Parallel Axis (Steiner) Theorem

Open any mechanics reference and the moment-of-inertia table is suspiciously short: one number per shape. A rod is $\tfrac{1}{12}ML^2$ , a disk $\tfrac12 MR^2$ , a sphere $\tfrac25 MR^2$ — always about an axis through the center of mass, and nothing else. Yet almost nothing in the real world spins about its center of mass. Doors swing on hinges at their edge, pendulums hang from pivots, wheels roll about a contact point, and a satellite's thruster torques act about whatever axis the mission demands. The table seems to answer none of these questions. It answers all of them, because of a single line of bookkeeping discovered in the age of pendulum clocks: $I_P = I_{\mathrm{cm}} + Md^2$ . Shift the axis a perpendicular distance $d$ away from the center of mass and the moment of inertia grows by exactly the inertia of a point particle carrying the whole mass $M$ at that distance. The body's shape never enters the correction. All the shape information lives in the tabulated $I_{\mathrm{cm}}$ ; the shift term treats the body as a structureless dot. That clean separation — intrinsic spin cost plus orbital cost of the center of mass — is the parallel axis theorem, and its tensor version moves all nine components of the inertia tensor in one stroke. Master it and a family of problems collapses: the physical pendulum and its curious minimum-period pivot, rolling bodies analyzed about their contact point, composite objects assembled piece by piece (holes included, as negative mass), and the reversible pendulum with which Kater measured $g$ to five figures in 1817. The whole thing rests on one vanishing sum.

Key takeaways

The parallel axis (Steiner) theorem transfers the inertia tensor from the center of mass to any other reference point: shape stays in $I^{(\mathrm{cm})}$ , the shift adds only a point-mass term.
Scalar form: $I_P = I_{\mathrm{cm}} + Md^2$ , with $d$ the perpendicular distance between the parallel axes — never negative, so the CM axis is always the easiest to spin about.
The full tensor form $I^{(P)}_{ij} = I^{(\mathrm{cm})}_{ij} + M\big(a^2\delta_{ij} - a_i a_j\big)$ moves all nine components — moments and products of inertia — at once.
The algebraic engine is the vanishing first moment $\sum_\alpha m_\alpha \mathbf{r}'_\alpha = \mathbf{0}$ : every cross term carries this factor, so the theorem connects an arbitrary axis to the CM axis — never directly to another off-CM axis.
It powers the physical pendulum ( $T = 2\pi\sqrt{(I_{\mathrm{cm}}+M\ell^2)/Mg\ell}$ , minimum period at $\ell = k$ , the radius of gyration), rolling about the contact point, and composite-body assembly.
Key idea: Move the reference point off the center of mass and the inertia tensor picks up exactly the inertia of the whole mass concentrated at the CM, orbiting the new point: $I^{(P)}_{ij} = I^{(\mathrm{cm})}_{ij} + M(a^2\delta_{ij} - a_i a_j)$ . The cross terms vanish only because moments are taken from the CM.

Where this lives today

Every CAD package computes Steiner shifts thousands of times before you click "export". SolidWorks, CATIA, and their kin store each part's inertia tensor about its own center of mass; when you assemble a gearbox or a drone, the software walks the assembly tree applying equation (2) and sums — the negative-mass bookkeeping of the plate example scaled to ten thousand parts. The same recursion runs live inside modern robots: recursive Newton–Euler algorithms in a humanoid's whole-body controller shift every link's inertia to the current joint axes at kilohertz rates, and a robot's URDF description file literally ships as "mass, CM position, inertia tensor about the CM" per link — a Steiner theorem waiting to be applied at every timestep. Spacecraft engineering lives by the theorem's other face: as propellant drains, a satellite's CM migrates, and mass-properties engineers maintain the vehicle's tensor as a function of fill fraction via time-dependent Steiner corrections. Closer to home, sports-equipment designers quote a bat's swing weight — its moment about the grip axis, $I_{\mathrm{cm}} + Md^2$ — because the grip, not the CM, is where humans pivot things. The 350-year-old bookkeeping rule is still the interface between the table and the world.

The physical picture

Spin a bicycle wheel about its axle: every gram of rim circles the center, and the cost is the tabulated $I_{\mathrm{cm}}$ . Now grab the wheel at a rim point and spin it about that. Something new happens — the center of mass itself sweeps a circle of radius $d$ about your hand. The wheel does two jobs at once: it spins about its CM, and its CM orbits the axis. The first job costs the old $I_{\mathrm{cm}}$ ; the second costs what a point mass $M$ on a string of length $d$ would cost, namely $Md^2$ . The theorem says the total is exactly the sum — no interference term, ever.

Figure

The price of an off-center axis is the orbit of the center of mass

The right-hand panel of the figure carries the deeper message. Plot the moment of inertia against the offset $d$ for the whole family of parallel axes and you get a parabola whose floor sits exactly at the center of mass: since $Md^2 \ge 0$ vanishes only at $d = 0$ , the CM is provably the easiest point to spin the body about in any direction, and every offset axis is strictly harder, with the penalty growing quadratically — double the offset, quadruple the surcharge. This is why a skater's spin axis passes through her center of mass and why a rotor on an off-center bearing shakes: the geometry taxes every millimeter of misalignment. There is also a conservation-law flavor to the split: the same algebra that separates $I_P$ into "spin about CM" plus "orbit of CM" separates kinetic energy into translation plus spin (König's theorem) and angular momentum into orbital plus spin. All three are one miracle — measure positions from the center of mass and the first moment $\sum_\alpha m_\alpha \mathbf{r}'_\alpha$ vanishes identically, taking every cross term with it.

The scalar law and the tensor law

Start with the version you will use daily. For rotation about an axis through point $P$ , parallel to an axis through the center of mass a perpendicular distance $d$ away:

I_P = I_{\mathrm{cm}} + M d^2 \qquad (1)

where: $I_P$ — moment of inertia about the axis through $P$ ; $I_{\mathrm{cm}}$ — moment about the parallel axis through the CM (the table entry); $M$ — total mass; $d$ — perpendicular distance between the two axes. Read it physically: the correction is what a point mass $M$ at distance $d$ would contribute, and it knows nothing about the body's shape — a rod, a disk, and a potato of equal mass pay the same surcharge for the same offset. Note also what the formula does not allow: both axes must be parallel, and one of them must pass through the CM. The full statement transfers the entire inertia tensor — all three moments and all three products of inertia — in a single tensor equation. Let $\mathbf{a}$ be the displacement vector from the CM to the new reference point $P$ :

I^{(P)}_{ij} = I^{(\mathrm{cm})}_{ij} + M\big(a^2\delta_{ij} - a_i a_j\big) \qquad (2)

where: $I^{(P)}_{ij}$ — components of the inertia tensor about $P$ ; $a_i$ — components of $\mathbf{a}$ ; $a^2 = a_x^2 + a_y^2 + a_z^2$ ; $\delta_{ij}$ — the Kronecker delta. The correction is precisely the inertia tensor of a single point particle of mass $M$ at displacement $\mathbf{a}$ — the one-particle case of the defining sum, verbatim. And because it is built only from $\mathbf{a}$ and $M$ , it is even in $\mathbf{a}$ : shifting to $+\mathbf{a}$ or $-\mathbf{a}$ costs the same, as symmetry demands. For reference, the tensor about the CM is the standard sum over particles — its construction and its two jobs (mapping $\boldsymbol{\omega}\to\mathbf{L}$ and pricing kinetic energy) belong to the moment of inertia tensor article and are not re-derived here:

I^{(\mathrm{cm})}_{ij} = \sum_\alpha m_\alpha\big(r_\alpha'^2\delta_{ij} - r'_{\alpha,i}r'_{\alpha,j}\big) \qquad (3)

where: $\mathbf{r}'_\alpha$ — position of particle $\alpha$ relative to the CM. The single fact about this sum that powers everything on this page is the definition of the center of mass itself:

\sum_\alpha m_\alpha\,\mathbf{r}'_\alpha = \mathbf{0} \qquad (4)

where: the sum is the first moment of the mass distribution about the CM. Equation (4) is not a theorem — it is what "center of mass" means. In the expansion below, every cross term between shape and displacement arrives multiplied by exactly this sum, and dies; measure from any other point and both survive. That single observation is the theorem's entire engine.

Figure

Why the cross terms die: the first moment about the CM is zero

Derivation AFull tensor Steiner theorem

We are deriving the nine-component shift rule (2) directly from the defining sum of the inertia tensor, and watching exactly where the center of mass earns its privileged role. The strategy is plain expansion: write each particle's position relative to $P$ as its CM-relative position minus the shift vector, square out, and sort the terms into three groups — pure shape, pure shift, and cross terms. The step to watch is the fate of the cross terms: they do not cancel by any clever identity, they cancel because each one carries the factor $\sum_\alpha m_\alpha \mathbf{r}'_\alpha$ , which equation (4) makes identically zero. That is also why the derivation fails between two arbitrary points — replace the CM by anything else and this factor survives.

Step 1. Write the definition of the tensor about $P$ . With $\mathbf{a}$ the displacement from the CM to $P$ , particle $\alpha$ sits at $\mathbf{r}'_\alpha - \mathbf{a}$ relative to $P$ :

I^{(P)}_{ij} = \sum_\alpha m_\alpha\Big[(\mathbf{r}'_\alpha-\mathbf{a})^2\,\delta_{ij} - (r'_{\alpha,i}-a_i)(r'_{\alpha,j}-a_j)\Big]

Why the sign convention is safe: the final correction is even in $\mathbf{a}$ , so CM-to-P versus P-to-CM cannot matter.

Step 2. Expand the square and the product:

(\mathbf{r}'_\alpha-\mathbf{a})^2 = r'^2_\alpha - 2\,\mathbf{r}'_\alpha\!\cdot\!\mathbf{a} + a^2

(r'_{\alpha,i}-a_i)(r'_{\alpha,j}-a_j) = r'_{\alpha,i}r'_{\alpha,j} - r'_{\alpha,i}a_j - a_i r'_{\alpha,j} + a_ia_j

Why nothing is lost: this is exact binomial algebra, valid particle by particle; all the physics enters when we sum.

Step 3. Sum over particles and sort into three groups. The quadratic-in- $\mathbf{r}'$ terms reassemble the CM tensor (3); the constant-in- $\mathbf{r}'$ terms factor out of the sum and give $\big(\sum_\alpha m_\alpha\big)\big(a^2\delta_{ij} - a_ia_j\big) = M\big(a^2\delta_{ij} - a_ia_j\big)$ ; and the terms linear in $\mathbf{r}'_\alpha$ collect into

-2\,a_k\Big(\sum_\alpha m_\alpha r'_{\alpha,k}\Big)\delta_{ij} + a_j\Big(\sum_\alpha m_\alpha r'_{\alpha,i}\Big) + a_i\Big(\sum_\alpha m_\alpha r'_{\alpha,j}\Big)

Why this grouping is the whole proof: shape talks to shift only through these linear terms — everything else is already either pure $I^{(\mathrm{cm})}$ or pure point-mass.

Step 4. Kill the linear group with the CM condition. Each bracket is a component of the first moment $\sum_\alpha m_\alpha \mathbf{r}'_\alpha$ , which vanishes by (4). The entire cross-term group is identically zero — not small, zero. Why this is the only place the CM enters: had we expanded about any other point $Q$ , these brackets would equal $M$ times the CM's displacement from $Q$ , and the "theorem" would sprout a messy linear correction. The clean two-term rule exists only relative to the center of mass.

Step 5. Read off the result:

I^{(P)}_{ij} = I^{(\mathrm{cm})}_{ij} + M\big(a^2\delta_{ij}-a_ia_j\big)

The correction is the inertia tensor of one point particle of mass $M$ at displacement $\mathbf{a}$ , it is positive semi-definite, and it vanishes only when $\mathbf{a} = \mathbf{0}$ — so every component-wise moment is minimized at the CM. As a quick consistency check: taking the trace of both sides yields $\mathrm{tr}\,I^{(P)} = \mathrm{tr}\,I^{(\mathrm{cm})} + 2Ma^2$ , matching equation (7) below, since $\delta_{ii} = 3$ gives $M(3a^2 - a^2) = 2Ma^2$ .

Extracting the working rules

The tensor statement (2) is the master copy; the blackboard rules are its shadows — contraction recovers the scalar law, double application relates off-CM axes, tracing gives a one-line audit.

Go deeper · Derivation B · 2 minFrom the tensor to the scalar rule and the two-axis relation

Derivation BFrom the tensor to the scalar rule and the two-axis relation

The goal is to compress the nine-component rule (2) into the three working identities of daily practice: the scalar law (1), the recognition that its $d$ is the perpendicular distance to the axis (not the distance between points), and the route-through-the-CM relation for two arbitrary parallel axes. The strategy is a single contraction with the unit vector $\hat{\mathbf{n}}$ along the rotation axis — the general recipe for extracting the scalar moment about any axis from the tensor — followed by using the result twice. Watch how the component of $\mathbf{a}$ along the axis drops out: sliding the reference point up and down the axis changes nothing, exactly as rotational geometry demands.

Step 1. Contract both sides of (2) with $\hat{\mathbf{n}}$ . Using $\hat n_i\,\delta_{ij}\,\hat n_j = 1$ and $\hat n_i a_i = \mathbf{a}\cdot\hat{\mathbf{n}}$ :

\hat{\mathbf{n}}^{\mathsf T} I^{(P)} \hat{\mathbf{n}} = \hat{\mathbf{n}}^{\mathsf T} I^{(\mathrm{cm})} \hat{\mathbf{n}} + M\big(a^2 - (\mathbf{a}\cdot\hat{\mathbf{n}})^2\big) \qquad (5)

Why contraction is the right move: $\hat{\mathbf{n}}^{\mathsf T} I\, \hat{\mathbf{n}}$ is precisely the scalar moment of inertia about the axis $\hat{\mathbf{n}}$ — the number in $L = I\omega$ when the rotation is locked to that axis.

Step 2. Recognize the geometry. Decompose $\mathbf{a}$ into components along and perpendicular to the axis: $a^2 - (\mathbf{a}\cdot\hat{\mathbf{n}})^2 = a_\perp^2 = d^2$ , the squared perpendicular distance between the two parallel axes. Equation (5) collapses to $I_P = I_{\mathrm{cm}} + Md^2$ — the scalar law (1), now with its $d$ properly defined. Why the parallel component is invisible: moving the reference point along the rotation axis does not move the axis. Only the sideways offset costs anything.

Step 3. Relate two arbitrary parallel axes by applying the scalar law twice and eliminating $I_{\mathrm{cm}}$ :

I_P = I_Q + M\big(d_P^2 - d_Q^2\big) \qquad (6)

where: $d_P$ and $d_Q$ — perpendicular distances from the CM axis to the axes through $P$ and $Q$ . This is the only legal hop between two off-CM axes: down to the CM, then back up. Why no direct rule can exist: relative to $Q$ the first moment does not vanish, so the cross terms of Derivation A survive and depend on the body's shape — precisely the dependence the CM route removes. The tempting shortcut $I_P = I_Q + M\,\overline{PQ}^2$ misses by 4% in the plate example below.

Step 4. Trace both sides of the tensor law for a coordinate-free check:

\mathrm{tr}\,I^{(P)} = \mathrm{tr}\,I^{(\mathrm{cm})} + 2Ma^2 \qquad (7)

Why the factor 2: $\delta_{ii} = 3$ in three dimensions, so the correction traces to $M(3a^2 - a_ia_i) = M(3a^2 - a^2) = 2Ma^2$ . Since the trace is rotation-invariant, this one number audits a whole tensor computation at a glance. *As a check: applying (1) to a uniform rod about its end gives $I = \tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac13 ML^2$ ; and the two-axis relation (6) with $Q$ = CM reduces to (1) identically.*

Which form answers which question

you want	use	notes
moment about an offset axis, table in hand	$I_P = I_{\mathrm{cm}} + Md^2$ , eq. (1)	$d$ = perpendicular distance between the parallel axes
the whole tensor at a new point (products included)	tensor form, eq. (2)	correction = point-mass tensor at $\mathbf{a}$
moment about a skew direction through $P$	contract: eq. (5)	only $a_\perp$ matters; axial offset is free
relate two off-CM axes	route through CM, eq. (6)	never a direct $Md^2$ hop
audit a tensor-shift computation	trace check, eq. (7)	$+2Ma^2$ , rotation-invariant
recover $I_{\mathrm{cm}}$ from a measured $I_P$	$I_{\mathrm{cm}} = I_P - Md^2$	"reverse Steiner": subtract to reach the floor

Where the theorem earns its living: the physical pendulum

Hang any rigid body — a meter stick, a leg, a church bell — from a horizontal pivot a distance $\ell$ from its CM and let it swing. The resisting inertia is the moment about the pivot, which Steiner delivers without a single new integral.

Go deeper · Derivation C · 2 minPhysical pendulum period and the minimum-period pivot

Derivation CPhysical pendulum period and the minimum-period pivot

We are deriving the small-oscillation period of a rigid body pivoted a distance $\ell$ from its center of mass, then extracting two payoffs the bare formula hides: the pivot location that minimizes the period, and the conjugate-pivot property that made Kater's reversible pendulum the world's standard gravimeter for a century. The strategy: use Steiner for the pivot inertia, write the torque equation for small angles, and then study the resulting equivalent length $L_{\mathrm{eq}}(\ell)$ as a function of where you drill the hole. Watch the competition: moving the pivot toward the CM shrinks the inertia but shrinks the gravity lever arm faster.

Step 1. Inertia about the pivot, by the scalar law (1):

I_{\text{pivot}} = I_{\mathrm{cm}} + M\ell^2

Why no new integral is needed: the table (or one CM computation) supplies $I_{\mathrm{cm}}$ ; the drill hole's position enters only through $M\ell^2$ .

Step 2. Torque equation about the pivot for a small tilt $\theta$ . Gravity $Mg$ acts at the CM with lever arm $\ell\sin\theta \approx \ell\theta$ :

\big(I_{\mathrm{cm}} + M\ell^2\big)\,\ddot\theta = -Mg\ell\,\theta

Why the pivot is the right reference: the unknown pivot reaction force acts at zero lever arm there, so it never enters — the same constraint-force elimination that powers all of rigid-body practice.

Step 3. Read off the frequency and period of simple harmonic motion:

T = 2\pi\sqrt{\dfrac{I_{\mathrm{cm}}+M\ell^2}{Mg\ell}} \qquad (8)

where: the ratio under the root has dimensions of time squared; the numerator is Steiner's pivot inertia and the denominator is the gravitational stiffness $Mg\ell$ .

Step 4. Define the equivalent simple-pendulum length and the radius of gyration $k$ (via $I_{\mathrm{cm}} = Mk^2$ ):

L_{\mathrm{eq}} = \dfrac{I_{\mathrm{cm}}+M\ell^2}{M\ell} = \ell + \dfrac{k^2}{\ell} \qquad (9)

where: $L_{\mathrm{eq}}$ — the length of the ideal point-mass pendulum with the same period, $T = 2\pi\sqrt{L_{\mathrm{eq}}/g}$ . The two terms compete: pivot far out and the $\ell$ term dominates (the body is nearly a point mass on a long arm); pivot close in and $k^2/\ell$ blows up (huge inertia, feeble torque). A pendulum pivoted exactly at its CM has infinite period — no restoring torque at all.

Step 5. Minimize. $\mathrm{d}L_{\mathrm{eq}}/\mathrm{d}\ell = 1 - k^2/\ell^2 = 0$ gives $\ell = k$ , where $L_{\mathrm{eq}}^{\min} = 2k$ : the fastest-swinging pivot sits exactly one radius of gyration from the CM. Why the minimum is physical, not formal: at $\ell = k$ the intrinsic inertia $Mk^2$ and the orbital inertia $M\ell^2$ contribute equally — the crossover point of Steiner's two terms.

Step 6. Conjugate pivots. Fix a period, i.e. fix $L_{\mathrm{eq}}$ ; equation (9) is a quadratic in $\ell$ with two roots satisfying $\ell_1\ell_2 = k^2$ and $\ell_1 + \ell_2 = L_{\mathrm{eq}}$ : every physical pendulum has pairs of pivots with identical periods, separated by exactly the equivalent length. Why this mattered for two centuries: measure the common period of two conjugate pivots and the distance between them, and $g = 4\pi^2(\ell_1+\ell_2)/T^2$ follows with no knowledge of the mass distribution — Kater's reversible pendulum (1817). *As a check: for the uniform rod of the worked example below, equation (9) gives $L_{\mathrm{eq}} = 0.45 + 0.0675/0.45 = 0.600\ \mathrm{m} = \tfrac23 L$ , the classic result.*

Figure

The equivalent length has a minimum at one radius of gyration from the CM

Limiting cases

Case	Result	Meaning
Axis through the CM	$d = 0 \Rightarrow I_P = I_{\mathrm{cm}}$	No displacement, no correction — the floor of the parabola $I(d)$ .
Point particle	$I_{\mathrm{cm}} = 0,\; I_P = Md^2$	No intrinsic spin inertia; all of $I_P$ is orbital. The correction is the point-particle tensor.
Far-away axis, $d \gg$ body size	$I_P \approx Md^2$	Every body looks like a point mass from far enough away.
Uniform rod about its end	$\tfrac{1}{12}ML^2 + M(L/2)^2 = \tfrac13 ML^2$	The classic table pair — the end axis costs exactly 4× the center axis.
Disk about an in-plane diameter shifted to the rim	$\tfrac14 MR^2 + MR^2 = \tfrac54 MR^2$	The perpendicular axis theorem result $\tfrac14 MR^2$ plus a Steiner shift — the theorems compose.
Rolling wheel, instantaneous contact axis	$I_{\text{contact}} = I_{\mathrm{cm}} + MR^2$	Rolling KE $= \tfrac12 I_{\text{contact}}\omega^2$ contains both translation and spin.
Tensor trace	$\mathrm{tr}\,I^{(P)} = \mathrm{tr}\,I^{(\mathrm{cm})} + 2Ma^2$	Coordinate-free audit; the 2 comes from $\delta_{ii} = 3$ minus one.

Common trapYou can hop directly between any two parallel axes with one Md-squared term

You cannot — and this is the single most common Steiner error. The theorem connects an arbitrary axis to the parallel axis through the center of mass, full stop: the cross terms vanish only because the first moment about the CM is zero, equation (4); relative to any other reference point they survive and no clean shift rule exists. To relate axes through two general points, route through the CM — $I_P - I_Q = M(d_P^2 - d_Q^2)$ , equation (6), both distances measured from the CM axis. The illegal shortcut $I_P = I_Q + M\,\overline{PQ}^2$ lands 4.2% low in the plate-with-a-hole example below — a large error by machine-design standards. Quick self-audit: the direct hop is legal only when $Q$ is the CM, in which case it is the theorem.

Worked examples

1 · A meter-stick pendulum, three numbers from one table entry (confidence builder)

A uniform rod of mass $M = 1.2\ \mathrm{kg}$ and length $L = 0.90\ \mathrm{m}$ has $I_{\mathrm{cm}} = \tfrac{1}{12}ML^2$ about a transverse axis through its center. Find its moment of inertia about a transverse axis through one end, its small-oscillation period when hung from that end, and the pivot position that would swing fastest.

Step 1. The table entry: $I_{\mathrm{cm}} = \tfrac{1}{12}(1.2)(0.90)^2 = \tfrac{1}{12}(1.2)(0.81) = 0.081\ \mathrm{kg\,m^2}$ .

Step 2. Shift to the end with equation (1). The end is $d = L/2 = 0.45\ \mathrm{m}$ from the CM: $I_{\text{end}} = 0.081 + 1.2\,(0.45)^2 = 0.081 + 0.243 = \boxed{0.324\ \mathrm{kg\,m^2}}$ — exactly $\tfrac13 ML^2$ , and exactly $4\times$ the center value: the orbital term $Md^2$ is three times the intrinsic one.

Step 3. Period about the end from equation (8), with $\ell = 0.45\ \mathrm{m}$ : the stiffness is $Mg\ell = 1.2 \times 9.81 \times 0.45 = 5.297\ \mathrm{N\,m}$ , so $T = 2\pi\sqrt{0.324/5.297} = \boxed{1.55\ \mathrm{s}}$ . Equivalently $L_{\mathrm{eq}} = I_{\text{end}}/(M\ell) = 0.324/0.54 = 0.600\ \mathrm{m} = \tfrac23 L$ : the rod swings like an ideal pendulum two-thirds its length.

Step 4. The fastest pivot. The radius of gyration is $k = \sqrt{I_{\mathrm{cm}}/M} = \sqrt{0.0675} = 0.260\ \mathrm{m}$ ; the minimum period occurs at $\ell = k$ , where $L_{\mathrm{eq}} = 2k = 0.520\ \mathrm{m}$ and $T_{\min} = 2\pi\sqrt{0.520/9.81} = \boxed{1.45\ \mathrm{s}}$ . Drilling the pivot hole anywhere else — nearer the CM or nearer the end — swings slower.

2 · A disk with an off-center hole: assembly, reverse Steiner, and the illegal hop (exam level)

A uniform plate is made by cutting a circular hole of radius $r = 0.10\ \mathrm{m}$ from a solid disk of radius $R = 0.40\ \mathrm{m}$ ; the hole's center sits $d = 0.20\ \mathrm{m}$ from the disk center $O$ . The finished plate has mass $M = 7.5\ \mathrm{kg}$ . Find the plate's moment of inertia about the axis through $O$ perpendicular to the plate, locate the plate's CM, find the moment about the parallel axis through the hole's center $H$ — and show what the illegal direct hop would have given.

Step 1. Reconstruct the pieces. The surface density is $\sigma = M/[\pi(R^2 - r^2)] = 7.5/[\pi(0.16 - 0.01)] = 15.92\ \mathrm{kg/m^2}$ , so the un-cut disk would have mass $m_{\text{full}} = \sigma\pi R^2 = 8.00\ \mathrm{kg}$ and the removed plug $m_{\text{hole}} = \sigma\pi r^2 = 0.50\ \mathrm{kg}$ . Treat the plate as full disk plus negative-mass plug.

Step 2. Assemble about $O$ , Steiner on the plug. Full disk: $\tfrac12(8.0)(0.16) = 0.640$ . Plug about its own center $\tfrac12(0.5)(0.01) = 0.0025$ , shifted to $O$ : $0.0025 + 0.5\,(0.20)^2 = 0.0225$ . Subtract: $I_O = 0.640 - 0.0225 = \boxed{0.6175\ \mathrm{kg\,m^2}}$ .

Step 3. Locate the CM: $x_{\mathrm{cm}} = (-m_{\text{hole}}\,d)/M = -0.5(0.20)/7.5 = -0.0133\ \mathrm{m}$ — 13.3 mm from $O$ , on the side away from the hole.

Step 4. Reverse Steiner to reach the floor. $O$ is itself an off-CM point, so $I_{\mathrm{cm}} = I_O - M x_{\mathrm{cm}}^2 = 0.6175 - 0.0013 = \boxed{0.6162\ \mathrm{kg\,m^2}}$ — barely below $I_O$ , because 13 mm is small and the penalty is quadratic.

Step 5. The axis through $H$ , legally, via route-through-CM. Distance from CM to $H$ : $0.20 + 0.0133 = 0.2133\ \mathrm{m}$ , so $I_H = 0.6162 + 7.5(0.2133)^2 = 0.6162 + 0.3413 = \boxed{0.9575\ \mathrm{kg\,m^2}}$ .

Step 6. The illegal hop, for contrast: $I_O + M\,\overline{OH}^2 = 0.6175 + 7.5(0.04) = 0.9175\ \mathrm{kg\,m^2}$ — low by $4.2\%$ , precisely because $O$ is not the CM. The warning box is not pedantry; it is four percent of your flywheel.

Origins · 1 minJakob Steiner and how this was discovered

Jakob Steiner

1796–1863 · Berlin

Steiner is the unlikeliest name in a mechanics course: a Swiss farmer's son who did not learn to write until he was fourteen, walked to Pestalozzi's experimental school at eighteen, and rose to the first geometry chair at the University of Berlin, created for him in 1834. He was the purest of pure geometers — he considered calculation a crutch, and his Systematische Entwickelung (1832) rebuilt geometry on projective principles; his isoperimetric arguments and symmetrization technique still carry his name in analysis. The axis-shift rule is a small corner of his work on second moments of figures — the physical statement was already implicit in Huygens' pendulum studies 150 years earlier — but Steiner's systematic geometric treatment attached his name to it permanently. A fitting irony: the man who distrusted formulas is cited daily through one line of algebra by engineers who have never read a word of projective geometry.

History

1673

— Huygens' Horologium Oscillatorium analyzes the compound pendulum and the center of oscillation; the parallel-axis shift and the conjugate-pivot property are implicit in his results.

1765

— Euler's Theoria motus corporum solidorum systematizes moments of inertia and principal axes, making axis-transfer rules part of a general rigid-body theory.

1817

— Kater's reversible pendulum turns the conjugate-pivot property into a precision instrument, measuring g to five significant figures with two knife edges and a clock.

1834

— Steiner takes the new geometry chair at Berlin; his synthetic treatment of second moments of figures ties his name to the axis-shift theorem.

1958

— Explorer 1's flat-spin surprise makes accurate inertia bookkeeping — CM tensors plus Steiner shifts for every appendage — a permanent discipline of spacecraft engineering.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 9 (Angular Momentum, Part II) · Goldstein, Classical Mechanics, 3rd ed., Ch. 5 · Landau & Lifshitz, Mechanics (Vol. 1), §32 · Taylor, Classical Mechanics, Ch. 10 Key numerical results on this page: rod end result $\tfrac13 ML^2$ from $\tfrac1{12}+\tfrac14$ ; hoop tangent-axis value $\tfrac32 MR^2$ ; trace identity factor $2Ma^2$ ; pendulum minimum at $\ell=k$ . Worked-example values: rod: $I_{\text{end}} = 0.324\ \mathrm{kg\,m^2}$ , $T = 1.55\ \mathrm{s}$ , $T_{\min} = 1.45\ \mathrm{s}$ ; plate: $I_O = 0.6175$ , $I_{\mathrm{cm}} = 0.6162$ , $I_H = 0.9575\ \mathrm{kg\,m^2}$ , illegal hop 4.2% low.

Builds on: Moment of Inertia Tensor · Products of Inertia and the Inertia Matrix Leads to: Principal Axes and Principal Moments of Inertia · Perpendicular Axis Theorem · Angular Momentum of a Rigid Body · Rotational Kinetic Energy

Why does the theorem only work from the center of mass and not between any two points?

Because its proof has exactly one non-trivial step: the cross terms between shape and shift arrive multiplied by the first moment $\sum_\alpha m_\alpha \mathbf{r}'_\alpha$ , and that sum is zero only when positions are measured from the CM — that is the definition of the CM. Expand about any other point $Q$ and the cross terms survive, proportional to the CM's displacement from $Q$ and dependent on the body's actual mass distribution — so no shape-blind rule can exist between two general points. The practical consequence is equation (6): to connect two off-CM axes, go down to the CM and back up.

Does the parallel axis theorem apply to the products of inertia too?

Yes — that is what the tensor form (2) buys. The off-diagonal components shift as $I^{(P)}_{xy} = I^{(\mathrm{cm})}_{xy} - M a_x a_y$ (and cyclic). Unlike the diagonal correction, this piece can have either sign: shifting the origin can create products of inertia where the CM tensor had none, or cancel existing ones. That is why a body whose CM tensor is diagonal generally has a non-diagonal tensor about a pivot — the $-Ma_ia_j$ term pollutes the off-diagonal slots unless $\mathbf{a}$ lies along a principal axis — and why "shift, then re-diagonalize" is the standard route to principal axes at a support point.

Is there a version of this theorem for kinetic energy or angular momentum?

The same first-moment cancellation, equation (4), produces a matching split for every quadratic quantity in mechanics. For kinetic energy it is König's theorem, $T = \tfrac12 M v_{\mathrm{cm}}^2 + T_{\mathrm{spin}}$ ; for angular momentum it is orbital-of-CM plus spin-about-CM. Steiner is the "static" member of the family — it decomposes the inertia tensor rather than a motion — but the engine is identical. A useful habit follows: if a cross term seems to survive in any such calculation, you almost certainly measured something from a point that is not the CM.

What exactly is the radius of gyration and why should I care about it?

The radius of gyration $k$ is defined by $I_{\mathrm{cm}} = Mk^2$ : the distance at which a point mass $M$ would replicate the body's intrinsic moment. It compresses the whole shape integral into one length and makes Steiner statements comparable across bodies: $I_P = M(k^2 + d^2)$ , so the offset matters relative to $k$ . It also owns two elegant facts from the pendulum derivation: the fastest pivot sits at $\ell = k$ exactly, and conjugate pivot pairs satisfy $\ell_1\ell_2 = k^2$ — the geometric mean of conjugate distances is the radius of gyration. For the 0.90 m rod, $k = L/\sqrt{12} = 0.26$ m; for a uniform disk about its symmetry axis, $k = R/\sqrt{2}$ .

Why can I treat a rolling wheel as rotating about its contact point, and does Steiner enter?

A wheel rolling without slipping has zero velocity at the contact point, so the velocity field is momentarily a pure rotation about the contact line — the instantaneous axis. The kinetic energy is then $\tfrac12 I_{\text{contact}}\,\omega^2$ with $I_{\text{contact}} = I_{\mathrm{cm}} + MR^2$ by Steiner, and this single term automatically contains both the translational $\tfrac12 Mv^2$ and the spin $\tfrac12 I_{\mathrm{cm}}\omega^2$ (substitute $v = \omega R$ and both reappear). Bonus: friction acts at the contact, so it exerts no torque about it. One caution — the contact point accelerates, so $\tau = I\dot\omega$ about it needs care in general, but for rolling on a plane the correction vanishes and the simple form survives.

How did Kater's pendulum measure g so precisely, and what does Steiner have to do with it?

It is the conjugate-pivot property of Derivation C, which is pure Steiner: $L_{\mathrm{eq}} = \ell + k^2/\ell$ takes the same value at two distances with $\ell_1\ell_2 = k^2$ and $\ell_1 + \ell_2 = L_{\mathrm{eq}}$ . Kater built a rod with two knife edges on opposite sides of the CM and tuned an adjustable mass until the periods about both edges matched. Then $g = 4\pi^2(\ell_1+\ell_2)/T^2$ , where $\ell_1+\ell_2$ is simply the measured distance between the edges — no mass distribution, CM location, or moment of inertia needs to be known; the conjugate condition eliminates everything hard about the body. Length and time were what 19th-century metrology measured superbly, so the design held the world gravimetry standard for over a century.

My CM tensor is diagonal. After a Steiner shift the tensor has off-diagonal entries. Did I make an error?

Almost certainly not — that is expected. The correction has off-diagonal entries $-Ma_ia_j$ whenever the displacement $\mathbf{a}$ has two or more nonzero components, i.e. whenever you shift diagonally relative to the CM principal axes. Physically: about the new point the mass is arranged asymmetrically even though it was symmetric about its own CM, so products of inertia genuinely appear and the principal axes at $P$ tilt relative to those at the CM. Only shifts along a CM principal axis preserve diagonality. If you need the principal frame at the new point: shift first, then re-diagonalize.

Parallel Axis (Steiner) Theorem

Where this lives today

The physical picture

The scalar law and the tensor law

Extracting the working rules

Which form answers which question

Where the theorem earns its living: the physical pendulum

Limiting cases

Worked examples

1 · A meter-stick pendulum, three numbers from one table entry (confidence builder)

2 · A disk with an off-center hole: assembly, reverse Steiner, and the illegal hop (exam level)

Going further

Before this

Where this leads