Year 12 Physics Module 5 ⏱ ~45 min 5 MC · 3 Short Answer Lesson 7 of 18

Uniform Circular Motion

A rider on a rotating platform can keep the same speedometer reading while their velocity changes every instant. This lesson connects period, frequency and angular speed to the inward acceleration and radial net force that continually redirect that velocity.

Today's hook: A car rounds a bend at constant speed. Its speed is unchanged, yet its velocity is not. What direction must its acceleration point, and which real force or combination of forces supplies the required radial net force?
0/5TASKS
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Connect and predict

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice
Before you read, predict

A car travels around a roundabout at constant speed. Is it accelerating? Explain your reasoning.

Consider what you know about velocity as a vector quantity. Does constant speed mean constant velocity?

Warm-up, velocity is a vector. Which of the following changes constitutes an acceleration?

Learning Intentions
goals

Know, Define Centripetal Acceleration and Force

  • Understand that circular motion requires a net inward force
  • State the direction of centripetal acceleration
  • Identify the force(s) that provide centripetal force in different situations

Understand, Solve Problems Using $a_c = v^2/r$ and $F_c = mv^2/r$

  • Select the appropriate formula for the given quantities
  • Rearrange equations to find unknown variables
  • Work with angular velocity and relate it to linear speed

Can Do, Relate $\omega$ to $v$ and Explain Zero Work

  • Use $v = \omega r$ to convert between angular and linear quantities
  • Calculate period and frequency from angular velocity
  • Explain why centripetal force does zero work in uniform circular motion
Scan these before reading
vocab
Uniform circular motionMotion in a circle at constant speed. The speed is constant but the velocity changes direction continuously.
Period ($T$)The time for one complete revolution, measured in seconds (s).
Angular velocity ($\omega$)The angular displacement per unit time: $\omega=\Delta\theta/\Delta t$. For uniform circular motion, $\omega = 2\pi/T = 2\pi f$, measured in rad s$^{-1}$.
Centripetal acceleration ($a_c$)Acceleration directed toward the centre of the circle. $a_c = v^2/r = \omega^2 r$.
Centripetal force ($F_c$)The net force directed toward the centre that maintains circular motion. $F_c = mv^2/r$. It is not a new type of force.
Centrifugal forceA fictitious (pseudo) force experienced in rotating reference frames. It does not exist in an inertial frame of reference.
Cross-lesson links: L06 introduced moments and equilibrium. L07 begins circular-motion kinematics, then applies Newton’s second law radially. Later lessons use this foundation for conical pendulums, banked bends, vertical circles and gravitational orbits.
Misconceptions to fix
✗ Wrong: "Centrifugal force pushes you outward on a roundabout."
✓ Right: Inertia wants you to travel in a straight line (Newton's First Law). Friction or tension provides an inward centripetal force that continuously changes your direction. There is no real outward force in an inertial reference frame.
✗ Wrong: "Centripetal force is a new type of force on the free-body diagram."
✓ Right: Centripetal force is the name for the net inward force. It can be provided by tension, friction, gravity, normal force, or any combination, it is not listed separately on a free-body diagram.
✗ Wrong: "If speed is constant, acceleration must be zero."
✓ Right: Acceleration measures rate of change of velocity (a vector). Even when speed is constant, a continuous change in direction means velocity is changing, so $a_c = v^2/r \neq 0$.

Centrifugal force is a real force that acts outward in all reference frames.

An object moving at constant speed always has zero acceleration.

Core Content

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Describe period, frequency and angular speed

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Describing Circular Motion
+5 XP

Period, frequency, angular velocity, and linear speed

Uniform circular motion diagram showing key quantities: period, angular velocity, linear speed and radius

Uniform circular motion, key relationships between period $T$, angular velocity $\omega$, linear speed $v$ and radius $r$.

A turntable completes repeated revolutions. One revolution sweeps through $2\pi$ radians; the outer edge travels farther than a point near the axis during the same time. Describing this motion requires:

  • Period ($T$): The time for one complete revolution. Measured in seconds (s).
  • Frequency ($f$): The number of revolutions per second. $f = 1/T$, measured in hertz (Hz).
  • Angular speed ($\omega$): The angular displacement per unit time, $\omega=\Delta\theta/\Delta t$. For uniform circular motion, one revolution is $2\pi$ radians, so $\omega = 2\pi/T = 2\pi f$, measured in rad s$^{-1}$. The radian is a dimensionless ratio of arc length to radius, but “rad” is retained to show an angular quantity.
  • Linear (tangential) speed ($v$): The distance travelled per unit time along the circular path. $v = 2\pi r/T = \omega r$.

The relationship $v = \omega r$ is particularly important. It tells us that for a given angular velocity, points farther from the centre travel faster. This is why the outer edge of a spinning DVD moves faster than the inner part.

Key Circular Motion Relationships

$f = \dfrac{1}{T}$ (Hz)

$\omega = \dfrac{2\pi}{T} = 2\pi f$ (rad/s)

$v = \dfrac{2\pi r}{T} = \omega r$ (m/s)

Worked example 1+5 XP on full reveal

Period, speed and angular speed. A car travels around a 50 m radius circular track in 12 s. Find $v$, $\omega$ and $f$.

1
Identify the cycle data
$r=50\text{ m}$ and $T=12\text{ s}$.
One lap is one period and corresponds to $2\pi$ radians.
2
Find tangential speed
$v=2\pi r/T=2\pi(50)/12=\mathbf{26.2\text{ m/s}}$.
The car covers one circumference during each period.
3
Find angular speed
$\omega=2\pi/T=\mathbf{0.524\text{ rad/s}}$.
“rad” records that this dimensionless ratio describes angular change.
4
Find frequency
$f=1/T=\mathbf{0.0833\text{ Hz}}$.
The reciprocal check $fT=1$ confirms the pair is consistent.
Step 1 of 4

For circular motion: $f = 1/T$ (Hz); $\omega = 2\pi/T = 2\pi f$ (rad/s); $v = 2\pi r/T = \omega r$ (m/s). At fixed $\omega$, larger $r$ means faster linear speed, outer edge of a spinning disc moves faster than the centre.

Pause, copy the highlighted circular motion relationships into your book before moving on.

A wheel completes 5 revolutions per second. Its angular velocity is:

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Represent tangential velocity and radial acceleration

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Centripetal Acceleration
+5 XP

Why constant speed does not mean zero acceleration

We just saw how to describe circular motion using $f$, $\omega$ and $v$. That raises a question: if speed is constant, how can there be acceleration? This card answers it → velocity is a vector; continuous change of direction means $a_c = v^2/r$ directed toward the centre.

In uniform circular motion, the speed is constant but the velocity is not. Velocity is a vector, it has both magnitude (speed) and direction. As the object moves around the circle, its direction continuously changes. A change in velocity means there is acceleration.

Consider the velocity vectors at two nearby points on the circle. The velocity is always tangent to the circle. When we subtract these vectors to find $\Delta \vec{v}$, the result points toward the centre of the circle. Taking the limit as the time interval approaches zero gives the instantaneous acceleration.

This acceleration is called centripetal acceleration ($a_c$), and it is always directed toward the centre of the circle.

Centripetal Acceleration

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

Direction: always toward the centre of the circle. Units: m s$^{-2}$.

The two forms are equivalent since $v = \omega r$:

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

Predict before substituting

From $a_c=v^2/r$: at fixed radius, $a_c\propto v^2$, so doubling speed makes acceleration four times larger. At fixed speed, $a_c\propto 1/r$, so doubling radius halves the acceleration. Use these relationships as a reasonableness check.

Key insight

At the peak of a projectile's flight, vertical velocity is zero but acceleration is still $-9.8\ \text{m s}^{-2}$. Similarly, in circular motion the speed may be constant but the acceleration is never zero, it continuously changes the direction of velocity. Constant speed $\neq$ constant velocity $\neq$ zero acceleration.

Worked example 2+5 XP on full reveal

Centripetal acceleration. Find $a_c$ for the same car, using $v=26.2\text{ m/s}$ and $r=50\text{ m}$.

1
Predict
$a_c\propto v^2/r$ and must point inward.
Direction is part of a vector answer, not an optional label.
2
Substitute
$a_c=(26.2)^2/50$.
Squaring m/s and dividing by m gives m/s².
3
Calculate and state direction
$a_c=\mathbf{13.7\text{ m/s}^2}$ toward the centre.
The tangential velocity and radial acceleration are perpendicular.
Step 1 of 3

Centripetal acceleration $a_c = v^2/r = \omega^2 r$ (m s⁻²), always directed toward the centre of the circle. Increasing $v$ or decreasing $r$ both increase $a_c$; constant speed does NOT mean zero acceleration.

Add the highlighted centripetal acceleration formula and direction rule to your notes before the check below.

An object in uniform circular motion has acceleration directed toward the centre.

If the speed of a circular-motion object doubles and radius stays the same, centripetal acceleration doubles.

Centripetal acceleration is always perpendicular to the velocity of the object.

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Derive and identify the radial net force

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Centripetal Force
+5 XP

The net force required to maintain circular motion

We just saw that centripetal acceleration is always directed toward the centre. That raises a question: what force causes this inward acceleration? This card answers it → by Newton's 2nd Law, $F_c = mv^2/r$; it is not a new force type, tension, friction or gravity can each provide it.

By Newton's Second Law, the inward acceleration requires an inward radial net force. “Centripetal force” names that resultant; it is not an additional force to draw beside tension, friction, gravity or a normal force.

Centripetal Force

$F_c = ma_c = \dfrac{mv^2}{r} = m\omega^2 r$

Direction: always toward the centre of the circle. Units: N.

Critical understanding: Centripetal force is not a new type of force. It is simply the name we give to the net force directed toward the centre that causes circular motion. This net force can be provided by:

  • Tension a ball on a string being whirled in a horizontal circle
  • Friction a car turning on a flat road
  • Gravity a satellite in orbit, or the Moon around Earth
  • Normal force a car on a banked curve
  • Any combination of these forces

If the net inward force is removed, the object travels in a straight line at constant velocity (Newton's First Law), it does not fly radially outward.

Worked example 3+5 XP on full reveal

Radial force on a car. A 1200 kg car rounds a level 40 m radius curve at 15 m/s. Find the static-friction force required.

1
Identify the real radial force
On the level curve, static friction supplies the inward resultant.
Do not draw a separate “centripetal force” in addition to friction.
2
Write the radial equation
$\sum F_r=F_f=mv^2/r$.
Take inward as positive for the radial component equation.
3
Substitute
$F_f=1200(15)^2/40$.
The required force scales with $v^2$, so speed strongly affects the demand on friction.
4
Calculate and report direction
$F_f=\mathbf{6.75\times10^3\text{ N}}$ toward the centre.
This is the required static friction, not automatically the maximum available friction.
Step 1 of 4

Centripetal force $F_c = mv^2/r = m\omega^2 r$ (N), directed toward the centre. It is the net inward force, provided by tension, friction, gravity or normal force (not a new type). Remove it and the object moves in a straight line (Newton's 1st Law).

Pause, copy the highlighted centripetal force formula and the "not a new force" rule into your book before moving on.

Speed doubles, radius stays constant. Centripetal force $F_c$ is multiplied by:

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Explain zero work in ideal uniform motion

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Work and Energy in Uniform Circular Motion
+5 XP

Syllabus 5.21, Why the centripetal force does no work

We just saw that centripetal force acts inward at every point. That raises a question: if a force is always acting, does it change the object's energy? This card answers it → $F_c$ is always perpendicular to displacement, so $W = Fd\cos90° = 0$ and $\Delta KE = 0$.

In uniform circular motion, speed is constant. Because $KE = \tfrac{1}{2}mv^2$, the kinetic energy never changes:

Work–Energy Theorem in Circular Motion

$\Delta KE = 0$

$W_{\text{net}} = \Delta KE = 0$

$W = F_c \, d \, \cos\theta = F_c \, d \, \cos 90° = \mathbf{0}$

$\theta = 90°$ between $F_c$ (inward) and displacement (tangent)

Why is the work zero? At every instant, the centripetal force points toward the centre, while the displacement is tangent to the circle. These two directions are perpendicular ($\theta = 90°$), so $\cos 90° = 0$ and no work is done.

Key insight: In the ideal uniform model, the radial net force continuously redirects the velocity vector without changing its magnitude. Therefore that radial resultant does zero instantaneous work. A separate tangential force would change speed and would no longer be uniform circular motion.

Real World, Roundabout

A 1500 kg car travels at 35 km/h ($9.72\ \text{m/s}$) around a roundabout of radius 15 m.

$a_c = \dfrac{v^2}{r} = \dfrac{(9.72)^2}{15} = \mathbf{6.30\ \text{m/s}^2}$

$F_c = \dfrac{mv^2}{r} = \dfrac{1500 \times (9.72)^2}{15} = \mathbf{9450\ \text{N}}$

This 9450 N centripetal force is provided entirely by friction between the tyres and the road. On a wet road, maximum friction is reduced. If friction is insufficient, the car follows Newton's First Law and continues in a straight line, sliding outward, not because of an outward force, but because the inward force is too weak.

✗ "Because there is a net force, the object must be gaining energy."
✓ Force only does work when it has a component parallel to displacement. In ideal uniform circular motion, the radial net force is perpendicular to motion, so its work is zero and kinetic energy remains constant.

In ideal uniform circular motion $\Delta KE = 0$ because the radial net force is perpendicular to instantaneous displacement ($\theta = 90°$, so $W = Fd\cos90° = 0$). It changes the direction of velocity but not its magnitude.

Add the highlighted work-energy result and the zero-work reason to your notes before moving on.

In uniform circular motion, the work done by the centripetal force is:

Investigate the force relationships and work
syllabus investigation

Use the controls to conduct a fair-test investigation. Change only one variable at a time and record at least three trials for each relationship.

  1. Hold $m$ and $r$ constant; vary $v$. Plot $F_c$ against $v^2$. A straight line through the origin supports $F_c\propto v^2$ and has gradient $m/r$.
  2. Hold $v$ and $r$ constant; vary $m$. Plot $F_c$ against $m$. The predicted gradient is $v^2/r$.
  3. Hold $m$ and $v$ constant; vary $r$. Plot $F_c$ against $1/r$. The predicted gradient is $mv^2$.
  4. Play the animation and compare the inward force direction with the tangential displacement. Explain why the radial force does zero work even when its magnitude changes between trials.

Evaluate: a simulator generates model data, so agreement checks internal mathematical consistency; it is not independent experimental validation. A physical investigation would measure force, speed and radius with uncertainty and repeated trials.

Open fullscreen ↗
Essential formulae, uniform circular motion

Frequency & period: $f = 1/T$ (Hz)

Angular velocity: $\omega=\Delta\theta/\Delta t = 2\pi/T = 2\pi f$ for uniform circular motion (rad s$^{-1}$)

Linear speed: $v = 2\pi r/T = \omega r$ (m/s)

Centripetal acceleration: $a_c = v^2/r = \omega^2 r$ (m s$^{-2}$), toward centre

Centripetal force: $F_c = mv^2/r = m\omega^2 r$ (N), toward centre

Work done by $F_c$: $W = 0$ (force perpendicular to displacement)

Fill the gap. If a wheel rotates at 3 revolutions per second, its angular velocity is $\omega = \_\_\_\_ \pi$ rad/s. (Give the coefficient of $\pi$ as a whole number.)

Activities

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Apply, diagnose and consolidate

Activity 1, Drill Problems
ApplyBand 3–4

Apply the formulas to standard circular motion problems. Show all working.

  1. An object moves in a circle of radius 20 m at a speed of 8 m/s. Calculate the centripetal acceleration.
  2. A 0.5 kg mass is attached to a string of length 0.8 m and whirled in a horizontal circle at 2 revolutions per second. Calculate the tension in the string (which provides the centripetal force).
  3. A laboratory rotor has radius 0.30 m and completes 4.0 revolutions per second. Calculate the speed of a point on its rim.

Drill check: for the rotor in Activity 1 Q3, the rim speed is approximately:

Activity 2, Force Identification
UnderstandBand 4

For each scenario, identify what real force(s) provide the centripetal force.

  1. A ball attached to a string and swung in a horizontal circle on a smooth table.
  2. A car travelling around a banked curve (no friction needed at design speed).
  3. An electron orbiting the nucleus in a simplified Bohr model.
  4. Wet clothes stuck to the inside wall of a spinning dryer drum.

Three of these correctly describe centripetal force. Pick the odd one out.

Wrap-up, Misconceptions & Summary

Misconceptions, final check

✗ "An object in uniform circular motion is in equilibrium because its speed is constant."
✓ The object is NOT in equilibrium. Equilibrium requires constant velocity (speed and direction). In circular motion the direction changes continuously, so there is centripetal acceleration $a_c = v^2/r$ and a nonzero net force $F_c = mv^2/r$. Newton's First Law rules out equilibrium wherever a net force exists.
✗ "Because centripetal force acts on the object, it must be doing work and changing the object's energy."
✓ Work requires a force component parallel to displacement. The centripetal force is always perpendicular to the tangential displacement ($\theta = 90°$), so $W = Fd\cos 90° = 0$. No energy change occurs.

Copy into your books

Key Definitions

  • Uniform circular motion: constant speed, changing direction
  • Centripetal acceleration: inward, $a_c = v^2/r$
  • Centripetal force: net inward force, not a new force type

Core Formulae

  • $v = \omega r = 2\pi r/T$
  • $a_c = v^2/r = \omega^2 r$
  • $F_c = mv^2/r = m\omega^2 r$

Work and Energy

  • $W = Fd\cos\theta$; $\theta = 90°$ so $W = 0$
  • $\Delta KE = 0$ (speed constant)
  • $F_c$ redirects, does not accelerate or decelerate

Key Principles

  • Speed constant $\neq$ acceleration zero
  • No centrifugal force in inertial frames
  • Remove $F_c$ → straight line (Newton's 1st)

Which set correctly describes uniform circular motion?

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Apply circular motion independently

Quick recall, uniform circular motion
+5 XP

A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 9 marks
+5 XP

ApplyBand 4(2 marks) 1. A 0.3 kg ball on a 0.6 m string makes 2.5 revolutions per second. Calculate the tension in the string.

1 mark: correct $\omega$ or $v$ with working · 1 mark: correct tension using $F_c = m\omega^2 r$

AnalyseBand 5(3 marks) 2. Two identical cars travel around the same circular track. Car B travels at twice the speed of car A. Compare their centripetal accelerations and radial net forces, and justify the comparison using equations.

1 mark: $a_B=4a_A$ · 1 mark: $F_B=4F_A$ · 1 mark: justification using the squared speed relationship

EvaluateBand 6(4 marks) 3. Evaluate the statement: "An object in uniform circular motion is in equilibrium because its speed is constant." Use Newton's laws and the concept of acceleration to justify your answer.

1 mark: identifies statement as false · 1 mark: defines equilibrium correctly (constant velocity, not speed) · 1 mark: centripetal acceleration $a_c = v^2/r \neq 0$ · 1 mark: links to Newton's 2nd Law, net force $F_c = mv^2/r$ exists

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (2 marks): $f = 2.5\ \text{Hz}$, so $\omega = 2\pi f = 2\pi \times 2.5 = 5\pi = 15.7\ \text{rad/s}$ (1 mark). $F_c = T = m\omega^2 r = 0.3 \times (15.7)^2 \times 0.6 = 0.3 \times 246.5 \times 0.6 = \mathbf{44.4\ \text{N}}$ (1 mark).

Q2 (3 marks): At fixed radius, $a_c=v^2/r$. Therefore $a_B=(2v_A)^2/r=4a_A$ (1 mark). For identical masses, $F_c=ma_c=mv^2/r$, so $F_B=4F_A$ (1 mark). Both results follow because the required acceleration and radial net force are proportional to the square of speed, not speed itself (1 mark).

Q3 (4 marks): The statement is false (1 mark). Equilibrium requires constant velocity both constant speed and constant direction (Newton's 1st Law). In uniform circular motion, the direction of velocity changes continuously, so the object is not in equilibrium (1 mark). The continuous change in direction produces a centripetal acceleration $a_c = v^2/r$ directed toward the centre (1 mark). By Newton's 2nd Law, this requires a nonzero net force $F_c = mv^2/r$. Since the net force is not zero, equilibrium cannot exist, even though speed is constant (1 mark).

Activity Answers

Activity 1 Q1: $a_c = v^2/r = 8^2/20 = 64/20 = \mathbf{3.2\ \text{m/s}^2}$

Activity 1 Q2: $\omega = 2\pi \times 2 = 4\pi = 12.57\ \text{rad/s}$. $F_c = m\omega^2 r = 0.5 \times (12.57)^2 \times 0.8 = \mathbf{63.2\ \text{N}}$.

Activity 1 Q3: $v=2\pi rf=2\pi(0.30)(4.0)=\mathbf{7.5\ \text{m/s}}$ to two significant figures.

Activity 2 (a): Tension in the string supplies the inward radial net force.

Activity 2 (b): The horizontal component of the normal force, on a banked curve the normal force is tilted inward, its horizontal component provides the centripetal force.

Activity 2 (c): Electrostatic (Coulomb) force, attraction between positive nucleus and negative electron provides centripetal force.

Activity 2 (d): Normal force from the drum wall, the wall pushes inward on the clothes, providing the centripetal force. Water escapes through the holes, not pushed out by a force.

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Retrieve and reflect

Check what actually stuck
Take the full module quiz
quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

Start the module quiz →
Arcade practice · Asteroid Blaster

Quick-fire circular motion questions in arcade format, locks in your mental model of centripetal force and direction before the next lesson.

How did your thinking change?

At the start you were asked about a car rounding a bend at constant speed. You can now separate the scalar and vector claims: speed is constant, but the tangential velocity direction changes continuously. The acceleration and radial net force point toward the centre, and on a level bend static friction supplies that resultant. If the inward force disappears, the car continues along the instantaneous tangent rather than moving radially outward.

Has your understanding of centripetal force changed?

← Previous: Torque and Rotational Equilibrium Next: Horizontal Circular Motion →

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