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If a rigid body's mass density is invariant under every rotation about an axis through its centre of mass, its inertia tensor in any orthonormal basis with third vector on that axis has the displayed form. In particular all transverse directions are principal axes with equal principal moments of inertia.

For a proof, let that axis be $z$. In the defining integrals for the inertia tensor, a half-turn $(x,y,z)\mapsto (-x,-y,z)$ makes the $xz$ and $yz$ products change sign, so their integrals vanish. A quarter-turn $(x,y,z)\mapsto (-y,x,z)$ makes the $xy$ integral vanish and equates the integrals of $x^2$ and $y^2$. Therefore all off-diagonal components vanish and $I_{11}=I_{22}$. The values are $I_{\perp }=\int \rho [(x^2+y^2)/2+z^2]d^{3}r$ and $I_{\parallel }=\int \rho (x^2+y^2)d^{3}r$. No reflection symmetry in $z$ is needed.

A rigid Skyrmion orientation is quantized on a cover of its Skyrmion collective-coordinate orbit. A lifted combined rotation in the static field's stabilizer subgroup identifies the same classical field and imposes ${\hat{D}}^J(R){\hat{D}}^I(A)\Psi ={\chi }_{\mathrm{FR}}\Psi$, where the sign is the chosen Finkelstein-Rubinstein constraints character of that configuration-space loop. Choosing the nontrivial character models unit baryons as fermions. A $2\pi$ spatial rotation or isorotation of charge $B$ then has sign $(-1{)}^B$: deforming the field to separated unit lumps adds the unit-lump loop classes in the fundamental group ${\mathbb{Z}}_2$, while the labelled orbital paths can be contracted in three dimensions. The resulting Finkelstein-Rubinstein constraints sign is the product of the unit-lump signs. Hence spin and isospin are half-integer for odd $B$ and integer for even $B$. Further stabilizer constraints restrict which pairs $(J,I)$ and which body-fixed states occur. The energy operator comes from the inertia tensors on the orbit; symmetry selects allowed states but does not by itself determine their energies. A bosonic choice of the trivial topological character is mathematically possible but does not model a fermionic nucleon.

If two principal moments of inertia coincide, for example $I_1=I_2\ne I_3$, every direction in their two-dimensional eigenspace is a principal axis. The perpendicular axis is determined, but there is no privileged orthogonal pair within that plane. Rotating the pair in the plane leaves the diagonal inertia tensor unchanged. If all three principal moments of inertia coincide, ${\mathbf{I}}_O=I_{0}1$ and every axis through $O$ is principal. This follows from the eigenvalue multiplicity and eigenspace, rather than from a failure of diagonalization. ^[ref]

Equal principal moments of inertia do not prove that a body's outline has continuous rotational symmetry. For example, six equal point masses at $(\pm a,0,0)$, $(0,\pm a,0)$ and $(0,0,\pm a)$ have ${\mathbf{I}}_O=4m{a}^{2}1$. This discrete layout has the same axial moment of inertia in every direction, although its actual mass distribution changes under most rotations. Conversely, even when all three principal moments of inertia are distinct, principal axes are unoriented lines: reversing an eigenvector's sign gives the same axis. Near repeated eigenvalues, small measurement errors can rotate fitted principal axes substantially within the nearly degenerate eigenspace, so quoted directions need an uncertainty assessment as well as quoted principal moments of inertia.

The inertia tensor packages the moments of inertia about every axis through a specified point $O$ into one symmetric matrix. An axis still needs a direction: if $\mathbf{n}$ is a unit vector, its scalar moment of inertia is $I_{O,\mathbf{n}}={\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$. The same mass layout can therefore be easy to spin about one direction and harder about another. The inertia tensor also determines rotational kinetic energy and relates angular velocity to angular momentum in the appropriate rigid body model.

Take position vectors $\mathbf{r}=(x,y,z)$ relative to $O$, expressed in an orthonormal basis. For point masses $m_a$, or a continuous distribution with mass density $\rho$ and $dm=\rho dV$,Here ${\delta }_{ij}$ is the Kronecker delta. Equivalently, using the identity matrix $1$ and an outer product,Every entry has SI units $\mathrm{kg}{\mathrm{m}}^2$. The diagonal entries are axial moments of inertia; the off-diagonal entries carry a minus sign in this convention. Some texts instead define the integrals without the prefixed minus sign $\int xydm$, $\int xzdm$, $\int yzdm$ as their named off-diagonal quantities and put the minus signs into the displayed matrix. Check that convention before combining formulas. ^{[ref] [ref]}

The symmetric matrix is positive semidefinite, because its quadratic form isThe dot product and cross product identify this as a sum of squared distances when $\mathbf{u}$ is a unit vector. A zero moment of inertia occurs precisely when all the mass lies on that axis, up to sets of zero mass. A genuine three-dimensional body with positive mass density in a volume has strictly positive axial moments of inertia. Do not confuse this inertia tensor with the mass second-moment tensor $\mathbf{S}=\int \mathbf{r}{\mathbf{r}}^{T}dm$: they obey ${\mathbf{I}}_O=\mathrm{tr}(\mathbf{S})1-\mathbf{S}$, where $\mathrm{tr}$ denotes the matrix trace.

If $O$ is fixed in an inertial frame and each velocity is $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$, summing the individual kinetic energies and angular momenta givesThese formulas also describe angular momentum and rotational kinetic energy relative to the centre of mass $G$ of a translating rigid body. In that case the total kinetic energy in the inertial frame is $K_{\mathrm{total}}=\frac{1}{2}M|{\mathbf{V}}_G{|}^2+\frac{1}{2}{\mathbf{\omega }}^T{\mathbf{I}}_G\mathbf{\omega }$. For a general moving point $O$, the body's actual velocities include ${\mathbf{V}}_O$, so ${\mathbf{I}}_O\mathbf{\omega }$ alone need not be its angular momentum about $O$. ^{[ref] [ref]}

Changing the orthonormal basis changes components, rather than the physical distribution: if the columns of an orthogonal matrix $Q$ are the new basis vectors expressed in the old basis, then ${\mathbf{I}}_O^{\prime }=Q^T{\mathbf{I}}_{O}Q$. Changing the reference point is a different operation, governed by the tensor parallel-axis theorem. Both principal axes and principal moments of inertia must therefore be quoted with their reference point.

For a uniform ball of radius $R>0$ and positive mass $M$, let $G$ be its centre of mass. Spherical symmetry gives the second-moment matrix $\int u{u}^{T}dm=(M{R}^2/5)I$, so the central inertia tensor is $(2M{R}^2/5)I$. The tensor parallel-axis theorem gives the displayed shifted tensor. For $X\ne 0$ its eigenvector along $X$ has eigenvalue $2M{R}^2/5$, while the two-dimensional orthogonal complement has repeated eigenvalue $M(2R^2/5+|X{|}^2)$. Thus it is an isotropic tensor exactly at $P=G$. Displacement along a chosen axis does not change that axis's moment; the transverse moments increase by $M|X{|}^2$.

For a uniform planar lamina with $|x_1-s{x}_2|\le a$ and $|x_2|\le b$, put $u=x_1-s{x}_2$, $v=x_2$. The Jacobian determinant is one, the mass is $M=4ab\sigma$, and symmetry puts the centre of mass at zero. Uniform rectangular moments give $⟨u^2⟩=a^2/3$, $⟨v^2⟩=b^2/3$, $⟨uv⟩=0$. Substitution into $I_{ij}=\int (|x{|}^2{\delta }_{ij}-x_i{x}_j)dm$ yields the displayed inertia tensor. The in-plane principal moments of inertia are $M[a^2+(1+s^2)b^2\pm \sqrt{(a^2+(s^2-1)b^2{)}^2+4s^2{b}^4}]/6$; the perpendicular moment is their sum. This separates the geometric shear from the elementary rectangular integrals.

For a uniform cube of side $s$ and mass $M$, reflection symmetry makes central off-diagonal inertia components zero. Equal coordinate second moments are $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩=s^2/12$, so each diagonal component is $M{s}^2/6$. Thus every central axis is a principal axis. A nonspherical shape can have an isotropic central inertia tensor: that tensor measures only quadratic moments of the mass distribution, not the whole shape.

An exact planar example makes the off-diagonal entries tangible. Join four equal point masses $m$ with negligible-mass rigid connectors at $a(2,1,0)$, $a(-2,-1,0)$, $a(1,-1,0)$ and $a(-1,1,0)$, where $a>0$ is a length. Their centre of mass is $O=(0,0,0)$. Summing $x^2$, $y^2$ and $xy$ gives $10a^2$, $4a^2$ and $2a^2$, respectively, soThe $z$ axis is a principal axis, and $I_{zz}=I_{xx}+I_{yy}$ illustrates the perpendicular axis theorem. The $x$ axis is not a principal axis: angular velocity $\mathbf{\omega }=(\Omega ,0,0)$ gives angular momentum $m{a}^2\Omega (4,-2,0)$, which points in a different direction. A bearing can enforce fixed-axis rotation despite this difference, supplying the necessary transverse torque.

For an in-plane unit vector $\mathbf{n}=(\cos \theta ,\sin \theta ,0)$, the planar model hasIts extrema are the principal moments of inertia $(7-\sqrt{13})m{a}^2$ and $(7+\sqrt{13})m{a}^2$; the remaining principal moment of inertia is $14m{a}^2$. These values follow from the two-by-two characteristic polynomial $(4-\lambda )(10-\lambda )-4$, so they can be checked without trusting a numerical eigenvalue routine. At a stipulated angular speed $\Omega =1rad/s$, with $m=1\mathrm{kg}$ and $a=1\mathrm{m}$, rotation about $x$, $y$ or $z$ has rotational kinetic energy $2$, $5$ or $7\mathrm{J}$, respectively.

A truly three-dimensional example adds two more equal point masses at $a(0,0,1)$ and $a(0,0,-1)$. Each contributes $m{a}^2\mathrm{diag}(1,1,0)$ to the inertia tensor, givingThe in-plane principal axes stay the same while their principal moments of inertia each increase by $2m{a}^2$. The $z$ moment of inertia stays unchanged because the added point masses lie on that axis. This comparison changes the total mass from $4m$ to $6m$; it holds the original four point masses fixed rather than redistributing a fixed total mass.

The mass second-moment tensor about a centre of mass is $I_{jk}=\int y_j{y}_{k}dm$. It measures the shape of a mass distribution and appears in tidal torque theory. It differs from the mechanical inertia tensor, which is ${\mathcal{I}}_{jk}=\mathrm{tr}(I){\delta }_{jk}-I_{jk}$. For a uniform ellipsoid of mass $M$ and semiaxes $a_1,a_2,a_3$, its principal values are $M{a}_j^2/5$.

The moment of inertia about a specified axis measures how a mass distribution resists changes of angular velocity about that axis. For point masses $m_i$ at perpendicular distances $r_{\perp i}$, or a continuous distribution with mass density $\rho$,Its SI unit is $\mathrm{kg}{\mathrm{m}}^2$. The distance is to the entire axis, not to one point on it: moving a point mass along the axis leaves its contribution unchanged. The moment of inertia depends on both the distribution and the chosen axis; it is not determined by total mass alone. About an axis fixed in an inertial frame, a rigid body with larger moment of inertia has smaller angular acceleration under the same net axial torque, and greater rotational kinetic energy at the same angular speed.^[ref]

For a simple model, put four equal point masses of total mass $M$ on light rigid spokes at radius $a$. Their moment of inertia is $M{a}^2$. Doubling $a$ quadruples the moment of inertia without changing $M$. The illustrated placements $a=R/3$ and $a=R$ give $M{R}^2/9$ and $M{R}^2$. Actual spokes and finite-sized weights contribute their own moments of inertia; the point mass model neglects them. Repositioning weights during rotation also breaks the fixed-distribution assumption, so it requires a separate work and angular momentum calculation.

The moment of inertia of a uniform disc follows by dividing it into concentric annuli: $dm=(2M/R^2)rdr$ and $I={\int }_0^R{r}^2(2M/R^2)rdr=M{R}^2/2$. For an ideal thin ring every mass element is at $R$, so $I=M{R}^2$. A ring of finite radial thickness instead uses the moment of inertia of a uniform annulus, $I=M(R^2+b^2)/2$, with inner radius $b$. A moment of inertia is thus a compact summary of radial placement rather than a complete description of shape. For several joined pieces, their moments of inertia add only when each is calculated about the same axis.^[ref]

A physical pendulum makes the moment of inertia measurable through timing. Martín Monteiro, Cecilia Stari, Cecilia Cabeza and Arturo C. Martí used a metal bar with suspension holes $1.0\mathrm{cm}$ apart and an attached smartphone. They report bar length $1.199\mathrm{m}$, bar mass $0.2518\mathrm{kg}$ and phone mass $0.1311\mathrm{kg}$. Changing the suspension hole changes the pivot axis. Their readouts were phone gyroscope angular velocity records, analyzed by autocorrelation or a fit to a sinusoid, and an independent photogate; the logged records for the sinusoid fits lasted $10\mathrm{s}$. They observed agreement between the three timing methods and the geometry-based model, including a minimum of oscillation period as the suspension point varied. These are reported apparatus details and observations, not measurements made for this article.^[ref]

To infer a moment of inertia from such a measurement, determine the total mass $M_{\mathrm{tot}}$ including every attached part, locate the assembly's centre of mass $G$, and measure its distance $h>0$ below the horizontal pivot $O$. Record successive crossings in the same direction, or the elapsed time $t_N$ for $N$ complete oscillations, giving oscillation period $P=t_N/N$. The physical pendulum model with a fixed pivot, a uniform gravitational field, negligible friction and the small-angle approximation givesThe crossing times, masses and geometric distances are the measured inputs; $I_{O,\mathrm{inferred}}$ is an inference conditional on this model, while an integral over the assumed shape supplies an independent model prediction. The parallel axis theorem converts the inferred pivot moment of inertia to one about a parallel axis through $G$. The controls and interpretation of that comparison are described there. An axial measurement determines one scalar moment of inertia, not the complete inertia tensor needed for general rotation.

An inertia tensor can be isotropic even when the body is not a sphere. An example is a uniform thin disc of radius $2a$ and mass $M$, with a thin rod of length $\sqrt{3}a$ and mass $2M$ attached normally at its center. The combined center of mass is one third of the rod length from the disc. Standard disc and rod moments and the parallel axis theorem give all three principal moments equal to $2M{a}^2$, so $J=2M{a}^{2}I$. The angular momentum is therefore a fixed scalar times the angular velocity. In torque-free motion, conservation of the inertial angular-momentum vector makes the angular-velocity vector constant for every initial spin. Rotational symmetry of the shape is not required; equality of its second moments is enough.

If a region is invariant under one coordinate reflection and an integrable function changes sign under that reflection, its multiple integral is zero. Apply the change of variables formula to the reflection, whose absolute Jacobian determinant is one: the integral equals its negative. This applies to odd coordinate powers in symmetric volumes and to the off-diagonal entries of an inertia tensor when the mass density shares the coordinate-reflection symmetries.

For a uniform square of side $a$, the principal inertia tensor is $\mathrm{diag}(m{a}^2/12,m{a}^2/12,m{a}^2/6)$. A rotation axis tilted $45^{\circ }$ to the surface has equal in-plane and normal angular-velocity components. Multiplying each by its own principal moment gives the stated magnitude. In general $I_{\mathrm{axis}}\omega$ is only the axial projection of angular momentum.

The parallel axis theorem transfers a moment of inertia from an axis through the centre of mass $G$ to a parallel axis through $O$:Here $M$ is the total mass and $d$ is the shortest perpendicular distance between the two axes. Both moments of inertia concern the same mass distribution and the same axis direction. The axis for $I_G$ must pass through the centre of mass; $I_O$ may refer to an axis outside the body. No uniform-density or planar-shape assumption is needed. Within any family of parallel axes, the centre of mass axis minimizes the moment of inertia.^[ref]

A short proof isolates why the centre of mass matters. Choose parallel axes perpendicular to an $xy$ plane, put $G$ at the origin, and place $O$ at $(d,0)$. Thenbecause $\int xdm=0$ in centre of mass coordinates. Individual mass elements can move closer to the new axis, but their weighted cross terms cancel. For arbitrary parallel axes $A$ and $B$, use $I_B-I_A=M(d_B^2-d_A^2)$, where each distance is measured from the parallel centre of mass axis; adding $M$ times the squared separation of $A$ and $B$ is generally incorrect. A displacement along the axis has $d=0$. Changing the axis direction instead requires the inertia tensor or, in its applicable planar setting, the perpendicular axis theorem.

For a uniform disc, mass $M$ and radius $R$, the central perpendicular moment of inertia is $I_G=M{R}^2/2$. Shifting that axis to the rim gives $d=R$ and $I_O=3M{R}^2/2$, three times $I_G$. In the computed $M=2\mathrm{kg}$, $R=0.30\mathrm{m}$ example, $I_G=0.090\mathrm{kg}{\mathrm{m}}^2$, $M{d}^2=0.180\mathrm{kg}{\mathrm{m}}^2$ and $I_O=0.270\mathrm{kg}{\mathrm{m}}^2$. At angular speed $10rad/s$ the rim-pivoted disc has rotational kinetic energy $13.5\mathrm{J}$. Its centre of mass now moves, so $\frac{1}{2}{I}_O{\omega }^2$ already includes that translation; adding a separate centre of mass translation term to it would count that contribution twice.

The physical pendulum measurement described under moment of inertia provides a practical check: infer $I_O=Mgh(P/2\pi {)}^2$ at several pivot positions, then subtract $M{h}^2$ to obtain $I_G$. Keep all pivot axes parallel and measure $h$ to the centre of mass of the entire moving assembly. The same assembly should yield the same $I_G$ within measurement precision. If a smartphone is attached, its mass and its own moment of inertia belong in both the geometric prediction and the centre of mass calculation. The published bar-and-phone study explicitly adds both contributions using the parallel axis theorem.^[ref]

For a careful replication, release small swings without a push, repeat timings at several small amplitudes, and check that the extrapolated oscillation period is stable as amplitude decreases. Time many complete oscillations and repeat the run; two crossings of a photogate per cycle must not be mistaken for two cycles. Inspect decay of amplitude and compare early and late timings to detect significant pivot friction or air drag. Mark the bar and phone mount to detect relative slip, tighten the attachment, and check that the pivot support does not rock or migrate between holes. Repeat with an independent photogate or video readout to check the phone's timing. These are proposed controls for the replication, not additional results reported by the authors. The ideal model excludes appreciable damping, large swings and flexing; subtraction of $M{h}^2$ also magnifies relative errors when $I_G$ is much smaller than $I_O$.

The oscillation period need not increase monotonically with pivot distance: the shift raises the moment of inertia but also increases the gravitational restoring torque. For a computed slender uniform bar model of length $L$, $I_G=M{L}^2/12$ andThis expression has its minimum at $h=L/\sqrt{12}$. With $L=1\mathrm{m}$, $h=0.50\mathrm{m}$ and stipulated $g=9.81m/{\mathrm{s}}^2$, the prediction is $P\simeq 1.638\mathrm{s}$; it is not a recorded timing or a prediction for the bar with a phone attached. The competing terms explain why a timing experiment can reveal a mass distribution rather than merely its weight.

Put $\mathbf{r}=(x,y,z)$ with ${\mathbf{e}}_3$ along $z$, and interpret the symmetry as axisymmetry of the whole mass density. A half-turn about ${\mathbf{e}}_3$ sends $(x,y,z)$ to $(-x,-y,z)$ and leaves the integration domain and mass density invariant. Therefore $\int \rho xz{d}^{3}r=\int \rho yz{d}^{3}r=0$. A quarter-turn sends $(x,y,z)$ to $(-y,x,z)$; it givesAll off-diagonal entries of the inertia tensor consequently vanish. Writing $s^2=x^2+y^2$, the principal moments of inertia satisfyThus every orthonormal basis with third vector along the symmetry axis is a principal body frame. This is axisymmetry of the inertia tensor. In general there is no requirement that $I_3=I_1$.

Take the origin at the centre of mass, so ${\int }_B\rho \mathbf{r}{d}^{3}r=0$, and write $r_i=\mathbf{r}\cdot {\mathbf{e}}_i$ in the specified orthonormal basis. The inertia tensor isHere ${\delta }_{ij}$ is the Kronecker delta. For a rigid body with angular velocity $\mathbf{\omega }$, this tensor gives $\mathbf{L}=\mathbf{I}\mathbf{\omega }$ and rotational kinetic energy $T=\mathbf{\omega }\cdot \mathbf{I}\mathbf{\omega }/2$.

The inertia tensor is a real symmetric matrix, so it admits an orthonormal basis of eigenvectors. The lines in these eigenvector directions through the centre of mass are its principal axes; the corresponding eigenvalues $I_1,I_2,I_3$ are its principal moments of inertia:If an eigenvalue repeats, the corresponding principal axes need not be unique.

In the quadratic form, the two equal off-diagonal contributions together supply the mixed term. Comparing coefficients givesFor a nontrivial physical moment of inertia scale take $I_0>0$; then this inertia tensor is positive definite.

Use coordinates $\mathbf{r}$ measured from the center of mass $G$, so $\int \mathbf{r}dm=0$. The displacement from $P$ is $\mathbf{r}-\mathbf{X}$. Expand the defining inertia tensor:All terms linear in $\mathbf{r}$ integrate to zero. The remaining constant terms give the tensor parallel-axis theorem,For the uniform cube, let $M=8\rho a^3$. Symmetry makes the off-diagonal inertia tensor entries zero. Direct integration gives $\int x_i^{2}dm=M{a}^2/3$ for each coordinate, and therefore $I(G)=(2M{a}^2/3)I_3$. With $\mathbf{X}=a(1,1,1{)}^T$, the tensor parallel-axis theorem givesThe unit eigenvector $u=(1,1,1{)}^T/\sqrt{3}$ has eigenvalue $2M{a}^2/3$. Every vector in the plane $x+y+z=0$ is annihilated by $11^T$ and hence has eigenvalue $11M{a}^2/3$. One orthonormal basis of that eigenspace isThe principal moments are $2M{a}^2/3$ along the body diagonal and $11M{a}^2/3$ twice in its perpendicular plane. Any orthonormal pair in that plane supplies the remaining principal axes.

In orthonormal Cartesian coordinates, a second-rank tensor has components transforming under a change of axes $Q$ as $M_{ij}^{\prime }=Q_{ik}{Q}_{j\ell }{M}_{k\ell }$, where $Q{Q}^T=I$. Its contraction is invariant:Thus $M_{ii}$ is a scalar. This is tensor contraction with the Euclidean metric. In general nonorthonormal coordinates the invariant expression is $g^{ij}{M}_{ij}$, not an unweighted sum of diagonal covariant components.

For the planar plate, every mass element has $z=0$. The inertia tensor is consequentlyIt follows directly that ${\mathbf{e}}_z$ is an eigenvector andThe real symmetric matrix has three real eigenvalues. Its trace is $2{\int }_D(x^2+y^2)\rho dS=2M_{\perp }$, so $M_1+M_2+M_{\perp }=2M_{\perp }$. Hence the perpendicular axis theorem gives $M_{\perp }=M_1+M_2$.

For a disc of constant areal density ${\rho }_0$, symmetry gives ${\int }_{D}xydS=0$, and polar integration gives ${\int }_D{x}^{2}dS={\int }_D{y}^{2}dS=\pi a^4/4$. Thereforewhere the scalar $M$ in the last expression denotes the total mass rather than the tensor.

For a uniform circular cross-section, averaging gives $⟨x^2⟩=⟨y^2⟩=R^2/4$; for a uniform axial coordinate in $[-h,h]$, $⟨z^2⟩=h^2/3$. Symmetry makes the coordinate axes principal axes, since all products of distinct coordinates integrate to zero. Thus the principal moments of inertia about the centre areThe inertia tensor is a scalar multiple of identity exactly when $h^2/3=R^2/4$. Thereforeis the height-to-radius ratio for sphere-like rotational inertia; the height-to-diameter ratio is $\sqrt{3}/2$. The geometric shape remains cylindrical, but its angular momentum is parallel to angular velocity for every axis, just as for a sphere.