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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 $I_z=\rho \pi l(r_0^4+r_0^3{r}_1+r_0^2{r}_1^2+r_0{r}_1^3+r_1^4)/10$.

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.

A homogeneous spherical bob rigidly fixed to a massless support has centre-of-mass distance $L$ from its pivot and intrinsic moment of inertia $2M{a}^2/5$. The parallel axis theorem and linear gravitational restoring torque give the title period for small oscillations. Relative to a point mass at the same centre, the leading fractional period increase is $a^2/(5L^2)$. If $L$ is instead measured to the sphere's surface, the centre distance must first be changed to $L+a$.

With a pivot at distance a from the centre of mass, the parallel axis theorem and small-angle restoring torque give $T^2=(4{\pi }^2/g)[a+I_{\mathrm{CM}}/(Ma)]$. Differentiation, or the arithmetic–geometric mean inequality, minimizes the period at the radius of gyration $\sqrt{I_{\mathrm{CM}}/M}$. The minimum is $T_{\min }=2\pi \sqrt{2a_{\min }/g}$. The allowed pivot locations must include this point; a constraint keeping the pivot outside the body can move the physical optimum to a boundary.

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.

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.

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.

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.

The parallel axis theorem states $I_O=I_G+m{d}^2$ for two parallel axes, one through the centre of mass and the other a perpendicular distance $d$ away. To see this, expand the squared distance to the translated axis in the mass integral; its mixed term vanishes because the centroidal first moment is zero.

For a uniform solid hemisphere, the moment about a diameter in its flat face is half the corresponding full-sphere moment. With full-sphere mass $2m$, it is $\frac{1}{2}[\frac{2}{5}(2m)a^2]=2m{a}^2/5$. The original PDF gives the centroidal displacement $d=3a/8$; the TeX's $3a/5$ is incorrect. Moving the diameter axis to its parallel centroidal axis therefore yieldsThis is the moment of inertia of a uniform solid hemisphere, not that of a thin hemispherical shell.

Take a cross-section perpendicular to the rotation axis. Let $O$ be the underlying sphere centre, $G$ the centre of mass and $P$ the contact point. During rounded-surface contact, $OP=a$ vertically and $OG=d$ points toward the curved face, with downward component $d\cos \theta$. ThereforeFor rolling without slipping, $P$ is instantaneously at rest and the body rotates about it with angular speed $|\dot{\theta }|$. Thus the centroid speed is $b|\dot{\theta }|$. Translation and rotation contributeThe initial vertical base has $\theta =\pi /2$ and $K=0$, so conserved energy is $mga$. Equating the later $K+V$ to this value givesOn the initial roll toward the lowest centroid position, $\dot{\theta }<0$. Static contact friction does no work because the contacting material point is at rest; its coefficient must be large enough for the assumed no-slip motion. The formula describes the rounded-contact phase of the rolling uniform solid hemisphere and should not be continued through a change of contact geometry without another model.

The parallel axis theorem says that for an axis through a body's centre of mass and a parallel axis at perpendicular distance $d$,where $M$ is the total mass. The displacement is the distance between the two parallel axes; neither axis may be arbitrary in place of the center-of-mass axis. Expanding the squared distances of mass elements gives this formula, since the cross term integrates to zero at the centre of mass.

The moment of inertia of the uniform disc about its central perpendicular axis isThe parallel axis theorem gives $I_A=I_O+m{a}^2=3m{a}^2/2$. For angle $\beta$ from the downward resting position, the gravitational torque is $-mga\sin \beta$, so $I_A\ddot{\beta }=-mga\sin \beta$. Linearizing at $\beta =0$ gives $\ddot{\beta }+(2g/(3a))\beta =0$. Thus the uniform disc pivoted at its rim is a physical pendulum with period

Let $\mathbf{r}$ be position relative to the center of mass, and let the new origin have vector $\mathbf{c}$ relative to it. The inertia tensor about this origin isExpanding and using $\int \mathbf{r}dm=0$ proves the tensor form of the parallel axis theorem:For a uniform cube of side $L$ and mass $M$, centered coordinate integrals give $I_0=(M{L}^2/6)I$. The displacement to a vertex is $(L/2,L/2,L/2)$ up to signs. Using axes along its incident edges givesThe principal moments of inertia are $\boxed{M{L}^2/6,11M{L}^2/12,11M{L}^2/12}$. The first principal axis is the body diagonal through the vertex; every direction perpendicular to it has the repeated moment.

Let $M$ be the mass of a complete uniform sphere of radius $a$. The intact marble has $I_1=2M{a}^2/5$. Each removed bubble has mass $M/8$, radius $a/2$, and centre a distance $a/2$ from the rolling axis. By the parallel axis theorem, each removesThus $M_2=3M/4$, $I_2=2M{a}^2/5-7M{a}^2/80=5M{a}^2/16$, and $I_2/(M_2{a}^2)=5/12$. The two accelerations are therefore $(5/7)g\sin \alpha$ and $(12/17)g\sin \alpha$. From rest, distance is proportional to acceleration, so

Let a change of orthonormal coordinates be represented by a rotation matrix $R$. Since both angular momentum and angular velocity are vectors,Using $L_i=I_{ij}{\omega }_j$ in both frames givesfor every vector $\mathbf{\omega }$. Multiplication by $R^{-1}=R^T$ yieldswhich is exactly the transformation law for a rank-two tensor. Thus $I_{ij}$ is a rank-two tensor.

Put $\mathbf{r}=\mathbf{x}-\mathbf{a}$. The vector triple product givesThereforeAt the centre of mass,Expanding the first formula, the terms linear in $x_i$ vanish because the centre of mass is the origin, while $\rho {\int }_{B}dV=M$. Hence the parallel axis theorem in tensor form is

The axis $y=0,z=h$ is parallel to the $x$-axis and lies a distance $|h|$ from it. The parallel axis theorem therefore givesto leading order in the thickness.

The parallel axis theorem says that for an axis through $x$, parallel to the centre-of-mass axis in direction $\hat{d}$,LetMoments about the axis through $x_C$ add, soUsing $M_A=M_C-M_B$ and solving giveswhen the denominator is nonzero. In the degenerate equal-distance case, this inertia equation alone does not determine $M_B$; the centre-of-mass equation $M_C{x}_C=M_A{x}_A+M_B{x}_B$ supplies the remaining information.

The disc is rigidly attached, so its rotation axis is perpendicular to its face. Its moment of inertia about its center is $M{a}^2/2$: with surface mass density $M/(\pi a^2)$, an annulus of radius $s$ contributes $s^2[M/(\pi a^2)]2\pi sds$, whose integral is $M{a}^2/2$. The parallel axis theorem adds $M{d}^2$ at the pivot. Adding the rod's moment of inertia gives

The inertia tensor is $I_{ij}=\int (r^2{\delta }_{ij}-x_i{x}_j)dm$. The mixed moments vanish by reflection in $x$ or $y$. Every diagonal integrand is even in $z$, so its integral over the upper hemisphere is half its integral over the corresponding full ball of mass $2M$. A uniform ball's coordinate moments are each $(2M)a^2/5$, obtained by spherical symmetry and $⟨r^2⟩=3a^2/5$. Hence the hemisphere has each coordinate second moment $M{a}^2/5$ andThe tensor parallel axis theorem gives $I_O=I_C+M(|d{|}^{2}I-d{d}^t)$ with $d=(0,0,3a/8)$. ThereforeTranslation along the symmetry axis changes the two transverse moments but leaves the axial one unchanged.

The inertia tensor about the centroid has entries $J_{ij}={\int }_B\rho (r^2{\delta }_{ij}-x_i{x}_j)dV$. Reflection symmetry makes every off-diagonal entry zero. Under $x=a\xi$, $y=a\eta$, $z=c\zeta$, the ellipsoid becomes the unit ball. Its three coordinate second moments are equal, andUsing the stated mass gives ${\int }_B\rho x^{2}dV=M{a}^2/5$, the same for $y$, and ${\int }_B\rho z^{2}dV=M{c}^2/5$. ThereforeFor the parallel coordinate axes based at $P=(0,0,c)$, the parallel axis theorem in tensor form is $J^{(P)}=J^{(G)}+M(|d{|}^{2}I-d{d}^T)$, where $d=P-G$. Thus

Write $r=(x_1,x_2,x_3{)}^T$ for position relative to the centroid. The inertia tensor there isIn particular its diagonal entries integrate the squares of distances from the corresponding axes, and its off-diagonal entries are the negative products of coordinates.

Insert ${\mathbf{v}}_Q={\mathbf{v}}_G+\mathbf{\omega }\times r$ into $T=\frac{1}{2}{\int }_A\rho |{\mathbf{v}}_Q{|}^{2}dV$. The cross term vanishes because ${\int }_A\rho rdV=0$. Using $|\mathbf{\omega }\times r{|}^2=|\mathbf{\omega }{|}^2|r{|}^2-(\mathbf{\omega }\cdot r{)}^2$ givesIf $a$ is the vector from $G$ to $O$, positions relative to $O$ are $r-a$. Expanding the defining tensor integral with this substitution again removes the terms linear in $r$. Thus the tensor parallel axis theorem isSince $O$ is fixed, ${\mathbf{v}}_G=\mathbf{\omega }\times (-a)$, so $m|{\mathbf{v}}_G{|}^2=m{\mathbf{\omega }}^T(|a{|}^{2}I-a{a}^T)\mathbf{\omega }$. Substitution into the energy decomposition yields