Centripetal Acceleration and Force

Overview

Centripetal Acceleration and Force explains why an object moving in a circular path requires an acceleration and a resultant inward force, even when its speed is constant.

A common misconception is that constant speed means zero acceleration. This is false because velocity is a vector quantity, and a change in direction means the velocity changes.

This page deepens the core ideas introduced in Circular Motion.

Why It Matters

This is the core physics of circular motion. Students must distinguish the inward acceleration from the real forces that produce it.

Definition

Centripetal acceleration is the inward acceleration required for motion along a circular path. Centripetal force is not an extra force; it is the resultant radial force provided by real physical forces.

Key Representations

$$a_c=\frac{v^2}{r}$$ $$a_c={\omega }^{2}r$$ $$\sum F_{\text{radial}}=\frac{m{v}^2}{r}$$ $$F_c=\frac{m{v}^2}{r}=m{\omega }^{2}r$$

Scalar vs Vector Distinction

Scalars

• speed
• mass
• radius
• time
• period
• frequency

Vectors

• velocity $\vec{v}$
• acceleration $\vec{a}$
• force $\vec{F}$

Key Idea

In uniform circular motion:

• speed may remain constant
• velocity changes continuously

Therefore:

• acceleration is non-zero
• resultant force is non-zero

Why Acceleration Exists in Circular Motion

Suppose a particle moves in a circle at constant speed.

At two nearby positions:

• the magnitudes of the velocity vectors are equal
• their directions are different

Hence:

$$\Delta \vec{v}\ne 0$$

So:

$$\vec{a}=\frac{\Delta \vec{v}}{\Delta t}$$

Therefore the particle accelerates.

For uniform circular motion, this acceleration always points toward the centre of the circle.

Centripetal Acceleration

The inward acceleration required to maintain circular motion is called centripetal acceleration.

• centripetal means centre-seeking

Magnitude:

$$a_c=\frac{v^2}{r}$$

Also:

$$a_c={\omega }^{2}r$$

Where:

• $v$ = speed
• $r$ = radius
• $\omega$ = angular speed

Direction

The vector acceleration is:

$${\vec{a}}_c$$

and points toward the centre of the circular path.

Why the Formula Makes Sense

Formula:

$$a_c=\frac{v^2}{r}$$

Interpretation

• larger speed → larger acceleration needed
• smaller radius → sharper turning → larger acceleration

Example

A racing car moving faster around the same bend needs a much larger inward acceleration.

Angular Form

Using:

$$v=r\omega$$

Substitute into:

$$a_c=\frac{v^2}{r}$$

Then:

$$a_c=\frac{(r\omega {)}^2}{r}={\omega }^{2}r$$

Useful when angular speed is given.

Centripetal Force

By Newton’s Second Law:

$${\vec{F}}_{\text{net}}=m\vec{a}$$

Therefore the inward resultant force needed is:

$$F_c=m{a}_c$$

So:

$$F_c=\frac{m{v}^2}{r}$$

or

$$F_c=m{\omega }^{2}r$$

Important Warning: Not a New Force Type

“Centripetal force” is not an extra physical force added to the diagram.

It is the resultant inward force produced by real forces such as:

• tension
• friction
• normal contact force
• gravitational force
• electric force

Example

For a stone tied to a string:

• tension provides centripetal force

For a car on flat road:

• friction provides centripetal force

For a satellite:

• gravity provides centripetal force

Direction and Sign Reasoning

If radial inward direction is chosen positive, then:

$$\sum F_{\text{inward}}=\frac{m{v}^2}{r}$$

If outward is chosen positive, signs must be handled consistently.

At H2 level, choosing inward as positive is usually simplest.

Worked Example 1: Centripetal Acceleration

A ball moves in a circle of radius $0.80\text{ m}$ at speed $4.0{\text{ m s}}^{-1}$.

Find centripetal acceleration.

Solution

$$a_c=\frac{v^2}{r}=\frac{4.0^2}{0.80}=\frac{16}{0.80}=20{\text{ m s}}^{-2}$$

Toward the centre.

Worked Example 2: Resultant Force

The ball in Example 1 has mass $0.25\text{ kg}$.

Find the required resultant inward force.

Solution

$$F=ma=0.25(20)=5.0\text{ N}$$

Toward the centre.

Worked Example 3: Angular Speed Form

A disc rotates at angular speed:

$$\omega =6.0{\text{ rad s}}^{-1}$$

A point is $0.30\text{ m}$ from the centre.

Find centripetal acceleration.

Solution

$$a_c={\omega }^{2}r=6.0^2(0.30)=36(0.30)=10.8{\text{ m s}}^{-2}$$

Comparison with Tangential Acceleration

Centripetal Acceleration

• changes direction of velocity
• points inward

Tangential Acceleration

• changes speed
• acts along tangent

In uniform circular motion:

• tangential acceleration = 0
• centripetal acceleration $\ne 0$

Common Exam Pitfalls

1. Saying Constant Speed Means No Acceleration

Wrong. Direction changes.

2. Drawing Centripetal Force as Extra Arrow

Wrong. Use only real forces.

3. Wrong Direction

Acceleration and resultant force point toward centre.

4. Using Radius as Diameter

Always check whether given value is radius or diameter.

5. Mixing Speed and Velocity

Use speed $v$ in magnitude formulas.

Summary

For circular motion:

• changing direction means changing velocity
• changing velocity means acceleration
• inward acceleration is centripetal acceleration

$$a_c=\frac{v^2}{r}={\omega }^{2}r$$

The required inward resultant force is:

$$F_c=\frac{m{v}^2}{r}=m{\omega }^{2}r$$

This “centripetal force” is not separate; it is provided by real forces.

Links

• Circular Motion
• Circular Motion Force Analysis
• Vertical Circular Motion
• Orbital Motion in Gravity
• Dynamics
• Vectors