Moment of inertia ($I$)

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.

A uniform disk of radius $R$ and area mass density $\rho$ has a circular hole of radius $r_o$, whose centre H is displaced by $r_o$ from the original centre C. Set $D=\pi \rho R^2$, $h=\pi \rho r_o^2$ and $W=D-h$. The parallel axis theorem gives $I_H=D{R}^2/2+D{r}_o^2-h{r}_o^2/2$. Removing the material shifts the remaining centre of mass G away from H by $CG=h{r}_o/W$, so $HG=D{r}_o/W$ and $I_G=I_H-D^2{r}_o^2/W$. Thus the rotational component of free rigid-body kinetic energy is smaller than fixed-axis kinetic energy about H at the same angular speed, although total rolling kinetic energy also includes translation.

Integrating $dC=(2/3)r^{2}dm$ over a uniform-density spherical layer gives the displayed result. It becomes $2M{R}^2/5$ for a solid sphere and tends to $2M{R}^2/3$ only in the thin-shell limit. Calling a layer hollow does not make its thickness negligible.

For a solid cylinder with mass density $\rho (r)=k{r}^p$ and $p>-2$, integrating cylindrical shells gives $M=2\pi Lk{R}^{p+2}/(p+2)$ and axial moment of inertia $I_z=2\pi Lk{R}^{p+4}/(p+4)$. Their ratio proves the displayed formula. It gives the uniform-cylinder result $M{R}^2/2$ at p=0 and $2M{R}^2/3$ at p=2. The radial density weighting must enter both mass and inertia integrals.

For a cylinder of uniform mass density $\rho$, length $L$, inner radius $R_i$ and outer radius $R_o$, an annular element has $dm=2\pi \rho Lrdr$. Thus the axial moment of inertia is $I=\int r^{2}dm=\pi \rho L(R_o^4-R_i^4)/2$. Dividing by $M=\pi \rho L(R_o^2-R_i^2)$ yields the displayed result. At fixed mass, the solid-cylinder limit is $M{R}_o^2/2$ and the thin-shell limit is $M{R}^2$.

A uniform solid with circular cross-sectional radius $r(z)$ can be sliced into thin discs. A slice of thickness $dz$ has mass $dm=\rho \pi r(z{)}^{2}dz$ and axial moment of inertia $d{I}_z=\frac{1}{2}r(z{)}^{2}dm$. ThusAll slice axes coincide, so no parallel axis theorem correction is needed. For a linearly tapered section of length $l$ with end radii $r_0,r_1$, direct integration gives $M=\rho \pi l(r_0^2+r_0{r}_1+r_1^2)/3$ and $I_z=\rho \pi l(r_0^4+r_0^3{r}_1+r_0^2{r}_1^2+r_0{r}_1^3+r_1^4)/10$.

For three mutually perpendicular axes through one point of a nonnegative mass distribution, $I_x+I_y-I_z=2\int z^{2}dm\ge 0$, with cyclic counterparts. Equality holds precisely when the mass distribution is supported in the plane $z=0$, giving the perpendicular axis theorem. The point need not be the centre of mass, and the axes need not be principal axes.

A uniform square of side $s$ has the displayed moment of inertia about the perpendicular axis through its centre. Integrating $x^2+y^2$ over $[-s/2,s/2{]}^2$ gives $s^4/6$; multiplying by the surface mass density $M/s^2$ gives the result. A square prism of any uniform thickness has the same axial formula when expressed in terms of its total mass. If the side is denoted $2a$ instead, the result is $2M{a}^2/3$, so the side-length convention must be specified before substituting numerical data into a normal mode calculation.

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.

A uniform thin square of side $l$ and mass $M$ has $I_z=M{l}^2/6$ about the perpendicular central axis. Any central in-plane axis and its perpendicular in-plane partner have equal moments by $90^{\circ }$ rotational symmetry. The perpendicular axis theorem says their sum is $I_z$, proving $I_{\varphi }=M{l}^2/12$ for every angle $\varphi$. A non-square rectangle generally lacks this angle independence.

The centroid of a uniform solid hemisphere of radius $a$ lies $3a/8$ from its flat face. About a diameter of the underlying sphere parallel to that face, its moment of inertia is $2m{a}^2/5$. The parallel axis theorem therefore gives $I_G=2m{a}^2/5-m(3a/8{)}^2=83m{a}^2/320$ about the parallel centroidal axis. A thin hemispherical shell has a different centroid and inertia.

For a uniform solid cylinder of radius $a$, length $L$ and mass density $\rho$, a coaxial shell carries mass $dm=2\pi \rho Lrdr$. The axial moment of inertia is $I={\int }_0^a{r}^{2}dm=\pi \rho L{a}^4/2=m{a}^2/2$. The axial moment of inertia is independent of the length once mass and radius are fixed. A thin cylindrical shell instead has $I=m{a}^2$, demonstrating why the uniform solid assumption matters.

For a uniform solid cylinder of length $l$, radius $r$ and mass $m$, a thin circular slice at axial coordinate $z$ has transverse moment of inertia $(z^2+r^2/4)dm$ about a diametral axis through the cylinder's midpoint. Integrating from $-l/2$ to $l/2$ gives $I_{\perp ,G}=m{l}^2/12+m{r}^2/4$. The parallel axis theorem gives $I_{\perp ,\mathrm{end}}=m{l}^2/3+m{r}^2/4$ about a parallel axis through an end-face centre. The transverse and symmetry-axis moments of inertia coincide exactly when $l=\sqrt{3}r$.

If a fixed-axis wheel of radius $R$ drives a fixed-axis roller of radius $S$ without slip, their angular-speed magnitudes satisfy $\sigma =(R/S)\omega$. Their kinetic energy is $\frac{1}{2}[I+J(R/S{)}^2]{\omega }^2$, so the displayed reflected moment of inertia governs the wheel coordinate. A roller resisting torque of magnitude $\gamma {\sigma }^2$ consumes power $\gamma {\sigma }^3$ and therefore reflects to wheel torque $\gamma (R/S{)}^3{\omega }^2$. The squared ratio belongs to inertia; the cubed ratio belongs to this quadratic resisting torque.

A uniform planar annulus of outer radius $a$, inner radius $b<a$ and mass $M$ has surface density $M/[\pi (a^2-b^2)]$. Integrating $r^2$ over its mass gives the axial moment of inertia $I=M(a^2+b^2)/2$. Subtracting the inner disc from a complete disc gives the same result, provided their surface densities match.

// Target: electromagnetism.bigb

For a uniform disc of radius $a$ and mass $M$, annuli of radius $r$ and thickness $dr$ have mass $(M/\pi a^2)2\pi rdr$. Their axial moment of inertia sums to $I={\int }_0^a{r}^{2}dm=M{a}^2/2$.

A uniform solid sphere of mass $M$ and radius $a$ has the indicated moment of inertia about any diameter. Integrating the squared perpendicular distance against uniform mass density gives $I=8\pi \rho a^5/15=2M{a}^2/5$. It is different from the value for a thin spherical shell and determines the rotational contribution to kinetic energy during rolling without slipping.

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 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.

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 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.

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.

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 nonnegative mass distribution relative to an origin, the second-moment matrix is $\mathsf{S}=\int \mathbf{r}{\mathbf{r}}^{\mathsf{T}}dm$ and the inertia tensor is $\mathsf{I}=\mathrm{tr}(\mathsf{S})1-\mathsf{S}$. Taking the trace gives the displayed inverse formula, which must be positive semidefinite. Conversely, any such $\mathsf{S}$ can be realized by pairs of point masses along its orthogonal principal directions. Thus this is also a realizability test when total mass and dimensions are unrestricted. In principal coordinates it is equivalent to the moment-of-inertia triangle inequalities: $I_2+I_3-I_1=2S_{11}\ge 0$, with cyclic counterparts. A symmetric positive definite matrix alone need not be a physical inertia tensor.

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.

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.

PASCO's PASPORT Rotary Motion Sensor Experiment Guide, Experiment 1, specifies a PS-2120A sensor, a rod carrying two equal-radius weights, a three-step drive wheel, a thread and hanging mass, a clamp-on guide wheel, a support, a balance and calipers. The thread leaves the drive wheel tangentially. The sensor records angular velocity against time; a linear regression over the accelerating interval supplies angular acceleration $\alpha$. The guide measures the drive radius with calipers and repeats the run with the weights removed to subtract the apparatus's own moment of inertia. Its tables are blank: it prescribes a measurement and does not report an observed numerical agreement. ^[ref]

With hanging mass $m_h$ including its hanger, effective drive radius $b$, no thread slip and negligible guide-wheel inertia, Newton's second law gives thread tension $F=m_h(g-b\alpha )$. The axial torque is $bF$. Under negligible friction, the inference is thereforeRepeat the inference for the unloaded assembly and subtract: $I_{\mathrm{weights}}=I_{\mathrm{loaded}}-I_{\mathrm{empty}}$. Omitting the $m_h{b}^2$ term incorrectly treats thread tension as the hanging weight. Comparing with $(M_1+M_2)R^2$ tests the point mass approximation; finite-sized weights also have their own moments of inertia about their centres of mass. ^[ref]

Useful proposed checks go beyond that ideal inference. Verify the rotary encoder's angle scale by turning the shaft through a marked full revolution; check the recorded time against an independent clock and the balance against a known mass. Measure $b$ at the thread's running surface, including thread thickness; avoid overlapping wraps that change $b$. Release without a push, inspect the angular velocity record for a straight interval, and stop before the hanger lands or the thread runs out. Repeat with several hanging masses, weight radii and rotation directions. A consistent inferred moment of inertia across drive torques supports the model; curvature, direction dependence or drift indicates a limitation rather than a new principal moment of inertia.

For approximately constant resisting torque ${\tau }_f\ge 0$ in one rotation direction, an original calibration model isFit the left-hand quantity against $\alpha$ across several drives: the slope estimates the axial moment of inertia and the intercept estimates ${\tau }_f$. This calibration is conditional on friction remaining constant over the fitted angular speeds; a freely slowing run can expose speed dependence. Sensor differentiation noise, guide-wheel moment of inertia, thread slip, flexing connectors and uncertain weight radii limit the inference. Removing weights can also change bearing loads, so empty-apparatus subtraction does not automatically cancel friction. These are proposed controls and a model extension, not results claimed by PASCO.

A fixed-axis readout measures ${\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$, not the full inertia tensor. In a proposed generalization, repeat about $x$, $y$, $z$ and the three diagonal directions $({\mathbf{e}}_x+{\mathbf{e}}_y)/\sqrt{2}$, $({\mathbf{e}}_x+{\mathbf{e}}_z)/\sqrt{2}$, $({\mathbf{e}}_y+{\mathbf{e}}_z)/\sqrt{2}$, all through the same $O$. For example,Six independent scalar readings then determine the six entries of a symmetric matrix. Each mounting needs its own apparatus correction, and different reference points require the tensor parallel-axis theorem before the readings can be combined.

A rotary encoder converts shaft rotation into a readout of angle or changes in angle. The PS-2120A uses interruptions of a light beam to encode shaft motion; its product specification gives 4000 divisions per revolution. That count is a resolution specification, not a guarantee of the precision of an inferred angular acceleration or moment of inertia. ^[ref]

If $G$ is the center of mass and $X=P-G$, expanding the defining inertia tensor about $P$ gives the displayed formula. The cross terms vanish because $\int rdm=0$ in coordinates centred at $G$. Contracting with a unit axis $u$ gives the parallel axis theorem, with the squared distance between the parallel axes equal to $|X{|}^2-(u\cdot X{)}^2$.

The principal moments of inertia about a specified point $O$ are the eigenvalues $I_1,I_2,I_3$ of the inertia tensor ${\mathbf{I}}_O$. Their associated unit vectors that are eigenvectors specify principal axes through $O$. The spectral theorem for real symmetric matrices guarantees an orthonormal basis of these eigenvectors, even for an irregular rigid body. Choose their signs to make a right-handed basis and collect them as the columns of an orthogonal matrix $Q$:Thus diagonalization removes the off-diagonal coupling in the rotational kinetic energy. The eigenvalues belong to the mass distribution and reference point, independently of which orthonormal basis initially describes them. ^[ref]

In this principal body frame, write angular velocity as $\mathbf{\omega }={\sum }_i{\omega }_i{\mathbf{e}}_i$. ThenUse these as total angular momentum and kinetic energy only under the reference-point conditions stated under inertia tensor. Along a single principal axis with $I_i>0$, angular momentum is parallel to angular velocity. For a mixture of directions with different principal moments of inertia, it generally is not. A principal axis need not be a geometric symmetry axis: it is selected by the complete mass distribution. A principal axis also need not give stable free rotation; for three distinct principal moments of inertia, the intermediate axis theorem supplies the stability distinction.

For any unit vector $\mathbf{n}={\sum }_i{n}_i{\mathbf{e}}_i$,The axial moment of inertia is a weighted average of the principal moments of inertia. At fixed angular speed, the least and greatest rotational kinetic energies occur in the corresponding eigenspaces. At fixed nonzero angular momentum magnitude $L$, instead, $K=\frac{1}{2}{\sum }_i{L}_i^2/I_i$, so the largest principal moment of inertia gives the smallest rotational kinetic energy. These comparisons hold the reference point fixed.

There is an additional physical restriction beyond positive semidefiniteness. In principal axes coordinates,with two analogous cyclic inequalities. Therefore three arbitrary positive numbers do not always represent a possible inertia tensor. For a distribution in the $xy$ plane through $O$, $I_3=I_1+I_2$, as in the planar inertia tensor of paired point masses. A uniform solid ellipsoid of mass $M$ and semiaxes $a,b,c$ supplies a continuous three-dimensional example: using its mass second-moment tensor, its central principal moments of inertia are $M(b^2+c^2)/5$, $M(a^2+c^2)/5$, $M(a^2+b^2)/5$. Setting $a=b=c=R$ recovers the moment of inertia of a uniform solid sphere.

For strictly positive principal moments of inertia and a fixed rotational kinetic energy $K_0>0$, the allowed angular velocity components form an ellipsoid:Its semiaxis lengths $s_i$ have units of angular speed. The longest semiaxis corresponds to the smallest principal moment of inertia: that direction permits the greatest angular speed for the same rotational kinetic energy. This is a surface in angular velocity space, not the physical outline of the body. If one principal moment of inertia vanishes in an ideal line-mass model, this constant-energy surface is unbounded in that direction and is no longer an ellipsoid.

The rotational energy ellipsoid gives the geometry behind Poinsot's ellipsoid, but a constant-energy surface alone does not determine a free-rotation trajectory. A torque-free trajectory must also satisfy the conserved angular momentum constraint and the Euler equations for a torque-free rigid body.

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.

For torque-free rigid-body motion, the kinetic-energy ellipsoid fixed in the body rolls without slipping on the invariable plane perpendicular to the conserved angular momentum. Its point of contact is the angular-velocity vector.

The herpolhode is the contact curve traced on the fixed invariable plane in Poinsot's torque-free construction. It is distinct from the polhode, which is described in body coordinates.

The polhode is the contact curve traced on the body's energy ellipsoid during Poinsot's torque-free rolling construction.

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.

For uniform density between concentric radii $R_1,R_2$, subtract the missing solid core's central moment of inertia and then shift the axis by $R_2$. Here $M=4\pi \rho (R_2^3-R_1^3)/3$. The thin-shell limit gives $I=5M{R}_2^2/3$, while a full solid sphere gives $7M{R}_2^2/5$.

For a planar lamina in the $xy$-plane, the moments about three mutually perpendicular axes through one point satisfyThis follows pointwise from $x^2+y^2=y^2+x^2$ in the defining mass integrals.

For an axisymmetric planar lamina of total mass $M$, let $I_z$ be its moment of inertia about the central axis normal to its plane. The perpendicular axis theorem and rotational symmetry give $I_x=I_y=I_z/2$ about central diameters. An in-plane tangent line a perpendicular distance $R$ from a parallel central diameter therefore has moment of inertia $I_{\tan }=I_z/2+M{R}^2$ by the parallel axis theorem. An axis normal to the lamina through the same rim point instead has $I_z+M{R}^2$; specifying the point without the axis direction is insufficient.

For a uniformly filled, negligibly light disc of fixed radius $R$, remaining mass $m(t)$, angular velocity $\omega (t)$, and tangential exhaust speed $u$ relative to its rim, angular-momentum balance givesAt zero relative exhaust speed, $\omega /m$ is constant.

For a uniform solid cone of height $l$, base radius $R$, and mass $M$, whose symmetry axis points from vertex to base, the centre of mass lies at $3l/4$ from the vertex. The principal moments at the vertex are

Consider the uniform solid of revolution about the $x_3$-axis whose boundary isFor total mass $M$, the center of mass lies a distance $3h/5$ from the pointed origin, and the principal moments about that origin are