This unit is ≈4% of the A-Level Physics, across 8 lessons. Full syllabus
Lesson 1 of 8 · Options and board-specific extensions
Rotational dynamics
8 min read · about 1 h with practice3 quick checks<1% of the testStretch: Stretch: harder material that separates the top grades
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Rotational dynamics is half of the AQA Engineering physics option, which is Paper 3 Section B (35 marks). Every quantity in linear mechanics has a rotational twin. Questions reward learners who move between the two confidently: angular SUVAT, torque, rotational kinetic energy, angular momentum and power, usually set in an engineering context such as a flywheel, a motor or a rolling object. Expect at least one multi-step calculation and a 6-mark question on flywheels or angular momentum.
By the end you’ll be able to
Use I = Σmr² and given moments of inertia; calculate rotational kinetic energy ½Iω²
Apply the rotational equations of motion (angular SUVAT) and T = Iα
Apply conservation of angular momentum (ice skaters, spinning platforms) and angular impulse
Explain flywheel design and uses (energy storage, smoothing torque) and calculate power P = Tω
What the exam asks
Moment of inertia:I=mr2 for a point mass and I= for an extended object. You need to know what affects . Formulae such as are given.
vii.Check your understanding
3 questions on rotational dynamics. Every option is explained once you answer.
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PromptCard 1 of 3
Define moment of inertia. What does it depend on?
Σ
m
r2
qualitatively
I
21mr2
Rotational kinetic energy:Ek=21Iω2. What limits a flywheel’s energy storage, and how flywheels are used for smoothing, energy storage in vehicles and production machines.
Angular kinematics: displacement, speed, velocity and acceleration; graphs for uniform and non-uniform angular acceleration; the four equations for uniform α.
Torque:T=Fr and T=Iα.
Angular momentum:Iω, its conservation, and angular impulse TΔt=Δ(Iω), with examples from sport.
Work and power:W=Tθ and P=Tω, including frictional torque in real machines.
All of these equations are on the AQA data sheet. The marks come from choosing the right one and handling units.
Core ideas
The linear–rotational analogy
Linear
Rotational
Link
displacement s (m)
angular displacement θ (rad)
s=rθ
velocity v (m s−1)
angular velocity ω (rad s−1)
v=rω
acceleration a (m s−2)
angular acceleration α (rad s)
mass m (kg)
moment of inertia I (kg m2)
I=Σm
force F (N)
torque T (N m)
T=Fr
F=ma
T=Iα
momentum mv
angular momentum Iω (kg m2s−1 or N m s)
impulse FΔt
angular impulse TΔt
Ek=21mv
W=Fs, P=Fv
W=Tθ,
Moment of inertia
I=Σmr2 measures how hard it is to change an object’s rotation. It depends on:
the mass;
how far the mass is from the axis, which matters most because of the r2;
which axis is chosen.
For the same mass and radius, a hoop (mr2) has a larger I than a uniform disc (21mr2), which has a larger I than a solid sphere (52mr2). Moving mass outwards increases I sharply.
Angular kinematics
ω=ΔtΔθ and α=ΔtΔω. Angles must be in radians. To convert revolutions per minute to rad s−1, multiply by 602π. For uniform angular acceleration:
On graphs, the gradient of θ against t is ω, the gradient of ω against t is α, and the area under an ω–t graph is θ. For non-uniform α the ω–t graph curves: find α from a tangent and θ by counting squares.
Torque and T=Iα
T in T=Iα is the resultant torque, which is the driving torque minus the frictional torque. A machine turning at constant ω has zero resultant torque, so its driving torque equals the frictional torque.
Rotational kinetic energy and rolling
A rolling object without slipping has v=rω and two kinds of kinetic energy:
mgh=21mv2+21Iω2⇒v2=1+I/(mr2)2gh
The smaller mr2I is, the faster the object rolls. In a race down a slope, a solid sphere beats a solid cylinder, which beats a hoop, whatever their masses and radii.
Angular momentum and angular impulse
Angular momentum L=Iω is conserved when no resultant external torque acts. A skater or diver who pulls in their arms or tucks reduces I, so ω rises. Their kinetic energy 2IL2increases, because they do work pulling the mass inwards. When two rotating objects lock together (a child landing on a roundabout, or clutch plates engaging), angular momentum is conserved but kinetic energy is lost, just like an inelastic collision.
Angular impulse TΔt=Δ(Iω) is the area under a torque–time graph.
Work, power and frictional torque
W=Tθ and P=Tω. Bearings and air resistance produce a frictional torque. To keep a machine at constant ω, the power input must equal Tfrictionω. A flywheel that spins down to rest transfers all its kinetic energy to work against friction: 21Iω2=Tfriction.
Flywheels
Storing energy. Since Ek=21Iω2, a flywheel of given mass stores more energy if its mass is concentrated in a heavy rim (spoked design) and, above all, if it spins faster, because Ek∝ω2. The limit is the tensile strength of the material: the stress in the rim grows with the rim speed, and too high a speed makes the flywheel burst. Strong, low-density materials such as carbon-fibre composites can spin much faster than steel and store more energy per kilogram. Energy losses are reduced with vacuum enclosures and low-friction or magnetic bearings.
Smoothing torque and speed. A single-cylinder engine drives only during its power stroke, so its torque arrives in pulses. A flywheel with a large I means the fluctuating resultant torque causes only a small α, so the speed stays nearly constant. The flywheel absorbs energy when the driving torque exceeds the load torque and gives it back when it is less.
Production machines. In presses, punches and forges, a small motor spins the flywheel up steadily. The flywheel then delivers a large amount of energy in a short time during the working stroke, at high power, while slowing only a little.
Vehicles. In regenerative braking (KERS), the vehicle’s kinetic energy goes into the flywheel through a variable-ratio transmission during braking and is returned during acceleration. The drawbacks are the extra mass, gyroscopic effects when the vehicle turns, and the need for a strong safety casing.
Worked examples
Exam technique
Convert first. Change rev min⁻¹ to rad s−1 and revolutions to radians before substituting anything.
Translate from linear. Ask which linear equation you would use, then swap in the rotational symbols. The four angular SUVAT equations behave exactly like the linear ones.
“Show that” questions: give the final value to at least one more significant figure than the value quoted, with every step shown.
Angular momentum questions: start with “no external torque, so angular momentum is conserved”, then write I1ω1=I2ω2. Comment on kinetic energy only after that.
Graphs: the gradient of an ω–t graph gives α and the area gives θ. For a straight line, use a large gradient triangle. The area of a trapezium is exact.
Flywheel 6-markers: cover the uses (smoothing, storage, production machines), then the design (Ek=21Iω, a rim-heavy shape, dominating, the strength limit), then losses and safety.
Common mistakes
Quick recap
I=Σmr2 depends on mass and, above all, on how far that mass is from the axis.
Ek=21Iω2. For rolling, add 21mv2 and use v=rω.
Angular SUVAT uses θ, ω1, ω2, and , with angles in radians.
T=Fr=Iα (resultant torque). W=Tθ and P=.
L=Iω is conserved with no external torque. Angular impulse is TΔt=Δ(Iω), the area under a T– graph.
Flywheels smooth torque and speed and store energy. For the most energy, put the mass in the rim and spin fast, up to the strength limit.