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Rotational motion and moment of inertia

Everything you learned about straight-line motion runs again, with angles in place of metres, and the four SUVAT equations come back wearing Greek letters. One new quantity replaces mass, and it cares where the mass is as much as how much there is.

Year 13AQA 3.11.1

Builds on Circular motion and Work, energy and power.

IN THIS TOPIC

Describe rotation with angular displacement, angular velocity and angular acceleration, in radians.
Solve uniform angular acceleration problems with the four angular equations of motion.
Say what moment of inertia measures and how mass distribution changes it, using values you are given.
Calculate rotational kinetic energy with half I omega squared, including for flywheels storing energy.

WHAT YOU PROBABLY THINK

How hard something is to set spinning depends only on its mass.

Angles instead of metres

Take any rigid body turning about a fixed axis, a wheel, a drum, a turbine, and every particle in it sweeps the same angle in the same time. That shared angle is what rotational motion measures. The angular displacement θ is the angle turned, in radians. The angular velocity ω is the rate it is swept out, $\omega = \Delta\theta / \Delta t$ , in rad s⁻¹, and the angular acceleration α is the rate ω changes, $\alpha = \Delta\omega / \Delta t$ , in rad s⁻².

A point at radius r links the two languages. It moves along its arc at speed $v = \omega r$ , straight from the circular motion work, and when the rotation rate changes its speed along the arc changes too:

a = \alpha r

NOT ON THE DATA SHEET: LEARN IT

so the rim of a wheel accelerates αr along its own circle while the axle goes nowhere. Every point shares one ω and one α. What differs from point to point is r, and with it the speed.

FIG. 1One disc, one angular velocity. The point at twice the radius moves twice as fast along its arc, because v = ωr, yet both points sweep the same angle in the same time.

For constant α, the four SUVAT equations return with their symbols swapped, s for θ, u and v for $\omega_{1}$ and $\omega_{2}$ , a for α. All four are printed in the booklet:

\omega_{2} = \omega_{1} + \alpha t

ON YOUR DATA SHEET

\omega_{2}^{2} = \omega_{1}^{2} + 2\alpha\theta

ON YOUR DATA SHEET

\theta = \omega_{1}t + \frac{\alpha t^{2}}{2}

ON YOUR DATA SHEET

\theta = \frac{(\omega_{1} + \omega_{2})t}{2}

ON YOUR DATA SHEET

Solve them exactly as you solved SUVAT. List the four quantities, mark what you know, pick the equation missing the one you do not want.

WORKED EXAMPLE

A drum spinning up

A washing machine drum accelerates uniformly from rest to its spin speed of 1200 revolutions per minute in 4.0 s. Find the angular acceleration and the number of revolutions made.

Convert first. 1200 rev min⁻¹ = 1200 × 2π/60 = 126 rad s⁻¹.

α = (ω₂ − ω₁)/t = 126/4.0 = 31 rad s⁻².

θ = (ω₁ + ω₂)t/2 = 126 × 4.0/2 = 251 rad, and 251/2π = 40 revolutions.

Revolutions to radians is the door into every one of these questions. One revolution is 2π radians, and rev min⁻¹ needs dividing by 60 as well.

Moment of inertia

In F = ma, mass measures the reluctance to accelerate. Rotation has its own reluctance, and it is not mass alone. Push a roundabout at its rim with children clinging to the edge and it starts stubbornly; sit the same children by the centre and the same push spins it up easily. The mass has not changed. Its distribution has.

The quantity doing mass's job is the moment of inertia I. Cut the body into particles, multiply each mass by the square of its distance from the axis, and add:

I = m_{1}r_{1}^{2} + m_{2}r_{2}^{2} + \ldots

ON YOUR DATA SHEET

which the booklet compresses with a capital sigma, standing for the sum over every particle. The unit is kg m². The square is the point to hold onto. Mass twice as far from the axis counts four times over, so pushing material out to the rim raises I far faster than adding material at the middle.

FIG. 2Equal masses, equal radii, different machines. The hoop carries all its mass at full radius and has I = mr²; the uniform disc spreads its mass inward and manages only half that.

That is the lie dealt with. A light hoop can resist spin-up more than a heavy compact disc, because r² rewards distance so heavily. You are never asked to derive I for a shape. Questions hand you the value, or the formula for it, and your job is to use it, plus one idea in words, that mass concentrated far from the axis means a large I and mass hugging the axis means a small one.

YOUR TURN

Two masses on a light rod

Two 0.40 kg masses sit at the ends of a light rod, each 0.30 m from the central axis, spinning at 5.0 rad s⁻¹. Find the moment of inertia and state what happens to I if the masses slide halfway in.

Show the working

I = 2 × 0.40 × 0.30² = 0.072 kg m². The rod is light, so only the masses count.

At 0.15 m each r² falls by a factor of four, so I drops to a quarter, 0.018 kg m². Halving the distance does far more than halving would suggest.

Rotational kinetic energy and the flywheel

A rotating body stores kinetic energy even though it goes nowhere, because every particle in it is moving. Adding the half m v squared of each particle, with v = ωr, gives

E_{k} = \frac{1}{2}I\omega^{2}

ON YOUR DATA SHEET

the exact twin of half m v squared with I for m and ω for v. A machine built to exploit this is the flywheel, a wheel made deliberately hard to spin up, with its mass pushed out to the rim where r² works hardest. Spin it fast and it becomes an energy store.

FIG. 3Energy stored against spin rate for one flywheel. The curve is a parabola, so doubling ω quadruples the stored energy, and the fastest spin earns most of the storage.

The ω² makes speed worth more than size. Doubling the spin rate quadruples the store, so modern flywheel batteries are modest discs spun to tens of thousands of revolutions per minute in a vacuum, on magnetic bearings, feeding braking energy back to trams and grid-scale stores smoothing out demand spikes.

Linear quantity	Rotational twin
displacement s	angular displacement θ
velocity v	angular velocity ω
acceleration a	angular acceleration α
mass m	moment of inertia I
kinetic energy = ½mv²	E_k = ½Iω²
force F	torque T (next lesson)
momentum mv	angular momentum Iω (next lesson)

The table is the map of the whole sub-topic. Every linear idea you own has a rotational twin, and the next lesson fills in the last two rows.

TRY IT UNSEEN

The flywheel as a battery

A flywheel is a uniform disc of mass 140 kg and radius 0.60 m, for which I = ½mr², spun at 3000 revolutions per minute. Find the energy it stores, and how long it could supply 5.0 kW.

Show the working

I = ½ × 140 × 0.60² = 25.2 kg m², and ω = 3000 × 2π/60 = 314 rad s⁻¹.

E_k = ½ × 25.2 × 314² = 1.2 × 10⁶ J.

At 5.0 kW that lasts t = 1.24 × 10⁶/5000 ≈ 250 s, about four minutes. A tonne-scale flywheel holds a phone-battery-scale energy, so real designs chase ω, not mass.

THE EXAM BIT

Convert revolutions before anything else. One revolution is 2π rad, so N rev min⁻¹ is N × 2π/60 rad s⁻¹. Leaving a speed in rev min⁻¹ poisons every later line.
Angular SUVAT is marked exactly like linear SUVAT. List θ, ω₁, ω₂, α, t, pick the equation missing the unwanted one, and quote it before substituting.
Moment of inertia in words earns marks without algebra. It measures resistance to angular acceleration, and it depends on how far the mass sits from the axis, with r squared. You will be given I or its formula, never asked to derive it.
For flywheel questions the examiner wants the design point stated, mass concentrated at the rim to maximise I, and the energy point, E_k ∝ ω², so spin rate beats mass.
If a question gives a diameter, halve it. Feeding a diameter into r² multiplies I and E_k by four, and it is this option's most reliable lost mark.

CHECK YOURSELF

A potter's grindstone of moment of inertia 1.7 kg m² accelerates uniformly from rest to 30 rad s⁻¹ in 12 s. Find the angular acceleration, the angle turned in that time, and the kinetic energy at full speed.

Show a hint

Two angular SUVAT equations, then the rotational kinetic energy formula.

Show the answer

α = (ω₂ − ω₁)/t = 30/12 = 2.5 rad s⁻².

θ = ω₁t + αt²/2 = 0 + 2.5 × 12²/2 = 180 rad, about 29 revolutions.

E_k = ½Iω² = ½ × 1.7 × 30² = 765 J, the work the motor did against the stone's inertia.

Rotation reruns SUVAT with θ, ω and α standing in for s, v and a.

I sums mr² over the body, so distance from the axis counts squared.

A spinning store holds ½Iω², and the ω² makes spin rate worth more than mass.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

17 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

6 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

CHECK YOUR PROGRESS

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Describe rotation with angular displacement, angular velocity and angular acceleration, in radians.
Solve uniform angular acceleration problems with the four angular equations of motion.
Say what moment of inertia measures and how mass distribution changes it, using values you are given.
Calculate rotational kinetic energy with half I omega squared, including for flywheels storing energy.

Open the full revision checklist to track your progress across the whole unit.

Rotational motion and moment of inertia

Angles instead of metres

Moment of inertia

Rotational kinetic energy and the flywheel

Question workbook

Question workbook

Answer book