A wheel rotates at 3.0\ rad s^-1. A point on the rim is 0.40\ m from the axis. Find its linear speed.

WJEC Eduqas Eduqas 2020-style practice: A 0.30\ kg ball on a string is whirled in a horizontal circle of radius 0.80\ m at 2.5\ revolutions per second. Calculate the angular velocity and the tension in the string.

Angular velocity: ω = 2π f = 2π × 2.5 = 15.7\ rad s^-1. The tension provides the centripetal force: F = mω^2 r = 0.30 × (15.7)^2 × 0.80 = 0.30 × 247 × 0.80 = 59\ N. Markers reward ω = 2π f, recognising the tension supplies the centripetal force F = mω^2 r, and the tension about 59\ N.

WJEC Eduqas Eduqas 2022-style practice: A car of mass 1100\ kg travels over a humpback bridge whose top forms an arc of radius 35\ m. Calculate the maximum speed at which the car can cross while staying in contact with the road. Take g = 9.81\ m s^-2.

At the top of the hump the weight provides the centripetal force when the normal force just reaches zero (the point of losing contact). Setting the normal force to zero: mg = mv^2r, so v^2 = gr = 9.81 × 35 = 343, giving v = 18.5\ m s^-1. Markers reward identifying that contact is lost when N = 0, equating the weight to the centripetal force, and the maximum speed about 19\ m s^-1.

EnglandPhysicsSyllabus dot point

What keeps an object moving in a circle, and how do we relate its speed, period and the centripetal force?

Circular motion: angular velocity and the period, the centripetal acceleration, the centripetal force, and applications such as banked tracks, vertical circles and the conical pendulum.

A focused answer to the Eduqas A-Level Physics Component 1 circular motion content, covering angular velocity and the period, the centripetal acceleration, the centripetal force that maintains circular motion, and applications including banked tracks, vertical circles and the conical pendulum.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

Eduqas wants you to define angular velocity and relate it to the period and frequency, derive and use the centripetal acceleration $a = \frac{v^2}{r} = \omega^2 r$ , explain that a resultant centripetal force is required for circular motion, and apply these ideas to horizontal circles, vertical circles and banked tracks.

The answer

Angular velocity and the period

Centripetal acceleration

Centripetal force

Applications

Speed at the bottom of a vertical circle

A $0.25\ \text{kg}$ ball on a string of length $0.90\ \text{m}$ is swung in a vertical circle. At the lowest point its speed is $4.0\ \text{m s}^{-1}$ . Find the tension in the string there. Take $g = 9.81\ \text{m s}^{-2}$ .

step 1: Identify the forces at the lowest point

At the bottom, the tension $T$ acts upward (towards the centre) and the weight $mg$ acts downward (away from the centre). The resultant towards the centre is the centripetal force.

step 2: Write Newton's second law towards the centre

$T - mg = \dfrac{mv^2}{r}$ , so $T = mg + \dfrac{mv^2}{r}$ .

step 3: Substitute

$T = (0.25)(9.81) + \dfrac{(0.25)(4.0)^2}{0.90} = 2.45 + \dfrac{4.0}{0.90} = 2.45 + 4.44 = 6.9\ \text{N}$ .

Examples in context

Circular motion governs satellite and planetary orbits (gravity as the centripetal force), the design of road and railway curves and the banking of velodrome and motor-racing tracks, and the operation of centrifuges that separate blood or isotopes. Theme-park rides such as rotors and loop-the-loops are designed around the minimum speed needed to keep riders safely against the wall or the track.

Try this

Q1. State the direction of the centripetal acceleration of an object in uniform circular motion. [1 mark]

Cue. Towards the centre of the circle.

Q2. A wheel rotates at $3.0\ \text{rad s}^{-1}$ . A point on the rim is $0.40\ \text{m}$ from the axis. Find its linear speed. [2 marks]

Cue. $v = \omega r = 3.0 \times 0.40 = 1.2\ \text{m s}^{-1}$ .

Q3. Explain why the centripetal force does no work on an object in uniform circular motion. [2 marks]

Cue. The force is always perpendicular to the velocity, so the work $W = Fx\cos 90^\circ = 0$ and the kinetic energy is unchanged.

Exam-style practice questions

Practice questions written in the style of WJEC Eduqas exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Eduqas 20204 marksA

0.30\ \text{kg}

ball on a string is whirled in a horizontal circle of radius

0.80\ \text{m}

2.5\ \text{revolutions per second}

. Calculate the angular velocity and the tension in the string.

Show worked answer →

Angular velocity: $\omega = 2\pi f = 2\pi \times 2.5 = 15.7\ \text{rad s}^{-1}$ .

The tension provides the centripetal force: $F = m\omega^2 r = 0.30 \times (15.7)^2 \times 0.80 = 0.30 \times 247 \times 0.80 = 59\ \text{N}$ .

Markers reward $\omega = 2\pi f$ , recognising the tension supplies the centripetal force $F = m\omega^2 r$ , and the tension about $59\ \text{N}$ .

Eduqas 20224 marksA car of mass

1100\ \text{kg}

travels over a humpback bridge whose top forms an arc of radius

35\ \text{m}

. Calculate the maximum speed at which the car can cross while staying in contact with the road. Take

g = 9.81\ \text{m s}^{-2}

Show worked answer →

At the top of the hump the weight provides the centripetal force when the normal force just reaches zero (the point of losing contact).

Setting the normal force to zero: $mg = \dfrac{mv^2}{r}$ , so $v^2 = gr = 9.81 \times 35 = 343$ , giving $v = 18.5\ \text{m s}^{-1}$ .

Markers reward identifying that contact is lost when $N = 0$ , equating the weight to the centripetal force, and the maximum speed about $19\ \text{m s}^{-1}$ .

Related dot points

Sources & how we know this

Eduqas GCE AS/A Level Physics specification (A720QS) — WJEC Eduqas (2015)