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

These notes condense sub-topic A.4, Rigid body mechanics, from the IB Diploma Programme Physics guide, first assessment 2025. A.4 is additional higher level (HL only) with no standard level content. For full explanations and longer worked examples, read the study guide, then test yourself with the practice questions.

Related pages: the IB Physics hub, the printable checklist, forces and momentum revision notes and kinematics revision notes. Not sure which areas are weakest? Take the free 10-minute diagnostics.

Key terms

• Torque: the turning effect of a force about an axis, τ = Fr sin θ, in N m.
• Rotational equilibrium: zero resultant torque.
• Angular displacement, Δθ: the angle turned through, in rad.
• Angular velocity, ω: rate of change of angle, in rad s⁻¹. The guide uses angular speed with a sense, not a vector treatment.
• Angular acceleration, α: rate of change of ω, in rad s⁻².
• Moment of inertia, I: how the distribution of mass about an axis resists a change in rotation, in kg m².
• Angular momentum, L: Iω, in kg m² s⁻¹.
• Angular impulse: τΔt, equal to the change in L.

Formulae

Quantity	Equation	Notes
Torque	τ = Fr sin θ	θ is the angle between F and the line from the axis
Point masses	I = Σmr²	r is the distance to the axis
Newton’s second law	τ = Iα	τ is the resultant (average) torque
Angular momentum	L = Iω	Conserved if no resultant torque acts
Angular impulse	ΔL = τΔt = Δ(Iω)
Rotational energy	Ek = ½Iω² = L²/2I
Link to linear motion	v = ωr	For a point at distance r from the axis

Uniform angular acceleration:

Equation	Missing quantity
Δθ = ((ωf + ωi)/2) t	α
ωf = ωi + αt	Δθ
Δθ = ωi t + ½αt²	ωf
ωf² = ωi² + 2αΔθ	t

Choose the equation that leaves out the quantity you neither know nor need.

Linear and rotational: the analogy

Linear idea	Rotational partner
s	Δθ
v	ω
a	α
m	I
F	τ
Newton’s second law, F = ma	τ = Iα
p = mv	L = Iω
FΔt (impulse)	τΔt (angular impulse)
Ek = ½mv² = p²/2m	Ek = ½Iω² = L²/2I

The guide’s linking idea is that the equations of motion and the conservation laws for rotation are analogous to those for linear motion. The analogy is a memory aid; always check that you are using a rotational quantity with a rotational one.

Methods in steps

Rotational equilibrium problem

1. Draw the body and mark every force at its point of application. A uniform body’s weight acts at its centre of mass.
2. Choose an axis, ideally through the point where an unknown force acts.
3. Choose a positive sense. Write clockwise torques equal to counter-clockwise torques.
4. Solve. If a force or angle is needed, add the condition that forces sum to zero.

Angular acceleration problem

1. Find the resultant torque, with signs.
2. Find I about the same axis, using Σmr² or the formula you are given.
3. α = τ/I, then use the angular equations of motion.

Coupled bodies (string over a pulley)

1. Write F = ma for the hanging mass.
2. Write τ = Iα for the pulley, using the tension.
3. Link them with a = αR, assuming the string does not slip.
4. Eliminate the tension.

Changing I or joining bodies

1. Check there is no external resultant torque.
2. Write L before = L after, with I and ω on both sides.
3. Calculate Ek before and after if energy is asked about. L is conserved, Ek need not be.

Rolling without slipping

1. Use v = ωr.
2. Write total Ek = ½mv² + ½Iω².
3. Substitute the given I and eliminate ω.

Small worked reminders

Torque. A force of 4.0 N perpendicular to a handle, 0.35 m from the axis: τ = 4.0 × 0.35 = 1.4 N m. The same force along the handle gives sin 0° = 0, so no torque.

Angular impulse. A wheel with I = 0.20 kg m² reaches 10 rad s⁻¹ from rest. Then L = 2.0 kg m² s⁻¹ and Ek = L²/2I = 4.0/0.40 = 10 J. The torque that did this over 4.0 s was on average ΔL/Δt = 0.50 N m.

Conservation. If a spinning body’s I falls to a third of its value and no external torque acts, ω triples and Ek triples. The extra energy comes from the work done on the body.

Rolling. A hoop (I = mr², given) rolls from rest down a drop of 0.80 m without slipping. Then mgh = ½mv² + ½mv² = mv², so v = √(gh) = √(9.81 × 0.80) = 2.80 m s⁻¹. Half the energy is rotational. With ω = v/r, a hoop of radius 0.20 m turns at 14.0 rad s⁻¹ at the bottom.

Sense. For a seesaw balanced about its midpoint, a 30 kg child at 1.5 m on the left gives a counter-clockwise torque. A second child on the right balances it only if their mass times distance is also 45 kg m.

Must-know distinctions

• Torque and energy: both have the unit N m, but torque is a turning effect with a sense and energy is a scalar. Write torque in N m and energy in J.
• Angular speed and linear speed: ω is the same for every point of a rigid body. v = ωr differs with r.
• Mass and moment of inertia: mass is fixed for a body, I changes with the axis and with the arrangement of mass.
• Conservation of L and of Ek: collisions where bodies join lose kinetic energy but keep L if no external torque acts.
• Equilibrium and rest: zero resultant torque means constant ω, which can be non-zero. Rotational equilibrium does not require the body to be stationary.
• Resultant and individual torques: τ = Iα uses the resultant torque, including friction at the axle.

Quick self-test

1. A force of 12 N acts perpendicular to a spanner at 0.50 m from the nut. What is the torque?
2. A torque of 2.0 N m acts on a body of I = 0.50 kg m². What is α?
3. A wheel at 5.0 rad s⁻¹ has α = 2.0 rad s⁻² for 6.0 s. Find ω at the end.
4. A body starts from rest with α = 5.0 rad s⁻². Find the angular displacement when ω = 8.0 rad s⁻¹.
5. Masses of 0.20 kg at 0.50 m and 0.30 kg at 0.20 m from an axis: find I.
6. A body with I = 0.40 kg m² turns at 3.0 rad s⁻¹. Find L.
7. A skater spinning at 3.0 rad s⁻¹ reduces I from 2.0 to 0.80 kg m². Find the new ω.
8. A body has L = 6.0 kg m² s⁻¹ and I = 0.90 kg m². Find Ek.
9. A torque of 3.0 N m acts for 2.5 s. What angular impulse is this?
10. A wheel of radius 0.15 m rolls without slipping at 3.0 m s⁻¹. Find ω.
11. State the condition for rotational equilibrium.
12. Why does a hoop have a larger I than a disc of the same mass and radius?

Answers

1. τ = 12 × 0.50 = 6.0 N m.
2. α = 2.0/0.50 = 4.0 rad s⁻².
3. ω = 5.0 + 2.0 × 6.0 = 17 rad s⁻¹.
4. Δθ = ωf²/2α = 64/10 = 6.4 rad.
5. I = 0.20 × 0.25 + 0.30 × 0.04 = 0.050 + 0.012 = 0.062 kg m².
6. L = 0.40 × 3.0 = 1.2 kg m² s⁻¹.
7. ω = 2.0 × 3.0/0.80 = 7.5 rad s⁻¹.
8. Ek = 36/(2 × 0.90) = 20 J.
9. ΔL = 3.0 × 2.5 = 7.5 kg m² s⁻¹ (equivalently N m s).
10. ω = v/r = 3.0/0.15 = 20 rad s⁻¹.
11. The resultant torque about any axis is zero.
12. More of the hoop’s mass is at the full radius, far from the axis, so Σmr² is larger.

Where marks are usually lost

• Using a force that is not perpendicular to r without the factor sin θ.
• Forgetting a torque, such as the weight of a uniform beam at its midpoint, or the sense of one torque.
• Reporting torque in J, or I in kg.
• Using one radius for every mass in Σmr² when the masses sit at different distances from the axis.
• Using revolutions or rev min⁻¹ in an equation that needs rad and rad s⁻¹.
• Writing L = Iω for the final state without adding the second body’s I after coupling.
• Claiming Ek is conserved when bodies join.
• Giving a final answer without a sense when the question asks for the direction of rotation.
• Not showing the substitution in a “show that” question, where the full route is needed.

Official syllabus

These notes follow sub-topic A.4 (Rigid body mechanics) of the IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization.