OxfordAQA International AS and A-level Physics: Circular motion: angular speed and centripetal force (9630)

This guide follows the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for exams from May/June 2020 onwards. It covers section 3.6.1 (Circular motion). The specification marks section 3.6 as International A-level only, so this content is not part of International AS. Use it with the course hub and tick points off on the printable checklist.

Other pages for this topic: revision notes and practice questions. If your force and acceleration work feels shaky, read motion and Newton’s laws first.

What the specification asks for

Section	Outcome
3.6.1	Explain why travelling round a circle at a steady speed involves an acceleration, and why a centripetal force is needed
3.6.1	Use angular speed ω = v/r = 2πf
3.6.1	Use centripetal acceleration a = v²/r = ω²r
3.6.1	Use centripetal force F = mv²/r = mω²r
3.6.1	Know that the derivation of a = v²/r will not be examined

The specification gives these as results to use. You are not asked to derive a = v²/r, so learn the result and what it means. The unit of angular speed is the radian per second, so you also need to be comfortable with the radian.

Why a circle needs a force

Velocity is a vector. It has a size (the speed) and a direction. An object moving round a circle at constant speed keeps changing direction, so its velocity keeps changing. A changing velocity is an acceleration, even though the speed never changes.

At every instant the velocity points along the tangent to the circle. The change in velocity points towards the centre. So:

• the acceleration is directed towards the centre of the circle
• it is perpendicular to the velocity
• by Newton’s second law, the resultant force is also directed towards the centre

This resultant force is the centripetal force. The word describes the direction and the job done, not a new kind of force. Something real has to provide it: tension, friction, a normal contact force, gravitational attraction, or a combination. When you draw a free-body diagram, draw only the real forces. Do not add an extra arrow labelled “centripetal force” or an outward “centrifugal force”. The resultant of the real forces, taken towards the centre, is what equals mv²/r.

The force always acts at 90° to the velocity, so nothing pushes the object faster or slower. It turns the velocity and leaves the speed alone. If the force vanishes, for example when the string snaps, the object leaves along the tangent at the speed it had, in a straight line.

Angles in radians

One radian is the angle formed at the centre when the arc along the edge is exactly as long as the radius. For an arc of length s on a circle of radius r:

θ = s/r

A full turn has arc length 2πr, so it is 2π rad. To convert, use degrees × π/180 for radians, and radians × 180/π for degrees.

Worked example 1. A bead slides 0.84 m along a circular wire of radius 0.70 m. Find the angle turned, in radians and degrees.

θ = s/r = 0.84/0.70 = 1.2 rad In degrees: 1.2 × 180/π = 68.8°

Also, 135° is 135 × π/180 = 3π/4 rad = 2.36 rad. Angular speed formulas only work if the angle is in radians, so convert first.

Angular speed

Angular speed ω is the angle turned per unit time. It is measured in rad s⁻¹. For motion at constant speed, with v the linear speed and r the radius:

ω = v/r = 2πf

where f is the frequency (revolutions per second, in hertz). The period T is the time for one revolution, so f = 1/T and ω = 2π/T. The link v = ωr follows by rearranging the first equality. It says that points further from the axis move faster, even though every point on a rigid turntable has the same ω.

Revolutions per minute need converting: divide by 60 for revolutions per second, then multiply by 2π.

Worked example 2. A playground roundabout makes one revolution in 5.2 s. A child sits 1.4 m from the axis. Find ω, the child’s speed, and the child’s centripetal acceleration.

1. f = 1/5.2 = 0.192 Hz
2. ω = 2πf = 2π/5.2 = 1.21 rad s⁻¹
3. v = ωr = 1.208 × 1.4 = 1.69 m s⁻¹
4. a = ω²r = 1.208² × 1.4 = 2.04 m s⁻²

Check with the other form: v²/r = 1.692²/1.4 = 2.04 m s⁻². The two forms must agree.

Worked example 3. A laboratory drum spins at 450 revolutions per minute. A point on its rim is 0.12 m from the axis. Find ω, v and a.

1. f = 450/60 = 7.5 Hz
2. ω = 2π × 7.5 = 47.1 rad s⁻¹
3. v = ωr = 47.12 × 0.12 = 5.65 m s⁻¹
4. a = ω²r = 47.12² × 0.12 = 266 m s⁻², which is about 27 times g

Centripetal acceleration

The specification gives two equivalent forms:

a = v²/r = ω²r

Choose the form that matches the data. If you are given speed and radius, use v²/r. If you are given frequency or period, find ω and use ω²r. Substituting v = ωr into v²/r gives ω²r, so you can always convert between them. The direction is towards the centre.

Two scaling results are worth knowing:

• At fixed radius, doubling the speed multiplies the acceleration by four.
• At fixed ω, doubling the radius doubles the acceleration. At fixed v, doubling the radius halves it.

Which comparison applies depends on what is held constant. On a rigid rotating disc, ω is the same everywhere, so use ω²r. For two cars on different bends at the same speed, v is the same, so use v²/r.

Centripetal force

Apply F = ma with the centripetal acceleration:

F = mv²/r = mω²r

F here is the resultant force towards the centre. Always start by asking which real forces point towards the centre, and which point away. Then set the resultant equal to mv²/r.

Situation	What provides the centripetal force
Object on a string on a smooth horizontal surface	Tension in the string
Small object resting on a spinning disc	Friction between object and disc
Car on a flat circular bend	Sideways grip of the tyres on the road (friction)
Satellite in a circular orbit	Gravitational attraction towards the planet
Rider held against the inner wall of a rotating cylinder	Push of the wall on the rider (normal contact force)

Worked example 4. A stone of mass 0.35 kg is swung on a light string of length 1.15 m, travelling at a constant 6.4 m s⁻¹ round a horizontal circle on a smooth surface. Find the acceleration and the tension.

1. a = v²/r = 6.4²/1.15 = 35.6 m s⁻²
2. T = ma = 0.35 × 35.62 = 12.5 N, directed along the string towards the centre

The angular speed is ω = v/r = 5.57 rad s⁻¹, and the check mω²r = 0.35 × 5.565² × 1.15 = 12.5 N agrees.

Worked example 5. A car of mass 1250 kg takes a flat bend of radius 48 m at 14 m s⁻¹. (a) What sideways force must the road exert? (b) The tyres can supply at most 7500 N. What is the greatest speed on this bend?

(a) F = mv²/r = 1250 × 14²/48 = 1250 × 196/48 = 5.1 × 10³ N (5104 N)

(b) Rearrange F = mv²/r to make v the subject: v = √(Fr/m) = √(7500 × 48/1250) = √288 = 17.0 m s⁻¹

The car is safe at 14 m s⁻¹ because 5104 N is less than the 7500 N limit.

Worked example 6. A small trolley of mass 0.15 kg circles at a radius of 0.32 m on a frictionless table. It is held on a string passing through a hole in the table to a hanging mass of 0.40 kg. The hanging mass stays at rest. Find the trolley’s speed, angular speed and frequency. Use g = 9.81 N kg⁻¹.

1. The hanging mass is in equilibrium, so the tension is T = Mg = 0.40 × 9.81 = 3.92 N.
2. This tension is the centripetal force on the trolley: T = mv²/r.
3. v = √(Tr/m) = √(3.924 × 0.32/0.15) = 2.89 m s⁻¹
4. ω = v/r = 2.893/0.32 = 9.04 rad s⁻¹
5. f = ω/2π = 1.44 Hz

Worked example 7. A satellite of mass 850 kg moves in a circular orbit of radius 7.2 × 10⁶ m with a period of 6.1 × 10³ s. Find its angular speed, its speed, its centripetal acceleration and the resultant force on it. (These figures are supplied data for this question.)

1. ω = 2π/T = 2π/6100 = 1.03 × 10⁻³ rad s⁻¹
2. v = ωr = 1.030 × 10⁻³ × 7.2 × 10⁶ = 7.4 × 10³ m s⁻¹
3. a = ω²r = (1.030 × 10⁻³)² × 7.2 × 10⁶ = 7.6 m s⁻²
4. F = ma = 850 × 7.639 = 6.5 × 10³ N, towards the planet’s centre

The force is the satellite’s weight in this field. It is the only force on the satellite, so it is the whole of the centripetal force.

Reading data and graphs

For a fixed mass and radius, F = mω²r means a graph of F against ω² is a straight line through the origin with gradient mr. A graph of F against v² has gradient m/r. If the gradient of an F against ω² graph is 0.027 kg m and the radius was 0.45 m, then m = 0.027/0.45 = 0.060 kg.

In experiments with a bung on a string, the centripetal force is found from the weight of hanging masses, and the period is found by timing many revolutions and dividing, which reduces the percentage uncertainty in T. Plotting F against 1/T² gives a straight line, because ω = 2π/T gives F = 4π²mr/T².

Common errors

• Using degrees in ω = v/r or ω = 2πf. Angular speed needs radians.
• Forgetting to convert revolutions per minute to revolutions per second before multiplying by 2π.
• Squaring the wrong quantity: it is v squared over r, not (v/r) squared.
• Drawing an outward “centrifugal” arrow on a free-body diagram. Only real forces appear, and the resultant points inward.
• Saying “constant speed means no acceleration”. The speed is constant; the velocity is not.
• Treating centripetal force as an extra force on top of tension or friction. It is their resultant towards the centre.
• Using the diameter instead of the radius.
• Claiming the centripetal force does work. It acts at right angles to the motion, so the kinetic energy stays constant.

Check yourself

Try the practice questions, then use the revision notes for the final weeks. If you are unsure where you stand across physics, take one of the free 10-minute diagnostics.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams from May/June 2020 onwards, published by OxfordAQA. Section 3.6.1 Circular motion, within section 3.6 Circular and periodic motion (International A-level only).