Δ v}{v} = \frac{Δ s}{r}

2

Divide by Time:

Introduce acceleration and note that $Δ$ s/ $Δ$ t is the speed v.

a = \frac{Δ v}{Δ t} = \frac{v}{r} \frac{Δ s}{Δ t}

3

Substitute \u0394s/\u0394t=v:

So centripetal acceleration is $v^{2}$ /r. Using v=r $ω$ gives $ω^{2}$ r.

a = \frac{v ^{2}}{r} = ω^{2} r

Result

a = \frac{v ^{2}}{r} = ω^{2} r

Source: Edexcel A-Level Physics — Further Mechanics

Free formulas

Rearrangements

Solve for $a$

Make a the subject

a = \frac{v ^{2}}{r}

a is already the subject of the formula.

Difficulty: 1/5

Solve for $v$

Make v the subject

v = a r

Start from Centripetal Acceleration. To make v the subject, clear r, then make $v^{2}$ the subject, then take the square root.

Difficulty: 4/5

Solve for $r$

Make r the subject

r = \frac{v ^{2}}{a}

Start from Centripetal Acceleration. To make r the subject, clear r, then divide by a.

Difficulty: 3/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

The graph follows a parabolic curve starting from the origin because acceleration is proportional to the square of velocity. For a physics student, this means that even small increases in velocity result in significantly larger requirements for acceleration to maintain circular motion. The most important feature of this curve is that the steepening slope demonstrates how rapidly the demand for centripetal acceleration grows as velocity increases.

Graph type: parabolic

Why it behaves this way

Intuition

An object moving in a circle is constantly being pulled or pushed towards the center, preventing it from flying off in a straight line tangent to the circle.

a

Centripetal acceleration

The acceleration an object experiences towards the center of its circular path, caused by the continuous change in its direction of motion.

v

Speed of the object

How fast the object is moving along its circular path. Higher speed means it covers more distance per unit time, requiring a greater change in direction to stay on the curve.

r

Radius of the circular path

The distance from the center of rotation to the object. A smaller radius means a tighter curve, demanding a more rapid change in direction for a given speed.

Signs and relationships

v^2: The acceleration depends on the square of the speed because both the magnitude of the velocity vector and the rate at which its direction changes are proportional to the speed.
/r: The acceleration is inversely proportional to the radius. For a given speed, a smaller radius implies a sharper curve, which necessitates a greater acceleration towards the center to keep the object on that path.

Free study cues

Insight

Canonical usage

In the International System of Units (SI), centripetal acceleration is expressed in meters per second squared (m/s2), velocity in meters per second (m/s), and radius in meters (m).

Common confusion

The most common confusion arises from using inconsistent units for velocity and radius, such as velocity in kilometers per hour and radius in meters, leading to incorrect acceleration values.

Unit systems

$a$ m/s2 · Centripetal acceleration is a vector quantity, but this formula calculates its magnitude.

$v$ m/s · Represents the magnitude of the tangential velocity or speed.

$r$ m · The radius of the circular path.

One free problem

Practice Problem

A race car travels around a circular track with a radius of 50 meters at a constant speed of 20 m/s. Calculate the centripetal acceleration experienced by the car.

Velocity20 m/s

Radius50 m

Solve for: $a$

Hint: Square the speed first before dividing by the radius.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Finding acceleration of a car on a roundabout.

Study smarter

Tips

Confirm that velocity is in m/s and radius is in meters to get acceleration in m/s².
Note that acceleration is proportional to the square of the speed; doubling speed quadruples the acceleration.
Always remember that centripetal acceleration vectors point directly toward the center of the circle.

Avoid these traps

Common Mistakes

Using diameter instead of radius.
Forgetting v is squared.

Keep going

Related Formulas

Common questions

Frequently Asked Questions

Derives the acceleration of an object moving in a circle at constant speed.

This equation is used for any object undergoing uniform circular motion or moving along a curved trajectory with a known radius. It assumes the object's path is part of a perfect circle at the moment of measurement.

It is fundamental for calculating the forces needed to keep satellites in orbit, designing safe highway curves, and engineering amusement park rides. Without understanding this acceleration, we could not predict the tension in mechanical rotors or the friction required for vehicles to turn.

Using diameter instead of radius. Forgetting v is squared.

Finding acceleration of a car on a roundabout.

Confirm that velocity is in m/s and radius is in meters to get acceleration in m/s². Note that acceleration is proportional to the square of the speed; doubling speed quadruples the acceleration. Always remember that centripetal acceleration vectors point directly toward the center of the circle.

References

Sources

Halliday, Resnick, Walker, Fundamentals of Physics
Wikipedia: Centripetal acceleration
NIST Guide for the Use of the International System of Units (SI) (NIST Special Publication 811, 2008 edition)
Fundamentals of Physics, 10th Edition by Halliday, Resnick, and Walker (2014)
Halliday, Resnick, and Walker, Fundamentals of Physics
Edexcel A-Level Physics — Further Mechanics

Centripetal Acceleration

Overview

Variables

Derivation

Relate Change in Velocity to Angle Swept:

Divide by Time:

Substitute \u0394s/\u0394t=v:

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Centripetal Force

Frequently Asked Questions

Sources