Dynamics of Angular Motion

Key idea: Learn torque, moment of inertia, and the rotational form of Newton’s second law, with exam tips and worked examples.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • show an understanding of and use the terms angular momentum and moment of inertia of a rotating rigid body
  • calculate the moment of inertia about an axis for simple bodies by using calculus, the parallel-axis theorem or otherwise (knowledge of the perpendicular-axis theorem and mathematical derivation of the moment of inertia for spheres are not required)
  • show an understanding of torque produced by a force relative to a reference point, and apply the principle that torque is related to the rate of change of angular momentum to solve problems, such as those involving point masses, rigid bodies, or bodies with a variable moment of inertia e.g. an ice-skater

Dynamics of angular motion is the forces side of rotation: external torque changes angular momentum. For a rigid body rotating about a fixed axis with constant moment of inertia, the working equation is: ∑τₑₓₜ = Iα

1. Definitions (Must Know)

  • Torque (moment of a force), τ (N m): turning effect of a force about a chosen point/axis.
    • Magnitude: τ = rF sin φ = Fd_⊥ where r is distance from axis to point of application, φ is the angle between vector r and vector F, and d_⊥ is the perpendicular moment arm.
  • Net external torque, ∑τₑₓₜ: the signed sum of external torques about one chosen axis.
  • Moment of inertia, I (kg m²): rotational “inertia” about an axis: I = ∑ mᵢrᵢ² for a set of discrete masses.
  • Fixed-axis, constant-I form: ∑τₑₓₜ = Iα
  • Symbols used in this lesson: τ torque (N m), F force (N), r radius (m), d_⊥ moment arm (m), φ angle (rad), I moment of inertia (kg m²), α angular acceleration (rad s⁻²).

2. Key Ideas (What Earns Marks)

  • Always state the axis (or pivot) you’re taking moments about; torque depends on it.
  • Use the perpendicular distance: τ = Fd_⊥ is often the safest form.
  • I depends on both the mass distribution and the chosen axis.
  • For a rigid body about a fixed axis: bigger I means smaller α for the same τₙₑₜ.

Torque formulas (choose the one that matches the diagram):

SituationSafe formNotes
You know angle to radiusτ = rF sin φUse φ between vector r and vector F
You know moment armτ = Fd_⊥d_⊥ is perpendicular distance to the line of action
A force acts at the end of a position vector from a pivot; its torque equals the force multiplied by the perpendicular distance to the line of action.
Torque geometry is defined about a chosen pivot: use either rF sin φ or the perpendicular moment arm Fd⊥.

3. Detailed Explanations

A. Torque: what matters physically

For a given force magnitude F:

  • torque increases if you apply it further from the axis (bigger r),
  • torque increases if the force is more perpendicular to the radius (bigger sin φ),
  • torque is zero if the line of action passes through the axis (d_⊥ = 0).

B. Obtaining ∑τₑₓₜ = Iα

For a point mass m constrained at distance r from the axis, tangential acceleration is aₜ = rα. Its tangential resultant force is: Fₜ = maₜ = mrα The torque about the axis from this tangential force is: τ = rFₜ = r(mrα) = mr²α

For a rigid body, the general starting point is ∑ vector τₑₓₜ = d vector L/dt. When the body rotates about a fixed axis and I about that axis is constant, L = Iω, so:

∑τₑₓₜ = d(Iω)/dt = Iα

This form is not valid unchanged when the rotation axis or moment of inertia varies; use the angular-momentum relation instead.

C. Choosing signs (1D rotation)

Pick a sign convention (e.g. anticlockwise is +). Then:

  • torques that tend to rotate the body anticlockwise are positive,
  • clockwise torques are negative, and you add them to get τₙₑₜ.

4. Common Mistakes

  • Using rF when the force is not perpendicular (should be rF sin φ or Fd_⊥).
  • Measuring r to the wrong point (use the distance from axis to where the force acts).
  • Mixing torques about different axes in the same equation.
  • Writing τ = Iω (incorrect; for a fixed axis and constant I, use ∑τₑₓₜ = Iα).

5. Exam Tips

  • Draw the axis/pivot and mark d_⊥; it prevents many sign/geometry errors.
  • In multi-force problems, compute each torque separately with its sign, then sum.
  • Unit check:
    • I in kg m²,
    • α in rad s⁻²,
    • τ in N m.

6. Worked Examples

Modelled example 1

Torque from a force at an angle

Core

Problem

A 20 N force acts 0.25 m from a bolt at 60° to the spanner. Find torque magnitude.
Study the worked solution
  1. Select the perpendicular component

    Method

    τ = rF sin φ.

    Reason

    Only force perpendicular to the radius produces torque.

    Working

    F_⊥ = F sin 60°
  2. Evaluate

    Method

    τ = 4.3 N m.

    Reason

    Multiply moment arm by perpendicular force.

    Working

    (0.25)(20) sin 60° = 4.3 N m

Guided practice 2

Using τₙₑₜ = Iα

About 4 min

Problem

A disc has I = 0.50 kg m² and experiences net torque 2.0 N m. Find α.

Try this before viewing the solution

Hints

Hint 1: rotational second law
Use τₙₑₜ = Iα.
View solution step by step
  1. Rearrange

    Method

    α = τₙₑₜ/I.

    Reason

    Moment of inertia is rotational resistance to angular acceleration.

    Working

    τₙₑₜ = Iα
  2. Evaluate

    Method

    α = 4.0 rad s⁻².

    Reason

    Divide torque by inertia.

    Working

    2.0/0.50 = 4.0 rad s⁻²

Common misconception 3

Two torques, one axis

Find and correct the mistake

Learner claim

Torques + 6.0 and -2.0 N m act on a body with I = 2.0 kg m². A learner adds magnitudes to get 8.0 N m. Explain the sign error and find α.

Try this before viewing the solution

Net torque

View solution step by step
  1. Sum signed torques

    Method

    τₙₑₜ = +4.0 N m.

    Reason

    The torques act in opposite senses.

    Working

    6.0 + (-2.0) = 4.0
  2. Find acceleration

    Method

    α = +2.0 rad s⁻².

    Reason

    α = τₙₑₜ/I retains the net-torque sign.

    Working

    4.0/2.0 = 2.0 rad s⁻²

Examiner practice 4

Force at a handle (moment arm form)

4 marks

Examination question

An 18 N perpendicular force acts 0.80 m from a door hinge. The door has I = 3.6 kg m². Find α. [4 marks]

Try this before viewing the solution

View solution step by step
  1. Identify moment arm

    1 mark

    Method

    d_⊥ = 0.80 m.

    Reason

    The force is perpendicular.

    Working

    τ = Fd_⊥
  2. Calculate torque

    1 mark

    Method

    τ = 14.4 N m.

    Reason

    Multiply force by perpendicular distance.

    Working

    (18)(0.80) = 14.4
  3. Apply dynamics

    1 mark

    Method

    α = τ/I.

    Reason

    Net torque produces angular acceleration.

    Working

    τ = Iα
  4. Report

    1 mark

    Method

    α = 4.0 rad s⁻².

    Reason

    14.4/3.6 = 4.0.

    Working

    4.0 rad s⁻²

Challenge 5

Multiple forces about a pivot (signs)

Minimal support

Independent transfer

A 10 N force at 0.30 m acts anticlockwise; a 6.0 N force at 0.50 m acts clockwise. For I = 0.80 kg m², find α.

Try this before viewing the solution

Hints

Hint 1: assign rotational signs first
Take anticlockwise positive before calculating each torque.
View solution step by step
  1. Calculate both torques

    Method

    τ₁ = +3.0 and τ₂ = -3.0 N m.

    Reason

    They act in opposite senses.

    Working

    τ₁ = (10)(0.30), τ₂ = -(6.0)(0.50)
  2. Find net torque

    Method

    τₙₑₜ = 0.

    Reason

    The equal opposing torques cancel.

    Working

    + 3.0-3.0 = 0
  3. Infer acceleration

    Method

    α = 0.

    Reason

    α = τₙₑₜ/I.

    Working

    0/0.80 = 0

7. Mind Stretchers

Mind stretcher 1: Same torque, different axisExtension

You apply the same force at the same point on a rigid body, but you choose a different pivot point.

Does the torque change? Explain briefly.

Answer

Yes. Torque depends on the perpendicular distance from the chosen axis/pivot to the line of action of the force, so changing the pivot generally changes d_⊥ and hence τ.

Mind stretcher 2: Can you have motion with zero net torque?Extension

If τₙₑₜ = 0 about an axis, does that guarantee the object is not rotating? Explain.

Answer

No. τₙₑₜ = 0 implies α = 0 (for fixed-axis rotation), so the angular velocity is constant. The object could be at rest (ω = 0) or rotating at a steady rate (ω ≠ 0).

8. Optional/Enrichment: Beyond Fixed-Axis Rotation

If the rotation axis changes direction (3D rigid body rotation) or the frame is non-inertial, you need more advanced tools (e.g. Euler’s equations / rotating-frame dynamics). This is beyond the fixed-axis H3 scope unless explicitly guided by the question.

Next step

Return to the Rotational Motion hub, or continue to the more general Torque and Angular Momentum relation.

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