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 reference point with common angular velocity \(\vec{\omega}\), the angular momentum about that point is a linear function of \(\vec{\omega}\), \(L_j=\sum_k I_{jk}\,\omega_k\), where the real symmetric array \(I_{jk}=\sum_i m_i\left(r_i^2\,\delta_{jk}-x_{i,j}x_{i,k}\right)\) is the moment-of-inertia tensor. Being real and symmetric it is orthogonally diagonalizable, possessing three mutually perpendicular principal axes with real non-negative principal moments \(I_1,I_2,I_3\) along which \(\vec{L}\) is parallel to \(\vec{\omega}\).

Why it matters

Rotational inertia is not a single number. A rigid body resists angular acceleration differently about different axes, and in general \(\vec{L}\) is not parallel to \(\vec{\omega}\). The inertia tensor is the object that encodes this directional dependence and turns Newtonian rotation into a clean linear-algebra problem: eigenvalues are the principal moments, eigenvectors the principal axes.

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 / Dzhanibekov effect), and the balancing of rotating machinery all follow from the structure of this one symmetric tensor.

Assumptions

Rigid body.If interparticle distances change, \(\vec{v}_i=\vec{\omega}\times\vec{r}_i\) fails, the mass geometry is time-dependent, and \(I_{jk}\) is no longer a fixed body-frame constant.

A single common angular velocity.Every particle shares one \(\vec{\omega}\); drop this and each element rotates independently, so no single tensor maps \(\vec{\omega}\mapsto\vec{L}\).

Reference point is fixed, or is the centre of mass.If the origin both accelerates and is not the centre of mass, translation–rotation cross terms appear and \(\vec{L}\) about that point is not simply \(\mathbf{I}\vec{\omega}\).

Non-relativistic classical mass points.At relativistic rim speeds the additive \(\vec{L}=\sum \vec{r}\times m\vec{v}\) with \(\vec{p}=m\vec{v}\) breaks; mass and momentum become velocity-dependent and the \(m_i\)-weighted Euclidean form is invalid.

Euclidean 3-space with a fixed orthonormal basis.The triple-product identity and the appearance of \(\delta_{jk}\) presuppose a flat metric; in curved or non-Cartesian coordinates the components carry extra metric factors.

Derivation

\[ \vec{L}=\sum_i \vec{r}_i\times\vec{p}_i=\sum_i m_i\,(\vec{r}_i\times\vec{v}_i) \]

Total angular momentum about the fixed point is the additive sum of single-particle terms (prior result: angular momentum and torque), with \(\vec{p}_i=m_i\vec{v}_i\). A

\[ \vec{v}_i=\vec{\omega}\times\vec{r}_i \]

For a rigid body pivoting about the origin, each particle's velocity is the rotational velocity field; rigidity forbids any radial term. This is where the rigid-body assumption enters. A

\[ \vec{L}=\sum_i m_i\,\vec{r}_i\times(\vec{\omega}\times\vec{r}_i) \]

Substitute step 2 into step 1. All the motion is now carried by the single vector \(\vec{\omega}\). A

\[ \vec{r}_i\times(\vec{\omega}\times\vec{r}_i)=\vec{\omega}\,(\vec{r}_i\cdot\vec{r}_i)-\vec{r}_i\,(\vec{r}_i\cdot\vec{\omega})=r_i^2\,\vec{\omega}-(\vec{r}_i\cdot\vec{\omega})\,\vec{r}_i \]

Apply the BAC–CAB vector triple-product identity \(\vec{a}\times(\vec{b}\times\vec{c})=\vec{b}(\vec{a}\cdot\vec{c})-\vec{c}(\vec{a}\cdot\vec{b})\) with \(\vec{a}=\vec{c}=\vec{r}_i,\ \vec{b}=\vec{\omega}\). B

\[ \vec{L}=\sum_i m_i\left[\,r_i^2\,\vec{\omega}-(\vec{r}_i\cdot\vec{\omega})\,\vec{r}_i\,\right] \]

Insert step 4 into step 3. The result is linear in \(\vec{\omega}\), but the second term drags in the direction of \(\vec{r}_i\) — the geometric reason \(\vec{L}\) generally tilts away from \(\vec{\omega}\). A

\[ L_j=\sum_i m_i\left[\,r_i^2\,\omega_j-x_{i,j}\Big(\sum_k x_{i,k}\,\omega_k\Big)\right] \]

Take the \(j\)-th Cartesian component and expand the dot product as \(\vec{r}_i\cdot\vec{\omega}=\sum_k x_{i,k}\,\omega_k\). B

\[ \omega_j=\sum_k \delta_{jk}\,\omega_k \;\Rightarrow\; L_j=\sum_k\left[\sum_i m_i\big(r_i^2\,\delta_{jk}-x_{i,j}x_{i,k}\big)\right]\omega_k \]

Rewrite \(\omega_j\) with the Kronecker delta so both terms carry a common \(\omega_k\), then factor it out. The bracket depends only on the mass geometry. C

\[ I_{jk}\equiv\sum_i m_i\big(r_i^2\,\delta_{jk}-x_{i,j}x_{i,k}\big)\;\xrightarrow{\text{continuum}}\;\int \rho(\vec{r})\big(r^2\,\delta_{jk}-x_j x_k\big)\,dV \]

Define the coefficient array as the inertia tensor, and replace the particle sum by a mass-density integral for a continuous body. Pure definition. A

\[ L_j=\sum_k I_{jk}\,\omega_k \qquad\Longleftrightarrow\qquad \vec{L}=\mathbf{I}\,\vec{\omega} \]

Combine steps 7 and 8: the map is manifestly linear, so doubling \(\vec{\omega}\) doubles \(\vec{L}\). A

\[ I_{kj}=\sum_i m_i\big(r_i^2\,\delta_{kj}-x_{i,k}x_{i,j}\big)=I_{jk} \]

Swap \(j\leftrightarrow k\): both \(\delta_{jk}\) and the product \(x_{i,j}x_{i,k}\) are symmetric, so \(\mathbf{I}\) is a real symmetric matrix. B

\[ \mathbf{I}=\mathbf{Q}\,\boldsymbol{\Lambda}\,\mathbf{Q}^{\mathsf T},\quad \boldsymbol{\Lambda}=\mathrm{diag}(I_1,I_2,I_3),\quad \mathbf{Q}^{\mathsf T}\mathbf{Q}=\mathbf{1} \]

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

\[ \hat{\mathbf n}^{\mathsf T}\mathbf{I}\,\hat{\mathbf n}=\sum_i m_i\left[\,r_i^2-(\hat{\mathbf n}\cdot\vec{r}_i)^2\,\right]=\sum_i m_i\,d_{i,\perp}^2\;\ge 0 \]

For any unit vector \(\hat{\mathbf n}\), the quadratic form equals \(\sum m_i\) times squared perpendicular distance to the axis — manifestly non-negative. Hence \(\mathbf{I}\) is positive semidefinite and every \(I_a\ge 0\). C

Result

\[ \vec{L}=\mathbf{I}\,\vec{\omega},\qquad I_{jk}=\sum_i m_i\big(r_i^2\,\delta_{jk}-x_{i,j}x_{i,k}\big)=\int \rho(\vec{r})\big(r^2\,\delta_{jk}-x_j x_k\big)\,dV,\qquad \mathbf{I}=\mathbf{I}^{\mathsf T} \]

Reading. Angular momentum is a linear map of angular velocity. The diagonal entries \(I_{xx}=\sum m(y^2+z^2)\) are ordinary moments of inertia about each axis; the off-diagonal entries \(I_{xy}=-\sum m\,xy\) are the products of inertia that measure mass imbalance and tilt \(\vec{L}\) off \(\vec{\omega}\). Because \(\mathbf{I}\) is symmetric there always exists a body-fixed orthonormal frame (the principal axes) in which the products vanish and \(L_a=I_a\omega_a\) decouples axis by axis.

Units check. Each term of \(I_{jk}\) is mass times length squared, so \([\mathbf{I}]=\mathrm{kg\,m^2}\). Then \([\mathbf{I}\vec{\omega}]=\mathrm{kg\,m^2}\cdot\mathrm{s^{-1}}=\mathrm{kg\,m^2\,s^{-1}}\), matching \([\vec{r}\times\vec{p}]=\mathrm{m}\cdot\mathrm{kg\,m\,s^{-1}}\). Consistent.

Limiting cases

Single point mass on the rotation axis: \(d_\perp=0\), so that axis contributes zero moment — the tensor is rank-deficient along it.
Rotation about a principal axis: products of inertia vanish, \(\vec{L}=I_a\vec{\omega}\) is parallel to \(\vec{\omega}\), and the body is dynamically balanced.
Spherical top (\(I_1=I_2=I_3=I\)): \(I_{jk}=I\,\delta_{jk}\), so \(\vec{L}=I\vec{\omega}\) for every axis — rotation behaves like the scalar case and every axis is principal.
Planar lamina in the \(xy\) plane: \(z=0\) gives the perpendicular-axis relation \(I_{zz}=I_{xx}+I_{yy}\).
Symmetric top (\(I_1=I_2\neq I_3\)): the two perpendicular principal moments are degenerate and any pair of orthogonal axes in that plane will serve as principal axes.

Breaks when

The body deforms. For a non-rigid or fluid body (a spinning water balloon, a rotating gas cloud, a figure skater pulling in their arms), the mass geometry changes with time, so \(I_{jk}\) is not a body-frame constant and \(d\vec{L}/dt=\mathbf{I}\,d\vec{\omega}/dt\) gains extra \((d\mathbf{I}/dt)\vec{\omega}\) 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; \(\vec{L}\) about that point is not \(\mathbf{I}\vec{\omega}\) and one must add the orbital piece \(\vec{R}_{\rm cm}\times M\vec{V}_{\rm cm}\).
Relativistic rotation. When rim speeds approach \(c\), the additive \(\sum \vec{r}\times m\vec{v}\) and rigidity itself (Born rigidity is over-constrained) fail; the Newtonian tensor is superseded by the relativistic angular-momentum tensor.
Quantum / sub-atomic angular momentum. For intrinsic spin there is no mass distribution to integrate over; spin angular momentum is not \(\mathbf{I}\vec{\omega}\) of any classical body, and for a molecule the tensor survives only as the rotational-constant structure \(B=\hbar^2/2I\) in the rigid-rotor Hamiltonian.

Failure modes

Assuming \(\vec{L}\parallel\vec{\omega}\) always. Students carry over the scalar \(L=I\omega\); off a principal axis \(\vec{L}\) tilts, which is the whole reason bearings feel a rotating reaction load.
Dropping the minus sign on products of inertia. Writing \(I_{xy}=+\sum m\,xy\) instead of \(-\sum m\,xy\) flips off-diagonal signs and corrupts the principal-axis directions.
Confusing \(I_{xx}=\sum m(y^2+z^2)\) with \(\sum m\,x^2\). The diagonal element excludes its own coordinate; using \(x^2\) is a classic 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 \(I_{jk}=I_{jk}^{\rm cm}+M(d^2\delta_{jk}-d_j d_k)\), not naive summation.
Treating \(\mathbf{I}\) as a fixed lab-frame matrix. The constant tensor lives in the body frame; in the lab it rotates as \(\mathbf{I}(t)=\mathbf{R}(t)\,\mathbf{I}_{\rm body}\,\mathbf{R}(t)^{\mathsf T}\).
Forgetting units on the tensor. Reporting a bare number without \(\mathrm{kg\,m^2}\) 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 \(\mathbf{I}\) that maps \(\vec{\omega}\mapsto\vec{L}\) also delivers the rotational kinetic energy as a quadratic form, \(T=\tfrac12\,\vec{\omega}^{\mathsf T}\mathbf{I}\,\vec{\omega}=\tfrac12\sum_a I_a\omega_a^2\) in principal axes. The level surfaces \(\vec{\omega}^{\mathsf T}\mathbf{I}\,\vec{\omega}=\text{const}\) are ellipsoids (the inertia / energy ellipsoid) whose semi-axes scale as \(1/\sqrt{I_a}\); this geometric picture is the basis of Poinsot's construction for torque-free motion.

Symmetry of the tensor is doing real work. Because \(\mathbf{I}=\mathbf{I}^{\mathsf T}\), the spectral theorem guarantees a real orthonormal eigenbasis — the principal axes are always mutually perpendicular and the principal moments always real and non-negative. This mathematical symmetry holds for any lump of matter. A geometric symmetry of the body (a mirror plane, or an \(n\)-fold axis with \(n\ge 3\)) does something sharper: it forces off-diagonal products to vanish and pins principal axes to the symmetry directions, which is why a well-designed flywheel spins without shaking its bearings.

Dynamically, feeding \(\vec{L}=\mathbf{I}\vec{\omega}\) into \(d\vec{L}/dt=\vec{\tau}\) and transforming to the rotating body frame yields Euler's equations, \(I_1\dot\omega_1-(I_2-I_3)\omega_2\omega_3=\tau_1\) and cyclic permutations. Their fixed points and stability follow entirely from the ordering of \(I_1,I_2,I_3\): torque-free rotation about the largest and smallest principal axes is stable, about the intermediate axis unstable — the tennis-racket theorem, a direct downstream consequence of the tensor derived here.

Coordinate-freely, \(\mathbf{I}\) is a symmetric second-rank tensor: under a rotation \(\mathbf{R}\) its components transform as \(\mathbf{I}'=\mathbf{R}\,\mathbf{I}\,\mathbf{R}^{\mathsf T}\), i.e. \(I'_{jk}=\sum_{pq}R_{jp}R_{kq}I_{pq}\), and its rotational invariants — the trace \(\mathrm{tr}\,\mathbf{I}=2\sum_i m_i r_i^2\) and determinant \(\det\mathbf{I}=I_1I_2I_3\) — are frame-independent. Under \(SO(3)\) a symmetric rank-2 tensor decomposes into a scalar trace (the isotropic spin-0 part) and a symmetric traceless spin-2 part carrying the ellipsoid's shape — the identical structure that recurs across physics in the stress tensor, the electric quadrupole moment, and the polarizability.

Common misconceptions. A body does not have "a" moment of inertia — it has a tensor, and the scalar \(I=\hat{\mathbf n}\cdot\mathbf{I}\hat{\mathbf n}\) you memorised is only its value about one chosen axis. Nonzero angular momentum off a principal axis does not imply an applied torque: a freely rotating body can carry a fixed lab-frame \(\vec{L}\) while \(\vec{\omega}\) 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

\[ \textbf{Example 1: } \vec{L}\ \text{not parallel to}\ \vec{\omega}.\quad \text{Masses } m \text{ at } \vec{r}_1=(a,0,h),\ \vec{r}_2=(-a,0,-h),\ \vec{\omega}=\omega\,\hat{\mathbf z} \]

Set up: two point masses spun about the \(z\)-axis; find \(\vec{L}\) and its tilt. A

\[ I_{zz}=\sum m(x^2+y^2)=m a^2+m a^2=2m a^2 \]

Diagonal element about the spin axis; both masses have \(|x|=a,\ y=0\). A

\[ I_{xz}=-\sum m\,x\,z=-\big[m(a)(h)+m(-a)(-h)\big]=-2m a h \]

Product of inertia; the two masses add with the same sign, so it does not cancel. Also \(I_{yz}=I_{xy}=0\) since \(y=0\). B

\[ \vec{L}=\mathbf{I}\vec{\omega}\ \Rightarrow\ L_x=I_{xz}\omega=-2m a h\,\omega,\quad L_y=0,\quad L_z=I_{zz}\omega=2m a^2\,\omega \]

Only the third column of \(\mathbf{I}\) acts, since \(\vec{\omega}\) points along \(z\). B

\[ m=0.5\ \mathrm{kg},\ a=0.30\ \mathrm{m},\ h=0.40\ \mathrm{m},\ \omega=10\ \mathrm{rad/s} \]

Now insert numbers. A

\[ L_x=-2(0.5)(0.30)(0.40)(10)=-1.2,\qquad L_z=2(0.5)(0.30)^2(10)=0.9 \]

Arithmetic in SI units. A

\[ \vec{L}=(-1.2,\,0,\,0.9)\ \mathrm{kg\,m^2\,s^{-1}},\qquad |\vec{L}|=1.5\ \mathrm{kg\,m^2\,s^{-1}} \]

Reading. Although \(\vec{\omega}\) points purely along \(z\), \(\vec{L}\) tilts by \(\arctan(1.2/0.9)=53^\circ\) 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. \(\mathrm{kg}\times\mathrm{m}\times\mathrm{m}\times\mathrm{s^{-1}}=\mathrm{kg\,m^2\,s^{-1}}\). Correct.

\[ \textbf{Example 2: principal axes by diagonalization.}\quad m=2.0\ \mathrm{kg}\ \text{at}\ (a,a,0),(-a,-a,0),\ a=0.50\ \mathrm{m} \]

Set up: find the principal moments and axes of this two-mass configuration. A

\[ I_{xx}=\sum m(y^2+z^2)=2m a^2,\qquad I_{yy}=\sum m(x^2+z^2)=2m a^2,\qquad I_{zz}=\sum m(x^2+y^2)=4m a^2 \]

Each mass has \(z=0\); both contribute \(m a^2\) to the transverse diagonals. A

\[ I_{xy}=-\sum m\,x y=-\big[m(a)(a)+m(-a)(-a)\big]=-2m a^2,\qquad I_{xz}=I_{yz}=0 \]

Off-diagonal term does not vanish; \(z\) is already a principal axis. B

\[ \mathbf{I}=m a^2\begin{pmatrix} 2 & -2 & 0\\ -2 & 2 & 0\\ 0 & 0 & 4\end{pmatrix} \]

Assemble the symmetric tensor in symbolic form. A

\[ \det\begin{pmatrix} 2-\lambda & -2\\ -2 & 2-\lambda\end{pmatrix}=(2-\lambda)^2-4=\lambda(\lambda-4)=0 \]

Diagonalize the upper \(2\times2\) block; roots \(\lambda=0\) and \(\lambda=4\) (in units of \(m a^2\)). C

\[ \lambda=0:\ \hat{\mathbf n}=\tfrac{1}{\sqrt2}(1,1,0);\quad \lambda=4:\ \hat{\mathbf n}=\tfrac{1}{\sqrt2}(1,-1,0);\quad \lambda=4:\ \hat{\mathbf n}=(0,0,1) \]

Eigenvectors are the principal axes: the line through both masses (\(y=x\)), its in-plane perpendicular, and \(z\). With \(m a^2=(2.0)(0.50)^2=0.50\ \mathrm{kg\,m^2}\). C

\[ I_1=0\ (\text{axis } y=x),\qquad I_2=4m a^2=2.0\ \mathrm{kg\,m^2},\qquad I_3=4m a^2=2.0\ \mathrm{kg\,m^2} \]

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

Units check. \(\mathrm{kg}\times\mathrm{m^2}=\mathrm{kg\,m^2}\). Correct.

Problems

A single point mass \(m=3.0\ \mathrm{kg}\) sits at \((2.0,0,0)\ \mathrm{m}\). Write its full inertia tensor about the origin.
Solution
With \(x=2,\ y=z=0\): \(I_{xx}=m(y^2+z^2)=0\); \(I_{yy}=m(x^2+z^2)=3(4)=12\); \(I_{zz}=m(x^2+y^2)=12\). All products vanish (each contains a zero coordinate). So \(\mathbf{I}=\mathrm{diag}(0,12,12)\ \mathrm{kg\,m^2}\). The \(x\)-axis moment is zero because the mass lies on it.
A thin uniform rod of mass \(M=1.2\ \mathrm{kg}\) and length \(\ell=1.0\ \mathrm{m}\) lies along the \(x\)-axis, centred at the origin. Find \(I_{xx},\ I_{yy},\ I_{zz}\).
Solution
Linear density \(\lambda=M/\ell\). \(I_{xx}=\int(y^2+z^2)\,dm=0\) (rod on axis). \(I_{yy}=I_{zz}=\int_{-\ell/2}^{\ell/2}\lambda x^2\,dx=\lambda\left[\frac{x^3}{3}\right]_{-\ell/2}^{\ell/2}=\frac{\lambda \ell^3}{12}=\frac{M\ell^2}{12}\). Numerically \(=\frac{(1.2)(1.0)^2}{12}=0.10\ \mathrm{kg\,m^2}\). Result: \(\mathbf{I}=\mathrm{diag}(0,\,0.10,\,0.10)\ \mathrm{kg\,m^2}\).
Four equal masses \(m=1.0\ \mathrm{kg}\) sit at the corners of a square of side \(2a\) (\(a=0.50\ \mathrm{m}\)) in the \(xy\) plane, at \((\pm a,\pm a,0)\). Show all products of inertia vanish, give the tensor, and state whether \(\vec{L}\parallel\vec{\omega}\) for spin about \(z\).
Solution
\(I_{xy}=-\sum m\,xy=-m a^2(+1-1-1+1)=0\); likewise \(I_{xz}=I_{yz}=0\) since \(z=0\). Diagonals: \(I_{xx}=\sum m\,y^2=4m a^2\), \(I_{yy}=4m a^2\), \(I_{zz}=\sum m(x^2+y^2)=8m a^2\). With \(m a^2=0.25\): \(\mathbf{I}=\mathrm{diag}(1.0,\,1.0,\,2.0)\ \mathrm{kg\,m^2}\). Diagonal in these axes, so yes — spinning about \(z\) gives \(\vec{L}=I_{zz}\omega\,\hat{\mathbf z}\), parallel to \(\vec{\omega}\).
For Worked Example 1's configuration, at what nonzero value of the ratio \(h/a\) would \(\vec{L}\) make a \(45^\circ\) angle with the \(z\)-axis?
Solution
\(\tan\theta=\dfrac{|L_x|}{L_z}=\dfrac{2m a h\,\omega}{2m a^2\,\omega}=\dfrac{h}{a}\). Setting \(\theta=45^\circ\) gives \(\tan 45^\circ=1\), so \(h/a=1\). The tilt of \(\vec{L}\) depends only on the geometric ratio \(h/a\), independent of \(m\) and \(\omega\).
A rigid body has principal moments \(I_1=2.0,\ I_2=5.0,\ I_3=6.0\ \mathrm{kg\,m^2}\) and spins with \(\vec{\omega}=(3,0,4)\ \mathrm{rad/s}\) expressed in the principal frame. Find \(\vec{L}\), the rotational kinetic energy \(T\), and the angle between \(\vec{L}\) and \(\vec{\omega}\).
Solution
In principal axes \(L_a=I_a\omega_a\): \(\vec{L}=(2\times3,\,5\times0,\,6\times4)=(6,0,24)\ \mathrm{kg\,m^2\,s^{-1}}\), magnitude \(|\vec{L}|=\sqrt{36+576}=24.7\). Energy \(T=\tfrac12\sum I_a\omega_a^2=\tfrac12[2(9)+0+6(16)]=\tfrac12(18+96)=57\ \mathrm{J}\). Angle: \(\cos\phi=\dfrac{\vec{L}\cdot\vec{\omega}}{|\vec{L}||\vec{\omega}|}\) with \(\vec{L}\cdot\vec{\omega}=6(3)+24(4)=114\), \(|\vec{\omega}|=5\), \(|\vec{L}|=24.7\), so \(\cos\phi=\dfrac{114}{24.7\times5}=0.923\), \(\phi\approx22.6^\circ\). Note \(\vec{L}\cdot\vec{\omega}=2T=114\), a useful check.