--- title: Rotational Inertia module: Rotation moduleNumber: 5 lessonNumber: 1 order: 501 summary: > Push a wheel and a merry-go-round with the same force and they speed up at wildly different rates: the same mass resists rotation differently depending on where it sits relative to the axis. That single fact is the moment of inertia, $I=\int r_\perp^2\,\d m$, and this lesson builds it from the ground up. We tie angular motion to linear through $s=r\theta$, $v=r\omega$, and $a_t=r\alpha$, derive $I$ for rods, disks, and spheres, and use the parallel- and perpendicular-axis theorems to move between axes — always naming the axis, because the same body has as many moments of inertia as it has lines to spin about. topics: [Rotation] draft: false sources: - book: Tipler & Mosca ref: "Ch. 9 — Rotation; §§9-1–9-3" --- ## Rotational coordinates, energy, and validation In fixed-axis rotation, every point of a rigid body shares one angular coordinate $\theta$. Arc length at radius $r$ is $s=r\theta$, with angle measured in radians. Differentiation gives $$ \omega=\frac{\d\theta}{\d t},\qquad \alpha=\frac{\d\omega}{\d t},\qquad v=r\omega. $$ Tangential acceleration is $a_t=r\alpha$; radial acceleration is $a_r=r\omega^2$. The components are perpendicular and have different physical roles: one changes speed, the other changes velocity direction. $$ % caption: A point on a rotating disk carries a tangential acceleration $a_t=r\alpha$ along its path and a radial (centripetal) acceleration $a_r=r\omega^2$ toward the axis. The two are perpendicular and combine into the total acceleration $a$. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black,thick] (0,0) circle (2); \draw[fill=white,draw=black] (0,0) circle (2pt); \coordinate (P) at (45:2); \draw[black] (0,0)--(P); \draw[fill=acc!12,draw=acc] (P) circle (2.4pt); \draw[->,acc,very thick] (P)--($(P)+(135:1.05)$) node[above right] {$a_t$}; \draw[->,black,very thick] (P)--($(P)+(225:1.05)$); \node[black] at (1.35,0.78) {$a_r$}; \draw[->,acc,thick,densely dashed] (P)--($(P)+(180:0.95)$) node[above] {$a$}; \node[black] at (-1.35,-1.35) {axis}; \end{tikzpicture} $$ **Inertia scaling and numerical validation.** Geometrically similar bodies made from the same material have mass proportional to linear size cubed and inertia proportional to linear size fifth power. If every length is multiplied by factor $c$, then $M\to c^3M$ and $I\to c^5I$. The ratio $I/(MR^2)$ remains unchanged. Dimensionless inertia ratios therefore depend on shape rather than size for similar bodies, while absolute rotational energy and motor requirements can change enormously with scale. An inertia axis is a line. A cylinder has different inertia about its long axis and a transverse diameter through the same centre. A formula selected without an axis direction can be numerically correct for the wrong physical problem. Component transfers also require each component's own centre of mass, not the centre of mass of an incomplete assembly. A numerical quadrature provides an independent check for a difficult shape. Divide the body into small cells, assign each cell mass $\Delta m$, and evaluate $$ I\approx\sum r_{\perp,i}^2\Delta m_i. $$ Refinement should converge toward the analytic result for a uniform simple shape. Failure to converge often identifies distance measured from the wrong axis or a density normalization error. For cutouts, treat removed cells as negative mass contributions relative to the original solid. > **Worked example.** A solid disk of mass $1.8\ \mathrm{kg}$ and radius > $0.15\ \mathrm m$ has $I=MR^2/2=0.0203\ \mathrm{kg\,m^2}$. A constant torque > $0.090\ \mathrm{N\,m}$ applied for $12\ \mathrm s$ gives > > $$ > \alpha=\frac\tau I=4.44\ \mathrm{rad\,s^{-2}}, > \quad \omega_f=53.3\ \mathrm{rad\,s^{-1}}, > \quad \theta=\tfrac12\alpha t^2=320\ \mathrm{rad}. > $$ > > Torque work is $\tau\theta=28.8\ \mathrm J$; rotational kinetic energy is > $I\omega_f^2/2=28.8\ \mathrm J$. The agreement checks the constant-torque > assumption, the inertia axis, the kinematic equations, and the fixed-axis > energy formula at once. A discrepancy would flag a resisting torque, variable > drive, or an inertia value that does not describe the rotating assembly. **Fixed-axis acceleration fields and stress relevance.** The acceleration field of a rigid rotor can be expressed in polar components. At radius $r$, tangential acceleration $r\alpha$ is proportional to radius and radial acceleration $r\omega^2$ is also proportional to radius. The outer rim therefore has the largest acceleration magnitude. This does not by itself give material stress, because stress depends on geometry and internal force distribution, but it identifies why high-speed rotor design is controlled by rim conditions. With constant angular acceleration during startup, radial acceleration grows as time squared because $\omega=\omega_i+\alpha t$. Tangential acceleration remains constant at a fixed radius. A vibration sensor mounted on a rotor consequently measures a changing combination of tangential and radial components during startup. Treating its output as one scalar acceleration without resolving geometry can lead to an erroneous inferred angular acceleration. **Work of variable torque.** Motor torque commonly depends on speed or angle. The rotational work relation remains an angular integral: $$ W=\int_{\theta_i}^{\theta_f}\tau(\theta)\,\d\theta. $$ A torsional spring with $\tau=-\kappa\theta$ has stored potential energy $$ U=\int_0^\theta\kappa\theta'\,\d\theta'=\frac12\kappa\theta^2. $$ The sign convention matters: spring torque is opposite displacement, while external work required to twist the spring is positive. Releasing the spring converts this energy to rotational kinetic energy if losses are negligible. The torsion-pendulum period follows from the same quadratic potential and rotational inertia. > **Worked example.** A rotor experiences torque > $\tau(\theta)=0.40-0.020\theta$ in SI units over $0\le\theta\le10\ \mathrm{rad}$. > The work is > > $$ > W=\int_0^{10}(0.40-0.020\theta)\,\d\theta=3.0\ \mathrm J. > $$ > > With rotor inertia $0.060\ \mathrm{kg\,m^2}$ and a start from rest, the final > speed is $\sqrt{2W/I}=10\ \mathrm{rad\,s^{-1}}$. Replacing the variable torque > by its initial value would overestimate the work; replacing it by its average > gives the same result here only because the torque is linear over the interval. **Product-of-inertia interpretation.** Products of inertia such as $I_{xy}=-\int xy\,\d m$ vanish when a body is symmetric under reflection in either coordinate plane. This is why rectangular plates aligned with their sides on coordinate axes have diagonal inertia tensors about their centre. An asymmetric attached mass makes a product nonzero and rotates the principal axes away from the original geometric axes. Principal-axis measurement is therefore especially important for assemblies rather than ideal single-shape components. The sign convention for products of inertia varies across engineering texts. The physical content does not: off-diagonal terms couple angular-velocity components to angular-momentum components. A calculation should state its tensor convention before comparing tabulated matrices or performing a coordinate rotation. **Inertia bounds and plausibility tests.** If every mass element lies between radii $r_{min}$ and $r_{max}$ from an axis, then $$ Mr_{min}^2\le I\le Mr_{max}^2. $$ The bounds are crude but bracket the allowable inertia. A thin hoop saturates both bounds because all mass lies at one radius. A solid disk lies below $MR^2$ because much mass is inside its rim. For a component assembly, bounding each component and summing bounds can expose a mistaken radius or forgotten parallel-axis term. Physical plausibility also requires nonnegative inertia. Any negative result from a hole-subtraction method indicates that mass, axis, or transferred term has been assigned incorrectly. A zero inertia is possible only when all mass lies exactly on the axis, an idealization approached by a thin rod rotating about its length. **Reported-result protocol.** An inertia result should state the body, mass model, axis line, coordinate origin, method, and uncertainty. A bare number in $\mathrm{kg\,m^2}$ cannot be reproduced without axis information. For an experimental result, report the period range, amplitude range, calibration object, and treatment of platform inertia. For an analytic result, report density assumptions and whether holes or attached hardware were included. This information is part of the physical result, not administrative detail. ## Composite inertia and pendulum methods Compound bodies are decomposed into components with simple geometry. Each component is assigned a local centre-of-mass inertia and a signed position relative to the requested common axis. The sign of position does not affect the parallel-axis term because distance is squared, but it matters when checking the overall centre of mass or when constructing a coordinate diagram. Holes are handled as negative-mass components when the original solid has uniform density; the removed mass and its inertia are subtracted about the same axis. The calculation is most transparent in a table: | Component | $I_{cm}$ | offset | transferred contribution | | --- | --- | --- | --- | | base plate | tabulated or integrated | $d_1$ | $I_{1,cm}+M_1d_1^2$ | | attached disk | $M_2R^2/2$ | $d_2$ | $I_{2,cm}+M_2d_2^2$ | | point mass | $0$ | $d_3$ | $M_3d_3^2$ | The table prevents addition of component inertias referred to different axes. Every entry must have units $\mathrm{kg\,m^2}$ before summation. $$ % caption: A compound body is broken into a plate, an offset disk, and a point mass; each contribution is transferred to the common dashed axis before the moments are summed. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black,dashed] (0,-1.7)--(0,1.9) node[above] {axis}; \draw[thick,fill=black!5] (-1.3,-0.55) rectangle (1.1,0.55); \node at (-0.7,0) {plate}; \draw[black,thick,fill=black!5] (2.4,0) circle (0.6); \node[black] at (2.4,0) {disk}; \draw[fill=black!20,draw=black] (3.9,0) circle (3pt); \node at (3.9,0.45) {mass}; \draw[<->,black] (0,-1.15)--(2.4,-1.15) node[midway,below] {$d_2$}; \draw[<->,black] (0,1.35)--(3.9,1.35) node[midway,above] {$d_3$}; \end{tikzpicture} $$ > **Worked example.** A rectangular plate of mass $2.0\ \mathrm{kg}$, width > $0.40\ \mathrm m$, and negligible height has centre-normal inertia > $I_{p,cm}=Mw^2/12=0.0267\ \mathrm{kg\,m^2}$. A $0.50\ \mathrm{kg}$ disk of > radius $0.10\ \mathrm m$ is attached with its centre $0.35\ \mathrm m$ from the > plate centre axis. Its transferred inertia is > > $$ > I_d=\frac12(0.50)(0.10)^2+(0.50)(0.35)^2 > =0.0638\ \mathrm{kg\,m^2}. > $$ > > The assembly inertia is $0.0905\ \mathrm{kg\,m^2}$. The disk's offset term is > about twenty-five times its intrinsic inertia: the geometry of attachment, not > the disk radius, controls its contribution here. **Physical-pendulum fitting.** A physical pendulum supplies a second experimental method. For a rigid object suspended at several pivot distances $d$ from its centre of mass, $$ T^2=\frac{4\pi^2}{Mg}\left(\frac{I_{cm}}{d}+Md\right). $$ The relation is not linear in $d$, but the equivalent length $\ell_{eq}=I_{cm}/(Md)+d$ has a minimum. Measuring periods at multiple pivot locations can locate this minimum and estimate $I_{cm}$. This procedure is suitable when a torsion wire is unavailable, but pivot friction and uncertainty in centre- of-mass location require careful control. The small-angle assumption can be checked by repeating a period measurement at two amplitudes. A systematic period increase with amplitude indicates nonlinear restoring torque. It cannot be absorbed into a random timing error. Similarly, a period that changes during a long run can indicate wire heating, clamp slip, or slow change of support geometry. **Error propagation and residual tests.** First-order fractional uncertainty for an inertia inferred from $I=\kappa T^2/(4\pi^2)$ uncertainty contains the squared-period coefficient: $$ \left(\frac{\sigma_I}{I}\right)^2 \approx\left(\frac{\sigma_\kappa}{\kappa}\right)^2 +\left(2\frac{\sigma_T}{T}\right)^2. $$ Independent random contributions add in quadrature. Systematic contributions such as a radius calibration error should be reported separately because repeated trials do not reduce them. A torsion-pendulum calibration based on point masses has $I\propto r^2$, so fractional radius uncertainty is doubled in the inferred contribution. Residual plots decide whether an uncertainty model is adequate. Random residuals scattered about zero support a fitted period model. Curvature against added mass or radius suggests an axis shift or nonlinear support. A sequence of residuals drifting in time suggests apparatus change. The diagnostics test the physical assumptions that convert a measured period into an inertia value. **Cross-checks with energy and torque.** A measured inertia should predict both stored rotational energy and angular acceleration under known torque. If a rotor accelerated by torque $\tau$ has measured $\alpha$, the dynamical estimate is $I=\tau/\alpha$ after bearing torque has been corrected. Agreement with a torsion-pendulum estimate tests the fixed- axis model. Disagreement can arise from unmeasured friction torque, axis wobble, or deformation that makes the body nonrigid. One wheel can rotate freely about an axle, translate without spin, or roll under a contact constraint. Each model has different energy and acceleration relations. Torque-free rotation conserves angular momentum but need not keep body orientation fixed. An asymmetric body can rotate differently about different principal axes. Measurements of inertia inherit uncertainty from mass, distance, and period. Logarithmic slopes identify which measurement dominates the fractional uncertainty. Composite-body inertia is the sum of all component inertias about one common axis. Each component must be transferred from its own centre axis when necessary. A planar lamina obeys $I_z=I_x+I_y$ for mutually perpendicular axes through one point. The result follows by adding squared Cartesian distances. It does not apply to a three-dimensional solid. $$ % caption: For a flat lamina, the moment of inertia about the normal axis equals the sum of the two in-plane moments through the same point, $I_z=I_x+I_y$. It holds only because all mass lies in one plane. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[thick,fill=black!5] (-1.8,-1.1)--(1.8,-0.7)--(1.3,1.2)--(-1.5,0.9)--cycle; \draw[->,black] (-2.4,0)--(2.5,0) node[right] {$x$}; \draw[->,black] (0,-1.7)--(0,1.8) node[above] {$y$}; \fill (0,0) circle (1.8pt); \node[acc] at (2.15,1.45) {$I_z=I_x+I_y$}; \end{tikzpicture} $$ A torsion pendulum can determine an unknown inertia once torsion constant is calibrated. Its period squared is proportional to total inertia, so added test masses give a linear calibration relation. For constant angular acceleration, $$ \omega=\omega_i+\alpha t, \qquad \theta=\theta_i+\omega_it+\frac12\alpha t^2, \qquad \omega^2=\omega_i^2+2\alpha(\theta-\theta_i). $$ These equations are valid only while angular acceleration is constant. Radial acceleration remains present whenever angular speed is nonzero, including uniform rotation. ## Inertia from mass distributions Every particle in a fixed-axis rigid body has speed $v_i=r_i\omega$. Summing particle kinetic energies gives $$ K=\frac12\sum_i m_ir_i^2\omega^2=\frac12I\omega^2, \qquad I=\sum_i m_ir_i^2. $$ A continuous body has $I=\int r_\perp^2\,\d m$. The axis is part of the definition. Mass farther from the axis contributes more strongly because distance is squared. $$ % caption: Two equal masses at radii $r$ and $2r$ contribute to the moment of inertia in proportion to distance squared, so the outer mass contributes four times as much as the inner one. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black,dashed] (0,-2)--(0,2) node[above] {axis}; \draw[fill=white,draw=black] (1.2,0) circle (3.2pt); \draw[fill=acc!12,draw=acc] (3.2,0) circle (3.2pt); \node at (1.2,0.5) {$m$}; \node at (3.2,0.5) {$m$}; \draw[<->,black] (0,-0.6)--(1.2,-0.6) node[midway,below] {$r$}; \draw[<->,black] (0,-1.3)--(3.2,-1.3) node[midway,below] {$2r$}; \node at (1.2,1.15) {$mr^2$}; \node[acc] at (3.2,1.15) {$4mr^2$}; \end{tikzpicture} $$ Common symmetry-axis values include $MR^2$ for a hoop, $MR^2/2$ for a solid disk, $2MR^2/5$ for a solid sphere, and $ML^2/12$ for a thin rod through its centre. They follow from the integral with the appropriate linear, area, or volume density. Radius of gyration $k_g$ is defined by $I=Mk_g^2$. It specifies a hypothetical ring with the same total mass and inertia about the chosen axis; it is not generally a physical radius of the body. **Inertia integrals from mass elements.** The defining sum $I=\sum m_ir_i^2$ becomes an integral only after the mass element and its perpendicular distance to the specified axis have been established. A rod uses $\d m=\lambda\,\d x$, a lamina uses $\d m=\sigma\,\d A$, and a three- dimensional body uses $\d m=\rho\,\d V$. The density need not be uniform. A nonuniform rod, for example, has $I=\int x^2\lambda(x)\,\d x$, so locating its centre of mass is not sufficient to determine its inertia. The uniform-rod figure identifies the geometry for a direct derivation. With its centre at the origin and length $L$, linear density is $M/L$. Therefore $$ I_{cm}=\int_{-L/2}^{L/2}x^2\frac{M}{L}\,\d x =\frac{M}{L}\left[\frac{x^3}{3}\right]_{-L/2}^{L/2} =\frac1{12}ML^2. $$ The limits are symmetric because the axis passes through the centre. Moving the axis to one end changes the distance of every element; simply changing integration limits without changing the coordinate distance gives the wrong result. Direct integration about the end gives $$ I_{end}=\int_0^Lx^2\frac{M}{L}\,\d x=\frac13ML^2. $$ The difference is consistent with the parallel-axis theorem, $I_{end}=I_{cm}+M(L/2)^2$. Agreement between the two calculations is a valuable check on both axis placement and density normalization. A uniform disk can be divided into concentric rings. A ring of radius $r$ and width $\d r$ has mass $\d m=\sigma(2\pi r\,\d r)$, where $\sigma=M/(\pi R^2)$. Each ring lies entirely at distance $r$ from the symmetry axis, so $$ I=\int_0^Rr^2\sigma(2\pi r\,\d r) =2\pi\sigma\left[\frac{r^4}{4}\right]_0^R =\frac12MR^2. $$ The ring method exposes why a disk has smaller inertia than a hoop of equal mass and radius: disk mass occupies radii below $R$, whereas hoop mass is entirely at the maximum radius. It also prevents a common dimensional error. The area element is $2\pi r\,\d r$, not $\pi r^2\,\d r$. **Composite bodies and transferred axes.** The composite-body figure represents an assembly whose total inertia about the dashed axis is the sum of contributions from every rigidly attached part. If a component's tabulated inertia uses its own centre axis, transfer it before adding: $$ I_{total}=\sum_j\left(I_{j,cm}+M_jd_j^2\right). $$ The theorem follows from the coordinate shift $x=x'+d$: $$ \int(x'+d)^2\,\d m=\int x'^2\,\d m +2d\int x'\,\d m+d^2\int\d m. $$ The middle term vanishes only because $x'$ is measured from that component's centre of mass. The remaining terms are $I_{cm}+Md^2$. The theorem applies to parallel axes only. A change of axis direction requires either a fresh integral or the perpendicular-axis relation when the object is a planar lamina. > **Worked example.** A uniform rod of mass $2.0\ \mathrm{kg}$ and length > $1.0\ \mathrm m$ has a $0.50\ \mathrm{kg}$ point mass attached at one end. > About the rod centre, the rod contributes $ML^2/12=0.167\ \mathrm{kg\,m^2}$ and > the point mass contributes $m(L/2)^2=0.125\ \mathrm{kg\,m^2}$, for a total of > $0.292\ \mathrm{kg\,m^2}$. Treating all $2.5\ \mathrm{kg}$ as though located at > one radius would discard the rod's distributed mass and give a different result. The perpendicular-axis theorem gives another consistency check for a flat plate. If $z$ is normal to the plane, $r_z^2=x^2+y^2$ gives $$ I_z=\int(x^2+y^2)\,\d m=I_x+I_y. $$ It cannot be applied to a solid cylinder or sphere because some mass lies away from the reference plane. Inertia calculations always begin with geometry and axis selection; formulas are shortcuts only after those choices match the body. **Detailed composite-body calculation.** A composite inertia is most reliable when organized in a component table before numerical substitution, one row per component listing its mass, centre-of-mass inertia about a parallel axis, offset from the requested axis, and transferred contribution. Draw the common axis on the physical assembly first. A component on the axis has zero offset but can still carry large intrinsic inertia; a point mass has zero intrinsic inertia but can carry a large offset contribution. > **Worked example.** A thin uniform rod of length $L=1.20\ \mathrm m$ and mass > $M_r=3.0\ \mathrm{kg}$ carries a solid disk of mass $M_d=1.5\ \mathrm{kg}$ and > radius $R=0.20\ \mathrm m$ fixed at its right end. The requested axis is > perpendicular to the rod through its left end. The rod contributes directly > > $$ > I_r=\frac13M_rL^2=1.44\ \mathrm{kg\,m^2}. > $$ > > The disk's centre is distance $d=L$ from the axis, and its symmetry-axis inertia > is $I_{d,cm}=M_dR^2/2=0.030\ \mathrm{kg\,m^2}$, so the parallel-axis theorem > gives > > $$ > I_d=I_{d,cm}+M_dd^2 > =0.030+(1.5)(1.20)^2 > =2.19\ \mathrm{kg\,m^2}, > $$ > > and $I_{total}=3.63\ \mathrm{kg\,m^2}$. The offset term dominates the disk's > contribution. Using only $M_dR^2/2$ would model the disk as though its centre > lay on the pivot; using only $M_dL^2$ would treat it as a point mass and drop > its finite radius. The two-term decomposition separates those effects. > > As a check, every disk mass element lies between $L-R$ and $L+R$ from the pivot, > so the disk contribution must lie between $M_d(L-R)^2=1.50$ and > $M_d(L+R)^2=2.94\ \mathrm{kg\,m^2}$. The value $2.19\ \mathrm{kg\,m^2}$ falls > inside those bounds, catching any misplaced squared distance before the result > feeds a dynamical calculation. **Torsion-pendulum measurement.** A torsion wire supplies a restoring torque $\tau=-\kappa\theta$. A rigid body with inertia $I$ obeys $$ I\ddot\theta+\kappa\theta=0, \qquad T=2\pi\sqrt{\frac I\kappa}. $$ Rearrangement gives an inertia measurement, $$ I=\frac{\kappa T^2}{4\pi^2}. $$ The torsion constant can be calibrated using a reference object of known inertia, or by placing known point masses at measured radii. If the apparatus itself has inertia $I_0$, a measured period corresponds to $I_0+I_{object}$; subtracting the empty-platform inertia is required. The calibration figure expresses this as a linear $T^2$ versus added-inertia relation whose intercept represents $I_0$. $$ % caption: A torsion-pendulum calibration plots period squared against added moment of inertia. The slope is $4\pi^2/\kappa$, fixing the torsion constant $\kappa$, and the intercept gives the bare-platform inertia $I_0$. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->,black] (0,0)--(5.8,0) node[right] {added $I$}; \draw[->,black] (0,0)--(0,3.1) node[above] {$T^2$}; \draw[acc,thick] (0.3,0.5)--(5.3,2.75); \foreach \x/\y in {0.6/0.55,1.4/0.92,2.2/1.28,3/1.63,3.8/1.98,4.6/2.35,5.2/2.62} {\draw[fill=white,draw=black] (\x,\y) circle (1.6pt);} \fill[acc] (0,0.36) circle (1.8pt); \node[black,left] at (0,0.55) {$I_0$}; \end{tikzpicture} $$ For small independent uncertainty, fractional propagation from the measurement formula is $$ \frac{\Delta I}{I}\approx\frac{\Delta\kappa}{\kappa} +2\frac{\Delta T}{T}. $$ Period uncertainty is doubled because inertia depends on $T^2$. Timing many cycles reduces random start-stop uncertainty: if total time for $N$ cycles is $t_N$, use $T=t_N/N$. This does not remove systematic error from amplitude-dependent torsion, wire creep, air damping, or an axis that is not fixed. The torsion model requires small angle so torque remains proportional to angle. > **Worked example.** A calibrated wire has > $\kappa=0.0800\ \mathrm{N\,m\,rad^{-1}}$. The object-plus-platform period is > $2.50\ \mathrm s$ and the bare-platform period is $1.00\ \mathrm s$. Their > inertias are $I=\kappa T^2/4\pi^2 = 0.0127\ \mathrm{kg\,m^2}$ and > $0.00203\ \mathrm{kg\,m^2}$, so the object inertia is > $0.0106\ \mathrm{kg\,m^2}$. The subtraction must be done in inertia, not > period, because period goes as the square root of inertia. **Limits of tabulated values.** Tabulated inertia formulas assume uniform density and ideal shape. A drilled disk, spoked wheel, or object with a heavy hub requires decomposition or direct integration. Manufacturing tolerances can matter because mass at large radius has disproportionate influence. An axis displaced slightly from its intended location changes inertia by approximately $Md^2$, so alignment error becomes important for large masses and long supports. ## Planar rigid-body motion Fixed-axis formulas are a special case of planar rigid-body motion. For any two points $A$ and $B$ fixed in one rigid body, $$ \vec v_B=\vec v_A+\vec\omega\times\vec r_{B/A}. $$ The separation vector is constant in body coordinates, so all velocity differences are perpendicular to that separation. This relation explains why a translating and rotating plate can have one point instantaneously at rest while another moves rapidly. It also gives a geometric construction of an instantaneous centre: lines drawn perpendicular to known point velocities intersect at a point whose instantaneous velocity is zero. Acceleration adds a tangential and a radial relative term: $$ \vec a_B=\vec a_A+\vec\alpha\times\vec r_{B/A} +\vec\omega\times(\vec\omega\times\vec r_{B/A}). $$ The first relative term changes speed and the second turns velocity inward. A point that is instantaneously stationary can still have nonzero acceleration, which is why instantaneous-centre geometry cannot replace a complete acceleration analysis. The formulas apply to every planar rigid body, including noncircular bodies. At radius $r$ in fixed-axis rotation, the vector expression reduces to $a_t=r\alpha$ and $a_r=r\omega^2$. The total magnitude is $$ a=\sqrt{(r\alpha)^2+(r\omega^2)^2}. $$ The radial and tangential components are orthogonal, so adding their magnitudes directly is incorrect. A rotating fan at constant angular speed has zero tangential acceleration but nonzero radial acceleration at every point except the axis. **Several integral derivations.** The disk derivation uses concentric rings because every point on a ring shares one axis distance. A thin circular hoop places all mass at $R$, immediately giving $I=MR^2$. A thick annular disk of inner radius $R_1$ and outer radius $R_2$ has $$ I=\int_{R_1}^{R_2}r^2\sigma(2\pi r\,\d r) =\frac12M(R_1^2+R_2^2). $$ The result approaches the hoop formula as annulus thickness becomes small. It approaches the solid-disk formula as $R_1$ approaches zero. These limits check the integral without repeating its calculation. A uniform rectangular plate of sides $a$ and $b$ has centre-normal inertia $$ I_z=\int_{-a/2}^{a/2}\int_{-b/2}^{b/2}(x^2+y^2)\frac{M}{ab}\,\d y\,\d x =\frac1{12}M(a^2+b^2). $$ The two terms are the in-plane contributions identified by the perpendicular-axis figure. Choose coordinates that give the axis distance a simple expression, then integrate over the physical region. Spherical shells or disks can be used for a solid sphere. With disks perpendicular to a diameter, radius is $\sqrt{R^2-x^2}$ and disk mass is $\rho\pi(R^2-x^2)\,\d x$. Each disk has inertia one half its mass times its disk radius squared. Integration produces $I=2MR^2/5$. A spherical shell instead gives $2MR^2/3$ because all of its mass lies farther from the diameter axis on average. ## Inertia tensors and principal axes For unrestricted three-dimensional rotation, one scalar inertia is insufficient. Angular momentum is related to angular velocity by $$ \vec L=\mathbf I\vec\omega, $$ where $\mathbf I$ is an inertia tensor. Its diagonal terms are moments about three coordinate axes; asymmetric mass placement produces off-diagonal products of inertia. About a principal axis, $\vec L$ is parallel to $\vec\omega$ and the scalar relation $L=I\omega$ is recovered. $$ % caption: The principal axes diagonalize the inertia tensor. Rotation about any one of them makes $\vec L$ parallel to $\vec\omega$ with $L=I_k\omega$ and no cross-coupling between angular-velocity components. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[thick,fill=black!5] (0,0) ellipse (1.9 and 0.85); \draw[->,black] (0,0)--(2.6,0.6) node[right] {$x_1$}; \draw[->,black] (0,0)--(-1.4,1.5) node[left] {$x_2$}; \draw[->,acc,thick] (0,0)--(0,2.3) node[above] {$x_3$}; \fill (0,0) circle (1.5pt); \node[black] at (2.15,-0.85) {$I_1$}; \node[black] at (-1.7,1.05) {$I_2$}; \node[acc] at (0.5,1.9) {$I_3$}; \end{tikzpicture} $$ Fixed-axis machines avoid the tensor complication because bearings constrain one axis. An irregular freely rotating object generally does not: its material axes can precess or tumble even while total angular momentum is conserved. The tensor description is not an optional refinement in that case; it is required to connect angular momentum to angular velocity correctly. **Principal-axis measurements.** Principal inertias can be measured by torsion or bifilar suspension about several orientations. A repeated period measurement gives inertia about each suspension axis after calibration. Rotating the object and finding orientations where cross- coupling vanishes identifies principal directions. The procedure depends on small oscillation angle and a rigid body; flexible deformation introduces additional modes that spoil a single-inertia fit. Data fitting should distinguish random timing scatter from systematic drift. A linear regression of $T^2$ against added inertia estimates slope and intercept, but an intercept uncertainty propagates into the inferred apparatus inertia. A residual pattern that curves with added mass suggests torsion constant changes, support deformation, or an axis shift. Quoting only a fitted slope without a residual check can conceal these model failures. **Failure modes in inertia experiments.** Bearing friction usually affects decay more than the small-amplitude period, but large friction can shift measured timing and make zero-crossing detection ambiguous. A suspension wire can have nonlinear torque at large angle or slow creep over repeated trials. Mass clamps may move relative to the intended radius. The radius-squared dependence means a small radial placement error can dominate a mass uncertainty. Apparatus must be level if gravity contributes an unintended restoring torque. Rigid-body assumptions can fail when a long rod bends, a disk wobbles, or attached masses oscillate relative to the platform. The observed motion then contains several frequencies. A single-period measurement is no longer an inertia measurement unless the desired mode has been isolated. Inspecting time traces and frequency spectra is part of experimental validation, not an optional display step after calculation. ## Energy and consistency checks Every mass element of a rigid body constrained to rotate about one fixed axis has speed $v=r_\perp\omega$. Substitution into particle kinetic energy gives $$ K=\int\frac12v^2\,\d m =\frac12\omega^2\int r_\perp^2\,\d m =\frac12I\omega^2. $$ The derivation requires one angular speed for every element. It does not apply unchanged to a deforming body, where distances from the axis vary in time, or to a body whose axis translates, where centre-of-mass kinetic energy must be included. The fixed-axis energy is positive regardless of rotation sense because it depends on $\omega^2$. Work by a torque follows from tangential displacement $\d s=r\,\d\theta$: $$ \d W=F_t\,\d s=\tau\,\d\theta, \qquad W=\int\tau\,\d\theta=\Delta K_{rot}. $$ For constant torque about a fixed axis, $\tau=I\alpha$ and $\omega^2-\omega_i^2=2\alpha\Delta\theta$ reproduce the same energy change. This is an internal consistency check linking kinematics, dynamics, and energy. > **Worked example.** A rotor with $I=0.25\ \mathrm{kg\,m^2}$ speeds up from > $20$ to $80\ \mathrm{rad\,s^{-1}}$. Its rotational-energy increase is > > $$ > \Delta K=\frac12(0.25)(80^2-20^2)=750\ \mathrm J. > $$ > > An ideal motor must supply this much mechanical work, on top of bearing and > windage losses. Its instantaneous output is $P=\tau\omega$, so the power rises > during constant-torque acceleration even though the torque is fixed. **Integral checks by limiting cases.** An inertia integral should be tested against physical limits. For an annulus, $$ I=\frac12M(R_1^2+R_2^2). $$ As $R_1\to0$, this becomes the solid-disk result $MR_2^2/2$. As $R_1\to R_2$, it becomes hoop inertia $MR^2$. Both limits are required; an expression satisfying only one can still contain an integration or mass-normalization error. $$ % caption: The annulus moment $\tfrac12 M(R_1^2+R_2^2)$ interpolates between a solid disk (inner radius shrinks to zero) and a thin hoop (the annulus narrows to its rim). \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[thick] (-2.6,0) circle (0.9); \fill[acc!12,even odd rule] (0,0) circle (0.9) (0,0) circle (0.48); \draw[acc,thick] (0,0) circle (0.9); \draw[acc,thick] (0,0) circle (0.48); \draw[thick] (2.6,0) circle (0.9); \draw[thick] (2.6,0) circle (0.76); \node[black] at (-2.6,-1.35) {disk}; \node[black] at (0,-1.35) {annulus}; \node[black] at (2.6,-1.35) {hoop}; \end{tikzpicture} $$ A numerical integral independently checks the analytic result. Divide a rod into $N$ equal point masses at their segment centres and compute $\sum m_ix_i^2$. As $N$ increases, the sum approaches $ML^2/12$ about the centre. The error decreases because the discrete distribution better approximates continuous mass; it does not vanish by merely adding points at the wrong radii. **Energy measurement and balance checks.** A spin-down experiment records angular speed after motor torque is removed. If bearing torque is approximately constant, angular speed decreases linearly; if drag torque is proportional to angular speed, decay is exponential. Comparing the measured energy decrease $I\omega^2/2$ with heat or electrical recovery estimates tests the loss model. Treating all spin-down curves as constant angular deceleration can bias an inertia estimate. ## Axis transformations and kinematic measurement Angular position can be measured directly with an encoder, inferred from a marked rim point, or obtained by integrating angular velocity. Each method has a distinct error model. An optical encoder counts angular increments and can accumulate missed counts; a video measurement converts pixel position to angle and is sensitive to camera perspective; an accelerometer estimates angular motion indirectly and can drift under numerical integration. Agreement among methods tests calibration rather than merely producing several versions of the same number. For constant angular acceleration, a plot of angular velocity against time has slope $\alpha$. A plot of angular displacement against time is parabolic. Fitting both traces independently identifies whether constant-$\alpha$ kinematics is appropriate. A short interval may look linear even when angular acceleration changes appreciably over a full rotation, so the measurement interval belongs in the model statement. **Relative velocities in a rigid body.** Rigid-body velocity differences are perpendicular to the separation vector. If a door rotates about a hinge with angular speed $\omega$, the handle at twice the radius has twice the speed. The material remains rigid because all points have one common angular speed, not one common linear speed. A speed limit at the rim therefore constrains angular speed more severely for a larger rotor. The acceleration expression separates the two effects. Tangential acceleration is zero when $\omega$ is constant, whereas radial acceleration increases as $\omega^2$. A high-speed rotor can therefore have large internal stress even when its angular velocity is not changing. The kinematics alone does not calculate stress, but it specifies the acceleration field that internal forces must supply. **Axis transformations and checks.** Two applications of the parallel-axis theorem can be composed only when each shift is between parallel axes. A convenient check is reversibility: shifting from a centre axis to an offset axis and then subtracting $Md^2$ must recover the original inertia. A result below $I_{cm}$ for a parallel offset axis violates the theorem. The perpendicular-axis relation has a different geometric origin. It is valid for a planar lamina and axes through one common point. Moving one in-plane axis away from that point breaks the direct relation. The relation also fails for a thick plate unless all mass can be approximated as lying in one plane. ## Experimental inertia methods A bifilar suspension uses two parallel strings supporting a platform. A small yaw rotation raises the centre of mass slightly, providing a gravitational restoring torque. With string length $\ell$, half separation $a$, and total supported mass $M$, the small-angle period has the form $$ T=2\pi\sqrt{\frac{I\ell}{Mga^2}}. $$ This method avoids uncertainty in a torsion constant, but requires accurately parallel strings and small angular displacement. Unequal string lengths couple yaw to translation; then the measured period no longer identifies one pure rotational inertia. **Tensor diagonalization and principal-axis dynamics.** The inertia tensor is a real symmetric matrix. Symmetry guarantees three mutually orthogonal eigenvectors, the principal axes, and three real eigenvalues, $I_1,I_2,I_3$, the principal moments. In matrix form, diagonalization is written $$ \mathbf R^{\mathsf T}\mathbf I\mathbf R= \begin{pmatrix} I_1&0&0\\0&I_2&0\\0&0&I_3 \end{pmatrix}, $$ where columns of the rotation matrix $\mathbf R$ are principal-axis unit vectors. The diagonal form removes products of inertia only in that body-fixed coordinate system. Rotating the coordinate axes away from the principal directions generally restores off-diagonal terms; diagonal values are not components that remain unchanged in every frame. In a principal-axis frame, angular momentum components are $$ L_1=I_1\omega_1,\qquad L_2=I_2\omega_2,\qquad L_3=I_3\omega_3. $$ For torque-free motion, body-frame components obey Euler's equations: $$ I_1\dot\omega_1+(I_3-I_2)\omega_2\omega_3=0, $$ with cyclic permutations for the other two equations. Rotation exactly about one principal axis is a solution because the other angular-velocity components vanish. Small perturbations about the largest and smallest principal moments are stable, whereas rotation about the intermediate principal axis is unstable. The familiar tennis-racket flip is a consequence of this result. It lies outside the fixed-axis scalar model and demonstrates why a freely rotating body requires the principal moments and axes, rather than one scalar moment of inertia. The rotational kinetic energy in principal coordinates is $$ K=\frac12(I_1\omega_1^2+I_2\omega_2^2+I_3\omega_3^2). $$ Together with fixed $L^2$, this energy constrains torque-free motion to intersections of two quadratic surfaces in angular-velocity space. A constrained rotor in a machine does not explore these intersections because bearings apply external forces and torques that maintain the selected axis. ## Worked methods and reporting Take a thin rod of length $L$ whose linear density increases from left to right as $$ \lambda(x)=\lambda_0\left(1+\frac{x}{L}\right), \qquad 0\le x\le L. $$ The total mass is $$ M=\int_0^L\lambda(x)\,\d x=\frac32\lambda_0L, $$ so $\lambda_0=2M/(3L)$. Its centre of mass is $$ x_{cm}=\frac1M\int_0^Lx\lambda(x)\,\d x=\frac{5L}{9}. $$ The inertia about the left end follows directly from the mass elements: $$ I_{left}=\int_0^Lx^2\lambda(x)\,\d x =\lambda_0\left(\frac{L^3}{3}+\frac{L^3}{4}\right) =\frac{7}{18}ML^2. $$ The centre-of-mass inertia is obtained either by a second integral using $(x-x_{cm})^2$ or by reverse application of the parallel-axis theorem: $$ I_{cm}=I_{left}-Mx_{cm}^2 =\left(\frac7{18}-\frac{25}{81}\right)ML^2 =\frac{13}{162}ML^2. $$ For comparison, a uniform rod has centre inertia $ML^2/12=13.5ML^2/162$. The nonuniform rod has smaller centre inertia because more mass lies near its shifted centre of mass than at extreme distances. A uniform-rod formula cannot follow from total mass alone because the density profile changes both centre location and squared-distance weighting. The calculation also gives two checks. First, $I_{left}$ exceeds $I_{cm}$ by $Mx_{cm}^2$, as required. Second, both results have units $ML^2$. A negative result after axis transfer would signal an algebraic or coordinate-origin error. **Rolling inertia versus laboratory inertia measurement.** The quantity appearing in rolling energy is moment of inertia about the centre of mass, $I_{cm}$. Translational kinetic energy is then written separately: $$ K=\frac12Mv_{cm}^2+\frac12I_{cm}\omega^2. $$ $$ % caption: A rigid wheel admits three ideal models: spinning on a fixed axle, sliding without spin, and rolling under a no-slip contact. Each carries its own relation between $v$ and $\omega$. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black,thick] (-0.3,0)--(2.3,0); \draw[thick,fill=black!5] (1,0.75) circle (0.55); \fill (1,0.75) circle (1.5pt); \draw[->] (0.62,0.9) arc (150:410:0.38); \node[black] at (1,-0.4) {axle}; \draw[black,thick] (3.2,0)--(5.8,0); \draw[thick,fill=black!5] (4.5,0.75) circle (0.55); \draw[->,black,thick] (4.5,0.75)--(5.7,0.75) node[above] {$v$}; \node[black] at (4.5,-0.4) {slide}; \draw[black,thick] (6.7,0)--(9.3,0); \draw[thick,fill=black!5] (8,0.75) circle (0.55); \draw[->,black,thick] (8,0.75)--(9.2,0.75) node[above] {$v$}; \draw[->] (7.62,0.9) arc (150:410:0.38); \node[black] at (8,-0.4) {roll}; \end{tikzpicture} $$ An alternative instantaneous-contact expression uses $I_P=I_{cm}+MR^2$, but the contact point is not a fixed axis through the motion. It packages translation and rotation into one value at one instant. A laboratory torsion pendulum, by contrast, measures inertia about its fixed suspension axis. A rolling experiment cannot use the torsion-pendulum value directly unless the suspension axis and rolling axis are the same or a parallel-axis transfer has been made. The distinction is significant for an off-centre wheel. Its inertia about the geometric axle may differ from inertia about the centre of mass, and its centre of mass may move vertically during rotation. A simple rolling-energy formula assumes the rotation axis through the centre of mass has fixed relation to the body and that the rolling radius is well defined. Eccentric wheels require a more general rigid-body energy model with orientation-dependent potential energy. Experimental inertia estimates from rolling use the measured acceleration on an incline. Rewriting the ideal acceleration relation gives $$ \frac{I_{cm}}{MR^2}=\frac{g\sin\theta}{a}-1. $$ The method requires verified no-slip contact, known angle, and negligible rolling resistance. It infers $I_{cm}$ from translational acceleration. A torsion pendulum instead infers inertia from an oscillation period and a calibrated torsion constant. Agreement between the two methods is a strong test of both the contact model and the apparatus calibration. **Physical-pendulum derivation.** A physical pendulum is any rigid body free to rotate about a horizontal fixed pivot under gravity. If its centre of mass is distance $d$ from the pivot, gravity produces torque $$ \tau=-Mgd\sin\theta. $$ The rotational equation of motion is therefore $$ I_P\ddot\theta+Mgd\sin\theta=0. $$ At small angle, $\sin\theta\approx\theta$, giving simple harmonic motion with $$ \omega_0=\sqrt{\frac{Mgd}{I_P}}, \qquad T=2\pi\sqrt{\frac{I_P}{Mgd}}. $$ The inertia is about the pivot, not the centre of mass. With $I_P=I_{cm}+Md^2$, the measured period contains both mass distribution and pivot offset. The equivalent simple-pendulum length is $\ell_{eq}=I_P/(Md)$. It has the same small-angle period as a point mass suspended at that length. The model fails at large amplitude because torque is no longer proportional to angle. Pivot friction produces damping, and a flexible body can have internal modes. The small-angle physical-pendulum period is nevertheless a common and precise way to measure inertia about a chosen pivot when these effects are controlled. > **Worked example.** A uniform rod of mass $0.80\ \mathrm{kg}$ and length > $0.60\ \mathrm m$ swings about one end. Its pivot inertia is > $I_P=ML^2/3=0.096\ \mathrm{kg\,m^2}$ and its centre-of-mass distance is > $d=L/2=0.30\ \mathrm m$, so the small-angle period is > > $$ > T=2\pi\sqrt{\frac{0.096}{(0.80)(9.81)(0.30)}}=1.27\ \mathrm s. > $$ > > Treating the rod as a point mass at its centre would use $I=Md^2$ and predict a > different period; the rod's distributed mass raises its pivot inertia above that > approximation. The equivalent simple-pendulum length checks out: > $\ell_{eq}=I_P/(Md)=2L/3=0.40\ \mathrm m$, and $2\pi\sqrt{0.40/g}$ returns the > same period. **Sign and time dependence.** Counterclockwise angular coordinate is commonly positive in a plane diagram. A negative angular velocity denotes clockwise rotation; the sign of angular acceleration identifies whether signed angular speed is increasing or decreasing. **Continuous distributions.** Moment of inertia becomes an integral when mass is distributed continuously. A thin rod of length $L$ and uniform linear density $\lambda=M/L$ has element $\d m=\lambda\,\d x$. About an axis through its centre, $$ I=\int_{-L/2}^{L/2}x^2\lambda\,\d x=\frac1{12}ML^2. $$ The squared-distance factor must be measured perpendicular to the stated axis. $$ % caption: A uniform rod is split into mass elements $\d m=\lambda\,\d x$ at perpendicular distance $x$ from the central axis; the integral $\int x^2\,\d m$ weights outer material more heavily than central material. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[very thick] (-3.2,0)--(3.2,0); \draw[black,dashed] (0,-1.1)--(0,1.4) node[above] {axis}; \draw[acc,fill=acc!20] (1.5,-0.16) rectangle (1.85,0.16); \draw[<->,black] (0,-0.65)--(1.67,-0.65) node[midway,below] {$x$}; \node[acc] at (1.67,0.55) {$dm$}; \draw[<->,black] (-3.2,-1.15)--(3.2,-1.15) node[midway,below] {$L$}; \end{tikzpicture} $$ **Parallel axes.** The parallel-axis theorem moves a centre-of-mass inertia to a parallel axis at distance $d$: $$ I=I_{cm}+Md^2. $$ It follows by writing each particle coordinate as $x_i=x_i'+d$ and using $\sum m_ix_i'=0$. The centre-of-mass axis has the smallest inertia among parallel axes. $$ % caption: Two parallel axes a distance $d$ apart have moments differing by $Md^2$; the centre-of-mass axis gives the smallest moment among all parallel axes. \begin{tikzpicture}[>=stealth,font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black,dashed] (0,-1.55)--(0,1.55) node[above] {cm axis}; \draw[black,dashed] (3.6,-1.55)--(3.6,1.55) node[above] {axis}; \draw[thick,fill=black!5] (0,0) ellipse (1.35 and 0.82); \draw[fill=white,draw=black] (0,0) circle (2pt); \draw[<->,acc] (0,-1.15)--(3.6,-1.15) node[midway,below] {$d$}; \node[acc] at (1.8,1.15) {$I=I_{cm}+Md^2$}; \end{tikzpicture} $$ **Stating the rotation axis in a measurement.** A moment of inertia is incomplete without a line of rotation. A laboratory report should name an axis direction, a point or feature through which it passes, and any offset from the centre of mass. “The inertia of the disk” is ambiguous: the same disk has one value about its symmetry axis, a different value about a diameter, and another value about a parallel tangent axis. A coordinate description such as “the z axis through the centre” removes that ambiguity for a body mounted in a known orientation. A body rotating about a principal symmetry axis has angular momentum parallel to angular velocity, and the scalar fixed-axis equation $\tau=I\alpha$ applies. An arbitrary three-dimensional axis through a nonsymmetric body can give angular momentum that points away from angular velocity. The full relation then uses the inertia tensor, $$ \vec L=\mathbf I\vec\omega. $$ A fixed-axis calculation uses the corresponding scalar projection of the inertia tensor. Freely tumbling motion requires the tensor relation. A book released while rotating about a corner can change its body-axis components, so one scalar $I$ cannot describe the complete motion. Axis uncertainty can dominate an inertia measurement when mass lies far from the axis. In a parallel-axis transfer, an offset uncertainty $\delta d$ changes the added term by approximately $2Md\,\delta d$. A small ruler error therefore has a large effect when a heavy component is mounted far from the reference axis. The same issue appears in composite assemblies: each part must be located relative to the chosen common line before its inertia contribution is summed. Photographing or sketching the fixture with the reported axis is often as important as recording the mass and dimensions.