PhysicsCBSE Class 11 · Systems of Particles and Rotational Motion

Torque, Moment of Inertia and Angular Momentum Class 11

Updated 17 October 2026 · Fact-checked

Torque τ = r × F = rF sin θ is the rotational effect of a force. For rotation about a fixed axis, τ = Iα, with I = Σmr². Angular momentum is L = r × p = Iω, and τ = dL/dt, so L is conserved when the net external torque is zero. Rotational kinetic energy is ½Iω², the work done by a torque is τθ and power is τω.

Only the part of the force perpendicular to the rod, F sin θ, produces torque; F cos θ points along the rod and does not turn it.

Illustration drawn for QuestionBuddy.

Understand Torque, Moment of Inertia and Angular Momentum

Torque. $τ = r \times F$ , with magnitude $r F sin θ = F \times$ (perpendicular distance of the line of action from the axis). The SI unit is N m. Equilibrium of a rigid body needs $\sum F = 0$ and $\sum τ = 0$ .

Moment of inertia. $I = \sum m_{i} r_{i}^{2}$ measures resistance to angular acceleration about a given axis; it depends on mass, its distribution and the axis. Standard values (no derivation needed): ring about its axis $M R^{2}$ ; uniform disc or solid cylinder about its axis $\frac{1}{2} M R^{2}$ ; solid sphere about a diameter $\frac{2}{5} M R^{2}$ ; thin hollow sphere $\frac{2}{3} M R^{2}$ ; rod of length $L$ about a perpendicular axis through its centre $\frac{1}{12} M L^{2}$ . The radius of gyration is $k = I / M$ .

Rotational analogues. Linear to rotational: $m \to I$ , $v \to ω$ , $a \to α$ , $F \to τ$ , $p \to L$ , $F = ma \to τ = I α$ , $K = \frac{1}{2} m v^{2} \to \frac{1}{2} I ω^{2}$ , $P = F v \to τ ω$ , $W = F s \to τ θ$ . Rotational kinematics for constant $α$ : $ω = ω_{0} + α t$ , $θ = ω_{0} t + \frac{1}{2} α t^{2}$ , $ω^{2} = ω_{0}^{2} + 2 α θ$ .

Angular momentum. For a particle $L = r \times p$ ; for a rigid body about a fixed axis $L = I ω$ . A particle moving in a straight line has angular momentum $m v d$ about a point at perpendicular distance $d$ from its path. The law $τ_{e x t} = \frac{d L}{d t}$ gives conservation of angular momentum when $τ_{e x t} = 0$ : $I_{1} ω_{1} = I_{2} ω_{2}$ . So a skater or diver spins faster by reducing $I$ , but kinetic energy then *increases* (internal work is done).

Connected systems. For a mass hanging from a rope wound on a pulley or drum, write $m g - T = ma$ for the mass and $T R = I α$ for the drum, with $a = R α$ .

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

Torque: $τ = r \times F, τ = r F sin θ$; N m.
Rotational Newton's law: $τ = I α$; Fixed axis.
Angular momentum: $L = r \times p, L = I ω$; kg m² s⁻¹.
Torque and L: $τ_{e x t} = \frac{d L}{d t}$; L conserved when zero.
Rotational KE, work, power: $K = \frac{1}{2} I ω^{2}, W = τ θ, P = τ ω$; Analogues.
Constant α kinematics: $ω = ω_{0} + α t, θ = ω_{0} t + \frac{1}{2} α t^{2}$; Same as linear.

How to solve Torque, Moment of Inertia and Angular Momentum questions

Choose the axis first; then write τ = Iα for each rotating part and F = ma for each translating part, linked by a = Rα.

1Fix the axis and compute I about it using the standard values.
2Find the torque of each force about the same axis (perpendicular distance × force).
3Use τ = Iα for the angular acceleration.
4Use rotational kinematics for ω, θ or time.
5For internal changes with no external torque, equate L before and after.
6Use K = ½Iω² and W = τθ for energy and work.

Torque, Moment of Inertia 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.

Opening a door

Suppose a 15 N push is applied at the handle, 0.8 m from the hinges, perpendicular to the door.

Find the torque, and the force needed at 0.2 m from the hinge for the same torque.

Try it first, then show the solutionSolution

τ = 15 × 0.8 = 12 N m.
F = 12/0.2 = 60 N.

Answer: 12 N m; 60 N.

Memory hook: The farther from the hinge, the smaller the force.

A spinning stool

Suppose a student on a frictionless stool with arms stretched has I = 5 kg m² and spins at 2 rad/s; pulling the arms in makes I = 2 kg m².

Find the new angular speed.

Try it first, then show the solutionSolution

I₁ω₁ = I₂ω₂, so ω₂ = 5 × 2/2 = 5 rad/s.

Answer: 5 rad/s.

Memory hook: Smaller I means larger ω.

Tricks to remember Torque, Moment of Inertia and Angular Momentum

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

Rotational ladder
τ = Iα, L = Iω, K = ½Iω², P = τω: replace m by I.
Use it when: Recalling formulae.
Perpendicular distance
Torque = force × perpendicular arm.
Use it when: Angled forces.
Arms in, spin up
I down, ω up, KE up.
Use it when: Skater and diver.
Straight line L
L = mvd for a straight-line mover about a point.
Use it when: Angular momentum of a particle.

Torque, Moment of Inertia and Angular Momentum: solved examples

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

Example 1: Torque on a wrench · 2 marks · Very short answerBoard level

A force of 20 N is applied perpendicular to a 0.3 m wrench. Find the torque, and the torque if the force acts at 30° to the wrench.

Try it first, then show the solutionSolution

τ = rF = 0.3 × 20 = 6 N m. (1)
At 30°: τ = rF sin 30° = 3 N m. (1)

Answer: 6 N m; 3 N m.

Example 2: Rotational kinematics of a fan · 3 marks · Short answerBoard level

A fan starting from rest speeds up uniformly to 300 rpm in 10 s. Find the angular acceleration and the number of revolutions made in the 10 s.

Try it first, then show the solutionSolution

ω = 300 rpm = 5 rev/s = 10π rad/s; α = 10π/10 = π ≈ 3.14 rad/s². (1)
θ = ½αt² = ½ × π × 100 = 50π ≈ 157 rad. (1)
Revolutions = 157/2π = 25. (1)

Answer: α = π rad/s² ≈ 3.14 rad/s²; 25 revolutions.

Example 3: Torque on a disc · 3 marks · Short answerHarder board level

A constant torque of 4 N m acts on a uniform disc of mass 2 kg and radius 0.5 m, initially at rest. Find the angular acceleration, the time to reach 80 rad/s, and the work done by the torque in that time.

Try it first, then show the solutionSolution

I = ½MR² = ½ × 2 × 0.25 = 0.25 kg m²; α = τ/I = 16 rad/s². (1)
t = 80/16 = 5 s. (1)
θ = ½ × 16 × 25 = 200 rad; W = τθ = 800 J = ½Iω² = ½ × 0.25 × 6400. (1)

Answer: 16 rad/s²; 5 s; 800 J.

Example 4: Mass hanging from a drum · 4 marks · Long answerHarder board level

A 2 kg bucket hangs from a light rope wound on a drum (a uniform disc) of mass 4 kg and radius 0.2 m which turns without friction (g = 10 m/s²). Find the acceleration of the bucket, the tension in the rope, and the time the bucket takes to fall 2.5 m from rest.

Try it first, then show the solutionSolution

Bucket: mg − T = ma. Drum: TR = Iα = ½MR² (a/R), so T = ½Ma = 2a. (1)
20 − 2a = 2a gives a = 5 m/s². (1)
T = 2 × 5 = 10 N (check: TR = 2.0 N m = Iα = 0.08 × 25). (1)
t = √(2 × 2.5/5) = 1 s. (1)

Answer: a = 5 m/s²; T = 10 N; t = 1 s.

Example 5: Man walking on a turntable · 4 marks · Long answerHOTS / competency

A turntable of moment of inertia 300 kg m² turns freely at 1 rad/s with a 60 kg man standing 2 m from the axis (treat him as a point mass). He walks to the centre. Find the new angular speed, the change in rotational kinetic energy and explain where the energy comes from.

Try it first, then show the solutionSolution

I₁ = 300 + 60 × 2² = 540 kg m²; L = 540 × 1 = 540. (1)
I₂ = 300, so ω₂ = 540/300 = 1.8 rad/s. (1)
K₁ = ½ × 540 × 1 = 270 J; K₂ = ½ × 300 × 1.8² = 486 J; increase = 216 J. (1)
The extra energy is supplied by the man's own muscular (internal) work as he moves inward; L is unchanged. (1)

Answer: 1.8 rad/s; +216 J, from the man's internal work.

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

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

Mistake: Using the distance from the point of application instead of the perpendicular distance.
Forgetting sin θ.
Do this: τ = rF sin θ.
Usually costs: The answer mark.
Mistake: Using the wrong I for the shape.
Mixing ring and disc values.
Do this: Ring MR², disc ½MR², solid sphere 2/5 MR².
Usually costs: 1 to 2 marks.
Mistake: Conserving kinetic energy when angular momentum is conserved.
Mixing two laws.
Do this: Conserve L; K changes if internal work is done.
Usually costs: Reasoning mark.
Mistake: Converting rpm incorrectly.
Forgetting 2π/60.
Do this: rpm × 2π/60 = rad/s.
Usually costs: Accuracy mark.

Exam tips

State the axis about which I and τ are taken.
Write the rotational analogue before substituting.
Convert rpm to rad/s first.
In conservation questions state "no external torque".
Mention the source of energy change if K is not conserved.

Practice questions on Torque, Moment of Inertia 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
A solid sphere and a thin hollow sphere have the same mass and radius. The ratio of their moments of inertia about a diameter (hollow : solid) is
(a) 5 : 3 (b) 3 : 5 (c) 2 : 3 (d) 1 : 1
Hint
(2/3)/(2/5).
Answer
(a) 5 : 3.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A particle moving in a straight line can have angular momentum about a point not on its path.
Reason (R): Angular momentum about a point is r × p, which is non-zero when the line of motion does not pass through the point.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
L = mvd.
Answer
(a).
Q3 · 3 marks · Short answerHarder board level
A wheel of moment of inertia 2 kg m² rotating at 120 rpm is brought to rest in 10 s by a constant torque. Find the torque and the number of revolutions it turns before stopping.
Hint
ω = 4π rad/s.
Answer
α = 4π/10 = 1.257 rad/s²; τ = Iα = 2.51 N m; θ = ½ × 4π × 10 = 20π rad = 10 revolutions.
Q4 · 3 marks · Short answerHarder board level
A particle of mass 2 kg has position r = 3i + 4j m and velocity v = 2j m/s. Find its angular momentum about the origin.
Hint
L = m(r × v).
Answer
r × v = 3 × 2 k = 6 k, so L = 12 k kg m²/s.
Q5 · 4 marks · Long answerHOTS / competency
A uniform solid cylinder (mass 10 kg, radius 0.2 m) at rest is turned by a constant torque of 5 N m for 4 s. Find (a) the angular speed reached, (b) the kinetic energy and the work done by the torque, and (c) the power at t = 4 s. A brake then acts at the rim with a friction force of 20 N; find the time to stop.
Hint
I = ½MR² = 0.2 kg m².
Answer
(a) α = 5/0.2 = 25 rad/s²; ω = 100 rad/s. (b) K = ½ × 0.2 × 10⁴ = 1000 J; θ = 200 rad, W = 5 × 200 = 1000 J. (c) P = τω = 500 W. Braking torque = 20 × 0.2 = 4 N m, α = −20 rad/s², t = 100/20 = 5 s.

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.Torque?Show answer
r × F = rF sin θ.
2.Moment of inertia of a ring?Show answer
MR².
3.Moment of inertia of a disc about its axis?Show answer
½MR².
4.Angular momentum of a rigid body?Show answer
Iω.
5.Rotational power?Show answer
τω.
6.When is L conserved?Show answer
When the net external torque is zero.

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

Why does a skater spin faster with arms in?

I decreases, and with L = Iω constant, ω increases.

Is kinetic energy conserved when a skater pulls her arms in?

No; it increases because the skater does internal work.

What is the radius of gyration?

The distance from the axis at which the whole mass could be placed to have the same I: k = √(I/M).

Why is a longer spanner easier to use?

A larger distance gives a larger torque for the same force.

Does I depend on the axis?

Yes, the same body has different I about different axes.