Is tan θ = v²/rg on the HSC data sheet?

No. The 2019 data sheet gives Fc = mv^2/r and ac = v^2/r, but not the banked-curve formula. You may be asked to derive it: resolve the normal force so N cos θ = mg and N sin θ = mv^2/r, then divide the two equations.

Is centrifugal force real?

Not as a force on the object. In HSC Physics you describe motion from the ground, and there the only forces are real ones like friction, tension, normal force and gravity, with a net force towards the centre. The outward 'push' you feel in a turning car is your body tending to keep moving in a straight line.

Does the centripetal force do work?

No. It always acts at 90° to the direction of motion, so W = Fs cos 90° = 0. The kinetic energy, and so the speed, stays constant in uniform circular motion.

What is the difference between angular velocity and linear velocity?

Angular velocity ω is the angle turned per second (rad s^−1) and is the same for every point on a rotating rigid object. Linear velocity v is the actual speed along the path (m s^−1), and v = ωr, so it depends on the distance from the axis.

What is the minimum speed at the top of a vertical circle?

v = √(gr). At that speed the string or track pushes with zero force at the top, so gravity alone provides the centripetal force: mg = mv^2/r. Any slower and the object leaves the circle.

A car rounds a flat (unbanked) curve at constant speed. What provides the centripetal force?

Friction between the tyres and the road

Two children sit on a spinning merry-go-round, one 1.0 m from the centre and one 3.0 m from the centre. Which statement is correct?

They have the same angular velocity, and the outer child's linear speed is 3 times as large.

They have the same linear speed, and the outer child has the larger angular velocity.

They have the same angular velocity and the same linear speed.

The inner child has 3 times the angular velocity of the outer child.

A car goes round a banked curve more slowly than the curve's design speed. Which way does friction on the tyres act?

There is no friction on a banked curve at any speed.

Horizontally away from the centre

HSC Physics · Module 5 · Advanced Mechanics · 14 min read

Circular motion

Anything moving in a circle is changing direction all the time, so it is accelerating, even at a steady speed. In HSC Physics every circular motion question comes down to one idea: some real force (or combination of forces) must point towards the centre, and its size must be mv²/r.

This guide covers centripetal force and acceleration, cars on flat and banked curves (including where tan θ = v²/rg comes from), vertical circles, angular velocity, torque and why no work is done in uniform circular motion. Then a quick quiz and a full worked example on a banked curve.

What you need to know

Investigate uniform circular motion to explain and evaluate how the centripetal force is related to the object's mass, its speed and the radius of the circle.
Analyse the forces on a body in uniform circular motion in many situations, for example a car on a flat (unbanked) curve, a mass swung on a string and an object on a banked track.
Use a_c = v²/r, v = 2πr/T, F_c = mv²/r and ω = Δθ/t to solve circular motion problems, build models and make quantitative predictions.
For an object in uniform circular motion, investigate how the work done relates to its total energy.
Investigate how an applied torque relates to the rotation of a mechanical system, using τ = r_⊥F = rF sin θ.

The equations (from the 2019 HSC data sheet unless marked):

a_c = v²/rCentripetal acceleration, always towards the centre of the circle.

F_c = mv²/rThe net force needed to keep mass m moving in a circle of radius r at speed v.

v = 2πr/TSpeed from the period T (the time for one full revolution).

ω = Δθ/tAngular velocity in radians per second (rad s⁻¹). One revolution is 2π rad.

τ = r_⊥F = rF sin θTorque, with θ the angle between the lever arm r and the force F.

v = ωr not on the data sheetLinks angular and linear speed. Points further from the axis move faster at the same ω.

tan θ = v²/rg not on the data sheetDesign speed of a banked curve with no friction needed. Derive it; it is not given to you.

Dot points: PH-M5-05, PH-M5-06, PH-M5-07, PH-M5-08, PH-M5-09 (2017 Physics syllabus, HSC 2026 and 2027), in our own words.

Mini lesson

Why a circle needs a force

Velocity is a vector. An object going round a circle at constant speed keeps changing the direction of its velocity, so it has an acceleration. That acceleration points towards the centre and has size a_c = v²/r. Because it is at right angles to the velocity, it changes the direction but not the speed.

By Newton's second law a centre-pointing acceleration needs a centre-pointing net force, F_c = mv²/r. Take that force away and the object carries on in a straight line along the tangent. Doubling the speed needs four times the force; doubling the radius at the same speed needs half the force.

Blue: the velocity is along the tangent. Orange: the net force (and the acceleration) points to the centre, at 90° to the velocity, so the speed stays the same.

Centripetal force is a job, not a new force

Never add a separate 'centripetal force' to a free-body diagram. F_c is the net force, and some real force has to do the job. Name it every time:

Car on a flat curve: friction between the tyres and the road, pointing sideways towards the centre. Too fast and friction can't supply mv²/r, so the car slides outwards (really, it carries on straighter).
Mass swung on a string in a horizontal circle: the tension, or the horizontal part of it. For a conical pendulum, T cos θ = mg vertically and T sin θ = mv²/r horizontally, with r = l sin θ, not the string length l.
Satellite or moon: gravity alone (see the gravitational fields guide).
Charge in a magnetic field: the magnetic force qvB.
There is no outward 'centrifugal' force on the object. The outward push you feel in a turning car is your body tending to keep going straight while the car turns under you.

Banked curves: where tan θ = v²/rg comes from

Roads, racetracks and velodromes are banked (tilted towards the centre) so that the normal force helps turn the vehicle. At the design speed no sideways friction is needed, and only two forces act: the weight mg straight down and the normal force N at right angles to the road surface.

The car moves in a horizontal circle, so the centre is level with the car and the net force must be horizontal. Resolve N (not the weight) into components. N is tilted from the vertical by the bank angle θ:

Vertical: no vertical acceleration, so N cos θ = mg.
Horizontal: the horizontal part of N is the centripetal force, so N sin θ = mv²/r.
Divide the second line by the first: tan θ = v²/rg. The mass cancels, so the same bank suits a small car and a truck.
Note N = mg ÷ cos θ, so on a bank the normal force is bigger than the weight.
Faster than the design speed: N alone gives too little inward force, so friction acts down the slope. Slower: the car tends to slide down, so friction acts up the slope.

At the design speed only N and mg act. The vertical part of N balances mg; the horizontal part, N sin θ, is the centripetal force. N makes angle θ with the vertical.

Vertical circles

For a ball on a string swung in a vertical circle, write F_net = mv²/r towards the centre at the point you care about. At the top, tension and weight both point down, towards the centre. At the bottom, tension points up (to the centre) and weight points down (away from it).

The slowest the ball can go at the top and still follow the circle is when the string just goes slack (T = 0). Then gravity alone supplies the centripetal force: mg = mv²/r, so v_min = √(gr). Strictly, a vertical circle on a string is not uniform: the ball is slower at the top and faster at the bottom, because kinetic and gravitational potential energy swap as it rises and falls (½mv² + mgh stays constant if no energy is lost).

Take towards the centre as positive at each point. The tension is greatest at the bottom, where it must hold up the weight as well as supply mv²/r.

Angular velocity, period and torque

Angular velocity ω is the angle swept out per second: ω = Δθ/t, in rad s⁻¹. One revolution is 2π rad, so ω = 2π/T. Every point on a rigid rotating object (a wheel, a merry-go-round) has the same ω, but the linear speed v = ωr is larger further from the axis.

Torque is the turning effect of a force: τ = rF sin θ, where r is the distance from the pivot to where the force acts and θ is the angle between r and F. It is largest when the force is at 90° to the lever arm and zero when the force points along it. For example, a 40 N push on a door handle 0.75 m from the hinges, at 60° to the door, gives τ = 0.75 × 40 × sin 60° = 26 N m. A net torque changes how fast something rotates, just as a net force changes how fast something moves.

Work and energy in uniform circular motion

Work is W = Fs cos θ, where θ is the angle between the force and the displacement. In uniform circular motion the centripetal force is always at 90° to the direction of motion, so cos 90° = 0 and the centripetal force does no work.

No work means no change in kinetic energy, which is why the speed stays constant. The total energy of an object in uniform circular motion is constant. Work is only done (and the speed only changes) when there is a force component along the direction of motion, such as gravity on the sides of a vertical circle.

Common mistakes

Drawing 'centripetal force' as an extra arrow on a free-body diagram, or adding an outward centrifugal force.Draw only real forces (weight, normal, friction, tension). The centripetal force is their net result towards the centre.
Assuming N = mg for a car on a banked curve.Resolve N: N cos θ = mg, so N = mg ÷ cos θ, which is larger than mg.
Using the string length as the radius of a conical pendulum.The radius is the horizontal distance to the axis: r = l sin θ, with θ measured from the vertical.
Treating a ball on a string in a vertical circle as having the same speed all the way round.It speeds up going down and slows going up. Use conservation of energy between points, then F_net = mv²/r at each point.
Using angular velocity and linear velocity as if they were the same quantity.Points on one rigid body share ω (rad s⁻¹); their linear speeds differ, v = ωr (m s⁻¹).
Finding torque with the wrong angle, or with the distance in centimetres.Use θ between the lever arm and the force, convert cm to m, and check: τ is largest at 90°.

Worked example

A highway exit ramp follows a horizontal circular curve of radius 90 m. It is banked so that a car travelling at 72 km h⁻¹ needs no sideways friction from the road.

Show that the bank angle θ satisfies tan θ = v²/rg, and calculate θ. Then calculate the size of the normal force on a 1300 kg car taking the ramp at this speed.

(5 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. With v = 20 m s⁻¹, v²/rg ≈ 400 ÷ 900 ≈ 0.44, and tan 25° ≈ 0.47, so I expect an angle a little under 25°. The normal force must be more than the weight (1300 × 9.8 ≈ 1.3 × 10⁴ N), because only its vertical part holds the car up. I predict about 1.4 × 10⁴ N.

K · Knowns

I sketch a cross-section of the road with the car on it and put every value on the sketch. The speed is in km h⁻¹, so I convert it to m s⁻¹ straight away. There's no friction at the design speed, so only the weight and the normal force act.

r = 90 m, m = 1300 kg, g = 9.8 m s⁻²
v = 72 km h⁻¹ = 72 ÷ 3.6 = 20 m s⁻¹
friction = 0 (design speed)

Knowns on the sketch: r = 90 m (horizontal, to the centre), m = 1300 kg, v = 72 km h⁻¹ = 20 m s⁻¹. Unknowns: θ and N.

N · Not known

Two things are asked for: an angle in degrees and a force in newtons. I also have to show the formula, so the derivation comes before any numbers.

θ = ? °, N = ? N

O · Organise

The car goes round a horizontal circle, so the centre is level with the car and the net force must be horizontal. I resolve the normal force, not the weight, because the weight is already vertical. N is tilted from the vertical by θ.

Vertical (no vertical acceleration): N cos θ = mg (1)
Horizontal (centripetal): N sin θ = mv²/r (2)
(2) ÷ (1): tan θ = (mv²/r) ÷ mg = v²/rg
Then from (1): N = mg ÷ cos θ

W · Work it out

I substitute into tan θ first and keep the full value of θ for the cos. Then I check the answer with equation (2), because the horizontal part of N should equal mv²/r.

tan θ = 20² ÷ (90 × 9.8) = 400 ÷ 882 = 0.4535
θ = tan⁻¹(0.4535) = 24.39°
N = (1300 × 9.8) ÷ cos 24.39° = 12 740 ÷ 0.9107 = 13 989 N
Check: N sin θ = 13 989 × 0.4130 = 5778 N and mv²/r = 1300 × 400 ÷ 90 = 5778 N ✓

U · Units

The angle is in degrees and the force is in newtons. The data are to two significant figures, so I round both to two.

θ ≈ 24°, N ≈ 1.4 × 10⁴ N

D · Direction

The normal force is a vector, so I give its direction: at right angles to the road surface. That means it is tilted 24° from the vertical, towards the centre of the curve. That tilt is exactly what gives it the inward horizontal part.

N = 1.4 × 10⁴ N, perpendicular to the road, 24° from the vertical towards the centre

The ramp is banked at about 24°, and the normal force on the car is about 1.4 × 10⁴ N, perpendicular to the road surface (24° from the vertical, towards the centre).

Check against the prediction. I predicted a little under 25° and about 1.4 × 10⁴ N, more than the weight of 1.27 × 10⁴ N. Both fit: 24° and 1.4 × 10⁴ N.

HSC advice

What markers keep saying

When a period changes, show the effect with numbers, not just words. Since v = 2πr/T, halving the period doubles the speed, and F_c = mv²/r then goes up four times. Strong answers drew a free-body diagram with every force labelled and dealt with both the horizontal and the vertical forces.NESA marking feedback, 2024, summarised
For the minimum speed at the top of a vertical circle, say why only gravity acts (the tension is zero) and show the algebra from mg = mv²/r to v = √(gr). For speeds at different heights, use conservation of energy: the object has both kinetic and potential energy at every point.NESA marking feedback, 2025, summarised
Keep angular velocity and linear velocity apart. Points on the same rotating object share ω, but v = ωr grows with radius. Back the comparison up with a calculation using the radii you are given.NESA marking feedback, 2021, summarised
Where a force is tilted, resolve it into vertical and horizontal parts and write each equation separately before combining them. Set out every step of the derivation, not just the start and the result.NESA marking feedback, 2019, summarised
For torque, check the angle is the one between the lever arm and the force, and convert centimetres to metres before substituting.NESA marking feedback, 2024, summarised
On a banked curve, resolve the normal force (not the weight), because the net force points horizontally to the centre. Then say which way friction acts if the speed is above or below the design speed.CortexHSC teaching notes

Summaries in our own words. Read NESA's own feedback with the exam packs: 2019 · 2021 · 2024 · 2025.

How often it is examined

Circular motion questions were worth 31 marks across the seven HSC Physics exams from 2019 to 2025 (about 4.4 marks a paper), and appeared in 7 of 7 papers.

Marks per paper (Module 5, Circular Motion). Source: CortexHSC's own tagging of the NESA HSC Physics exams 2019–2025, each question counted under its main topic, so treat the numbers as close, not exact.

FAQ

Is tan θ = v²/rg on the HSC data sheet?: No. The 2019 data sheet gives F_c = mv²/r and a_c = v²/r, but not the banked-curve formula. You may be asked to derive it: resolve the normal force so N cos θ = mg and N sin θ = mv²/r, then divide the two equations.
Is centrifugal force real?: Not as a force on the object. In HSC Physics you describe motion from the ground, and there the only forces are real ones like friction, tension, normal force and gravity, with a net force towards the centre. The outward 'push' you feel in a turning car is your body tending to keep moving in a straight line.
Does the centripetal force do work?: No. It always acts at 90° to the direction of motion, so W = Fs cos 90° = 0. The kinetic energy, and so the speed, stays constant in uniform circular motion.
What is the difference between angular velocity and linear velocity?: Angular velocity ω is the angle turned per second (rad s⁻¹) and is the same for every point on a rotating rigid object. Linear velocity v is the actual speed along the path (m s⁻¹), and v = ωr, so it depends on the distance from the axis.
What is the minimum speed at the top of a vertical circle?: v = √(gr). At that speed the string or track pushes with zero force at the top, so gravity alone provides the centripetal force: mg = mv²/r. Any slower and the object leaves the circle.

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