Kinematics of Angular Motion

Key idea: Learn angular displacement, angular velocity, angular acceleration, and the constant-α rotational SUVAT equations, with worked examples and exam tips.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • show an understanding of and use the terms angular displacement, angular velocity, and angular acceleration of a rigid body with respect to a fixed axis
  • solve problems using the equations of motion for uniform angular acceleration that are analogous to the equations of motion for uniform linear acceleration

Kinematics of angular motion describes how a rigid body rotates about a fixed axis. The payoff is that the rotational equations under uniform angular acceleration are direct analogues of SUVAT.

1. Definitions (Must Know)

  • Angular displacement, θ (rad): angle turned about a fixed axis.
  • Angular velocity, ω (rad s⁻¹): rate of change of angular displacement, ω = dθ/dt.
  • Angular acceleration, α (rad s⁻²): rate of change of angular velocity, α = dω/dt = d²θ/dt².
  • Fixed axis rotation: every point on the rigid body moves in a circle centred on the axis, with the same θ, ω, α at any instant.
  • Radian: θ = s/r where s is arc length and r is radius.
  • Symbols used in this lesson: θ,θ₀ angle (rad), ω,ω₀ angular speed (rad s⁻¹), α angular acceleration (rad s⁻²), t time (s), r radius (m), s arc length (m), v tangential speed (m s⁻¹), aₜ tangential acceleration (m s⁻²).

2. Key Ideas (What Earns Marks)

  • Rotational kinematics mirrors linear kinematics:
    • x ↔ θ, v ↔ ω, a ↔ α.
  • For a point at distance r from the axis: s = rθ, v = rω, aₜ = rα
  • Under uniform angular acceleration (α constant), use the rotational “SUVAT” set:
    ω = ω₀ + α t; θ = θ₀ + ω₀ t + (1/2)α t²; ω² = ω₀² + 2α(θ-θ₀)

Quick map (use a table when asked to compare):

Linear (translation)Rotation (fixed axis)Units
displacement sangle θm, rad
velocity vangular speed ωm s⁻¹, rad s⁻¹
acceleration aangular acceleration αm s⁻², rad s⁻²
Two points on one rotating disc share angular velocity, but the point farther from the fixed axis has a larger tangential speed.
A rigid body shares θ, ω, and α at every radius; linear tangential quantities scale with distance from the axis.

3. Detailed Explanations

If a point is at radius r from the axis:

  • The arc length travelled is s = rθ.
  • Differentiating gives tangential speed: v = ds/dt = rdθ/dt = rω
  • Differentiating again gives tangential acceleration: aₜ = dv/dt = rdω/dt = rα

These are especially useful when a question gives a linear speed/acceleration and asks for ω or α (or vice versa).

B. Deriving the constant-α equations (one clean route)

If α is constant: α = dω/dt ⇒ ω = ω₀ + α t

Also ω = dθ/dt, so (use a dummy variable τ):

θ-θ₀ = ∫₀^t ω dτ = ∫₀^t(ω₀ + α τ) dτ = ω₀ t + (1/2)α t²

To eliminate t, use α = dω/dt = (dω/dθ)dθ/dt = ωdω/dθ:

α = ωdω/dθ ⇒ ∫_ω₀^ωω dω = α∫_θ₀^θdθ ⇒ ω² = ω₀² + 2α(θ-θ₀)

4. Common Mistakes

  • Using degrees in calculations without converting to radians (units of ω and α require radians).
  • Confusing tangential acceleration aₜ = rα with centripetal acceleration aᵣ = rω² (different directions, different causes).
  • Forgetting that r can differ for different points on the same rigid body (so different points have different v and aₜ, even though ω and α are shared).
  • Substituting into the constant-α equations when α is not constant (the question must justify “uniform angular acceleration”).

5. Exam Tips

  • Write a one-line map before you start: “use θ–ω–α as the rotational analogues of s–v–a.”
  • Keep a clear sign convention (clockwise vs anticlockwise) and stick to it.
  • Use radians by default; only convert to degrees at the end if the question asks for it.
  • If a question mixes linear and rotational quantities, immediately write v = rω and aₜ = rα for the point of interest.

6. Worked Examples

Modelled example 1

Using constant angular acceleration (direct)

Core

Problem

A wheel starts from rest with constant α = 2.0 rad s⁻² for 5.0 s. Find its final ω and angular displacement.
Study the worked solution
  1. Set the initial condition

    Method

    ω₀ = 0.

    Reason

    The wheel starts from rest.

    Working

    ω₀ = 0
  2. Find angular speed

    Method

    ω = 10 rad s⁻¹.

    Reason

    Use the constant-α velocity relation.

    Working

    ω = 0 + (2.0)(5.0) = 10
  3. Find displacement

    Method

    Δθ = 25 rad.

    Reason

    Use Δθ = ω₀t + (1/2)α t².

    Working

    Δθ = (1/2)(2.0)(5.0)² = 25 rad

Guided practice 2

Converting between linear speed and angular speed

About 4 min

Problem

A disc rim at radius 0.20 m moves at 3.0 m s⁻¹. Find ω.

Try this before viewing the solution

Hints

Hint 1: link rim distance to angle
Use v = rω.
View solution step by step
  1. Rearrange

    Method

    ω = v/r.

    Reason

    Tangential speed increases with distance from the axis.

    Working

    v = rω
  2. Evaluate

    Method

    ω = 15 rad s⁻¹.

    Reason

    Divide the tangential speed by radius.

    Working

    ω = 3.0/0.20 = 15 rad s⁻¹

Common misconception 3

Eliminating time (the ω² equation)

Find and correct the mistake

Learner claim

A fan changes from 50 to 150 rad s⁻¹ at constant α = 100 rad s⁻². A learner substitutes angular quantities into v² = u² + 2as but reports metres. Explain the notation and unit errors, then find Δθ.

Try this before viewing the solution

Angular-displacement unit

View solution step by step
  1. Use the angular relation

    Method

    ω² = ω₀² + 2αΔθ.

    Reason

    Time is absent and all variables are angular.

    Working

    Δθ = (ω²-ω₀²)/(2α)
  2. Evaluate

    Method

    Δθ = 100 rad.

    Reason

    Substitute angular speed and acceleration consistently.

    Working

    (150²-50²)/[2(100)] = 100 rad

Examiner practice 4

Tangential acceleration of a point on the rim

3 marks

Examination question

A rim point is 0.30 m from the axis while α = 4.0 rad s⁻². Find its tangential acceleration. [3 marks]

Try this before viewing the solution

View solution step by step
  1. State relation

    1 mark

    Method

    aₜ = rα.

    Reason

    Angular acceleration maps to tangential acceleration through radius.

    Working

    aₜ = rα
  2. Substitute

    1 mark

    Method

    aₜ = (0.30)(4.0).

    Reason

    Use SI quantities.

    Working

    aₜ = 1.2
  3. Report

    1 mark

    Method

    aₜ = 1.2 m s⁻².

    Reason

    Tangential acceleration is linear.

    Working

    1.2 m s⁻²

Challenge 5

Stopping time (sign convention matters)

Minimal support

Independent transfer

A wheel initially rotates at + 12 rad s⁻¹ and stops uniformly in 3.0 s. Find signed α and interpret its sign.

Try this before viewing the solution

Hints

Hint 1: set the final state
At rest, final ω = 0.
View solution step by step
  1. Apply the velocity relation

    Method

    0 = 12 + 3.0α.

    Reason

    Use the initial rotation direction as positive.

    Working

    ω = ω₀ + α t
  2. Solve and interpret

    Method

    α = -4.0 rad s⁻².

    Reason

    The negative direction opposes the positive initial angular velocity.

    Working

    α = (0-12)/3.0 = -4.0 rad s⁻²

7. Mind Stretchers

Mind stretcher 1: Same ω, different vExtension

Two points A and B are on the same rigid body rotating about a fixed axis, with r_A = 0.10 m and r_B = 0.30 m.

If the body has angular speed ω, compare v_A and v_B.

Answer

Using v = rω: v_B/v_A = r_B/r_A = 0.30/0.10 = 3 So v_B = 3v_A even though both points share the same ω.

Mind stretcher 2: Tangential vs centripetal accelerationExtension

A wheel spins at constant angular speed ω (so α = 0). Does a point on the rim have zero acceleration? Explain.

Answer

No. If ω is constant, the tangential acceleration is zero (aₜ = rα = 0), but the point still has centripetal (radial) acceleration: aᵣ = rω² directed toward the centre, because the velocity direction is continuously changing.

8. Optional/Enrichment: Variable Angular Acceleration (Calculus Form)

If α is not constant, the constant-α equations do not apply. The safe starting points are: ω = dθ/dt, α = dω/dt and you integrate/differentiate according to what the question gives.

Next step

Return to the Rotational Motion hub, or continue in sequence to Moment of Inertia.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027