Mini Physics
Part of Mini EducationSwitch subject
Search

Rotational Kinetic Energy

Key idea: Derive and use rotational kinetic energy, connect it to moment of inertia, and practise exam-style calculations.

  • GCE A-Level H3 Physics 2027
On this page

Learning objectives

  • Apply Eₖ,rot = ½Iω² to the rotational kinetic energy of a rigid body.
  • Derive Eₖ,rot = ½Iω² for a rigid body from the equations of motion.
  • recall and apply the result that the motion of a rigid body can be regarded as translational motion of its centre of mass with rotational motion about an axis through the centre of mass to solve problems, including the use of F ⩽ µN for solid surfaces in no-slip contact (no distinction is made between the coefficient of static and kinetic friction)

Rotational kinetic energy is the energy of a rigid body due to its rotation. The H3 result is: E_(k,rot) = (1/2)Iω²

1. Definitions (Must Know)

  • Rotational kinetic energy, E_(k,rot) (J): kinetic energy due to rotation about an axis.
  • Moment of inertia, I (kg m²): depends on axis and mass distribution.
  • Angular speed, ω (rad s⁻¹).
  • Tangential speed at radius r: v = rω.
  • Symbols used in this lesson: E_(k,rot) (J), I (kg m²), ω (rad s⁻¹), r (m), v (m s⁻¹), M (kg), R (m).

2. Key Ideas (What Earns Marks)

  • A rotating rigid body’s total kinetic energy can be written as the sum over its parts: Eₖ = ∑ (1/2)mᵢvᵢ² and with vᵢ = rᵢω, this becomes 1/2 Iω².
  • I is about a specific axis; if the axis changes, the energy formula still works but you must use the correct I.
  • For rolling without slipping, total kinetic energy is split into translation + rotation: Eₖ = 1/2 Mv_cm² + 1/2 I_cmω² (details are in the Rolling Motion lesson).

3. Detailed Explanations

A. Derivation of E_(k,rot) = (1/2)Iω²

Model the rigid body as many small masses mᵢ at distances rᵢ from the axis. Each has speed: vᵢ = rᵢω

Total kinetic energy:

E_(k,rot) = ∑ 1/2 mᵢ vᵢ²; = ∑ 1/2 mᵢ (rᵢ²ω²); = 1/2 ω² ∑ mᵢ rᵢ²; = 1/2 Iω²

B. Interpreting the formula

  • If I is large (mass further from axis), the same ω stores more rotational energy.
  • For a given L = Iω, decreasing I increases ω and changes E_(k,rot); energy is not generally conserved unless the system is isolated and no work is done internally/externally.

4. Common Mistakes

  • Treating I as a universal constant for the object (it changes with axis).
  • Using degrees in ω (always use rad s⁻¹).
  • Forgetting that a rolling object has both translational and rotational kinetic energy.

5. Exam Tips

  • If the question gives ω, reach for 1/2 Iω² immediately.
  • If the question gives linear speed at radius R, convert using ω = v/R.
  • Unit check: Iω² has units kg m² s⁻² = J.

6. Worked Examples

Modelled example 1

Rotational kinetic energy of a disc

Core

Problem

A solid disc has M = 2.0 kg, R = 0.30 m and ω = 20 rad s⁻¹. Find its rotational kinetic energy.
Study the worked solution

  1. Find disc inertia

    Method

    I = 0.090 kg m².

    Reason

    A solid disc about its central axis has I = (1/2)MR².

    Working

    I = (1/2)(2.0)(0.30)² = 0.090

  2. Find energy

    Method

    E_(k,rot) = 18 J.

    Reason

    Use (1/2)Iω².

    Working

    (1/2)(0.090)(20)² = 18 J

Guided practice 2

Same M, R, ω: ring vs disc

About 4 min

Problem

A thin ring and solid disc have the same M, R and ω. Compare their rotational energies.

Try this before viewing the solution

Hints

Hint 1: compare their inertias first
Use I_ring = MR² and I_disc = (1/2)MR².
View solution step by step

  1. Write both energies

    Method

    E_ring = (1/2)MR²ω² and E_disc = (1/4)MR²ω².

    Reason

    E = (1/2)Iω².

    Working

    I_ring = 2I_disc

  2. Compare

    Method

    The ring stores twice the rotational energy.

    Reason

    Angular speeds are equal, so energy ratio equals inertia ratio.

    Working

    E_ring/E_disc = 2

Common misconception 3

Convert linear speed to rotational energy

Find and correct the mistake

Learner claim

A wheel rim moves at 6.0 m s⁻¹ at radius 0.20 m, with I = 0.50 kg m². A learner substitutes 6.0 directly for ω. Explain the unit error and find the energy.

Try this before viewing the solution

Required angular speed

View solution step by step

  1. Convert speed

    Method

    ω = 30 rad s⁻¹.

    Reason

    v = Rω.

    Working

    ω = 6.0/0.20 = 30

  2. Calculate energy

    Method

    E = 225 J.

    Reason

    Use angular speed in (1/2)Iω².

    Working

    E = (1/2)(0.50)(30)² = 225 J

Examiner practice 4

Using angular momentum to find energy

4 marks

Examination question

A flywheel has I = 2.0 kg m² and L = 10 kg m² s⁻¹. Find ω and rotational energy. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Use angular momentum

    1 mark

    Method

    ω = L/I.

    Reason

    L = Iω for fixed-axis rotation.

    Working

    ω = 10/2.0

  2. Find speed

    1 mark

    Method

    ω = 5.0 rad s⁻¹.

    Reason

    Divide momentum by inertia.

    Working

    5.0 rad s⁻¹

  3. Use energy

    1 mark

    Method

    E = (1/2)Iω².

    Reason

    Use the calculated angular speed.

    Working

    E = (1/2)(2.0)(5.0)²

  4. Report

    1 mark

    Method

    E = 25 J.

    Reason

    Equivalently L²/(2I) = 25 J.

    Working

    25 J

Challenge 5

Work done by a constant torque

Minimal support

Independent transfer

A wheel with I = 0.80 kg m² starts from rest. Net torque 2.0 N m acts through 6.0 rad. If all work becomes rotational energy, find final ω.

Try this before viewing the solution

Hints

Hint 1: link torque work to energy
Use W = τΔθ = Δ Eₖ.
View solution step by step

  1. Find work

    Method

    W = 12 J.

    Reason

    Constant torque does work τΔθ.

    Working

    (2.0)(6.0) = 12 J

  2. Set energy balance

    Method

    12 = (1/2)(0.80)ω².

    Reason

    Initial rotational energy is zero.

    Working

    W = Δ Eₖ

  3. Solve

    Method

    ω = 5.5 rad s⁻¹.

    Reason

    ω² = 30 and speed magnitude uses the positive root.

    Working

    ω = square root of 30 = 5.5

7. Mind Stretchers

Mind stretcher 1: Energy change when I changes (link to ice skater)Extension

If external torque is negligible, angular momentum L is conserved. If a skater pulls arms in, I decreases and ω increases.

Does E_(k,rot) increase, decrease, or stay the same? Explain.

Answer

It generally increases. With L = Iω constant, ω = L/I, so: E_(k,rot) = 1/2 Iω² = 1/2 I(L/I)² = L²/2I If I decreases, E_(k,rot) increases. The extra energy comes from work done by the skater’s muscles.

Mind stretcher 2: Same angular momentum, different energyExtension

Two objects have the same angular momentum magnitude L about their spin axis. Object A has moment of inertia I_A and object B has I_B, with I_A < I_B.

Which has larger rotational kinetic energy? Explain.

Answer

Using E_(k,rot) = L²/2I (from L = Iω), smaller I gives larger energy for the same L: I_A < I_B ⇒ L²/2I_A > L²/2I_B So object A has the larger rotational kinetic energy.

8. Optional/Enrichment: Work–Energy in Rotation

There is a rotational work relation (fixed axis): W = ∫ τ dθ For constant torque, W = τΔθ, and it matches the change in rotational kinetic energy. This is useful, but H3 questions often let you use 1/2 Iω² directly.

Next step

Return to the Rotational Motion hub, or combine translation and rotation in Rolling Motion.

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