The Moment-of-Inertia Tensor

Derivation

The Moment-of-Inertia Tensor

D-035 Home PU-101 Threads matter · symmetry Depends on Angular Momentum, Torque, and Central Forces, eigenvalue-eigenvector-decomposition

Statement

For a rigid body rotating about a fixed point with angular velocity ω, the angular momentum about that point is a linear function of ω: L_j = I_jk ω_k, where the 3×3 array I_jk = Σ_i m_i(r_i²δ_jk − x_i,jx_i,k) is the real, symmetric moment-of-inertia tensor. Being symmetric, it possesses three mutually orthogonal principal axes in which it is diagonal, with real non-negative principal moments I₁, I₂, I₃.

Why it matters

Rotational inertia is not a single number. A rigid body resists angular acceleration differently about different axes, and in general its angular momentum L is not parallel to its angular velocity ω. The inertia tensor is the object that encodes this directional dependence and turns Newtonian rotation into a clean linear-algebra problem.

It is the gateway to rigid-body dynamics: the tumbling of satellites and asteroids, the wobble of an unbalanced wheel, gyroscopic precession, the free-rotation instability of the intermediate axis (the “tennis-racket theorem”), and the normal modes of rotating machinery all follow from the structure of this one symmetric tensor.

Assumptions

Rigid body.If interparticle distances change, v_i = ω×r_i fails, the mass distribution is time-dependent, and I_jk is no longer a fixed body-frame constant.

Common instantaneous angular velocity.Every particle shares one ω; drop this and each element rotates independently — the body is no longer rigid and no single tensor maps ω→L.

Reference point is fixed (or is the centre of mass).If the origin both moves and is not the centre of mass, cross terms couple translation to rotation and L about that point is not simply Iω.

Non-relativistic, classical mass points.At relativistic speeds the additive L = Σ r×p with p = m v breaks; mass and momentum acquire velocity dependence and the bilinear form is no longer m_i-weighted Euclidean geometry.

Euclidean 3-space with a fixed orthonormal basis.The identity a×(b×c) = b(a·c) − c(a·b) and the appearance of δ_jk presuppose a flat metric; in curved or non-Cartesian coordinates the components carry extra metric factors.

Derivation

L = Σ_i r_i × p_i = Σ_i m_i (r_i × v_i)

Total angular momentum about the fixed point is the additive sum of single-particle terms (prior result: angular-momentum-and-torque). A

v_i = ω × r_i

For a rigid body pivoting about the origin, each particle's velocity is the rotational velocity field; interparticle distances are constant so no radial term survives. A

L = Σ_i m_i r_i × (ω × r_i)

Substitute step 2 into step 1. A

r_i × (ω × r_i) = ω(r_i·r_i) − r_i(r_i·ω) = r_i²ω − (r_i·ω) r_i

Apply the BAC–CAB vector triple-product identity a×(b×c) = b(a·c) − c(a·b) with a = c = r_i, b = ω. B

L = Σ_i m_i [ r_i²ω − (r_i·ω) r_i ]

Insert step 4 into step 3. Note L is linear in ω but the second term drags in the direction of r_i — this is why L generally tilts away from ω. A

L_j = Σ_i m_i [ r_i²ω_j − x_i,j(Σ_k x_i,kω_k) ]

Take the j-th Cartesian component and write the dot product as a sum over k: r_i·ω = Σ_k x_i,kω_k. B

ω_j = Σ_k δ_jkω_k ⇒ L_j = Σ_k [ Σ_i m_i(r_i²δ_jk − x_i,jx_i,k) ] ω_k

Rewrite ω_j with the Kronecker delta so both terms carry a common ω_k, then factor it out. The bracket depends only on the mass geometry. C

L_j = Σ_k I_jk ω_k , I_jk ≡ Σ_i m_i(r_i²δ_jk − x_i,jx_i,k)

Define the coefficient array as the inertia tensor. For a continuous body replace Σ_i m_i → ∫ ρ dV. A

I_kj = Σ_i m_i(r_i²δ_kj − x_i,kx_i,j) = I_jk

Swap j↔k: δ_jk and the product x_i,jx_i,k are both symmetric, so I is a real symmetric matrix. B

I = Q Λ Q^⊤ , Λ = diag(I₁, I₂, I₃) , Q^⊤Q = 𝟙

A real symmetric matrix is orthogonally diagonalizable with real eigenvalues (prior result: eigenvalue–eigenvector decomposition / spectral theorem). The eigenvectors are the principal axes; the eigenvalues are the principal moments. C

n^⊤ I n = Σ_i m_i[ r_i² − (n·r_i)² ] = Σ_i m_i d_i,⊥² ≥ 0

For any unit vector n, the quadratic form equals the sum of m_i times squared perpendicular distance to the axis n — manifestly non-negative. Hence I is positive semidefinite and every principal moment I_a ≥ 0. C

Result

L_j = Σ_k I_jk ω_k , I_jk = Σ_i m_i(r_i²δ_jk − x_i,jx_i,k) = ∫ ρ(r)(r²δ_jk − x_jx_k) dV

Reading. Angular momentum is a linear map of angular velocity. The diagonal entries I_xx = Σm(y²+z²) are ordinary moments of inertia about each axis; the off-diagonal I_xy = −Σm xy are the products of inertia that measure mass imbalance and tilt L off ω. Because I is symmetric there always exists a body-fixed orthonormal frame (the principal axes) in which the products vanish and L_a = I_aω_a decouples axis by axis.

Units check. I_jk is mass × length²: [kg][m²] = kg·m². Then Iω has units kg·m² × s⁻¹ = kg·m²·s⁻¹, the units of angular momentum. Consistent.

Limiting cases

Single point mass on the rotation axis: d_⊥ = 0 so that axis contributes zero moment — the tensor is rank-deficient along it.
Rotation about a principal axis: products of inertia vanish, L = I_aω is parallel to ω, and the body is dynamically balanced.
Spherical top (I₁=I₂=I₃=I): I_jk = Iδ_jk, so L = Iω for every axis — rotation behaves like the scalar case.
Planar (lamina) body in the xy plane: z=0 gives the perpendicular-axis relation I_zz = I_xx + I_yy.
Thin symmetric top (I₁=I₂≠I₃): the perpendicular principal axes are degenerate and any pair orthogonal to the symmetry axis will serve.

Breaks when

The body deforms. For a non-rigid or fluid body (a spinning water balloon, a rotating gas cloud) the mass geometry changes with time, so I_jk is not a body-frame constant and dL/dt = I dω/dt gains extra (dI/dt)ω terms.
The reference point is arbitrary and accelerating. Taking moments about a point that is neither fixed nor the centre of mass introduces translation–rotation coupling; L about that point is no longer Iω and the parallel-axis shortcut does not apply.
Relativistic rotation. When rim speeds approach c, additive Σ r×mv and rigidity itself (Born rigidity is over-constrained) fail; the Newtonian tensor is superseded.
Quantum / sub-atomic angular momentum. For spin there is no mass distribution to integrate over; intrinsic angular momentum is not Iω of any classical body.

Failure modes

Assuming L ∥ ω always. Students carry over the scalar L = Iω; off a principal axis L tilts, which is the whole reason bearings feel a rotating reaction load.
Dropping the minus sign on products of inertia. Writing I_xy = +Σm xy instead of −Σm xy flips off-diagonal signs and corrupts the principal-axis directions.
Confusing I_xx = Σm(y²+z²) with Σm x². The diagonal element excludes its own coordinate; using x² is a classic sign/index slip.
Adding tensors computed about different origins. Inertia tensors only add when referred to the same point; combine sub-bodies via the tensor parallel-axis theorem, not naive summation.
Treating I as a fixed lab-frame matrix. The constant tensor lives in the body frame; in the lab it rotates as I(t) = R(t) I_body R(t)^⊤.
Forgetting units on the tensor. Reporting a bare number without kg·m² hides factor-of-mass or length errors.

Discussion

The deep content of this result is that rotational inertia is a rank-2 tensor, not a scalar. The same object I that maps ω→L also delivers the rotational kinetic energy as a quadratic form, T = ½ ω^⊤ I ω = ½ Σ_a I_aω_a² in principal axes. The surfaces ω^⊤ I ω = const are ellipsoids (the inertia/energy ellipsoid), whose semi-axes are set by 1/√I_a; this geometric picture is the basis of Poinsot's construction for torque-free motion.

Symmetry of the tensor is doing real work. Because I = I^⊤, the spectral theorem guarantees a real orthogonal eigenbasis — the principal axes are always mutually perpendicular and the principal moments always real and non-negative. Any geometric symmetry of the body (a mirror plane, an n-fold axis with n≥3) forces off-diagonal products to vanish and pins principal axes to symmetry directions, which is why a well-designed flywheel spins without shaking its bearings.

Dynamically, feeding L = Iω into dL/dt = τ and transforming to the rotating body frame yields Euler's equations, I₁ỏ₁ = (I₂−I₃)ω₂ω₃ and cyclic. Their fixed points and stability follow entirely from the ordering of I₁,I₂,I₃: rotation about the largest and smallest principal axes is stable, about the intermediate axis unstable — the tennis-racket / Dzhanibekov effect, a direct downstream consequence of the tensor derived here.

Coordinate-freely, I is a symmetric second-rank tensor: under a rotation R its components transform as I' = R I R^⊤, and its two rotational invariants — the trace tr I = 2Σm_ir_i² and determinant det I = I₁I₂I₃ — are frame-independent. This is the same tensorial structure that recurs across physics (stress, quadrupole moment, polarizability): a symmetric bilinear form whose eigen-decomposition names the “natural” directions of the system.

Common misconceptions. A body does not have “a” moment of inertia — it has a tensor, and the scalar I you memorised is only its value about one chosen axis. Nonzero angular momentum off a principal axis does not mean a torque is acting: a freely rotating body can carry a fixed lab-frame L while ω traces a cone. And “balanced” means the spin axis is a principal axis (products of inertia zero), which is stronger than merely passing through the centre of mass.

Worked examples

Example 1 — L is not parallel to ω. Two point masses m sit at r₁ = (a,0,h) and r₂ = (−a,0,−h), and the pair is spun about the z-axis, ω = ωẑ. Find L.

I_zz = Σm(x²+y²) = m a² + m a² = 2m a²

Diagonal element about the spin axis. A

I_xz = −Σm x z = −[ m(a)(h) + m(−a)(−h) ] = −2m a h

Product of inertia; the two masses add with the same sign, so it does not cancel. B

I_yz = −Σm y z = 0 , I_xy = −Σm x y = 0

Both masses have y = 0. A

L = Iω ⇒ L_x = I_xzω = −2m a h ω , L_y = 0 , L_z = I_zzω = 2m a²ω

Only the third column of I acts, since ω points along z. B

m = 0.5 kg, a = 0.30 m, h = 0.40 m, ω = 10 rad/s

Now insert numbers. A

L_x = −2(0.5)(0.30)(0.40)(10) = −1.2 , L_z = 2(0.5)(0.30)²(10) = 0.9

Arithmetic in SI. A

L = (−1.2, 0, 0.9) kg·m²·s⁻¹ , |L| = 1.5 kg·m²·s⁻¹

Reading. Although ω points purely along z, L tilts by arctan(1.2/0.9) = 53° from the axis. The nonzero product of inertia I_xz is exactly the imbalance a bearing would feel as a rotating side-load once per revolution.

Units check. kg × m × m × s⁻¹ = kg·m²·s⁻¹. Correct.

Example 2 — Finding principal axes by diagonalization. Two masses m = 2.0 kg lie at (a,a,0) and (−a,−a,0) with a = 0.50 m. Find the principal moments and axes.

I_xx = Σm(y²+z²) = 2m a² , I_yy = Σm(x²+z²) = 2m a²

Each mass has z = 0; both contribute m a². A

I_xy = −Σm x y = −[ m(a)(a) + m(−a)(−a) ] = −2m a² , I_zz = Σm(x²+y²) = 4m a²

Off-diagonal term does not vanish; z is already a principal axis (both I_xz=I_yz=0). B

I = m a² [ [2, −2, 0], [−2, 2, 0], [0, 0, 4] ]

Assemble the symmetric tensor in symbolic form. A

det( [ [2−λ, −2], [−2, 2−λ] ] ) = (2−λ)² − 4 = λ(λ−4) = 0

Diagonalize the upper 2×2 block; roots λ = 0 and λ = 4 (in units of m a²). C

λ=0: n = (1,1,0)/√2 ; λ=4: n = (1,−1,0)/√2 ; λ=4: n = (0,0,1)

Eigenvectors give the principal axes: the line through both masses (y=x), its in-plane perpendicular, and z. C

m a² = (2.0)(0.50)² = 0.50 kg·m²

Insert numbers into the common factor. A

I₁ = 0 (axis y=x) , I₂ = 4m a² = 2.0 kg·m² , I₃ = 4m a² = 2.0 kg·m²

Reading. One principal moment is exactly zero — the axis y=x passes through both point masses, so their perpendicular distance is zero. The other two moments are degenerate, making this a symmetric top; any axis perpendicular to y=x is also principal.

Units check. kg × m² = kg·m². Correct.

Problems

A single point mass m = 3.0 kg sits at (2.0, 0, 0) m. Write its full inertia tensor about the origin.
Solution
x=2, y=z=0. Diagonal: I_xx=m(y²+z²)=0; I_yy=m(x²+z²)=3(4)=12; I_zz=m(x²+y²)=12. All products vanish (any two coordinates include a zero). So I = diag(0, 12, 12) kg·m². The x-axis moment is zero because the mass lies on it.
A thin uniform rod of mass M = 1.2 kg and length L = 1.0 m lies along the x-axis, centred at the origin. Find I_xx, I_yy, I_zz.
Solution
Linear density λ = M/L. I_xx = ∫(y²+z²)dm = 0 (rod on axis). I_yy = I_zz = ∫_−L/2^L/2 x²λ dx = λ [x³/3]_−L/2^L/2 = λL³/12 = ML²/12. Numerically = (1.2)(1.0)²/12 = 0.10 kg·m². Result: I = diag(0, 0.10, 0.10) kg·m².
Four equal masses m = 1.0 kg sit at the corners of a square of side 2a (a = 0.50 m) in the xy plane, at (±a, ±a, 0). Show all products of inertia vanish and give the tensor. Is L ∥ ω for spin about z?
Solution
I_xy = −Σm xy: the four terms are m a²(+, −, +, −) summing to zero; likewise I_xz=I_yz=0 since z=0. Diagonals: I_xx=Σm y²=4m a², I_yy=4m a², I_zz=Σm(x²+y²)=8m a². With m a²=0.25: I=diag(1.0, 1.0, 2.0) kg·m². Diagonal in these axes, so yes — spinning about z gives L=I_zzωẑ, parallel to ω.
For Worked Example 1's configuration, at what non-zero value of the ratio h/a would L make a 45° angle with the z-axis?
Solution
tanθ = |L_x|/L_z = (2m a h ω)/(2m a² ω) = h/a. Setting θ=45° gives tan45°=1, so h/a = 1. The tilt of L depends only on the geometric ratio h/a, independent of m and ω.
A rigid body has principal moments I₁=2.0, I₂=5.0, I₃=6.0 kg·m² and spins with ω = (3, 0, 4) rad/s expressed in the principal frame. Find L, the rotational kinetic energy T, and the angle between L and ω.
Solution
In principal axes L_a=I_aω_a: L=(2×3, 5×0, 6×4)=(6, 0, 24) kg·m²·s⁻¹, magnitude |L|=√(36+576)=24.7. Energy T=½ΣI_aω_a²=½[2(9)+0+6(16)]=½(18+96)=57 J. Angle: cosφ=(L·ω)/(|L||ω|), with L·ω=6(3)+24(4)=18+96=114, |ω|=5, |L|=24.7, so cosφ=114/(24.7×5)=0.923, φ≈22.6°. Note L·ω=2T=114, a useful check.