PhysicsISC Class 11 · System of Particles and Rotational Motion

Moment of Inertia, Torque and Angular Momentum: Rotational Dynamics

Updated 17 October 2026 · Fact-checked

The moment of inertia I = Σmr² measures resistance to a change in rotation. Torque τ = Iα is the rotational form of Newton's second law. Rotational kinetic energy is ½Iω² and angular momentum L = Iω, which is conserved when the net external torque is zero.

With no external torque, Iω is constant, so pulling the masses in lowers I and the spin rate ω rises.

Illustration drawn for QuestionBuddy.

Understand Moment of Inertia, Torque and Angular Momentum

Moment of inertia. I = Σmᵢrᵢ², with r measured from the axis. It depends on the mass, how it is distributed and the axis chosen. Standard values: ring about its axis MR²; solid disc or cylinder about its axis ½MR²; solid sphere about a diameter (2/5)MR²; thin rod about its centre, perpendicular to it, ML²/12. The radius of gyration k is defined by I = Mk².

Axis theorems (statements). Parallel axes: I = I_cm + Md², where d is the distance between the parallel axes. For a rod, about its end I = ML²/12 + M(L/2)² = ML²/3. Perpendicular axes (for plane laminas): I_z = I_x + I_y. For a ring, I_diameter = ½MR² by symmetry.

Rotational dynamics. τ = Iα, ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ. Work done by a torque is τθ, and power is τω. Rotational kinetic energy is ½Iω².

Angular momentum. L = Iω (for a rigid body) and τ = dL/dt. If τ_ext = 0, L is constant: Iω is constant, so when I decreases, ω increases (a skater pulling in the arms), and the kinetic energy increases as the skater does work.

Rolling without slipping. v = Rω. Total K = ½Mv² + ½Iω² = ½Mv²(1 + k²/R²). A body rolling down an incline of height h reaches the foot with v² = 2gh/(1 + k²/R²), so a solid cylinder (k²/R² = ½) is slower than a sliding block.

Moment of Inertia, Torque and Angular Momentum: formulas and key facts

Moment of inertia: $I = Σ m r^{2} ∣ I = M k^{2}$; k = radius of gyration.
Standard bodies: Ring MR²
Disc MR²/2
Solid sphere 2MR²/5
Rod (centre) ML²/12; Axes: through the centre, perpendicular to the plane for ring and disc; a diameter for the sphere.
Theorems: $I = I_{c} m + M d^{2} ∣ I_{z} = I_{x} + I_{y}$; Parallel and perpendicular axes.
Rotational laws: $τ = I α ∣ K = ½ I ω^{2} ∣ L = I ω$; L is conserved if τ_ext = 0.
Rolling down an incline: $v^{2} = 2 g h / (1 + k^{2} / R^{2})$; Without slipping, from rest at height h.

Ring	MR²
Disc	MR²/2
Solid sphere	2MR²/5
Rod (centre)	ML²/12

Moment of Inertia, Torque and Angular Momentum in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A skater pulls in her arms

Suppose a skater spinning at 2 rad s⁻¹ has a moment of inertia 4 kg m² with arms out and 1.6 kg m² with arms pulled in.

Find her new angular speed and the work she does.

Try it first, then show the solutionSolution

L is conserved: 4 × 2 = 1.6 × ω₂ → ω₂ = 5 rad s⁻¹.
K₁ = ½ × 4 × 4 = 8 J; K₂ = ½ × 1.6 × 25 = 20 J.
Work done = 20 − 8 = 12 J.

Answer: 5 rad s⁻¹; 12 J of work.

Memory hook: Smaller I, larger ω; the extra energy comes from her muscles.

Tricks to remember Moment of Inertia, Torque and Angular Momentum

Short memory aids. Say each one aloud once, then use it on a question the same day.

Mass far from axis, larger I
A ring has more I than a disc of the same mass and radius.
Use it when: Comparing bodies and the order of arrival down an incline.
L conserved, so Iω constant
Whenever I changes with no outside torque, ω changes inversely.
Use it when: Skater, diver and merry-go-round problems.

Moment of Inertia, Torque and Angular Momentum: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Moment of inertia of a ring · 1 mark · MCQBoard level

The moment of inertia of a ring of mass M and radius R about an axis through its centre and perpendicular to its plane is:
(a) MR²/2
(b) MR²
(c) 2MR²/5
(d) 2MR²

Try it first, then show the solutionSolution

All the mass is at distance R from the axis, so I = MR².

Answer: (b) MR²

Example 2: Torque on a turbine wheel · 3 marks · Short answerBoard level

A wheel of moment of inertia 3 kg m² starts from rest under a steady torque of 12 N m. Find the angular acceleration, the angular velocity after 5 s and the kinetic energy then.

Try it first, then show the solutionSolution

α = τ/I = 12/3 = 4 rad s⁻². (1 mark)
ω = αt = 4 × 5 = 20 rad s⁻¹. (1 mark)
K = ½Iω² = ½ × 3 × 400 = 600 J. (1 mark)

Answer: α = 4 rad s⁻²; ω = 20 rad s⁻¹; K = 600 J.

Example 3: Parallel axis theorem for a bar · 3 marks · Short answerHarder board level

A uniform bar of mass 4 kg and length 3 m lies in a workshop. Find its moment of inertia about (i) an axis through the centre perpendicular to its length and (ii) a parallel axis through one end. Verify with ML²/3.

Try it first, then show the solutionSolution

I_cm = ML²/12 = 4 × 9/12 = 3 kg m². (1 mark)
I_end = I_cm + Md² = 3 + 4 × 1.5² = 3 + 9 = 12 kg m². (1 mark)
Check: ML²/3 = 4 × 9/3 = 12 kg m². (1 mark)

Answer: I_cm = 3 kg m²; I_end = 12 kg m².

Example 4: Pulley and falling mass · 4 marks · Short answerHarder board level

A string wrapped round a disc-shaped pulley of mass 4 kg and radius 0.1 m carries a 2 kg mass at its end (g = 10 m s⁻²). The mass is released. Find its acceleration, the tension and the angular acceleration of the pulley.

Try it first, then show the solutionSolution

Mass: mg − T = ma. Pulley: TR = Iα = ½MR² (a/R), so T = ½Ma. (1 mark)
20 − T = 2a and T = 2a → 20 = 4a. (1 mark)
a = 5 m s⁻²; T = 10 N. (1 mark)
α = a/R = 5/0.1 = 50 rad s⁻². (1 mark)

Answer: a = 5 m s⁻²; T = 10 N; α = 50 rad s⁻².

Example 5: Rolling cylinder · 5 marks · Long answerHOTS / competency

A solid cylinder of mass 2 kg and radius 0.2 m rolls without slipping from rest down an incline, falling through a height of 1.2 m (g = 10 m s⁻²). Find its speed at the bottom and the fractions of its energy that are translational and rotational. Compare with a block sliding without friction.

Try it first, then show the solutionSolution

For a solid cylinder k²/R² = ½, so v² = 2gh/(1 + ½) = 2 × 10 × 1.2/1.5 = 16. (2 marks)
v = 4 m s⁻¹. (1 mark)
K_trans = ½ × 2 × 16 = 16 J; K_rot = ½ × (½ × 2 × 0.04) × (4/0.2)² = 8 J; total 24 J = mgh = 24 J. Fractions 2/3 and 1/3. (1 mark)
A frictionless block: v = √(2gh) = √24 = 4.90 m s⁻¹, faster than the cylinder. (1 mark)

Answer: v = 4 m s⁻¹; translational 2/3, rotational 1/3; the block would reach 4.9 m s⁻¹.

Where students lose marks in Moment of Inertia, Torque and Angular Momentum

The slips examiners see most often, and what to write instead.

Mistake: Using the wrong axis for I
Using MR²/2 for a ring about its axis, or ML²/12 for a rod about its end.
Do this: Quote the standard result for the stated axis; use the parallel axis theorem to shift it.
Usually costs: 1 to 2 marks.
Mistake: Forgetting rotational energy in rolling
Equating mgh to ½Mv² only.
Do this: Rolling without slipping: mgh = ½Mv² + ½Iω².
Usually costs: 2 marks.
Mistake: Conserving energy instead of angular momentum
Using K constant when the skater changes I.
Do this: Kinetic energy is not conserved; angular momentum is.
Usually costs: 1 to 2 marks.

Exam tips

State the axis for every moment of inertia you use.
Write τ = Iα for the pulley and F = ma for the hanging mass, then link a = Rα.
Use L = Iω conservation whenever I changes with no external torque.
Give units: kg m², N m, rad s⁻¹.

Practice questions on Moment of Inertia, Torque and Angular Momentum

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If no external torque acts on a rotating body, which quantity remains constant?
(a) angular velocity
(b) kinetic energy
(c) angular momentum
(d) moment of inertia
Hint
τ = dL/dt.
Answer
(c) angular momentum
Q2 · 2 marks · Short answerBoard level
A uniform disc of mass 4 kg and radius 0.5 m rotates about its axis at 10 rad s⁻¹. Find its moment of inertia and angular momentum.
Hint
I = ½MR².
Answer
I = ½ × 4 × 0.25 = 0.5 kg m². L = Iω = 5 kg m² s⁻¹.
Q3 · 3 marks · Short answerHarder board level
A wheel of moment of inertia 0.5 kg m² speeds up from 20 rad s⁻¹ to 40 rad s⁻¹ in 5 s under a constant torque. Find the torque and the work done.
Hint
τ = Iα; work = change in K.
Answer
α = 4 rad s⁻²; τ = 2 N m. Work = ½ × 0.5 × (1600 − 400) = 300 J.
Q4 · 3 marks · Short answerHarder board level
A ring of mass 2 kg and radius 0.3 m is considered. Find I about a diameter using the perpendicular axes theorem, and then about a tangent in its plane.
Hint
I_z = I_x + I_y with I_x = I_y; then the parallel axis theorem.
Answer
I_z = MR² = 0.18, so I_diameter = 0.09 kg m². Tangent: 0.09 + 2 × 0.09 = 0.27 kg m².
Q5 · 3 marks · Short answerHarder board level
A student sits on a stool rotating at 3 rad s⁻¹ with arms out (I = 5 kg m²). With arms in, I = 3 kg m². Find the new angular speed and the change in kinetic energy.
Hint
Conserve L.
Answer
ω₂ = 5 × 3/3 = 5 rad s⁻¹. K₁ = ½ × 5 × 9 = 22.5 J; K₂ = ½ × 3 × 25 = 37.5 J; increase 15 J.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a playground merry-go-round (a disc of mass 200 kg and radius 2 m) turns at 2 rad s⁻¹. A 40 kg child at rest jumps on at the rim.
(i) Find the moment of inertia of the system after she lands. (1 mark)
(ii) Find the new angular speed. (1 mark)
(iii) Find the loss of kinetic energy. (1 mark)
(iv) What happens to ω if she then walks to the centre? (1 mark)
Hint
The child adds mr² to the moment of inertia.
Answer
(i) I₁ = ½ × 200 × 4 = 400; I = 400 + 40 × 4 = 560 kg m². (ii) ω = 400 × 2/560 = 1.43 rad s⁻¹. (iii) K₁ = 800 J; K₂ = ½ × 560 × (1.4286)² = 571 J; loss 229 J. (iv) At the centre her contribution to I is about zero, so ω returns to 2 rad s⁻¹.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the parallel axis theorem.Show answer
I = I_cm + Md².
2.I of a solid disc about its axis?Show answer
½MR².
3.Rotational form of Newton's second law?Show answer
τ = Iα.
4.Condition for conservation of angular momentum?Show answer
Net external torque is zero.

Moment of Inertia, Torque and Angular Momentum: frequently asked questions

Why does a ring roll down slower than a disc?

A ring has all its mass at the rim, so a larger share of the energy goes into rotation (k²/R² = 1 against ½), leaving less for translation.

Is moment of inertia a fixed property of a body?

No. It depends on the axis. The same body has different moments of inertia about different axes.

Does the perpendicular axes theorem apply to a solid?

No. It holds only for planar (lamina-type) bodies.