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IB DP Physics · Theme A Space, time and motion · A.4 (HL only) · ~9 min read

A.4 Rigid body mechanics

Torque and rotational equilibrium, angular displacement, velocity and acceleration, the rotational suvat equations, moment of inertia, Newton's second law for rotation, angular momentum and rotational kinetic energy.

TorqueRotational equilibriumAngular kinematicsMoment of inertiaτ = IαAngular momentumRotational kinetic energy
RSby Riyaz Sir
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No tag = SL and HLHL ONLY HL students onlyDATA BOOKLET given in the examM1A1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • A.4 is HL only. It is the rotational version of A.1–A.3: every linear quantity has a rotational partner (s → θ, v → ω, a → α, m → I, F → τ, p → L).
  • Torque τ = Fr sin θ turns a body. Zero resultant torque AND zero resultant force means equilibrium.
  • Moment of inertia I = Σmr2 measures how hard a body is to spin up. It depends on where the mass is, not just how much.
  • τ = Iα, L = Iω (conserved when no resultant torque acts) and Ek = ½Iω2 complete the set.

By the end you can

  • Calculate torques and solve rotational equilibrium problems. HL
  • Use angular displacement, velocity and acceleration, and the equations for uniform angular acceleration. HL
  • Calculate moments of inertia for point masses and use given values for other shapes. HL
  • Apply τ = Iα, angular impulse, conservation of angular momentum and rotational kinetic energy. HL

1Torque and equilibrium HL

pivotFθrτ = Fr sin θunit: N mOnly the component F sin θ, at 90° to r, turns the bar.
A force pushing straight along the bar towards the pivot cannot turn it. Only the perpendicular component counts.
τ = Fr sin θ DATA BOOKLET
θ is the angle between F and r; τ in N m
  • Same unit as energy, different idea: write torque in N m, never in J.
  • Couple: two equal, opposite, parallel forces. Zero resultant force but a non-zero torque: pure rotation.

Rotational equilibrium — the resultant torque about ANY axis is zero. Full equilibrium also needs the resultant force to be zero (translational equilibrium).

300 N1.5 mW = ?1.25 mClockwise moments = anticlockwise moments: 300 × 1.5 = W × 1.25, so W = 360 NEquilibrium also needs zero resultant force: the pivot pushes up with 660 N.
Take moments about the pivot: the unknown pivot force then drops out of the torque equation.
Worked example

A uniform 4.0 m plank of weight 200 N rests on two supports, one at each end. A 600 N person stands 1.0 m from the left end. Find the force from each support.

  1. moments about the left support: R × 4.0 = 600 × 1.0 + 200 × 2.0 M1
  2. R = 1000/4.0 = 250 N (right support) A1
  3. vertical forces: L + 250 = 800, so L = 550 N A1
Exam tipChoose the axis through the force you know least about. Its torque is zero there, so it disappears from the equation.

2Describing rotation HL

Angular displacement θ — the angle turned through, in radians.

Angular velocity ω — rate of change of angle, ω = Δθ/Δt, in rad s−1. Linked to linear speed by v = ωr.

Angular acceleration α — rate of change of ω, in rad s−2. Linked to the tangential acceleration by a = αr.

LinearRotational (uniform α) DATA BOOKLET
v = u + atωf = ωi + αt
s = ut + ½at2Δθ = ωit + ½αt2
v2 = u2 + 2asωf2 = ωi2 + 2αΔθ
s = ½(u + v)tΔθ = ½(ωf + ωi)t
Worked example

A wheel spinning at 30 rad s−1 slows uniformly to rest in 12 s. Find α and the number of turns.

  1. α = (0 − 30)/12 = −2.5 rad s−2 A1
  2. Δθ = ½(30 + 0)(12) = 180 rad M1
  3. turns = 180/2π = 28.6 A1
Common mistakeUsing revolutions or degrees in the equations. Convert to radians first: 1 rev = 2π rad, and 600 rpm = 600 × 2π/60 = 62.8 rad s−1.

3Moment of inertia and Newton's second law for rotation HL

I = Σmr2 DATA BOOKLET
r is the distance of each mass from the AXIS; I in kg m2
masses near the axisI = 2m(0.10)² smalleasy to spin upmasses far outI = 2m(0.25)² largehard to spin upSame mass, same axis: the distribution decides I = Σmr²
Moving the same masses further from the axis raises I sharply, because r is squared.
  • Other shapes: the exam gives the formula when needed, for example a solid cylinder about its axis I = ½MR2, a hoop I = MR2, a solid sphere I = 25MR2.
  • I depends on the axis. The same rod is easier to spin about its centre than about one end.
τ = Iα DATA BOOKLET
the resultant torque gives the angular acceleration
Worked example

Two 0.50 kg masses sit at the ends of a light rod, each 0.40 m from the axis. A torque of 2.0 N m acts for 5.0 s from rest. Find I, ω and the angle turned.

  1. I = 2 × 0.50 × 0.402 = 0.16 kg m2 A1
  2. α = τ/I = 2.0/0.16 = 12.5 rad s−2 M1
  3. ω = 0 + 12.5 × 5.0 = 62.5 rad s−1 A1
  4. Δθ = ½ × 12.5 × 5.02 = 156 rad A1
Worked example

A string wound round a pulley (I = 0.020 kg m2, radius 0.10 m) is pulled with a constant 5.0 N force. Find α and the linear acceleration of the string.

  1. τ = Fr = 5.0 × 0.10 = 0.50 N m M1
  2. α = 0.50/0.020 = 25 rad s−2 A1
  3. a = αr = 25 × 0.10 = 2.5 m s−2 A1

4Angular momentum HL

L = Iω DATA BOOKLET
in kg m2 s−1
ΔL = τΔt = Δ(Iω) DATA BOOKLET
angular impulse equals the change in angular momentum
  • Conservation: with no resultant external torque, total angular momentum stays constant. Divers, skaters, collapsing stars and spinning chairs all show it.
pullarms inI = 3.0 kg m², ω = 2.0 rad s⁻¹I = 1.2 kg m², ω = 5.0 rad s⁻¹L = Iω is conserved: smaller I, faster spin
Pulling her arms in cuts I, so ω rises to keep Iω constant.
Worked example

In the diagram, find the final angular velocity and the change in kinetic energy.

  1. L conserved: 3.0 × 2.0 = 1.2ω M1
  2. ω = 5.0 rad s−1 A1
  3. Ek: ½(3.0)(2.02) = 6.0 J → ½(1.2)(5.02) = 15 J M1
  4. Ek increases by 9.0 J: her muscles do work pulling her arms in R1
Common mistakeAssuming rotational kinetic energy is conserved when I changes. Angular momentum is conserved; kinetic energy usually is not.

5Rotational kinetic energy HL

Ek = ½Iω2 = L22I DATA BOOKLET
energy stored in rotation
rolls without slipping: v = ωrmgh = ½mv² + ½Iω²part of the energy goes intorotation, so it arrives slowerthan a frictionless sliding block
A rolling object shares its energy between moving along and spinning.
Worked example

A solid cylinder (I = ½MR2) rolls without slipping from rest down a slope of vertical height 1.0 m. Find its speed at the bottom.

  1. Mgh = ½Mv2 + ½(½MR2)(v/R)2 M1M1
  2. gh = ¾v2 A1
  3. v = √(4gh/3) = √(4 × 9.8 × 1.0/3) = 3.6 m s−1 (a sliding block would reach 4.4 m s−1) A1
Exam tipFlywheels store energy as ½Iω2. A heavy rim (large I) at high ω stores the most, which is why flywheels are rim-heavy.
Linking questionEvery result in this topic mirrors one from A.1–A.3. Can you write the rotational version of "impulse equals change in momentum" and "work equals force × distance" (W = τΔθ)?
Idea for your IA / explorationRoll cylinders with the same mass and radius but different mass distributions (solid, hollow) down a ramp. Time them and deduce I from the energy equation.

Test yourself — cover the answers first

Quick check

  1. What is the unit of torque?
  2. State the two conditions for a rigid body to be in equilibrium.
  3. Convert 300 rpm to rad s−1.
  4. Why does a hoop have a larger I than a disc of the same mass and radius?
  5. When is angular momentum conserved?
  6. A skater halves her moment of inertia. What happens to ω and Ek?

Exam-style

  1. A door 0.90 m wide is pushed at its edge with 40 N at 60° to the plane of the door. Find the torque about the hinges.
  2. A flywheel (I = 2.5 kg m2) spins at 120 rad s−1. A brake applies a torque of 15 N m. How long does it take to stop, and how much energy is dissipated?
  3. A 40 kg child runs at 3.0 m s−1 along a tangent and jumps onto the edge of a stationary roundabout (I = 500 kg m2, radius 2.0 m). Find the angular velocity afterwards.
Show answers
  1. N m (not J)
  2. zero resultant force and zero resultant torque
  3. 300 × 2π/60 = 31.4 rad s−1
  4. all its mass is at the maximum distance from the axis
  5. when no resultant external torque acts
  6. ω doubles; Ek doubles
  7. τ = Fr sin θ = 40 × 0.90 × sin 60°
    τ = 31 N m
  8. α = −15/2.5 = −6.0 rad s−2
    t = 120/6.0 = 20 s
    energy = ½ × 2.5 × 1202 = 1.8 × 104 J
  9. L before = mvr = 40 × 3.0 × 2.0 = 240 kg m2 s−1
    I after = 500 + 40 × 2.02 = 660 kg m2
    ω = 240/660 = 0.36 rad s−1
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by the International Baccalaureate Organization. Syllabus codes refer to the IB DP subject guides.