Why is "Using the full string length as the radius in a conical pendulum" a common mistake in Horizontal circular motion?

Why it happens: Confusing the slant length of the string with the horizontal radius of the circle. How to avoid it: The radius is the horizontal distance from the axis: r = , where is the angle to the vertical. Draw the cone and mark r horizontally. Source: 9231 Paper 3 Examiner Reports

Why is "Writing $F = mr\omega$ instead of $F = mr\omega^2$" a common mistake in Horizontal circular motion?

Why it happens: Carelessly copying v = r and forgetting the square in the acceleration. How to avoid it: Centripetal acceleration is r^2 (or v^2/r) — the angular speed is SQUARED. Check the units: r^2 gives m s⁻², r gives m s⁻¹. Source: 9231 Paper 3 mark schemes

Why is "Adding the centripetal force as an extra arrow on the force diagram" a common mistake in Horizontal circular motion?

Why it happens: Treating 'centripetal force' as a separate physical force rather than the resultant of the real forces. How to avoid it: Draw only the real forces (weight, tension, reaction, friction). Their RESULTANT toward the centre equals mv^2/r. Never draw a centripetal force as well. Source: 9231 Paper 3 Examiner Reports

Why is "Getting the direction of friction wrong on a banked bend" a common mistake in Horizontal circular motion?

Why it happens: Not thinking about which way the car tends to slip at the speed in question. How to avoid it: At the MAXIMUM speed the car tends to slide UP the bank, so friction acts DOWN the slope; at the minimum speed it is the reverse. Decide the tendency first. Source: 9231 Paper 3 Examiner Reports

Why is "Thinking a heavier car has a lower maximum speed on a flat bend" a common mistake in Horizontal circular motion?

Why it happens: Assuming a heavier car needs more friction, without noticing R = mg also grows. How to avoid it: On a flat bend v^2 = g r — the mass cancels. Heavier cars need more friction but the road provides proportionally more ( R = mg). Source: 9231 Paper 3 Examiner Reports

Why is "Putting the centripetal force equal to $R\cos\theta$ instead of $R\sin\theta$" a common mistake in Horizontal circular motion?

Why it happens: Mixing up which component of the normal reaction is horizontal (toward the centre). How to avoid it: On a bank at angle to the horizontal, the HORIZONTAL component of R is R (toward the centre) and the VERTICAL component is R. Match horizontal to the centripetal equation. Source: 9231 Paper 3 mark schemes

Cambridge International A Levels Further Mathematics (9231)

Horizontal circular motion

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Detailed Study Notes

Detailed notes on Circular motion for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Horizontal Circular Motion — Cambridge International AS & A Level Further Mathematics 9231 Paper 3 Further Mechanics (2026-2027 syllabus)

A particle moving in a horizontal circle always accelerates toward the centre: $a = r\omega^2 = \frac{v^2}{r}$ . The resultant of the real forces (tension, friction, normal reaction) supplies this centripetal force $F = mr\omega^2 = \frac{mv^2}{r}$ . Learn the four standard models — conical pendulum, particle on a string on a smooth table, and cars on flat and banked bends. Take $g = 10$ m s⁻².

At a glance

Centripetal acceleration is always toward the centre: $a = r\omega^2 = \frac{v^2}{r}$ .
Centripetal force $F = mr\omega^2 = \frac{mv^2}{r}$ is the RESULTANT of the real forces — not an extra force. (Given in MF19.)
Conical pendulum: resolve tension vertically ( $T\cos\theta = mg$ ) and horizontally ( $T\sin\theta = mr\omega^2$ ); radius $r = \ell\sin\theta$ .
Flat bend: friction provides the centripetal force; maximum speed $v^2 = \mu g r$ (mass cancels).
Banked bend: resolve the normal reaction (and friction if present); no-friction speed gives $\tan\theta = \frac{v^2}{rg}$ .

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

3.3a — Recall and use the magnitude of the centripetal acceleration $r\omega^2$ or $\frac{v^2}{r}$ , directed toward the centre.
3.3b — Apply Newton's second law along the radius to a particle in horizontal circular motion (conical pendulum, particle on a string on a smooth table).
3.3c — Model a vehicle on a flat or banked circular bend, resolving the normal reaction and friction to find speeds and angles.

Centripetal acceleration and force

Circular motion needs an acceleration toward the centre; the real forces must supply it.

The key fact. Even at constant speed, a particle going round a circle is accelerating, because the direction of its velocity keeps changing. This acceleration points straight toward the centre and has magnitude

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

(the two forms are equal because $v = r\omega$ ). By Newton's second law there must be a resultant force toward the centre — the centripetal force:

$F = mr\omega^2 = \frac{mv^2}{r}$

This expression is given in the MF19 formula list, but you must know how to use it.

Crucial point. "Centripetal force" is not a new force. It is the name for the resultant of the ordinary forces — tension, friction, weight components, normal-reaction components — once you add them up toward the centre. Never draw a separate "centripetal force" arrow on a diagram.

The method for every problem:

Draw the real forces only.
Resolve perpendicular to the motion that is in equilibrium (usually vertical) — set components equal.
Resolve toward the centre and set the resultant $= \frac{mv^2}{r}$ (or $mr\omega^2$ ).

Velocity v is tangent to the circle; the centripetal acceleration a points toward the centre O, of magnitude rω² = v²/r.

Cambridge tip. Pick $r\omega^2$ if the question gives an angular speed (or a period/frequency), and $\frac{v^2}{r}$ if it gives a linear speed. Choosing the right form saves a conversion.

$a = r\omega^2 = \frac{v^2}{r}$ , always toward the centre.
$F = mr\omega^2 = \frac{mv^2}{r}$ is the resultant of the real forces (in MF19).
Resolve perpendicular (balance) and toward the centre (Newton's 2nd law).

See the full worked example for horizontal circular motion →

The conical pendulum

A string sweeping out a cone: resolve the tension vertically and horizontally.

A particle of mass $m$ hangs from a light inextensible string of length $\ell$ fixed at the top. It moves in a horizontal circle so that the string traces out a cone, making angle $\theta$ with the vertical.

Only two forces act: the weight $mg$ (straight down) and the tension $T$ (along the string, up toward the fixed point).

Resolve vertically — there is no vertical acceleration, so the vertical part of the tension balances the weight:

$T\cos\theta = mg$

Resolve horizontally toward the centre — this is the centripetal direction, so the horizontal part of the tension equals $mr\omega^2$ :

$T\sin\theta = mr\omega^2 = \frac{mv^2}{r}$

The radius is not the string length. The particle sits at horizontal distance

$r = \ell\sin\theta$

from the central axis. Forgetting this is the single most common error.

A useful shortcut. Dividing the horizontal equation by the vertical one cancels $T$ and $m$ :

$\tan\theta = \frac{r\omega^2}{g} = \frac{v^2}{rg}$

The string makes angle θ with the vertical. The radius of the horizontal circle is r = ℓ sin θ. Resolve T vertically (= mg) and horizontally (= mrω²).

Cambridge tip. Write both equations before substituting numbers. If the question gives the speed of the particle, use $\frac{mv^2}{r}$ on the right of the horizontal equation; if it gives $\omega$ , use $mr\omega^2$ .

Vertical: $T\cos\theta = mg$ ; horizontal: $T\sin\theta = mr\omega^2$ .
Radius of the circle is $r = \ell\sin\theta$ , not the string length $\ell$ .
Dividing the equations gives $\tan\theta = \frac{r\omega^2}{g} = \frac{v^2}{rg}$ .

See the full worked example for horizontal circular motion →

Cars and cyclists on a flat bend

Friction alone provides the centripetal force; maximum speed v² = μgr.

When a car or cyclist rounds a flat (level) circular bend, the only horizontal force available is friction between the tyres and the road. This friction, directed toward the centre, supplies the entire centripetal force.

Particle on a string on a smooth horizontal table is the same idea with tension instead of friction: the table is smooth so the weight is balanced by the normal reaction and plays no part horizontally; the string tension provides $\frac{mv^2}{r}$ .

Flat bend — the equations. Resolve vertically (no vertical acceleration):

$R = mg$

Resolve toward the centre. The friction $F$ supplies the centripetal force:

$F = \frac{mv^2}{r}$

Maximum speed without skidding. Friction cannot exceed its limiting value $\mu R = \mu mg$ . At the greatest speed $F = \mu R$ , so:

$\frac{mv^2}{r} = \mu mg \quad\Longrightarrow\quad v^2 = \mu g r$

Notice the mass cancels — the maximum speed on a flat bend depends only on $\mu$ , $g$ and $r$ , not on how heavy the vehicle is.

On a flat bend the friction F points toward the centre and supplies the centripetal force. R = mg balances the weight.

Cambridge tip. State $R = mg$ first, then the centripetal equation. When the question asks for the greatest speed, put friction at its limiting value $\mu R$ — and remember to take the square root at the end.

Flat bend / smooth table: one horizontal force (friction or tension) gives $\frac{mv^2}{r}$ .
Vertical balance $R = mg$ ; centripetal $\frac{mv^2}{r} = F$ .
Maximum speed $v^2 = \mu g r$ — the mass cancels.

See the full worked example for horizontal circular motion →

The banked bend

Tilting the road lets the normal reaction help turn the car; resolve R (and friction).

A banked bend is tilted inward at angle $\theta$ to the horizontal so that part of the normal reaction helps push the car toward the centre — reducing or removing the need for friction.

Case 1: design speed (no friction needed). Only the weight $mg$ (down) and the normal reaction $R$ (perpendicular to the road surface) act. Resolve:

Vertically: $R\cos\theta = mg$
Toward the centre (horizontally): $R\sin\theta = \dfrac{mv^2}{r}$

Dividing cancels $R$ and $m$ :

$\tan\theta = \frac{v^2}{rg}$

This single speed is the one at which the bend "works perfectly" with no sideways friction.

Case 2: with friction. Above the design speed the car tends to slip up the bank, so friction $\mu R$ acts down the slope (and vice versa below it). Resolve both $R$ and the friction into vertical and horizontal parts, then apply the same two principles. At the maximum speed:

Vertically: $R\cos\theta - \mu R\sin\theta = mg$
Toward the centre: $R\sin\theta + \mu R\cos\theta = \dfrac{mv^2}{r}$

On a bank, R is perpendicular to the surface. Its horizontal component R sin θ points toward the centre; its vertical component R cos θ balances the weight (when no friction is needed).

Cambridge tip. Decide the direction of friction FIRST — at the upper speed limit the car slides up, so friction is down the slope. Get this wrong and every sign flips. For a "Show that", always write both resolved equations before dividing.

Normal reaction is perpendicular to the banked surface: components $R\cos\theta$ (vertical) and $R\sin\theta$ (horizontal, toward centre).
No-friction speed: $\tan\theta = \frac{v^2}{rg}$ (divide the two equations).
With friction at max speed: friction acts DOWN the slope ( $\mu R$ down).

See the full worked example for horizontal circular motion →

Quick recap

Centripetal acceleration $a = r\omega^2 = \frac{v^2}{r}$ is always toward the centre.
Centripetal force $F = mr\omega^2 = \frac{mv^2}{r}$ is the resultant of the real forces (in MF19).
Conical pendulum: $T\cos\theta = mg$ , $T\sin\theta = mr\omega^2$ , $r = \ell\sin\theta$ .
Flat bend / smooth table: friction or tension gives the centripetal force; $v_{\max}^2 = \mu g r$ .
Banked bend: resolve $R$ (and friction); no-friction speed $\tan\theta = \frac{v^2}{rg}$ .
Never draw a separate 'centripetal force' — use the resultant of the real forces.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$a = r\omega^2 = \frac{v^2}{r}$ (centripetal acceleration, toward centre).
$F = mr\omega^2 = \frac{mv^2}{r}$ (centripetal force — in MF19, but know how to use it).
Conical pendulum: $T\cos\theta = mg$ , $T\sin\theta = mr\omega^2$ , $r = \ell\sin\theta$ .
Flat bend maximum speed: $v^2 = \mu g r$ (mass cancels).
Banked bend (no friction): $\tan\theta = \frac{v^2}{rg}$ .

How it’s examined

Horizontal circular motion is a core Paper 3 (Further Mechanics) topic, typically worth 6-10 marks. Questions present a standard model — conical pendulum, particle on a smooth table, or a vehicle on a flat or banked bend — and expect two resolved equations: one perpendicular to the motion (balance) and one toward the centre (Newton's 2nd law). Marks are most often lost by using the string length as the radius in a conical pendulum, drawing a spurious 'centripetal force' arrow, forgetting to square $\omega$ , or choosing the wrong direction for friction on a bank. Method marks require both equations written before substitution; unsupported answers score nothing. Take $g = 10$ m s⁻² and give non-exact answers to 3 significant figures.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 3 Further Mechanics); 9231/31, 9231/32, 9231/33 Further Mechanics — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Horizontal circular motion

Step-by-step solutions to past-paper-style questions on horizontal circular motion, written exactly the way a tutor would explain them at the board.

Question type:

1Centripetal force on a particle in a horizontal circle
Getting startedDirect calculation• centripetal-force, definition
Question
A particle of mass $0.5$ kg moves in a horizontal circle of radius $0.8$ m with constant angular speed $3$ rad s⁻¹. Find the centripetal force acting on it.
Step-by-step solution
1. Step 1
  Choose the right form. Angular speed is given, so use $F = mr\omega^2$ .
  $F = mr\omega^2$
2. Step 2
  Substitute $m = 0.5$ kg, $r = 0.8$ m, $\omega = 3$ rad s⁻¹.
  $F = 0.5 \times 0.8 \times 3^2 = 0.5 \times 0.8 \times 9$
3. Step 3
  Evaluate.
  $F = 3.6$
Answer
$F = 3.6$ N, directed toward the centre.
Examiner tip
The centripetal force always points toward the centre. Use $\omega^2$ (square the angular speed) — squaring is the step beginners skip.
2Centripetal acceleration from linear speed
Getting startedDirect calculation• centripetal-acceleration, v-form
Question
A car travels round a circular bend of radius $40$ m at a constant speed of $12$ m s⁻¹. Find the magnitude of its (centripetal) acceleration.
Step-by-step solution
1. Step 1
  Choose the right form. Linear speed is given, so use $a = \frac{v^2}{r}$ .
  $a = \frac{v^2}{r}$
2. Step 2
  Substitute $v = 12$ m s⁻¹, $r = 40$ m.
  $a = \frac{12^2}{40} = \frac{144}{40}$
3. Step 3
  Evaluate.
  $a = 3.6$
Answer
$a = 3.6$ m s⁻², directed toward the centre of the bend.
Examiner tip
When the speed $v$ along the path is given, $\frac{v^2}{r}$ is faster than first finding $\omega$ . Both forms are equivalent via $v = r\omega$ .
3Conical pendulum — finding the tension and angular speed
Building confidenceMulti-step problem• conical-pendulum, resolving
Question
A particle of mass $0.2$ kg is attached to one end of a light inextensible string of length $0.5$ m. The other end is fixed and the particle moves as a conical pendulum in a horizontal circle, with the string making an angle of $30°$ with the vertical. Taking $g = 10$ m s⁻², find (a) the tension in the string and (b) the angular speed of the particle.
Step-by-step solution
1. Step 1
  Resolve vertically (no vertical acceleration): the vertical component of tension balances the weight.
  $T\cos 30° = mg$
2. Step 2
  (a) Solve for $T$ . With $m = 0.2$ , $g = 10$ , so $mg = 2$ N.
  $T = \frac{2}{\cos 30°} = \frac{2}{0.8660} \approx 2.31$
3. Step 3
  Radius of the circle. The particle is at horizontal distance $r = \ell\sin 30°$ from the axis.
  $r = 0.5\sin 30° = 0.5 \times 0.5 = 0.25$
4. Step 4
  Resolve horizontally toward the centre. The horizontal component of tension provides the centripetal force $mr\omega^2$ .
  $T\sin 30° = mr\omega^2$
5. Step 5
  (b) Solve for $\omega$ . Substitute $T = 2.309$ , $m = 0.2$ , $r = 0.25$ .
  $\omega^2 = \frac{T\sin 30°}{mr} = \frac{2.309 \times 0.5}{0.2 \times 0.25} = \frac{1.1547}{0.05} = 23.09$
6. Step 6
  Take the square root.
  $\omega = \sqrt{23.09} \approx 4.81$
Answer
$T \approx 2.31$ N and $\omega \approx 4.81$ rad s⁻¹.
Examiner tip
Two equations: vertical balance ( $T\cos\theta = mg$ ) and horizontal Newton's 2nd law ( $T\sin\theta = mr\omega^2$ ). Dividing them gives $\tan\theta = \frac{r\omega^2}{g}$ , a clean shortcut.
4Car on a flat bend — maximum speed before skidding
Building confidenceMulti-step problem• flat-bend, friction
Question
A car of mass $900$ kg travels round a flat (unbanked) circular bend of radius $50$ m. The coefficient of friction between the tyres and the road is $0.4$ . Taking $g = 10$ m s⁻², find the greatest speed at which the car can take the bend without skidding.
Step-by-step solution
1. Step 1
  Vertical balance. The road is horizontal, so the normal reaction equals the weight.
  $R = mg = 900 \times 10 = 9000$
2. Step 2
  Friction provides the centripetal force. At the greatest speed, friction is at its maximum value $\mu R$ .
  $F_{\max} = \mu R = 0.4 \times 9000 = 3600$
3. Step 3
  Newton's 2nd law toward the centre. This maximum friction equals $\frac{mv^2}{r}$ .
  $\mu R = \frac{mv^2}{r}$
4. Step 4
  Solve for $v^2$ . Substitute the numbers.
  $v^2 = \frac{\mu R \, r}{m} = \frac{3600 \times 50}{900} = 200$
5. Step 5
  Take the square root.
  $v = \sqrt{200} \approx 14.1$
Answer
The greatest speed is $\sqrt{200} \approx 14.1$ m s⁻¹.
Examiner tip
Notice $v^2 = \mu g r$ — the mass cancels. On a flat bend the maximum speed depends only on $\mu$ , $g$ and $r$ , never on how heavy the car is.
5Particle on a string on a smooth horizontal table
StretchMulti-step problem• string-on-table, tension
Question
A particle of mass $0.3$ kg lies on a smooth horizontal table, attached to a fixed point $O$ on the table by a light inextensible string of length $0.6$ m. It moves in a horizontal circle of radius $0.6$ m about $O$ at constant speed, completing one revolution every $0.8$ s. Find the tension in the string. Take $g = 10$ m s⁻².
Step-by-step solution
1. Step 1
  The table is smooth and horizontal. The weight is balanced by the normal reaction; they play no part in the horizontal circular motion. Only the string tension acts toward the centre.
2. Step 2
  Find the angular speed from the period $T = 0.8$ s.
  $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.8} = 2.5\pi \approx 7.854$
3. Step 3
  Newton's 2nd law toward the centre. The tension supplies the entire centripetal force.
  $T = mr\omega^2$
4. Step 4
  Substitute $m = 0.3$ , $r = 0.6$ , $\omega = 2.5\pi$ .
  $T = 0.3 \times 0.6 \times (2.5\pi)^2 = 0.18 \times 61.685 \approx 11.1$
Answer
$T \approx 11.1$ N.
Examiner tip
On a smooth horizontal table the weight is irrelevant to the circular motion — it is cancelled vertically. The whole centripetal force comes from the tension. Using $T = mr\omega^2$ avoids ever needing $v$ .
6Banked bend — the design speed (no friction needed)
StretchShow that / prove• banked-bend, resolving
Question
A bend of radius $80$ m is banked at an angle $\theta$ to the horizontal. A car can travel round the bend at $20$ m s⁻¹ with no tendency to slip, so no friction is needed. Taking $g = 10$ m s⁻², show that $\tan\theta = \frac{1}{2}$ and find $\theta$ .
Step-by-step solution
1. Step 1
  Forces on the car. With no friction, only the weight $mg$ (down) and the normal reaction $R$ (perpendicular to the banked surface) act. $R$ has a vertical component $R\cos\theta$ and a horizontal component $R\sin\theta$ toward the centre.
2. Step 2
  Resolve vertically (no vertical acceleration).
  $R\cos\theta = mg$
3. Step 3
  Newton's 2nd law toward the centre (horizontal): the horizontal component of $R$ is the centripetal force.
  $R\sin\theta = \frac{mv^2}{r}$
4. Step 4
  Divide the horizontal equation by the vertical one — $R$ and $m$ both cancel.
  $\tan\theta = \frac{v^2}{rg}$
5. Step 5
  Substitute $v = 20$ , $r = 80$ , $g = 10$ .
  $\tan\theta = \frac{20^2}{80 \times 10} = \frac{400}{800} = \frac{1}{2}$
6. Step 6
  Find the angle.
  $\theta = \arctan\tfrac{1}{2} \approx 26.6°$
Answer
$\tan\theta = \frac{1}{2}$ , so $\theta \approx 26.6°$ .
Examiner tip
The banking formula $\tan\theta = \frac{v^2}{rg}$ is found by dividing the two resolved equations, which cancels both $R$ and $m$ . For a 'Show that', both starting equations must be written explicitly before dividing.

Model Answers — Horizontal circular motion

High-scoring sample answers for horizontal circular motion on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
A particle of mass $0.25$ kg moves in a horizontal circle of radius $1.2$ m with constant angular speed $4$ rad s⁻¹. Find the centripetal force on the particle.
Model answer
Use $F = mr\omega^2$ (angular speed given): $F = 0.25 \times 1.2 \times 4^2 = 0.25 \times 1.2 \times 16 = 4.8 \text{ N}$

The force of $4.8$ N is directed toward the centre of the circle.
Why this scores
M1 for $F = mr\omega^2$ , M1 for substituting with $\omega^2 = 16$ , A1 for $4.8$ N. State the direction (toward the centre) for full credit.
Question 2
4 marks
A cyclist rides round a flat circular track of radius $25$ m at a steady speed of $7.5$ m s⁻¹. The combined mass of cyclist and bicycle is $80$ kg. Find (a) the centripetal acceleration and (b) the horizontal frictional force needed. Take $g = 10$ m s⁻².
Model answer
(a) Centripetal acceleration (speed given, so use the $v$ -form): $a = \frac{v^2}{r} = \frac{7.5^2}{25} = \frac{56.25}{25} = 2.25 \text{ m s}^{-2}$

(b) The friction provides the centripetal force: $F = ma = 80 \times 2.25 = 180 \text{ N}$

(Equivalently $F = \frac{mv^2}{r} = \frac{80 \times 56.25}{25} = 180$ N.) The friction acts horizontally toward the centre of the track.
Why this scores
(a) M1 for $a = v^2/r$ , A1 for $2.25$ ; (b) M1 for $F = ma$ (or $mv^2/r$ ), A1 for $180$ N. The friction is the centripetal force here — the road is flat so the normal reaction is purely vertical.
Question 3
6 marks
A particle of mass $0.4$ kg is attached to a fixed point by a light inextensible string of length $0.75$ m. It moves as a conical pendulum in a horizontal circle, the string making an angle of $40°$ with the vertical. Take $g = 10$ m s⁻². Find (a) the tension in the string and (b) the speed of the particle.
Model answer
Set up the two equations. Let $T$ be the tension and $\theta = 40°$ .

Resolve vertically (no vertical acceleration): $T\cos 40° = mg = 0.4 \times 10 = 4$

(a) Solve for the tension: $T = \frac{4}{\cos 40°} = \frac{4}{0.7660} \approx 5.22 \text{ N}$

Radius of the circle: $r = \ell\sin 40° = 0.75 \times 0.6428 = 0.4821 \text{ m}$

Resolve horizontally toward the centre (Newton's 2nd law), using $\frac{mv^2}{r}$ for the centripetal force: $T\sin 40° = \frac{mv^2}{r}$

(b) Solve for $v$ : $v^2 = \frac{T\sin 40° \, r}{m} = \frac{5.221 \times 0.6428 \times 0.4821}{0.4} = \frac{1.6179}{0.4} = 4.045$ $v = \sqrt{4.045} \approx 2.01 \text{ m s}^{-1}$
Why this scores
Vertical: M1 A1 for $T\cos\theta = mg$ giving $T \approx 5.22$ N. Horizontal: M1 for $T\sin\theta = mv^2/r$ with $r = \ell\sin\theta$ , M1 for solving, A1 for $v \approx 2.01$ m s⁻¹. The radius is $\ell\sin\theta$ , NOT $\ell$ — a very common slip.
Question 4
5 marks
A car of mass $1200$ kg travels round a flat circular bend of radius $60$ m. The coefficient of friction between the tyres and the road is $0.5$ . Taking $g = 10$ m s⁻², find the maximum speed at which the car can negotiate the bend without slipping, and state how the answer would change for a heavier car.
Model answer
Vertical balance (flat road): $R = mg = 1200 \times 10 = 12000 \text{ N}$

Maximum friction provides the centripetal force at the greatest speed: $\mu R = \frac{mv^2}{r}$

Since $R = mg$ , the mass cancels: $\mu mg = \frac{mv^2}{r} \;\Rightarrow\; v^2 = \mu g r = 0.5 \times 10 \times 60 = 300$

$v = \sqrt{300} \approx 17.3 \text{ m s}^{-1}$

Because $v^2 = \mu g r$ does not contain $m$ , a heavier car has exactly the same maximum speed.
Why this scores
M1 for $R = mg$ , M1 for $\mu R = mv^2/r$ , M1 for cancelling $m$ to get $v^2 = \mu g r$ , A1 for $17.3$ m s⁻¹, A1 for the correct statement that mass has no effect. The cancellation of $m$ is the examined insight.
Question 5
8 marks
A bend of radius $90$ m is banked at $20°$ to the horizontal. A car of mass $1000$ kg travels round it. The coefficient of friction between the tyres and the road is $0.3$ . Taking $g = 10$ m s⁻², find the maximum speed at which the car can travel round the bend without slipping up the slope.
Model answer
Forces. At the maximum speed the car tends to slip up the bank, so friction $F = \mu R$ acts down the slope. The normal reaction $R$ is perpendicular to the road. Resolve into horizontal and vertical components, with $\theta = 20°$ .

Resolve vertically (no vertical acceleration). The vertical component of $R$ is up, the vertical component of the down-slope friction is downward: $R\cos 20° - \mu R\sin 20° = mg$ $R(\cos 20° - 0.3\sin 20°) = 1000 \times 10 = 10000$ $R(0.9397 - 0.3 \times 0.3420) = R(0.8371) = 10000 \;\Rightarrow\; R = 11946 \text{ N}$

Newton's 2nd law toward the centre (horizontal). Both the horizontal component of $R$ and of the friction point toward the centre: $R\sin 20° + \mu R\cos 20° = \frac{mv^2}{r}$ $11946(0.3420 + 0.3 \times 0.9397) = \frac{1000\,v^2}{90}$ $11946(0.6239) = \frac{1000\,v^2}{90}$ $7453 = \frac{1000\,v^2}{90} \;\Rightarrow\; v^2 = \frac{7453 \times 90}{1000} = 670.8$ $v = \sqrt{670.8} \approx 25.9 \text{ m s}^{-1}$
Why this scores
M1 for identifying friction acting down the slope at maximum speed; M1 A1 for the vertical equation giving $R$ ; M1 for the centripetal equation with both components toward the centre; M1 A1 for solving; A1 for $v \approx 25.9$ m s⁻¹. The direction of friction (down-slope for the upper limit) is the key decision and the most common error.
Question 6
7 marks
A conical pendulum consists of a particle of mass $0.2\,\text{kg}$ attached to one end of a light inextensible string of length $0.5\,\text{m}$ ; the other end is fixed. The string will break if its tension exceeds $4\,\text{N}$ . Taking $g = 10\,\text{m s}^{-2}$ , find the maximum angular speed at which the particle can move in a horizontal circle, and the angle the string then makes with the vertical.
Model answer
Let the string make angle $\theta$ with the vertical and tension be $T$ . The radius of the circle is $r = l\sin\theta = 0.5\sin\theta$ .

Resolve vertically (no vertical acceleration): $T\cos\theta = mg \;\Rightarrow\; \cos\theta = \frac{mg}{T}.$

The tension is greatest, and is the limiting value $T = 4\,\text{N}$ , at the maximum angular speed. Then $\cos\theta = \frac{0.2 \times 10}{4} = 0.5 \;\Rightarrow\; \theta = 60^\circ.$

Resolve horizontally (centripetal): $T\sin\theta = m r \omega^2 = m(l\sin\theta)\omega^2$ , so $\sin\theta$ cancels: $T = m l \omega^2 \;\Rightarrow\; \omega^2 = \frac{T}{ml} = \frac{4}{0.2 \times 0.5} = 40.$

$\omega = \sqrt{40} = 6.32\,\text{rad s}^{-1}\ (3\text{ s.f.}), \qquad \theta = 60.0^\circ.$
Why this scores
M1 A1 for the vertical equation giving $\cos\theta = mg/T$ , M1 for setting $T = 4$ at the breaking point and finding $\theta = 60^\circ$ , M1 for the horizontal/centripetal equation with $r = l\sin\theta$ , A1 for $\omega^2 = T/(ml)$ , A1 for $\omega = 6.32\,\text{rad s}^{-1}$ . Recognising that maximum tension ⇒ maximum $\omega$ is the key insight.

Key Formulae — Horizontal circular motion

The formulae you need to memorise for horizontal circular motion on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Centripetal acceleration
$a = r\omega^2 = \frac{v^2}{r}$
$a$
acceleration directed toward the centre (m s⁻²)
$r$
radius of the circle (m)
$\omega$
angular speed (rad s⁻¹)
$v$
linear speed along the path (m s⁻¹)
When to use
Use $r\omega^2$ when angular speed is given, $\frac{v^2}{r}$ when linear speed is given. They are equal because $v = r\omega$ .
Centripetal force
$F = mr\omega^2 = \frac{mv^2}{r}$
$F$
net force toward the centre (N)
$m$
mass of the particle (kg)
When to use
The resultant force directed toward the centre. This result is given in MF19. Equate it to the actual force (tension, friction, component of reaction).
Conical pendulum equations
$T\cos\theta = mg, \qquad T\sin\theta = mr\omega^2$
$T$
tension in the string (N)
$\theta$
angle of the string to the vertical
$r$
$\ell\sin\theta$ , the radius of the horizontal circle (m)
When to use
Resolve vertically (balance) and horizontally (Newton's 2nd law). Dividing gives $\tan\theta = \frac{r\omega^2}{g} = \frac{v^2}{rg}$ .
Flat (unbanked) bend — maximum speed
$v^2 = \mu g r$
$\mu$
coefficient of friction between tyres and road
$g$
$10$ m s⁻²
$r$
radius of the bend (m)
When to use
Greatest speed on a flat bend where friction alone supplies the centripetal force. The mass cancels.
Banked bend — design (no-friction) speed
$\tan\theta = \frac{v^2}{rg}$
$\theta$
angle of banking to the horizontal
$v$
the speed at which no friction is required (m s⁻¹)
When to use
The single speed at which a banked bend can be taken with no sideways friction. Found by dividing the resolved vertical and centripetal equations.

Key Definitions and Keywords — Horizontal circular motion

Definitions to memorise and the exact keywords mark schemes credit for horizontal circular motion answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Centripetal acceleration
Examiner keyword
The acceleration of a particle in circular motion, directed toward the centre, of magnitude $a = r\omega^2 = \frac{v^2}{r}$ . It changes the direction of velocity, not its magnitude.
Centripetal force
Examiner keyword
The resultant force directed toward the centre that produces the centripetal acceleration: $F = mr\omega^2 = \frac{mv^2}{r}$ . It is not a new kind of force — it is supplied by tension, friction or a component of the normal reaction.
Conical pendulum
Examiner keyword
A particle on a string moving in a horizontal circle so that the string sweeps out a cone. The tension is resolved vertically ( $T\cos\theta = mg$ ) and horizontally ( $T\sin\theta = mr\omega^2$ ).
Banking
Examiner keyword
Tilting a road or track inward on a bend so that a component of the normal reaction contributes to the centripetal force, reducing reliance on friction. The no-friction speed satisfies $\tan\theta = \frac{v^2}{rg}$ .
Normal reaction ( $R$ )
The contact force exerted by a surface perpendicular to itself. On a flat road $R = mg$ ; on a banked road $R$ is perpendicular to the inclined surface, so it has both vertical and horizontal components.
Limiting friction
The maximum friction a surface can provide, equal to $\mu R$ . The maximum (or minimum) speed on a bend occurs when friction reaches this limiting value.

Common Mistakes and Misconceptions — Horizontal circular motion

The traps other students keep falling into on horizontal circular motion questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Using the full string length as the radius in a conical pendulum
9231 Paper 3 Examiner Reports
Why it happens
Confusing the slant length $\ell$ of the string with the horizontal radius of the circle.
How to avoid it
The radius is the horizontal distance from the axis: $r = \ell\sin\theta$ , where $\theta$ is the angle to the vertical. Draw the cone and mark $r$ horizontally.
✕Writing $F = mr\omega$ instead of $F = mr\omega^2$
9231 Paper 3 mark schemes
Why it happens
Carelessly copying $v = r\omega$ and forgetting the square in the acceleration.
How to avoid it
Centripetal acceleration is $r\omega^2$ (or $\frac{v^2}{r}$ ) — the angular speed is SQUARED. Check the units: $r\omega^2$ gives m s⁻², $r\omega$ gives m s⁻¹.
✕Adding the centripetal force as an extra arrow on the force diagram
9231 Paper 3 Examiner Reports
Why it happens
Treating 'centripetal force' as a separate physical force rather than the resultant of the real forces.
How to avoid it
Draw only the real forces (weight, tension, reaction, friction). Their RESULTANT toward the centre equals $\frac{mv^2}{r}$ . Never draw a centripetal force as well.
✕Getting the direction of friction wrong on a banked bend
9231 Paper 3 Examiner Reports
Why it happens
Not thinking about which way the car tends to slip at the speed in question.
How to avoid it
At the MAXIMUM speed the car tends to slide UP the bank, so friction acts DOWN the slope; at the minimum speed it is the reverse. Decide the tendency first.
✕Thinking a heavier car has a lower maximum speed on a flat bend
9231 Paper 3 Examiner Reports
Why it happens
Assuming a heavier car needs more friction, without noticing $R = mg$ also grows.
How to avoid it
On a flat bend $v^2 = \mu g r$ — the mass cancels. Heavier cars need more friction but the road provides proportionally more ( $\mu R = \mu mg$ ).
✕Putting the centripetal force equal to $R\cos\theta$ instead of $R\sin\theta$
9231 Paper 3 mark schemes
Why it happens
Mixing up which component of the normal reaction is horizontal (toward the centre).
How to avoid it
On a bank at angle $\theta$ to the horizontal, the HORIZONTAL component of $R$ is $R\sin\theta$ (toward the centre) and the VERTICAL component is $R\cos\theta$ . Match horizontal to the centripetal equation.

Horizontal circular motion

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Centripetal force on a particle in a horizontal circle

2Centripetal acceleration from linear speed

3Conical pendulum — finding the tension and angular speed

4Car on a flat bend — maximum speed before skidding

5Particle on a string on a smooth horizontal table

6Banked bend — the design speed (no friction needed)

Centripetal acceleration

Centripetal force

Conical pendulum equations

Flat (unbanked) bend — maximum speed

Banked bend — design (no-friction) speed

Centripetal acceleration

Centripetal force

Conical pendulum

Banking

Normal reaction (RRR)

Limiting friction

✕Using the full string length as the radius in a conical pendulum

✕Writing F=mrωF = mr\omegaF=mrω instead of F=mrω2F = mr\omega^2F=mrω2

✕Adding the centripetal force as an extra arrow on the force diagram

✕Getting the direction of friction wrong on a banked bend

✕Thinking a heavier car has a lower maximum speed on a flat bend

✕Putting the centripetal force equal to Rcos⁡θR\cos\thetaRcosθ instead of Rsin⁡θR\sin\thetaRsinθ

Normal reaction ( $R$ )

✕Writing $F = mr\omega$ instead of $F = mr\omega^2$

✕Putting the centripetal force equal to $R\cos\theta$ instead of $R\sin\theta$