A 0.50\,kg ball on a string moves in a horizontal circle of radius 0.80\,m at 4.0\,m s^-1. Find the tension in the string. [3 marks]

WalesPhysicsSyllabus dot point

Why is an object moving in a circle at constant speed accelerating, and what provides the force?

Angular velocity, the link between speed and angular velocity, centripetal acceleration and centripetal force.

A focused answer to WJEC A-Level Physics Unit 3 circular motion, covering angular velocity, the link v = r omega between speed and angular velocity, centripetal acceleration, and the centripetal force needed to keep a body moving in a circle.

Generated by Claude Opus 4.811 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

WJEC wants you to define angular velocity, link it to linear speed, and use the equations for centripetal acceleration and centripetal force, explaining why circular motion at constant speed is accelerated motion. Circular motion is also the gateway to orbits and to simple harmonic motion, so the ideas here recur throughout Unit 3 and Unit 4.

The answer

Angular velocity

A point on the rim of a wheel and a point near the hub share the same angular velocity but have different linear speeds, because $v = r\omega$ grows with radius.

Centripetal acceleration

Centripetal force

The centripetal force is the resultant force towards the centre. It is not a new kind of force; it is provided by a real force such as tension, gravity or friction. Without it the object would move off in a straight line (Newton's first law).

Angular and linear speed of a fairground ride

A fairground ride spins riders at a radius of $6.0\,\text{m}$ , completing one rotation every $4.0\,\text{s}$ . Find the angular velocity, the linear speed, and the centripetal acceleration.

step Angular velocity

$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4.0} = 1.57\,\text{rad s}^{-1}$ .

step Linear speed

$v = r\omega = 6.0 \times 1.57 = 9.4\,\text{m s}^{-1}$ .

step Centripetal acceleration

$a = r\omega^2 = 6.0 \times (1.57)^2 = 6.0 \times 2.47 = 14.8\,\text{m s}^{-2}$ , about $1.5g$ , which is why riders feel pressed into their seats.

Examples in context

Example 1. A banked race track: On a banked bend the track is tilted so that a component of the normal contact force points toward the centre, providing the centripetal force without relying on friction. At the design speed a car can corner even on a frictionless (icy) surface, because the geometry alone supplies $mv^2/r$ . This is why velodromes and motor-racing circuits bank their corners steeply.
Example 2. A centrifuge: A laboratory centrifuge spins samples at thousands of revolutions per minute, generating centripetal accelerations of many thousands of $g$ . Denser particles require a larger inward force to follow the circular path than the surrounding fluid can supply, so they migrate outward to the bottom of the tube. The huge $r\omega^2$ acceleration is what separates blood cells from plasma in seconds.
Example 3. A satellite orbit: A satellite circling the Earth needs a centripetal force, and here gravity supplies it: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ . There is no string or contact force, only the inward pull of gravity bending the satellite's path into a circle. This is the same centripetal condition, with gravity playing the role that tension or friction plays in earthbound examples, and it leads directly into the orbits topic studied later in the course.

Try this

Q1. A $0.50\,\text{kg}$ ball on a string moves in a horizontal circle of radius $0.80\,\text{m}$ at $4.0\,\text{m s}^{-1}$ . Find the tension in the string. [3 marks]

Cue. $F = \frac{mv^2}{r} = \frac{0.50\times4.0^2}{0.80} = 10\,\text{N}$ .

Q2. Explain why an object moving in a circle at constant speed is accelerating. [2 marks]

Cue. Its velocity direction is continuously changing, and a change in velocity is an acceleration.

Exam-style practice questions

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

WJEC 20195 marksA car of mass

1200\,\text{kg}

rounds a flat circular bend of radius

50\,\text{m}

15\,\text{m s}^{-1}

. Calculate the centripetal force required, and determine the minimum coefficient of friction between the tyres and the road that allows the car to corner without skidding. Take

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

Show worked answer →

Centripetal force needed: $F = \dfrac{mv^2}{r} = \dfrac{1200 \times (15)^2}{50} = \dfrac{1200 \times 225}{50} = 5400\,\text{N}$ .

This force is provided by friction between the tyres and road. The maximum available friction is $\mu mg$ , so for no skidding $\mu mg \ge F$ .

$\mu \ge \dfrac{F}{mg} = \dfrac{5400}{1200 \times 9.81} = \dfrac{5400}{11772} = 0.46$ .

The minimum coefficient of friction is about $0.46$ . Markers reward the centripetal force from $mv^2/r$ , recognising friction as the source, and the friction condition giving $\mu \approx 0.46$ .

WJEC 20213 marksA stone is whirled on a string in a vertical circle. Explain why the tension in the string is greatest at the bottom of the circle.

Show worked answer →

At every point the net force toward the centre must equal the centripetal force $mv^2/r$ .

At the bottom of the circle the centre is directly above the stone, so the upward tension must both support the weight and provide the centripetal force: $T - mg = mv^2/r$ , giving $T = mg + mv^2/r$ .

At the top, gravity acts toward the centre and helps provide the centripetal force, so less tension is needed: $T = mv^2/r - mg$ . Since the tension at the bottom has the weight added rather than subtracted, and the speed is also greater at the bottom, the tension is largest there. Markers reward resolving forces toward the centre at the bottom and showing the weight adds to the required tension.

Related dot points

Sources & how we know this

WJEC A-level Physics specification — WJEC (2015)