Specification: AQA A-level Physics (7408) 3.11.1.1 Engineering physics option | Paper: 3 Section B | Data and formulae booklet: I = Σmr² is given in the exam; the moments of inertia of particular shapes are not, so they are given in the question
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Moment of inertia: a measure of an object’s resistance to a change in its angular velocity about a given axis. It is defined as I = Σmr², the sum of mass × (perpendicular distance from the axis)² for every particle in the object. Unit: kg m².
Mass distribution: I depends on the total mass and on how that mass is distributed about the axis. Mass further from the axis contributes much more, because r is squared. Moving the axis changes I, so always state the axis.
Rotational analogue: moment of inertia plays the same part in rotation that mass plays in linear motion (T = Iα compared with F = ma).
The equation
moment of inertia = sum of (mass × distance from axis²) I = Σmr² and, for a single point mass, I = mr²
| Quantity | Symbol | Unit |
|---|---|---|
| moment of inertia about the axis | I | kilogram metres squared (kg m²) |
| mass of each particle (or point mass) | m | kilograms (kg) |
| perpendicular distance of the particle from the axis | r | metres (m) |
Rearranged: for a point mass: m = I ÷ r² | r = √(I ÷ m) | for several masses, add the mr² terms: I = m₁r₁² + m₂r₂² + … | Convert g to kg and cm or mm to m before squaring r. The moments of inertia of particular shapes (for example ½MR² for a uniform disc) are not in the booklet; the exam gives them when you need them.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A small 0.50 kg mass is fixed to a light rod 0.40 m from the axis of rotation. Calculate its moment of inertia about the axis.
Show solution
F: I = mr²
I: I = 0.50 × 0.40²
F: I is already the subject
A: I = 0.50 × 0.16 = 0.080 kg m²
Q2 (Core) Two small 2.0 kg masses are fixed to a light rod, one 0.30 m either side of the axis of rotation. Calculate the moment of inertia of the system about the axis.
Show solution
F: I = Σmr²
I: I = (2.0 × 0.30²) + (2.0 × 0.30²)
F: I is already the subject
A: I = 0.18 + 0.18 = 0.36 kg m²
Watch out: each mass contributes its own mr². Do not add the masses and the distances first: 4.0 × 0.60² = 1.44 kg m² is wrong.
Q3 (Core) A 250 g mass on the end of a light arm has a moment of inertia of 4.0 × 10−3 kg m² about the axis. Calculate its distance from the axis in cm, to 2 significant figures.
Show solution
Convert first: 250 g = 0.250 kg
F: I = mr²
I: 4.0 × 10−3 = 0.250 × r²
F: r² = 4.0 × 10−3 ÷ 0.250 = 0.016, so r = √0.016
A: r = 0.1264… m = 13 cm (2 s.f.)
Watch out: remember the square root. 0.016 m is r², not r.
Q4 (Core) A bicycle wheel is modelled as a thin ring (hoop) of radius 340 mm. Its moment of inertia about its axle is 0.15 kg m². Explain why I = MR² for a thin ring, and calculate the mass of the ring, to 2 significant figures.
Show solution
Why I = MR²: every particle of a thin ring is the same distance R from the axle, so Σmr² = (Σm)R² = MR²
Convert first: 340 mm = 0.340 m
F: I = MR²
I: 0.15 = M × 0.340²
F: M = 0.15 ÷ 0.1156
A: M = 1.297… kg = 1.3 kg (2 s.f.)
Q5 (Exam) A flywheel is a uniform solid disc of mass 25 kg and diameter 60 cm. Calculate its moment of inertia about its central axis, to 2 significant figures. (For a uniform disc, I = ½MR².)
Show solution
Convert first: diameter 60 cm, so R = 30 cm = 0.30 m
F: I = ½MR²
I: I = ½ × 25 × 0.30²
F: I is already the subject
A: I = 1.125 kg m² = 1.1 kg m² (2 s.f.)
Watch out: using the diameter instead of the radius gives 4.5 kg m², four times too big, because R is squared.
Q6 (Exam) A flywheel is made from a uniform disc of mass 12 kg and radius 0.20 m with a heavy rim of mass 18 kg fixed around its edge. The rim can be modelled as a thin ring of radius 0.20 m. Calculate the total moment of inertia, and the percentage of it provided by the rim. (Disc: I = ½MR²; thin ring: I = MR².)
Show solution
Step 1 – disc
Idisc = ½ × 12 × 0.20² = 0.24 kg m²
Step 2 – rim
Irim = 18 × 0.20² = 0.72 kg m²
Step 3 – total
F: I = ΣI (moments of inertia about the same axis add)
I: I = 0.24 + 0.72
A: 0.96 kg m², and the rim provides 0.72 ÷ 0.96 = 75%
This is why flywheels have most of their mass in a heavy rim: it gives a large I for a given mass.
Q7 (Exam) Three small masses are fixed to a light rod that rotates about one end: 0.20 kg at 0.10 m, 0.30 kg at 0.25 m and 0.50 kg at 0.40 m from the axis. Calculate the moment of inertia of the system, to 2 significant figures, and the percentage contributed by the 0.50 kg mass.
Show solution
F: I = Σmr²
I: I = (0.20 × 0.10²) + (0.30 × 0.25²) + (0.50 × 0.40²)
F: I = 0.002 + 0.01875 + 0.080 = 0.10075 kg m²
A: 0.10 kg m² (2 s.f.)
the 0.50 kg mass contributes 0.080 ÷ 0.10075 = 0.794…, so 79% (2 s.f.)
Watch out: the outermost mass dominates because r is squared. Halving its distance would cut its contribution to a quarter.
Q8 (Exam) Two small 1.5 kg masses are joined by a light rod 0.80 m long. Calculate the moment of inertia of the system (a) about an axis through the centre of the rod and (b) about an axis through one of the masses, both perpendicular to the rod. Comment on your answers.
Show solution
Step 1 – (a) axis through the centre
each mass is 0.40 m from the axis
F: I = Σmr²
I: I = (1.5 × 0.40²) + (1.5 × 0.40²)
A: 0.48 kg m²
Step 2 – (b) axis through one mass
that mass has r = 0, so contributes nothing; the other is 0.80 m away
I: I = (1.5 × 0²) + (1.5 × 0.80²)
A: 0.96 kg m²
Comment: same object, same mass, but I has doubled. Moment of inertia depends on the position of the axis, not just on the mass.
Q9 (Stretch) A helicopter rotor has three blades, each modelled as a uniform rod of mass 45 kg and length 5.0 m rotating about one end (I = ⅓ML²). The hub is a solid cylinder of mass 60 kg and radius 0.25 m (I = ½MR²). (a) Calculate the total moment of inertia of the rotor, to 2 significant figures. (b) A small 2.0 kg balance weight is added to the tip of each blade. Calculate the percentage increase in the moment of inertia, to 2 significant figures.
Show solution
Step 1 – blades
each blade: ⅓ × 45 × 5.0² = 375 kg m², so three blades = 1125 kg m²
Step 2 – hub
½ × 60 × 0.25² = 1.875 kg m² (tiny: its mass is close to the axis)
Step 3 – (a) total
I = 1125 + 1.875 = 1126.875 kg m²
A: 1.1 × 103 kg m² (2 s.f.)
Step 4 – (b) tip weights
each is a point mass: mr² = 2.0 × 5.0² = 50 kg m², so three add 150 kg m²
percentage increase = 150 ÷ 1126.875 × 100 = 13.3…
A: 13% (2 s.f.)
Watch out: 6.0 kg of tip weights is only about 3% of the rotor’s mass, yet adds 13% to I, because the weights are at the largest possible r.
Q10 (Stretch) A steel flywheel is an annulus (a thick ring) with inner radius 0.15 m, outer radius 0.25 m and thickness 80 mm. The density of steel is 7800 kg m⁻³. For an annulus, I = ½M(R₁² + R₂²). (a) Calculate the mass of the flywheel, to 2 significant figures. (b) Calculate its moment of inertia, to 2 significant figures. (c) Calculate how many times larger this is than the moment of inertia of a solid disc of the same mass and outer radius (I = ½MR²).
Show solution
Convert first: 80 mm = 0.080 m
Step 1 – (a) volume and mass
V = π(R₂² − R₁²) × thickness = π × (0.25² − 0.15²) × 0.080 = π × 0.040 × 0.080 = 0.01005… m³
M = ρV = 7800 × 0.01005… = 78.41… kg
A: 78 kg (2 s.f.)
Step 2 – (b) moment of inertia
F: I = ½M(R₁² + R₂²)
I: I = ½ × 78.41… × (0.15² + 0.25²) = ½ × 78.41… × 0.085
A: I = 3.332… kg m² = 3.3 kg m² (2 s.f.)
Step 3 – (c) comparison
solid disc: I = ½ × 78.41… × 0.25² = 2.450… kg m²
ratio = 3.332… ÷ 2.450… = 1.36
A: about 1.4 times larger (2 s.f.), because none of the annulus’s mass is close to the axis
Watch out: use the unrounded mass in part (b). Using 78 kg is fine here, but rounding early in longer problems can shift the final digit.
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