Circular Motion

What you'll learn

How to describe circular motion using period, frequency, radians and angular velocity.
How angular motion connects to linear speed using $\omega r$ .
Why an object moving at constant speed in a circle is still accelerating.
How to use centripetal acceleration and force equations in calculations and practical contexts.

Starting point: vectors and resultant force

A scalar has magnitude only. A vector has magnitude and direction. Speed is a scalar, but velocity is a vector because it includes direction.

Acceleration is the rate of change of velocity. So even if an object keeps the same speed, it can accelerate if its direction changes.

The resultant force is the single overall force that has the same effect as all the real forces acting on an object combined. By Newton’s second law, a non-zero resultant force causes acceleration.

Period and frequency

Circular motion repeats. One complete trip around the circle is one revolution or one cycle.

Definition

Period and frequency

The period of rotation, $T$ , is the time for one complete revolution, measured in seconds (s). The frequency, $f$ , is the number of revolutions per second, measured in hertz (Hz). They are related by $\frac{1}{T}$ .

Example

Finding period and frequency

A motor completes 48 revolutions in 12.0 s. Find its frequency and period.

The frequency is the number of revolutions per second:
$\frac{48}{12.0\ \text{s}} = 4.00\ \text{Hz}$ .
The period is the reciprocal of frequency:
$\frac{1}{4.00\ \text{Hz}} = 0.250\ \text{s}$ .
Check the pair is consistent:
$4.00\ \text{s}^{-1} \times 0.250\ \text{s} = 1.00$ .

Radians: the natural angle unit for circles

At A Level, circular motion uses angles measured in radians, not degrees. The key idea is that an angle can be defined by the arc length it cuts out on a circle.

Definition

The radian

One radian is the angle at the centre of a circle when the arc length is equal to the radius. More generally, for arc length $s$ and radius $r$ , the angle in radians is $θ=sr\theta = \frac{s}{r}$ , so $r\theta$ .

The diagram shows why radians fit naturally with circular motion: angle is directly linked to distance travelled around the circle.

A circle showing radian measure as arc length divided by radius

A full circle has circumference $2πr2\pi r$ , so a full turn is:

rad\theta = \frac{2\pi r}{r} = 2\pi\ \text{rad}

Tip

Radians are dimensionless

Because $θ=sr\theta = \frac{s}{r}$ is length divided by length, radians are technically dimensionless. Still, write rad in answers to show clearly that you are using radians.

Example

Using radians to find angular displacement

A point on the rim of a wheel of radius 0.18 m moves through an arc length of 0.72 m. Find the angular displacement in radians.

Use the radian definition:
$θ=sr\theta = \frac{s}{r}$ .
Substitute the arc length and radius:
$rad\theta = \frac{0.72\ \text{m}}{0.18\ \text{m}} = 4.0\ \text{rad}$ .
Compare with a full turn if useful:
$4.02π=0.64\frac{4.0}{2\pi} = 0.64$ , so the point has moved about 0.64 of a revolution.

Common Mistake

Using degrees in circular motion formulae

Equations such as $r\theta$ , $\omega r$ and $ω=2πT\omega = \frac{2\pi}{T}$ assume angles are in radians. Do not put degrees directly into these equations.

Angular velocity

Angular displacement is the angle swept out by the radius joining the centre of the circle to the object. The symbol is usually $θ\theta$ .

Definition

Angular velocity

Angular velocity, $ω\omega$ , is the rate of change of angular displacement: $ω=ΔθΔt\omega = \frac{\Delta \theta}{\Delta t}$ . Its unit is radians per second, written $s−1\text{rad s}^{-1}$ .

For steady circular motion, one complete revolution is $rad2\pi\ \text{rad}$ , so:

ω=2πT=2πf\omega = \frac{2\pi}{T} = 2\pi f

The same symbol $ω\omega$ is also used in simple harmonic motion (SHM). In SHM, it is often called angular frequency: it is the rate at which the phase angle changes, equivalent to the angular velocity of a matching reference circle.

Connecting angular velocity to linear speed

The linear speed, $v$ , is the speed along the circular path. Since arc length is $r\theta$ :

\frac{\Delta s}{\Delta t} = \frac{r\Delta \theta}{\Delta t} = \omega r

So the further you are from the centre, the greater your linear speed for the same angular velocity.

Example

Finding angular velocity and speed

A centrifuge spins at 1200 revolutions per minute. A sample is 0.24 m from the centre. Find $ω\omega$ and $v$ .

Convert revolutions per minute to hertz:
$\frac{1200}{60.0\ \text{s}} = 20.0\ \text{Hz}$ .
Find the angular velocity:
$s−1\omega = 2\pi f = 2\pi \times 20.0\ \text{s}^{-1} = 126\ \text{rad s}^{-1}$ .
Use $\omega r$ :
$126\ \text{s}^{-1} \times 0.24\ \text{m} = 30\ \text{m s}^{-1}$ .

Centripetal acceleration and centripetal force

If an object moves at constant speed in a circle, its velocity is constantly changing direction. Therefore it has an acceleration.

Definition

Centripetal acceleration and force

Centripetal means centre-seeking. In uniform circular motion, the acceleration is directed towards the centre of the circle, and the resultant force towards the centre is called the centripetal force.

The diagram shows the key directions: velocity is tangential, while acceleration and resultant force point inwards.

Uniform circular motion with tangent velocity and inward centripetal acceleration and force

Key Idea

Constant speed does not mean zero acceleration

In circular motion, the speed can stay constant while the velocity changes because its direction changes. The acceleration is towards the centre, not in the direction of motion.

For a small time interval, the velocity vector turns through the same small angle as the radius. This leads to:

\frac{v^2}{r}

Using $\omega r$ , this can also be written as:

\omega^2 r

Then using Newton’s second law, $F = ma$ , the centripetal force equations are:

\omega r, \quad a = \omega^2 r, \quad a = \frac{v^2}{r}, \quad F = \frac{mv^2}{r}, \quad F = m\omega^2 r

Here $r$ is radius in metres, $m$ is mass in kilograms, $v$ is speed in metres per second, $a$ is acceleration in metres per second squared, and $F$ is force in newtons.

Example

Calculating centripetal acceleration and force

A 0.150 kg object moves in a horizontal circle of radius 0.80 m with period 1.20 s. Find its centripetal acceleration and centripetal force.

Find the angular velocity:
$s−1\omega = \frac{2\pi}{T} = \frac{2\pi}{1.20\ \text{s}} = 5.24\ \text{rad s}^{-1}$ .
Find the centripetal acceleration using $\omega^2 r$ :
$\left(5.24\ \text{s}^{-1}\right)^2 \times 0.80\ \text{m} = 22.0\ \text{m s}^{-2}$ .
Calculate the centripetal force using $F = ma$ :
$0.150\ \text{kg} \times 22.0\ \text{m s}^{-2} = 3.30\ \text{N}$ .
Check with the direct equation:
$m\omega^2 r = 0.150\ \text{kg} \times \left(5.24\ \text{s}^{-1}\right)^2 \times 0.80\ \text{m} = 3.30\ \text{N}$ .

Common Mistake

Centripetal force is not an extra force

Do not draw “centripetal force” as an additional real force. The centripetal force is the inward resultant of real forces such as tension, friction, weight or normal contact force.

Common Mistake

Uniform circular motion only

These equations apply when the object has constant speed and fixed radius. If the speed is changing, there is also tangential acceleration, so the resultant force is not purely towards the centre.

Practical and data-analysis links

A common practical context is a bung or mass moving in a horizontal circle on a string. You measure the radius, time many revolutions, and calculate $T$ , $f$ and $ω\omega$ .

Timing many revolutions reduces percentage uncertainty. For example, timing 20 revolutions gives a smaller percentage uncertainty in $T$ than timing just one revolution.

If the centripetal force is provided by a hanging mass, the tension can be estimated using $F = M g$ , provided the hanging mass is stationary. To test $m\omega^2 r$ , keep $m$ and $r$ constant and plot $F$ against $ω2\omega^2$ . A straight line through the origin supports the model, with gradient $m r$ .

Example

Interpreting a centripetal-force graph

A rotating mass has $0.050\ \text{kg}$ and $0.80\ \text{m}$ . A graph of $F$ against $ω2\omega^2$ has gradient $m0.038\ \text{kg m}$ . Compare this with the expected gradient.

From $m\omega^2 r$ , the gradient of a graph of $F$ against $ω2\omega^2$ should be $m r$ .
Calculate the expected gradient:
$0.050\ \text{kg} \times 0.80\ \text{m} = 0.040\ \text{kg m}$ .
Compare with the measured gradient:
$0.040−0.0380.040×100%=5.0%\frac{0.040 - 0.038}{0.040} \times 100\% = 5.0\%$ , so the result is close if the experimental uncertainties are of that order.

Exam technique

In the exam

Convert first: revolutions per minute to hertz, and degrees to radians if an angle calculation is involved.
Draw the real forces, then identify which components give the inward resultant force.
Choose the equation that matches the data given: use $\frac{v^2}{r}$ if you know $v$ , or $\omega^2 r$ if you know $ω\omega$ .
Watch proportionality carefully: for fixed radius, $\propto v^2$ ; for fixed angular velocity, $\propto r$ .
Carry units through calculations and round final answers to a sensible number of significant figures.

Self review

Check yourself

Why can an object moving at constant speed in a circle still have acceleration?
How are $T$ , $f$ and $ω\omega$ related for one complete revolution?
If the speed of an object doubles while its radius stays the same, what happens to the centripetal force?

What you'll learn

How to describe circular motion using period, frequency, radians and angular velocity.
How angular motion connects to linear speed using $\omega r$ .
Why an object moving at constant speed in a circle is still accelerating.
How to use centripetal acceleration and force equations in calculations and practical contexts.

Starting point: vectors and resultant force

A scalar has magnitude only. A vector has magnitude and direction. Speed is a scalar, but velocity is a vector because it includes direction.

Acceleration is the rate of change of velocity. So even if an object keeps the same speed, it can accelerate if its direction changes.

Period and frequency

Circular motion repeats. One complete trip around the circle is one revolution or one cycle.

Definition

Period and frequency

Example

Finding period and frequency

A motor completes 48 revolutions in 12.0 s. Find its frequency and period.

The frequency is the number of revolutions per second:
$\frac{48}{12.0\ \text{s}} = 4.00\ \text{Hz}$ .
The period is the reciprocal of frequency:
$\frac{1}{4.00\ \text{Hz}} = 0.250\ \text{s}$ .
Check the pair is consistent:
$4.00\ \text{s}^{-1} \times 0.250\ \text{s} = 1.00$ .

Radians: the natural angle unit for circles

At A Level, circular motion uses angles measured in radians, not degrees. The key idea is that an angle can be defined by the arc length it cuts out on a circle.

Definition

The radian

The diagram shows why radians fit naturally with circular motion: angle is directly linked to distance travelled around the circle.

A circle showing radian measure as arc length divided by radius

A full circle has circumference $2πr2\pi r$ , so a full turn is:

rad\theta = \frac{2\pi r}{r} = 2\pi\ \text{rad}

Tip

Radians are dimensionless

Because $θ=sr\theta = \frac{s}{r}$ is length divided by length, radians are technically dimensionless. Still, write rad in answers to show clearly that you are using radians.

Example

Using radians to find angular displacement

A point on the rim of a wheel of radius 0.18 m moves through an arc length of 0.72 m. Find the angular displacement in radians.

Use the radian definition:
$θ=sr\theta = \frac{s}{r}$ .
Substitute the arc length and radius:
$rad\theta = \frac{0.72\ \text{m}}{0.18\ \text{m}} = 4.0\ \text{rad}$ .
Compare with a full turn if useful:
$4.02π=0.64\frac{4.0}{2\pi} = 0.64$ , so the point has moved about 0.64 of a revolution.

Common Mistake

Using degrees in circular motion formulae

Equations such as $r\theta$ , $\omega r$ and $ω=2πT\omega = \frac{2\pi}{T}$ assume angles are in radians. Do not put degrees directly into these equations.

Angular velocity

Angular displacement is the angle swept out by the radius joining the centre of the circle to the object. The symbol is usually $θ\theta$ .

Definition

Angular velocity

For steady circular motion, one complete revolution is $rad2\pi\ \text{rad}$ , so:

ω=2πT=2πf\omega = \frac{2\pi}{T} = 2\pi f

Connecting angular velocity to linear speed

The linear speed, $v$ , is the speed along the circular path. Since arc length is $r\theta$ :

\frac{\Delta s}{\Delta t} = \frac{r\Delta \theta}{\Delta t} = \omega r

So the further you are from the centre, the greater your linear speed for the same angular velocity.

Example

Finding angular velocity and speed

A centrifuge spins at 1200 revolutions per minute. A sample is 0.24 m from the centre. Find $ω\omega$ and $v$ .

Convert revolutions per minute to hertz:
$\frac{1200}{60.0\ \text{s}} = 20.0\ \text{Hz}$ .
Find the angular velocity:
$s−1\omega = 2\pi f = 2\pi \times 20.0\ \text{s}^{-1} = 126\ \text{rad s}^{-1}$ .
Use $\omega r$ :
$126\ \text{s}^{-1} \times 0.24\ \text{m} = 30\ \text{m s}^{-1}$ .

Centripetal acceleration and centripetal force

If an object moves at constant speed in a circle, its velocity is constantly changing direction. Therefore it has an acceleration.

Definition

Centripetal acceleration and force

The diagram shows the key directions: velocity is tangential, while acceleration and resultant force point inwards.

Uniform circular motion with tangent velocity and inward centripetal acceleration and force

Key Idea

Constant speed does not mean zero acceleration

In circular motion, the speed can stay constant while the velocity changes because its direction changes. The acceleration is towards the centre, not in the direction of motion.

For a small time interval, the velocity vector turns through the same small angle as the radius. This leads to:

\frac{v^2}{r}

Using $\omega r$ , this can also be written as:

\omega^2 r

Then using Newton’s second law, $F = ma$ , the centripetal force equations are:

\omega r, \quad a = \omega^2 r, \quad a = \frac{v^2}{r}, \quad F = \frac{mv^2}{r}, \quad F = m\omega^2 r

Here $r$ is radius in metres, $m$ is mass in kilograms, $v$ is speed in metres per second, $a$ is acceleration in metres per second squared, and $F$ is force in newtons.

Example

Calculating centripetal acceleration and force

A 0.150 kg object moves in a horizontal circle of radius 0.80 m with period 1.20 s. Find its centripetal acceleration and centripetal force.

Find the angular velocity:
$s−1\omega = \frac{2\pi}{T} = \frac{2\pi}{1.20\ \text{s}} = 5.24\ \text{rad s}^{-1}$ .
Find the centripetal acceleration using $\omega^2 r$ :
$\left(5.24\ \text{s}^{-1}\right)^2 \times 0.80\ \text{m} = 22.0\ \text{m s}^{-2}$ .
Calculate the centripetal force using $F = ma$ :
$0.150\ \text{kg} \times 22.0\ \text{m s}^{-2} = 3.30\ \text{N}$ .
Check with the direct equation:
$m\omega^2 r = 0.150\ \text{kg} \times \left(5.24\ \text{s}^{-1}\right)^2 \times 0.80\ \text{m} = 3.30\ \text{N}$ .

Common Mistake

Centripetal force is not an extra force

Do not draw “centripetal force” as an additional real force. The centripetal force is the inward resultant of real forces such as tension, friction, weight or normal contact force.

Common Mistake

Uniform circular motion only

These equations apply when the object has constant speed and fixed radius. If the speed is changing, there is also tangential acceleration, so the resultant force is not purely towards the centre.

Practical and data-analysis links

A common practical context is a bung or mass moving in a horizontal circle on a string. You measure the radius, time many revolutions, and calculate $T$ , $f$ and $ω\omega$ .

Timing many revolutions reduces percentage uncertainty. For example, timing 20 revolutions gives a smaller percentage uncertainty in $T$ than timing just one revolution.

Example

Interpreting a centripetal-force graph

A rotating mass has $0.050\ \text{kg}$ and $0.80\ \text{m}$ . A graph of $F$ against $ω2\omega^2$ has gradient $m0.038\ \text{kg m}$ . Compare this with the expected gradient.

From $m\omega^2 r$ , the gradient of a graph of $F$ against $ω2\omega^2$ should be $m r$ .
Calculate the expected gradient:
$0.050\ \text{kg} \times 0.80\ \text{m} = 0.040\ \text{kg m}$ .
Compare with the measured gradient:
$0.040−0.0380.040×100%=5.0%\frac{0.040 - 0.038}{0.040} \times 100\% = 5.0\%$ , so the result is close if the experimental uncertainties are of that order.

Exam technique

In the exam

Convert first: revolutions per minute to hertz, and degrees to radians if an angle calculation is involved.
Draw the real forces, then identify which components give the inward resultant force.
Choose the equation that matches the data given: use $\frac{v^2}{r}$ if you know $v$ , or $\omega^2 r$ if you know $ω\omega$ .
Watch proportionality carefully: for fixed radius, $\propto v^2$ ; for fixed angular velocity, $\propto r$ .
Carry units through calculations and round final answers to a sensible number of significant figures.

Self review

Check yourself

Why can an object moving at constant speed in a circle still have acceleration?
How are $T$ , $f$ and $ω\omega$ related for one complete revolution?
If the speed of an object doubles while its radius stays the same, what happens to the centripetal force?

What you'll learn

Starting point: vectors and resultant force

Period and frequency

Radians: the natural angle unit for circles

Angular velocity

Connecting angular velocity to linear speed

Centripetal acceleration and centripetal force

Practical and data-analysis links

Recap questions

Review Circular Motion by teaching Genie

Remember key concepts with flashcards

Circular Motion

What you'll learn

Starting point: vectors and resultant force

Period and frequency

Radians: the natural angle unit for circles

Angular velocity

Connecting angular velocity to linear speed

Centripetal acceleration and centripetal force

Practical and data-analysis links

Recap questions

Review Circular Motion by teaching Genie

Remember key concepts with flashcards

Circular Motion

What you'll learn

Starting point: vectors and resultant force

Period and frequency

Radians: the natural angle unit for circles

Angular velocity

Connecting angular velocity to linear speed

Centripetal acceleration and centripetal force

Practical and data-analysis links

Recap questions

Review Circular Motion by teaching Genie

Remember key concepts with flashcards

Circular Motion Revision Guide

Circular Motion

What you'll learn

Starting point: vectors and resultant force

Period and frequency

Radians: the natural angle unit for circles

Angular velocity

Connecting angular velocity to linear speed

Centripetal acceleration and centripetal force

Practical and data-analysis links

Recap questions

Review Circular Motion by teaching Genie

Remember key concepts with flashcards

Circular Motion Revision Guide