Horizontal Circular Motion

Now that we understand centripetal force and acceleration, we can analyse specific scenarios where objects move in horizontal circles. The key skill is identifying which real force provides the centripetal force in each situation, and then applying Newton's second law towards the centre.

Car on a Flat Curve

When a car travels around a flat (unbanked) circular curve, friction between the tyres and the road provides the centripetal force.

The forces on the car are:

Weight (mg) acting downward
Normal reaction (N) acting upward: N = mg (since there is no vertical acceleration)
Friction (f) acting horizontally towards the centre of the curve

For the car not to skid: $f \leq \mu N = \mu mg$

The centripetal force required is: $\frac{mv^2}{r} \leq \mu mg$

Cancelling m: $\frac{v^2}{r} \leq \mu g$

Maximum speed on a flat curve: $v_{\max} = \sqrt{\mu g r}$

Worked Example 1

A car enters a flat roundabout of radius 30 m. The coefficient of friction between the tyres and the road is 0.70. What is the maximum safe speed?

$v_{\max} = \sqrt{0.70 \times 9.81 \times 30} = \sqrt{206} = 14.4 \text{ m s}^{-1}$

This is about 52 km/h — consistent with typical roundabout speed limits.

Notice that the maximum speed does not depend on the mass of the car. A heavy lorry and a light car have the same maximum speed on the same curve (assuming the same friction coefficient).

Condition	μ	v_max for r = 30 m	v_max for r = 100 m
Dry tarmac	0.70	14.4 m s⁻¹ (52 km/h)	26.2 m s⁻¹ (94 km/h)
Wet tarmac	0.40	10.8 m s⁻¹ (39 km/h)	19.8 m s⁻¹ (71 km/h)
Icy road	0.10	5.4 m s⁻¹ (20 km/h)	9.9 m s⁻¹ (36 km/h)
Gravel	0.35	10.1 m s⁻¹ (37 km/h)	18.5 m s⁻¹ (67 km/h)

Banked Curves

On a banked curve, the road is tilted inward at an angle θ. This allows a component of the normal reaction force to contribute to the centripetal force, reducing (or eliminating) the reliance on friction.

Forces on a Banked Curve (No Friction)

For a car on a banked curve with no friction (e.g. an icy banked road), the only forces are weight and the normal reaction:

Vertical equilibrium: N cos θ = mg Horizontal (centripetal): N sin θ = mv²/r

Dividing the second equation by the first:

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

This gives the design speed — the speed at which no friction is needed:

$v = \sqrt{rg\tan\theta}$

Worked Example 2

A banked curve has radius 200 m and is banked at 15°. What is the design speed?

$v = \sqrt{200 \times 9.81 \times \tan 15°} = \sqrt{200 \times 9.81 \times 0.268} = \sqrt{525} = 22.9 \text{ m s}^{-1}$

This is about 82 km/h. At this speed, a car can navigate the curve even on ice (no friction at all). At higher speeds, friction must act down the slope to provide additional centripetal force. At lower speeds, friction must act up the slope to prevent the car sliding inward.

Worked Example 3: Banked Curve WITH Friction (Maximum Speed)

A banked curve at 20° with radius 80 m and μ = 0.30. Find maximum speed.

At maximum speed, friction acts down the slope (towards centre):

Perpendicular to slope: N cos θ − mg − μN sin θ = 0 $N(\cos 20° - 0.30\sin 20°) = mg$ $N \times 0.837 = mg$

Centripetal (horizontal): N sin θ + μN cos θ = mv²/r $N(\sin 20° + 0.30\cos 20°) = \frac{mv^2}{r}$ $N \times 0.624 = \frac{mv^2}{r}$

Dividing: $v^2 = \frac{0.624 \times g \times r}{0.837} = \frac{0.624 \times 9.81 \times 80}{0.837} = 585$ $v = 24.2 \text{ m s}^{-1} \approx 87 \text{ km/h}$

Compare to design speed (no friction): v = √(80 × 9.81 × tan 20°) = 16.9 m s⁻¹. Friction allows 43% higher speed.

The Conical Pendulum

A conical pendulum consists of a mass on a string that swings in a horizontal circle, with the string tracing out a cone shape.

The forces on the bob are:

Weight (mg) downward
Tension (T) along the string, at angle θ to the vertical

Vertical equilibrium: T cos θ = mg Horizontal (centripetal): T sin θ = mω²r

Since r = L sin θ (where L is the string length):

T sin θ = mω²L sin θ

Cancelling sin θ (assuming θ ≠ 0):

$T = m\omega^2 L$

From the vertical equation: $\cos\theta = \frac{g}{\omega^2 L}$

Worked Example 4

A conical pendulum has string length 0.80 m and the bob makes a horizontal circle with the string at 25° to the vertical. Find the angular velocity and the period.

$\cos 25° = \frac{g}{\omega^2 L}$ $\omega^2 = \frac{g}{L\cos 25°} = \frac{9.81}{0.80 \times 0.906} = \frac{9.81}{0.725} = 13.53$ $\omega = 3.68 \text{ rad s}^{-1}$

$T = \frac{2\pi}{\omega} = \frac{2\pi}{3.68} = 1.71 \text{ s}$

Notice: as ω increases, cos θ decreases, so θ increases — the bob swings wider. As ω → ∞, θ → 90° (the bob approaches a horizontal circle).

Roundabouts and Fairground Rides

Many fairground rides involve horizontal circular motion. A chair-o-plane (wave swinger), for example, is essentially a large conical pendulum. As the ride spins faster:

The angle θ of the chains increases
The radius of the circle increases
The riders experience greater centripetal acceleration

The "apparent weight" felt by the rider is the tension in the chains, which is always greater than their actual weight (since T cos θ = mg means T = mg/cos θ > mg).

Worked Example 5

A fairground chair-o-plane has chains of length 6.0 m. A rider of mass 70 kg is at 40° to the vertical. Find the tension in the chains and the rider's speed.

Tension: $T = \frac{mg}{\cos 40°} = \frac{70 \times 9.81}{0.766} = \frac{686.7}{0.766} = 896 \text{ N}$

Radius of circle: r = L sin 40° = 6.0 × 0.643 = 3.86 m

Centripetal force: T sin 40° = 896 × 0.643 = 576 N

Speed: $576 = \frac{70 \times v^2}{3.86}$ $v^2 = \frac{576 \times 3.86}{70} = 31.8$ $v = 5.6 \text{ m s}^{-1}$

Cyclists and Motorbikes on Curves

When a cyclist or motorbike goes around a curve, they lean inward. This is because the friction force at the tyre contact creates a torque about the centre of mass that would topple the rider outward. By leaning, the normal force creates an opposing torque that maintains rotational equilibrium.

The angle of lean θ to the vertical is given by:

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

This is identical to the banked curve formula — the cyclist effectively banks their body.

Common Exam Mistakes

Mistake 1: Drawing centripetal force as a separate arrow on free-body diagrams. On a flat curve, friction IS the centripetal force — do not draw both.

Mistake 2: Forgetting that v_max = √(μgr) is independent of mass. Both a lorry and a sports car have the same v_max on the same curve (same μ and r).

Mistake 3: On banked curves, resolving forces horizontally and vertically instead of along and perpendicular to the slope. Both approaches work, but the along/perpendicular method is simpler when friction is present.

Mistake 4: Assuming friction always acts towards the centre on a banked curve. At the design speed, friction = 0. Below the design speed, friction acts UP the slope. Above it, friction acts DOWN the slope.

Summary

Flat curve: friction provides centripetal force; v_max = √(μgr)
Banked curve: component of normal force provides centripetal force; design speed v = √(rg tan θ)
Conical pendulum: tension provides both vertical equilibrium and centripetal force
On banked curves, the design speed is independent of mass
Cyclists lean into curves at an angle given by tan θ = v²/(rg)
At the design speed on a banked curve, friction is zero; above it, friction acts down the slope; below it, friction acts up the slope

A-Level Deep Dive: Horizontal Circular Motion

Spec mapping

Edexcel 9PH0 specification Topic 6 — Further Mechanics, sub-strand on circular motion covers angular displacement and angular velocity, the relationship $v = r\omega$ , the centripetal acceleration $a = v^2/r = r\omega^2$ , and the centripetal force expressions $F = mv^2/r = mr\omega^2$ as applied to motion in a horizontal circle (refer to the official specification document for exact wording). Horizontal circular motion is examined principally on 9PH0 Paper 2 (Advanced Physics II) but the underlying force-resolution skills resurface on Paper 1 in the mechanics review and on Paper 3 in the synoptic 'Practical Skills' section. The Edexcel formula booklet supplies $a = v^2/r$ , $a = r\omega^2$ and $F = mv^2/r$ but does not state the banked-curve or conical-pendulum equations — these must be derived in the answer booklet from a free-body diagram.

Worked example with full mark scheme

Question (8 marks):

A racing-car test track has a curve banked at $\theta = 18°$ to the horizontal with radius $r = 95\,\text{m}$ . The track surface is treated as frictionless for design purposes.

(a) Show, using a labelled free-body diagram and Newton's second law, that the design speed at which a car can negotiate the curve with no friction is given by $v = \sqrt{rg\tan\theta}$ . (4)

(b) Calculate the design speed for this curve. (2)

(c) Explain how, and in which direction, friction must act if a car attempts to take the curve at a speed higher than the design speed. (2)

Solution with mark scheme:

(a) Step 1 — free-body diagram and forces.

The forces on the car are weight $mg$ acting vertically downward and the normal contact force $N$ acting perpendicular to the banked surface. With the bank tilted inward at angle $\theta$ , $N$ has a vertical component $N\cos\theta$ and a horizontal component $N\sin\theta$ directed toward the centre of the circle.

M1 — correct free-body diagram showing only $mg$ and $N$ (no spurious "centripetal force" arrow, no friction arrow).

Step 2 — vertical equilibrium.

The car has no vertical acceleration, so the vertical components balance:

$N\cos\theta = mg$

M1 — correct vertical equilibrium statement.

Step 3 — Newton's second law toward the centre.

The horizontal component of $N$ supplies the centripetal force:

$N\sin\theta = \dfrac{mv^2}{r}$

M1 — correct horizontal equation with $mv^2/r$ (not $mv^2$ or $v^2/r$ alone).

Step 4 — eliminate $N$ .

Dividing the second equation by the first:

$\tan\theta = \dfrac{v^2}{rg} \implies v = \sqrt{rg\tan\theta}$

A1 — printed result reached with division step shown explicitly. Common error: students multiply rather than divide, producing $v = \sqrt{rg/\tan\theta}$ . That loses the A1.

(b) Step 1 — substitute.

$v = \sqrt{95 \times 9.81 \times \tan 18°} = \sqrt{95 \times 9.81 \times 0.3249}$

M1 — correct substitution with consistent units.

Step 2 — evaluate.

$v = \sqrt{302.7} = 17.4\,\text{m s}^{-1}$

A1 — answer to 3 s.f. with units.

(c) Above the design speed, the required centripetal force exceeds $N\sin\theta$ from the bank alone, so the additional inward force must come from friction acting down the slope (toward the centre of the circle). B1 — direction "down the slope" or "toward the centre". B1 — reasoning that friction supplies the extra centripetal force, not the total.

Total: 8 marks (M3 A2 B2 plus diagram M1).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A conical pendulum consists of a small bob of mass $0.150\,\text{kg}$ attached to a light inextensible string of length $L = 0.90\,\text{m}$ . The bob travels in a horizontal circle with the string at angle $\theta$ to the vertical and angular velocity $\omega$ .

(a) Show that $\cos\theta = \dfrac{g}{\omega^2 L}$ . (3)

(b) The bob is observed to make 50 complete revolutions in 60 s. Calculate the angle $\theta$ . (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — vertical equilibrium $T\cos\theta = mg$ and centripetal equation $T\sin\theta = m\omega^2 r$ both stated, with $r = L\sin\theta$ .
M1 (AO2.1) — substitution of $r = L\sin\theta$ followed by cancellation of $\sin\theta$ on the centripetal equation.
A1 (AO1.1b) — division of equations to obtain the printed result $\cos\theta = g/(\omega^2 L)$ .

(b)

M1 (AO1.1b) — angular velocity from frequency: $f = 50/60 = 0.833\,\text{Hz}$ , $\omega = 2\pi f = 5.236\,\text{rad s}^{-1}$ .
M1 (AO1.1b) — substitution $\cos\theta = 9.81/(5.236^2 \times 0.90) = 9.81/24.67 = 0.3976$ .
A1 (AO2.5) — $\theta = \arccos(0.3976) = 66.6°$ to 3 s.f.

Total: 6 marks split AO1 = 4, AO2 = 2. This is an AO1-dominated derivation question. Edexcel reserves AO2 marks for the substitution insight ( $r = L\sin\theta$ ) and the unit-aware final answer (degrees, 3 s.f., recognising the answer must lie strictly between 0° and 90°).

Synoptic links

Connects to:

Vertical circular motion (Topic 6 continuation): the next lesson lifts the constraint that the plane of motion is horizontal, introducing weight components along the direction of travel. Energy conservation must be added to the force analysis because speed varies around the loop. The horizontal-circle skill of identifying which real force supplies the centripetal component carries over directly — but now tension and the radial weight component must be combined.
Civil and highway engineering (banked curves in practice): real motorway interchanges and railway curves are banked using the same $\tan\theta = v^2/(rg)$ design rule. Designers choose $\theta$ for the expected operating speed and rely on tyre/rail friction to handle deviations. Spirals (clothoid transitions) interpolate the bank angle smoothly to avoid jolts in the centripetal force.
Conical pendulum and rotating reference frames (Topic 6 + further-physics enrichment): in the rotating frame of the bob, the apparent "centrifugal" effect balances the horizontal component of tension. Treating the conical pendulum in both the inertial (Newton's-second-law) frame and the rotating (pseudo-force) frame is excellent preparation for Coriolis effects in atmospheric science.
Satellite orbits (Topic 6 + Topic 8 Gravitational Fields): circular orbits are horizontal-circle problems where gravity replaces friction or tension as the centripetal-force agent. The orbital-speed equation $v = \sqrt{GM/r}$ is derived by setting $GMm/r^2 = mv^2/r$ — exactly the same Newton-2 step used here, with a different real force on the left-hand side.
Charged particle in a magnetic field (Topic 7): the Lorentz force $F = Bqv$ acts perpendicular to the velocity of a moving charge, so a charged particle in a uniform field traces a horizontal circle of radius $r = mv/(Bq)$ . The "which force is the centripetal force?" identification is the same skill — here the magnetic force is the centripetal force.

Mark-scheme literacy

Horizontal-circular-motion questions on 9PH0 distribute AO marks broadly: