IB DP Physics: Rotational dynamics: torque, moment of inertia and angular momentum (DP Physics)

This guide teaches sub-topic A.4, Rigid body mechanics (torque, moment of inertia and angular momentum), from Topic A of the IB Diploma Programme Physics guide, first assessment 2025. A.4 is additional higher level (HL only): the guide lists no standard level content here, and the sub-topic is allotted 7 hours. If you take Physics at SL, you can skip it. Everything below is HL only.

Use the IB Physics hub to see the other sub-topics and the printable checklist to tick off each statement. Two companion pages follow this one: the revision notes and the practice questions. For the linear ideas that rotation copies, see forces and momentum and kinematics. To find out where you stand first, try the free 10-minute diagnostics.

What A.4 covers

Guide statement	What you must be able to do	Tier
Torque, τ = Fr sin θ	Find a force’s turning effect about an axis, with its clockwise or counter-clockwise sense	HL only
Rotational equilibrium	Use zero resultant torque to find unknown forces or positions	HL only
Unbalanced torque	Link a resultant torque on a rigid body to angular acceleration	HL only
Angular displacement, velocity, acceleration	Describe rotation with θ, ω and α	HL only
Equations for uniform angular acceleration	Use the four equations	HL only
Moment of inertia, I = Σmr²	Find I for point masses; explain how mass distribution matters	HL only
Newton’s second law for rotation, τ = Iα	Solve problems with a resultant torque	HL only
Angular momentum, L = Iω	Calculate L and use conservation of L	HL only
Angular impulse, ΔL = τΔt = Δ(Iω)	Link torque acting over time to a change in L	HL only
Rotational kinetic energy, Ek = ½Iω² = L²/2I	Calculate and compare energies	HL only

The guidance adds limits. The vector direction of a torque is not needed, but you must state whether it is clockwise or counter-clockwise. You do not calculate a centre of mass. The moment of inertia for a specific shape will be given when needed. Combined rotation and translation is limited to rolling without slipping. Expect situations where I changes and where two bodies are coupled.

Torque and rotational equilibrium

A force changes how a body turns only if its line of action misses the axis. The torque of a force F about an axis is

τ = Fr sin θ

where r is the distance from the axis to the point where the force acts, and θ is the angle between the force and the line from the axis to that point. Only the part of the force at right angles to that line, F sin θ, turns the body. Torque is measured in N m.

Sense. Choose clockwise or counter-clockwise as positive. A torque the other way is negative.

Centre of mass. When you treat the whole of a body’s weight as one force, it acts at the centre of mass. For a uniform beam that is the midpoint. You are not asked to calculate its position.

Rotational equilibrium. A body in rotational equilibrium has zero resultant torque about any axis. For a body at rest, the forces must also sum to zero. If the torques do not cancel, the unbalanced torque gives a rigid body an angular acceleration.

Worked example 1. A force of 18 N acts at 0.75 m from a hinge, at 40° to the line joining the hinge to the point of application. Find the torque.

τ = Fr sin θ = 18 × 0.75 × sin 40°
  = 8.68 N m  (3 s.f.)

Worked example 2. A uniform beam of length 5.0 m and mass 30 kg is balanced horizontally on a support placed 1.0 m in from its left end, with a block of mass m hanging at the left end. Find m, and the force from the pivot. Take g = 9.81 N kg⁻¹.

The beam’s weight acts at its midpoint, 2.5 m from the left end, so 1.5 m from the pivot. Take torques about the pivot.

counter-clockwise (block):  m × 9.81 × 1.0
clockwise (beam):           30 × 9.81 × 1.5
equilibrium: m × 1.0 = 30 × 1.5, so m = 45 kg
pivot force = (45 + 30) × 9.81 = 736 N upward

Taking torques about the pivot removed the unknown pivot force from the equation.

Describing rotation: angle, speed and acceleration

The guide describes rotation using angular displacement Δθ (in radians), angular velocity ω and angular acceleration α. Angular speed is used rather than a formal vector treatment of angular velocity, so you need magnitudes and a sense.

• ω = Δθ/Δt, in rad s⁻¹.
• α = Δω/Δt, in rad s⁻².
• For a point at distance r from the axis, its speed is v = ωr. This link is needed later for rolling.

One full turn is 2π rad. Convert revolutions and degrees to radians before using any equation here.

For uniform angular acceleration the four equations mirror the linear ones:

Rotational	Linear counterpart
Δθ = ((ωf + ωi)/2) t	s = ((v + u)/2) t
ωf = ωi + αt	v = u + at
Δθ = ωi t + ½αt²	s = ut + ½at²
ωf² = ωi² + 2αΔθ	v² = u² + 2as

Worked example 3. A flywheel turning at 12 rad s⁻¹ speeds up uniformly at 3.5 rad s⁻² for 8.0 s. Find the final angular speed and the number of turns.

ωf = 12 + 3.5 × 8.0 = 40 rad s⁻¹
Δθ = 12 × 8.0 + ½ × 3.5 × 8.0² = 96 + 112 = 208 rad
turns = 208 / 2π = 33.1
check: ωf² = 144 + 2 × 3.5 × 208 = 1600, and 40² = 1600

Moment of inertia

The moment of inertia I is the rotational counterpart of mass: it sets the resistance to any change in a body’s spin about a chosen axis. It depends on the mass and on how that mass is distributed about the axis: a kilogram placed far out contributes much more than one placed close in. For a system of point masses

I = Σmr²

where each r is the perpendicular distance from the axis. Unit: kg m². The same body has different values of I for different axes.

Worked example 4. Masses of 0.30 kg, 0.50 kg and 0.20 kg are fixed to a light rod at 0.40 m, 0.25 m and 0.80 m from an axis. Find I.

I = 0.30 × 0.40² + 0.50 × 0.25² + 0.20 × 0.80²
  = 0.048 + 0.03125 + 0.128 = 0.207 kg m²

Same masses, different axis. Two 0.60 kg masses sit at the ends of a light rod 0.80 m long. About the middle, I = 2 × 0.60 × 0.40² = 0.192 kg m². About one mass, I = 0.60 × 0.80² = 0.384 kg m². Moving the axis doubled I.

For extended bodies such as discs, rods and spheres, the question will supply the formula for I. Do not rely on remembering one.

Newton’s second law for rotation

The rotational form of Newton’s second law is

τ = Iα

where τ is the resultant torque (the average torque if it varies) and I is about the same axis. For the same torque, doubling I halves α.

Worked example 5. A cylinder of mass 4.0 kg and radius 0.25 m has I = ½MR² (given). A string wound around it is pulled with a steady tangential force of 6.0 N. Find α.

I = ½ × 4.0 × 0.25² = 0.125 kg m²
τ = 6.0 × 0.25 = 1.5 N m
α = τ/I = 1.5 / 0.125 = 12 rad s⁻²

Coupled bodies. When a hanging mass turns a pulley through a string that does not slip, the string links the two motions: a = αR. Apply F = ma to the mass and τ = Iα to the pulley, and join them with the string tension.

Worked example 6. A 0.50 kg mass hangs from a string wound on a disc pulley of mass 3.0 kg and radius 0.10 m, with I = ½MR² (given). Released from rest, find the acceleration of the mass and the tension.

I = ½ × 3.0 × 0.10² = 0.015 kg m²
mass:   mg − T = ma
pulley: T R = I α = I a / R, so T = (I/R²) a = 1.5 a
add:    mg = (m + 1.5) a, so a = 0.50 × 9.81 / 2.0 = 2.45 m s⁻²
T = 1.5 × 2.4525 = 3.68 N

The tension is less than mg = 4.91 N, which is why the mass accelerates.

Angular momentum and its conservation

A body turning with angular speed ω has angular momentum

L = Iω

in kg m² s⁻¹. Angular momentum stays constant unless a resultant torque acts. This applies to a system of bodies: if no external resultant torque acts, the total L before equals the total L after.

Worked example 7: coupling to a second body. A disc with I = 0.060 kg m² spins at 15 rad s⁻¹. A 0.20 kg lump of clay is dropped onto it at 0.15 m from the axis and sticks. Find the new angular speed and the change in kinetic energy.

I_final = 0.060 + 0.20 × 0.15² = 0.0645 kg m²
0.060 × 15 = 0.0645 × ω, so ω = 14.0 rad s⁻¹
Ek before = ½ × 0.060 × 15² = 6.75 J
Ek after  = ½ × 0.0645 × 13.95² = 6.28 J

L is conserved but kinetic energy is not. About 0.47 J is transferred to other forms during the sticking, as in an inelastic collision.

Worked example 8: changing I. A person on a freely turning stool holds weights out, with I = 5.2 kg m² and ω = 1.5 rad s⁻¹. Pulling the weights in reduces I to 3.4 kg m².

ω = 5.2 × 1.5 / 3.4 = 2.29 rad s⁻¹
Ek before = ½ × 5.2 × 1.5² = 5.85 J
Ek after  = ½ × 3.4 × 2.294² = 8.95 J

The kinetic energy rises because the person does work pulling the weights inwards.

Angular impulse and rotational energy

A resultant torque acting for a time Δt produces an angular impulse, which equals the change in angular momentum:

ΔL = τΔt = Δ(Iω)

This is the rotational version of impulse equals change in momentum.

The kinetic energy of rotation is

Ek = ½Iω² = L²/2I

The second form follows from L = Iω. For fixed L, a smaller I means more energy.

Worked example 9. A torque of 0.80 N m acts on a wheel of I = 0.040 kg m², initially at rest, for 5.0 s. Find ω and Ek.

ΔL = 0.80 × 5.0 = 4.0 kg m² s⁻¹
ω = 4.0 / 0.040 = 100 rad s⁻¹
Ek = L²/2I = 16 / 0.080 = 200 J   (check: ½ × 0.040 × 100² = 200 J)

Rolling without slipping

A round body can move forward and turn at once. If it rolls without slipping, the speed of its centre is v = ωr, and its kinetic energy has two parts: ½mv² for translation and ½Iω² for rotation.

Worked example 10. A uniform cylinder, with I = ½mr² (given), rolls from rest down a slope, falling through a height of 0.60 m. Find its speed at the bottom.

mgh = ½mv² + ½Iω² = ½mv² + ½(½mr²)(v/r)² = ¾mv²
v = √(4gh/3) = √(4 × 9.81 × 0.60 / 3) = 2.80 m s⁻¹

A block sliding without friction would reach √(2gh) = 3.43 m s⁻¹. The cylinder is slower because part of the energy goes into rotation. Note that mass cancels.

Common errors

• Using the wrong angle in τ = Fr sin θ. θ is measured from the line joining the axis to the point of application.
• Mixing sense: a clockwise torque must carry the opposite sign to a counter-clockwise one.
• Measuring r in Σmr² from the wrong point. It is the distance to the axis, not to the other mass.
• Leaving ω in rev min⁻¹ or degrees.
• Applying conservation of L when an external torque, such as friction at a bearing, acts during the process.
• Assuming kinetic energy is conserved whenever L is.

Official syllabus

This page follows sub-topic A.4 (Rigid body mechanics) of the IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. A.4 is additional higher level content only.