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Moment of inertia and rotational kinetic energy

Mass measures reluctance to accelerate along a line, and the moment of inertia does that job in rotation, summing mr² over the body. Distance from the axis counts squared, so a flywheel carries its mass at the rim. A turning body stores rotational kinetic energy ½Iω², so doubling the spin rate quadruples the store.

Year 13AQA 3.11.1.1, 3.11.1.2

Rotational motion and moment of inertia, part 2 of 2. Part 1 is Rotational motion and the angular equations.

Builds on Rotational motion and the angular equations and Work, energy and power.

IN THIS TOPIC

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.

COMMON MISCONCEPTION

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

Mass alone does not decide it. The moment of inertia sums mr², so where the mass sits counts squared: a light hoop with everything at the rim can resist spin-up more than a heavier disc whose mass hugs the axis.

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 sitting at the edge and it is hard to start turning; move the same children close to 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 THE AQA 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 key point. 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.

Two wheels of equal mass and equal radius side by side, each on its axis. The hoop carries all of its mass as dots on the rim; the uniform disc has its dots spread in towards the centre. Bars beneath compare the sums of m r squared: the hoop's is exactly twice the disc's, because distance from the axis counts squared. — FIG. 1Equal 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.

A light hoop can resist spin-up more than a heavy compact disc, because r² weights distance so heavily. You are never asked to derive I for a shape. Questions give 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 concentrated close to the axis means a small one.

GUIDED PRACTICE

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 its centre of mass does not move, 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 THE AQA 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.

A graph of stored energy against angular velocity for a 25 kilogram metre squared flywheel: a parabola through the origin, with dashed guides marking 0.31 megajoules at 157 radians per second and 1.24 megajoules at 314 radians per second. Twice the spin rate, four times the energy. — FIG. 2Energy stored against spin rate for one flywheel. The curve is a parabola, so doubling ω quadruples the stored energy, and most of the stored energy comes from the highest spin rates.

Because $E_{k} \propto \omega^{2}$ , spin rate matters more than mass. 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
momentum mv	angular momentum Iω

Every linear quantity has a rotational analogue. The last two rows, torque and angular momentum, are covered in the lessons on torque and angular acceleration and on angular momentum and rotational power.

INDEPENDENT PRACTICE

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 140 kg flywheel holds only about 0.35 kW h, so real designs increase ω rather than mass.

ASSESSMENT FOCUS

Moment of inertia can be answered in words, 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.
A flywheel answer states the design point, mass concentrated at the rim to maximise I, and the energy point, E_k ∝ ω², so spin rate matters more than mass.
If a question gives a diameter, halve it. Feeding a diameter into r² multiplies both I and E_k by four.
Energy answers still need ω in rad s⁻¹, so the 2π/60 conversion comes first whenever a spin rate arrives in rev min⁻¹.

CHECK YOURSELF

Two 1.2 kg masses sit at the ends of a light rod, each 0.25 m from the central axis, and the arrangement spins at 8.0 rad s⁻¹. Find the moment of inertia and the kinetic energy stored, and state what happens to that energy if the spin rate doubles.

Show a hint

Sum mr² over the two masses, then half I omega squared, then look at the square.

Show the answer

I = Σmr² = 2 × 1.2 × 0.25² = 0.15 kg m². The rod is light, so only the masses count.

E_k = ½Iω² = ½ × 0.15 × 8.0² = 4.8 J.

Doubling ω quadruples the store to 19.2 J, because E_k ∝ ω² while I has not changed.

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

A spinning body stores ½Iω², so doubling the spin rate quadruples the stored energy.

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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.

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