Rotational Inertia: Study Guide (Physics C: Mechanics 5.4)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 5.4 called Rotational Inertia. This guide is the one for the calculus-based course. As well as adding up point masses, you will derive rotational inertias by integration. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise.

Resistance to changes in rotation

Mass measures how hard it is to change an object’s velocity. Rotational inertia I does the same job for rotation: it measures how hard it is to change a rigid system’s angular velocity about a given axis. Topic 5.6 will show that the same net torque gives a smaller angular acceleration when I is larger.

I depends on two things:

• the mass of the system, and
• how that mass is spread out relative to the axis. Mass far from the axis counts much more than mass close to it.

So I is not a fixed property of an object. It belongs to an object and an axis. Always say which axis.

Point masses: I = mr²

A small object of mass m, a perpendicular distance r from the axis, has

I = mr² (unit: kg·m²)

For a collection of small objects about the same axis, add their contributions:

I = Σ mᵢrᵢ²

Here r is the perpendicular distance to the axis, not the distance to the origin. If the axis is the y-axis, a mass at (x, y) has r = |x|. If the axis is the z-axis, r = √(x² + y²). A mass on the axis adds nothing.

Because r is squared, doubling a mass’s distance from the axis makes its contribution four times as big.

Continuous objects: I = ∫r² dm

A solid object is a collection of tiny masses dm. The sum becomes an integral:

I = ∫ r² dm

where r is the perpendicular distance from dm to the axis. The method is always the same:

1. Choose an element dm whose parts are all the same distance r from the axis. For a thin rod, a short slice of length dx. For a disk about its centre, a thin ring of radius r.
2. Write dm in terms of a density and a coordinate. Rod: dm = λ dx, with λ the mass per unit length (kg/m). Flat disk: dm = σ dA = σ(2πr dr), with σ the mass per unit area (kg/m²).
3. Integrate r² dm over the whole object, with limits that cover it once.
4. Use the total mass (M = ∫dm) to replace λ or σ by M.

Deriving the standard results

Uniform rod, axis perpendicular to the rod at any point. Put the rod along x from 0 to L, with λ = M/L, and the axis at x = a. A slice at x is a distance |x − a| from the axis, and r² = (x − a)²:

I = ∫₀ᴸ (x − a)² (M/L) dx = M(L²/3 − La + a²)

• Axis at the centre, a = L/2: I = ML²/12.
• Axis at one end, a = 0: I = ML²/3, four times larger.

Thin hoop or thin cylindrical shell, axis along its centre line. Every bit of mass is at the same distance R from the axis, so I = ∫R² dm = R²∫dm = MR². The length of a cylindrical shell along the axis does not matter; only the distance from the axis does.

Uniform disk (or solid cylinder) about its central axis. Use rings. σ = M/(πR²), and a ring of radius r and width dr has dm = σ(2πr dr):

I = ∫₀ᴿ r² σ 2πr dr = 2πσ R⁴/4 = ½MR²

Annular ring (a disk with a hole), inner radius R₁, outer radius R₂. Same rings, from R₁ to R₂, with σ = M/[π(R₂² − R₁²)]:

I = 2πσ (R₂⁴ − R₁⁴)/4 = ½M(R₁² + R₂²)

Check the limits: R₁ → 0 gives the disk, ½MR₂². R₁ → R₂ gives the hoop, MR₂². Figure 2 shows how I changes between the two.

Why a hoop beats a disk. A hoop and a disk of the same mass and radius do not have the same I. In the hoop all the mass sits at R. In the disk much of it is near the centre, where r² is small. So I_hoop = MR² is twice I_disk = ½MR².

What you derive and what is given. In this course, calculus derivations are expected for thin rods (uniform or not) about any axis perpendicular to the rod, and for thin cylindrical shells, disks and other shapes built from coaxial rings or shells, about their central axis. For other shapes, such as a solid sphere, expect the rotational inertia to be given; you only need to reason about it qualitatively.

The parallel axis theorem

If you know I_cm about an axis through the centre of mass, you can find I about any parallel axis a distance d away:

I = I_cm + Md²

Check with the rod: ML²/12 + M(L/2)² = ML²/3, which matches the integral. In fact the rod result M(L²/3 − La + a²) is exactly ML²/12 + M(a − L/2)².

Two consequences:

• Md² is never negative, so I is smallest about an axis through the centre of mass. Every parallel axis gives a larger I.
• The theorem always starts from the centre-of-mass axis. To go from one off-centre axis to another, go through I_cm.

Worked example 1: a three-arm spinner

Question. A spinner toy has three small 0.050 kg weights on light arms, each 0.12 m from the centre, spaced 120° apart. Find I about (a) an axis through the centre, perpendicular to the toy, and (b) a parallel axis through one of the weights. (c) Check (b) with the parallel axis theorem.

1. (a) All three weights are 0.12 m from the axis: I = 3 × 0.050 × 0.12² = 2.16 × 10⁻³ kg·m².
2. (b) The axis weight has r = 0. The other two are one side of the equilateral triangle away: s = 0.12√3 = 0.208 m, so s² = 3(0.12)² = 0.0432 m². I = 2 × 0.050 × 0.0432 = 4.32 × 10⁻³ kg·m².
3. (c) The centre of mass is at the centre (symmetry). M = 0.15 kg and d = 0.12 m: I = 2.16 × 10⁻³ + 0.15 × 0.12² = 2.16 × 10⁻³ + 2.16 × 10⁻³ = 4.32 × 10⁻³ kg·m². ✓

Interpretation. Moving the axis from the centre to a weight doubles I. The centre-of-mass axis gives the smaller value, as it must.

Worked example 2: a nonuniform rod

Question. A tapered rod of length L lies along x from 0 to L. Its mass per unit length is λ(x) = λ₀(1 + x/L), so the end at x = L is twice as dense as the end at x = 0. (a) Find its mass M. (b) Derive I about a perpendicular axis through x = 0, in terms of M and L. (c) Find the centre of mass and I_cm. (d) Find I about the dense end, x = L. (e) Evaluate for M = 0.90 kg and L = 1.2 m.

1. (a) M = ∫₀ᴸ λ₀(1 + x/L) dx = λ₀(L + L/2) = 3λ₀L/2, so λ₀ = 2M/(3L).
2. (b) I₀ = ∫₀ᴸ x² λ₀(1 + x/L) dx = λ₀(L³/3 + L³/4) = 7λ₀L³/12. Substituting λ₀: I₀ = 7ML²/18 ≈ 0.389ML².
3. (c) x_cm = (1/M)∫₀ᴸ x λ dx = λ₀(L²/2 + L²/3)/M = 5L/9. Then I_cm = I₀ − M(5L/9)² = 7ML²/18 − 25ML²/81 = 13ML²/162 ≈ 0.0802ML².
4. (d) The dense end is L − 5L/9 = 4L/9 from the centre of mass: I_L = 13ML²/162 + M(4L/9)² = 13ML²/162 + 32ML²/162 = 5ML²/18 ≈ 0.278ML². (Integrating ∫(L − x)² λ dx directly gives the same.)
5. (e) I₀ = 0.504 kg·m², I_cm = 0.104 kg·m², I_L = 0.360 kg·m², with x_cm = 0.667 m.

Check. A uniform rod of the same M and L has I = 0.432 kg·m² about either end. Spinning about the light end, the tapered rod has more mass far away, so its I is larger (0.504). Spinning about the dense end, the heavy part is close, so its I is smaller (0.360). Its I_cm is just below ML²/12 = 0.108 kg·m², because its mass is slightly more bunched around the centre of mass.

Worked example 3: an annular flywheel

Question. A flywheel is a flat steel ring of mass 4.0 kg, inner radius 0.10 m and outer radius 0.25 m. Find I about its central axis and compare it with a solid disk and a thin hoop of the same mass and outer radius.

1. Annulus: I = ½M(R₁² + R₂²) = ½ × 4.0 × (0.010 + 0.0625) = 0.145 kg·m².
2. Solid disk: ½MR₂² = ½ × 4.0 × 0.0625 = 0.125 kg·m².
3. Hoop: MR₂² = 4.0 × 0.0625 = 0.250 kg·m².

Interpretation. Cutting out the centre and keeping the mass the same pushes mass outwards, so I rises from 0.125 to 0.145 kg·m², a ratio R₁/R₂ = 0.4 on Figure 2. Flywheel designers put as much of the mass as possible near the rim for this reason.

Linking to experiments

You cannot see I directly. In Topic 5.6 you will measure it from how a known torque changes the rotation. A clear test of I = Σmr² is to clamp two equal masses at the same distance r on either side of a light rotating bar, measure I for several values of r, and plot I against r². The prediction is a straight line: slope 2m, intercept equal to I of the bar and hub alone. Practice Question 6 uses data like this.

Common misconceptions

• “I depends only on mass.” Mass distribution and the axis matter as much (Worked example 1).
• Using the distance to the origin. r is the perpendicular distance to the axis.
• Using r instead of r². A mass twice as far contributes four times as much.
• Choosing dm badly. All of dm must be the same distance from the axis; a thin slice across a disk does not work for its central axis.
• Forgetting to replace λ or σ by M. Answers should be in terms of M and the lengths.
• Jumping between two off-centre axes with Md². The parallel axis theorem starts from I_cm (practice Question 7).
• “Same mass and radius means same I.” A hoop has twice the I of a disk.
• Adding I values about different axes. Sum only about one axis.

Where this leads

Topic 5.5, Rotational Equilibrium and Newton’s First Law in Rotational Form, uses net torque to describe steady rotation, and Topic 5.6 combines net torque and I to find angular acceleration. Rotational inertia also returns in Unit 6, for rotational kinetic energy. Try the practice questions now, then use the revision notes and the checklist. You can also go back to Topic 5.3, Torque, or the course roadmap.