Uniform circular-motion kinematics and force

Key idea: Uniform circular motion has constant speed but changing velocity. Radians link angle to arc length, and the inward resultant force supplies the centripetal acceleration.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 1 of 3

Check your understanding

By the end of this lesson, you should be able to

Use radians, angular displacement and angular velocity.
Connect angular and linear speed using v = rω.
Calculate centripetal acceleration and resultant force using v or ω.

Learn the idea

Big question: Why does an object accelerating towards a circle's centre not move towards the centre?

Connect radians, arc length and speed

An angle in radians is θ = s/r, so one full turn is 2π rad. Angular velocity ω = Δθ/Δt describes how quickly the radius line sweeps angle.

Since s = rθ, linear speed is v = rω. Points on a rigid rotating object share ω, but a point farther from the axis has greater linear speed.

Check your understanding: Two points on one rigid wheel are at radii r and 2r. Compare their angular and linear speeds.

Their angular speeds are equal; the outer point's linear speed is twice as large.

Identify the inward resultant

In uniform circular motion, speed is constant but velocity continually changes direction. The change points towards the centre, giving a = v²/r = rω².

Centripetal force is not an extra force. It is the name for the inward resultant of real forces such as tension, friction, gravity or a normal force. The instantaneous velocity remains tangential because acceleration changes its direction continuously.

Check your understanding: If the inward force suddenly disappears, which way does the object move?

Along the tangent at that instant, not radially outward.

Scroll diagram horizontally to read all labels.

Radians connect rotation to distance directly: θ = s/r, so an arc equal to the radius subtends 1 rad.

Scroll diagram horizontally to read all labels.

At every point, velocity is tangent to the path while centripetal acceleration and the resultant force point towards the centre.

Key ideas

Constant speed does not mean zero acceleration.
Velocity is tangent; acceleration and resultant force are radial inward.
Identify the real interaction supplying the inward resultant.

Relationships to know

θ = s/r
ω = Δθ/Δt
v = rω
a = v²/r = rω²
Fresultant = mv²/r = mrω²

Follow the reasoning

Worked example

Find the real force supplying circular motion

Question: A 0.40 kg mass on a 0.80 m string moves in a horizontal circle at 2.5 revolutions per second. Find angular speed, linear speed and string tension, assuming tension is the inward resultant.

Step 1: Convert frequency to angular speed
Why: One revolution is 2π radians.
Working: ω = 2πf = 2π(2.5) = 15.7 rad s⁻¹.
Step 2: Find linear speed
Why: The mass travels at radius 0.80 m.
Working: v = rω = 0.80(15.7) = 12.6 m s⁻¹.
Step 3: Find the inward resultant
Why: Tension supplies the required centripetal acceleration.
Working: T = mrω² = 0.40(0.80)(15.7²) = 79.0 N.

Answer: ω = 15.7 rad s⁻¹, v = 12.6 m s⁻¹ and tension is about 79 N inward.

Check: Using mv²/r gives the same tension; the units are newtons.

Now try it with support

Practise with support

A point moves in a circle of radius 0.40 m at angular speed 6.0 rad s⁻¹. Find its linear speed and centripetal acceleration.

Hints

Use v = rω.
Then use a = rω² directly.

View the guided answer

v = 0.40(6.0) = 2.4 m s⁻¹. a = 0.40(6.0²) = 14.4 m s⁻² inward.

Your turn

Practise independently

A car rounds a flat curve of radius 45 m at 12 m s⁻¹. Find its centripetal acceleration and identify the real force that supplies it.

Check your answer

a = 12²/45 = 3.2 m s⁻² towards the curve’s centre. On a flat road, static friction from the road on the tyres supplies the inward resultant force.

Common mistakes and exam guidance

Watch out for

Drawing a separate inward ‘centripetal force’ in addition to tension or friction.
Pointing acceleration along the tangent because that is the direction of motion.

In an exam

Draw a radial arrow towards the centre before writing the force equation.
Convert revolutions or degrees to radians before using angular equations.

Put the ideas together

Exam-style practice [7 marks]

A 950 kg car rounds a flat curve of radius 55 m at 14 m s⁻¹. Find its centripetal acceleration and the friction force required. If the maximum friction is 4200 N, calculate the greatest safe speed and explain what happens above it.

Plan before you answer

Treat friction as the real inward force.
For the limit, set mv²/r equal to maximum friction.
Describe the initial motion if insufficient force is available.

View the marking points and model answer

Marking points

Finds a = 3.56 m s⁻².
Finds friction about 3.39 × 10³ N inward.
Uses v = √(Fr/m).
Obtains maximum speed about 15.6 m s⁻¹.
States required force rises with v².
States tyres cannot provide enough inward resultant above the limit.
Describes motion as following a less curved path/tangent rather than an outward force.

Model answer

a = 14²/55 = 3.56 m s⁻², so friction = 950(3.56) = 3.39 × 10³ N inward. At the limit, v = √(4200×55/950) = 15.6 m s⁻¹. Above this, the required inward force exceeds available friction, so the car cannot follow the curve and initially follows a less curved, near-tangential path; no extra outward force is needed.

Finish from memory

Three-question recap

How many radians are in one revolution?
Check
2π rad.
Where do velocity and acceleration point in uniform circular motion?
Check
Velocity is tangential; acceleration is towards the centre.
What is centripetal force?
Check
The inward resultant of the real forces acting on the object.

Try this next

Apply the same inward-resultant model to gravity-driven satellite orbits.

Continue to Inverse-square gravitation and near-Earth field Return to Circular Motion and Gravitation

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027