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Two uniform discs of mass m and radius a stick at first edge contact. One initially translates at speed v and the other rests, with neither initially spinning. If the incoming velocity makes angle $\theta$ to the line of centres, the composite axial moment of inertia about its centre of mass is $3m{a}^2$. Conserved angular-momentum magnitude is $mav|\sin \theta |$, so the angular velocity magnitude is $v|\sin \theta |/(3a)$. The centre-of-mass speed is v/2. Splitting kinetic energy into translation and rotation gives $K_f=m{v}^2(1/4+{\sin }^2\theta /6)$. The difference from $m{v}^2/2$ is dissipated internally; it cannot be removed from angular-momentum conservation.

Angular acceleration is the time derivative of angular velocity, $\mathbf{\alpha }=d\mathbf{\omega }/dt$. For rotation about a fixed axis, its signed scalar component obeys $\tau =I\alpha$ when the axial moment of inertia is constant and $\tau$ is the total applied torque about that axis.

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

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

A wheel that oscillates about a spindle and regulates a watch through an escapement. In the ideal balance-spring model, a torsion spring supplies torque $-\kappa \varphi$ and the wheel has moment of inertia $I$, giving $I\ddot{\varphi }+\kappa \varphi =0$. The period follows from simple harmonic motion. Real spring, bearing and escapement effects disturb this ideal.

Exchange of angular momentum between a stellar bar and a dark matter halo through orbital resonances. If the unperturbed halo distribution function depends only on energy and decreases with it, the resonant torque from a rotating perturbation is nonnegative for positive pattern angular velocity. Strict transfer requires nonzero resonant overlap. The reaction torque slows a rigid bar of constant positive moment of inertia; evolving bar structure and halo redistribution require updating the response.

Let a bell have axial moment of inertia $M{k}^2$ and centre of mass h from its axle. A clapper of point mass m at radius l touches the bell at fixed relative angle $\beta$. While they remain in contact, the clapper angle $\theta$ from the upward vertical obeys $(M{k}^2+m{l}^2)\ddot{\theta }=Mgh\sin (\theta -\beta )+mgl\sin \theta$. The contact force opposing increasing angle is $N=mg\sin \theta -ml\ddot{\theta }$. Eliminating the acceleration givesThe first zero on the physical trajectory marks detachment, because a nonadhesive contact cannot support negative normal force. Selecting the correct trigonometric branch is part of the contact problem.

An original reduced model makes the calibration explicit. Let the torsion pendulum have restoring torque $-\kappa \theta$, with torsion constant $\kappa$, and viscous damping torque $-b\dot{\theta }$, with $b>0$. Allow an unwanted axial torque ${\tau }_p(t)$:Here ${\omega }_0$ is the undamped natural angular frequency and $Q$ the quality factor. The forced harmonic oscillator response must be included before a twist can be interpreted as a torque.

Using the complex amplitude convention $m_z(t)=\mathrm{Re}(\tilde{m}{e}^{\mathrm{i}\omega t})$ and similarly for $\theta$ and ${\tau }_p$, differentiation multiplies by $\mathrm{i}\omega$. Substitution givesThis is the torque-to-twist transfer function. With ${\tilde{\tau }}_p=0$ and real positive $\tilde{m}=m_1$, the twist amplitude isFor $\gamma <0$, the twist phase relative to $m_z$ is $\pi /2-atan2(b\omega ,\kappa -I{\omega }^2)$. At $\omega ={\omega }_0$ it is zero, although the reaction torque leads $m_z$ by $\pi /2$. A phase measured at the mirror is not automatically the phase of the applied torque.

A fixed-amplitude reference torque ${\tau }_c\cos \omega t$ measures $|H_{\tau }|$ directly. The Einstein-de Haas effect instead contains an extra factor $\omega$ when $m_1$ stays fixed. Consequently its twist amplitude peaks exactly at ${\omega }_0$ in this model, whereas the fixed-torque peak is at ${\omega }_0\sqrt{1-1/(2Q^2)}$ when $Q>1/\sqrt{2}$. To verify the first statement, divide the squared denominator of $A_{\mathrm{EdH}}$ by ${\omega }^2$: the quantity to minimize is $(\kappa /\omega -I\omega {)}^2+b^2$, minimized at ${\omega }_0$. Both response curves tend to zero at high frequency; only the magnetomechanical curve also tends to zero at zero frequency.

For a fully stipulated example, choose $I=2.00\times 10^{-10}\mathrm{kg}{\mathrm{m}}^2$, $\kappa =8.00\times 10^{-10}\mathrm{N}m/rad$, $b=8.00\times 10^{-11}\mathrm{N}\mathrm{m}s/rad$, $\gamma =-1.76\times 10^{11}{\mathrm{s}}^{-1}{\mathrm{T}}^{-1}$, and $m_1=1.76\times 10^{-4}\mathrm{A}{\mathrm{m}}^2$. These teaching inputs give ${\omega }_0=2.00rad/s$, $Q=5.00$, reaction torque amplitude $2.00\times 10^{-15}\mathrm{N}\mathrm{m}$ at ${\omega }_0$, andAn optical lever at distance $D=1.00\mathrm{m}$ gives small-displacement spot amplitude $x\simeq 2DA=25.0\mu \mathrm{m}$; its full peak-to-peak width is $4DA=50.0\mu \mathrm{m}$. For a fast reversal from $+m_1$ to $-m_1$, the same inputs give $\Delta L_{\mathrm{body}}=-2.00\times 10^{-15}\mathrm{kg}{\mathrm{m}}^2/s$ and initial $\Delta \dot{\theta }=-1.00\times 10^{-5}rad/s$, provided the external angular impulse is negligible.

One independent mechanical calibration adds known moment of inertia $\Delta I$, keeping $\kappa$ unchanged. If $T_a,T_b$ are the undamped oscillation periods before and after adding it, thenFor $\Delta I=I$ in the stipulated example, $T_a=\pi \mathrm{s}$ and $T_b=\sqrt{2}\pi \mathrm{s}$, recovering the chosen $I$ and $\kappa$. Real damped periods require a damping correction: a free decay has angular envelope $e^{-bt/(2I)}$ and oscillation angular frequency $\sqrt{\kappa /I-(b/(2I){)}^2}$. Measuring that decay determines $b/(2I)$. Calibrating the optical scale, mechanical response and change of magnetic dipole moment separately is what makes the inferred gyromagnetic ratio meaningful.

A Cavendish torsion balance is a torsion balance arranged to measure gravitational torque from nearby source masses. Its suspended horizontal rod rotates about a vertical fiber. In the linear elastic regime, a twist angle $\theta$, measured from the fiber's untwisted orientation, gives a restoring torque ${\tau }_{\mathrm{fiber}}=-\kappa \theta$, where $\kappa >0$ is the torsion constant. The minus sign is essential: the fiber resists the twist. Under an approximately constant gravitational torque, the rotational equation using moment of inertia $I$ and angular acceleration $\ddot{\theta }$ isAt equilibrium, the two torques balance. The angle is a displacement from the fiber's reference orientation, not the largest excursion during an oscillation. ^[ref]

For the toy parameters in the Cavendish experiment, the predicted equilibrium twist isHere $L|{\theta }_{\mathrm{eq}}|/r\simeq 0.0107$, consistent with using a small geometric displacement in the toy model, but not an estimate of its total measurement error. Reversing the source positions ideally takes the equilibrium angle from $+{\theta }_{\mathrm{eq}}$ to $-{\theta }_{\mathrm{eq}}$: the magnitude of the change is $2|{\theta }_{\mathrm{eq}}|\simeq 0.6119^{\circ }$. A stable zero offset then cancels in the difference. A real reading may track a reflected light spot; the law of reflection doubles a mirror's rotation in the outgoing beam, so the spot's angular change must not be confused with the rod's angle. ^[ref]

How can the fiber's torsion constant be found without already knowing the gravitational force? Measure the oscillation period and calculate the moment of inertia. For the same toy rod, neglecting its mass and treating both test masses as point masses, $I=m{L}^2+m{L}^2=2m{L}^2$. With the source torque absent or treated as constant, let $\varphi =\theta -{\theta }_{\mathrm{eq}}$ be a small departure from equilibrium. Neglect damping and any change of gravitational torque with angle. The rotational equation becomes simple harmonic motion:Here $A$ is the angular amplitude, $\delta$ specifies the initial phase, and $\omega$ is the angular frequency. A softer fiber gives a larger equilibrium twist but also a longer oscillation period: greater sensitivity costs observation time. An undamped oscillation does not settle at equilibrium; damping is needed for settling. ^[ref]

The toy moment of inertia and undamped oscillation period areTiming this oscillation period would recover the assumed torsion constant. The units check explains the result: torque divided by torsion constant is an angle in radians, and $I/\kappa$ has units of time squared when radians are treated as dimensionless in SI.

Combining the equilibrium twist with the period calibration, the two-near-pair model would infer the gravitational constant asThe test mass cancels in the second expression because it enters both the gravitational torque and the toy moment of inertia. If the measured quantity is the magnitude $|\Delta \theta |$ of the change between opposite source positions, use $|{\theta }_{\mathrm{eq}}|=|\Delta \theta |/2$. Substituting the toy twist and oscillation period returns the input $G$; it is a consistency check, not independent experimental evidence.

The sensitivity of a Cavendish torsion balance also explains its practical difficulty. Air motion can exert unwanted torque; unequal temperatures can drive that motion. Cavendish discusses this effect on p. 471 and describes observing from outside a closed room to avoid disturbing the instrument. ^[ref] Mechanical vibrations can excite the same slow oscillations used to calibrate the torsion constant, while changes in the fiber or source separations alter the inferred force. Precision therefore depends on resolving the equilibrium angle, calibrating the full suspended moment of inertia and torsion constant, and modeling the actual source geometry. The two-near-pair model omits opposite-source attraction, rod contributions, finite-size effects, damping, and the variation of gravitational torque as the rod turns. ^[ref]

Real chemical bonds can stretch as molecular rotation increases. The resulting larger moment of inertia lowers the energy eigenvalues relative to a rigid rotor. Within one vibrational quantum state, a leading effective correction isBoth $B_v$ and $D_v$ are in frequency units here; $v$ labels the vibrational quantum state. For positive $D_v$, the adjacent-line spacing becomesso the departures grow with $J$. Molecular vibration also changes the effective rotational constant: even the lowest vibrational quantum state does not have motionless atomic nuclei.^[ref]

The quartic correction is a limited expansion, not a model to extrapolate to arbitrary $J$. It is small relative to the rigid contribution only while $(D_v/B_v)J(J+1)\ll 1$; eventually higher terms, changed internal motion, or chemical bond breaking matter. In the synthetic plot, the largest upper-state label is $J=5$, giving a stipulated correction ratio of $0.03$. No universal maximum valid $J$ follows without a particular molecule and precision target.

The quantum numbers also have limits of meaning. Here $J$ describes spin-free rotation. In a molecule with electronic spin, rotation alone is commonly labelled $N$, while $J$ includes electronic angular momentum; including nuclear spin introduces additional total-angular momentum labels and resolved splittings. An applied electric field can mix different field-free $J$ states through the Stark effect. For identical atomic nuclei, the Pauli exclusion principle can restrict combinations of rotational and nuclear-spin states. The simple $2J+1$ degeneracy and every-integer-$J$ ladder therefore describe the stated idealization, not all observed sublevels.^[ref]

Classical mechanics describes the motion of bodies by assigning positions and velocities and specifying how their interactions change those quantities. A point mass is the simplest model: its size and internal structure are ignored, leaving a mass and a moving position. A rigid body retains orientation and rotational kinetic energy. Choosing a model means deciding which features affect the question being asked; a sliding block may be treated as a point mass, whereas predicting its rolling without slipping requires its moment of inertia.

Three complementary questions organize the subject. Newton's second law asks how the net force determines the instantaneous acceleration. The work-energy theorem asks how that force, acting through a displacement, changes kinetic energy. Conservation of mechanical energy asks which combinations of position and velocity are accessible when the interactions admit a time-independent potential energy and no other force supplies net power. These descriptions are compatible, but answer different questions: an energy calculation may find a maximum speed without first solving for position at every time.

The harmonic oscillator below makes this unity visible. A restoring force produces a repeating position trace; the corresponding phase plane has a closed ellipse; kinetic energy and potential energy exchange while their sum stays fixed. Lagrangian mechanics and Hamiltonian mechanics provide further ways to organize the same dynamics, especially when coordinates or constraints make individual forces awkward to track.

The scope of classical mechanics also matters. Newtonian mechanics approximates motion at speeds small compared with the speed of light; special relativity changes that description when this approximation fails. Quantum mechanics is needed for phenomena for which a classical assignment of simultaneous position and momentum is inadequate. These limitations do not prevent classical mechanics from being a useful predictive model within its regime. ^[ref]

Conservation of angular momentum says that total angular momentum stays fixed when the net external torque about the chosen fixed inertial origin vanishes. Individual angular momenta and angular velocities can change. The radius change at constant angular momentum below makes this distinction visible: pulling two rotating point masses inward lowers the moment of inertia, raises the angular velocity, and supplies work that increases the kinetic energy.

Follow the same constant-mass point masses throughout, in an inertial frame, with all positions measured from a fixed origin $O$. If the internal pair forces are opposite and central, meaning directed along each pair's separation, thenThus zero net external torque throughout an interval implies constant total angular momentum throughout that interval. A vanishing external torque component, such as ${\tau }_z^{\mathrm{ext}}=0$, conserves the corresponding $L_z$ even when the other components change. This is the balance law for a fixed collection; angular-momentum flux must also be accounted for when matter crosses its boundary. ^[ref]

The central qualification is essential. Write ${\mathbf{F}}_{ij}$ for the internal force on member $i$ from member $j$. Under Newton's third law, ${\mathbf{F}}_{ji}=-{\mathbf{F}}_{ij}$, but the sum of their torques isThis vanishes when ${\mathbf{F}}_{ij}$ is parallel to ${\mathbf{r}}_i-{\mathbf{r}}_j$. Equal-and-opposite forces alone do not suffice: two opposite transverse forces with separated lines of action form a couple with nonzero torque. Cancellation of the internal pair torques, together with Newton's second law, proves the total balance above.

For one point mass under a central force about $O$, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ gives ${\mathbf{\tau }}_O=\mathbf{r}\times \mathbf{F}=0$. The force may still change the speed and kinetic energy; its zero torque preserves angular momentum. With nonzero ${\mathbf{L}}_O$, $\mathbf{r}\cdot {\mathbf{L}}_O=0$ keeps the motion in a fixed plane. Choosing the $z$ axis perpendicular to that plane, $r^2\dot{\theta }=L_z/m$ governs its angular progress. This removes an unknown from the central-force radial turning point calculation and helps reduce the Kepler orbit to a radial problem.

There is a historical geometric version of this result. In Book I, Section II, Proposition I of the Principia, Newton proves that motion under a central force toward a fixed centre stays in a plane and sweeps equal areas in equal times, using successive central impulses and equal triangle areas. ^[ref] In the modern notation above, constant areal velocity is $\dot{A}=L_z/(2m)$: the old geometric proof and the angular momentum calculation express the same invariant.

A rigid body with moment of inertia $I$, constant driving power $P$ and linear resisting torque $-\gamma \omega$ has an exponential approach of ${\omega }^2$ to $P/\gamma$. The ideal drive requires torque $P/\omega$, so starting from rest is a limiting energy model with singular initial torque.

Let a horizontal gyroscope have signed spin angular momentum $L_x$ along the axle. The disk's centre of mass is at $+l\hat{\mathbf{x}}$ and a counterweight is at $-l_c\hat{\mathbf{x}}$. Gravity supplies torque $g(ml-m_c{l}_c)\hat{\mathbf{y}}$. Since $\mathbf{\tau }=\mathbf{\Omega }\times \mathbf{L}$, steady horizontal precession gives the displayed signed vertical angular velocity. Neglect the axle's mass and the counterweight's axial spin moment of inertia. The zero-torque choice $m_c{l}_c=ml$ allows the axle to remain fixed. Moving the counterweight through this position reverses the gravitational torque and the associated slow precession.

Two identical torsional loads of moment of inertia $I$ hang in series from massless fibres of lengths ${\ell }_1,{\ell }_2$ and common filament torsional stiffness $K_t$. Put $R={\ell }_2/{\ell }_1$ and ${\omega }_0^2=K_t/(I{\ell }_1)$. The elastic energy is $K_t{\varphi }_1^2/(2{\ell }_1)+K_t({\varphi }_2-{\varphi }_1{)}^2/(2{\ell }_2)$ and the kinetic energy is $I({\dot{\varphi }}_1^2+{\dot{\varphi }}_2^2)/2$. Their normal modes have the displayed squared angular frequencies. Writing $\lambda ={\omega }^2/{\omega }_0^2$, their amplitude ratio is $A_2/A_1=1/(1-R\lambda )$. The lower mode twists in phase and approaches a combined inertia $2I$ as $R\to 0$, giving ${\omega }_{-}^2\to {\omega }_0^2/2$. The upper mode approaches the original single-load oscillation as $R\to \infty$, giving ${\omega }_{+}^2\to {\omega }_0^2$.

For a spool of outer radius $R$, axle radius $r$, mass $M$ and axial moment of inertia $I$, a lower-tangent pull of tension $T$ at upward angle $\theta$ produces horizontal acceleration $a=TR(R\cos \theta -r)/(I+M{R}^2)$ while rolling without slipping. This follows by combining $Ma=T\cos \theta +f$, $I\alpha =Tr+Rf$ and $a=-R\alpha$, with rightward translation and counterclockwise rotation positive. The acceleration reverses at ${\theta }_c$; zero acceleration means equilibrium only if initially at rest. At that angle the static friction limit gives $T_{\max }=\mu Mg/(\cos {\theta }_c+\mu \sin {\theta }_c)$.

A damped pendulum is a simple pendulum whose mechanical energy is removed by a resisting torque. Here the model is undriven, with a fixed pivot, a point mass $m>0$ at distance $\ell >0$, uniform gravitational acceleration $g>0$, and resisting torque $-bd\theta /dt$ with constant $b>0$. The angle $\theta$ is measured from the downward vertical in radians. An ideal viscous rotary dashpot realizes this torque law; tangential linear drag $-\alpha v$ on the bob gives the same equation with $b=\alpha {\ell }^2$. The moment of inertia is $m{\ell }^2$, soHere $t$ is physical time, ${\omega }_0$ is the undamped small-amplitude angular frequency, and $\gamma$ has units of inverse time. In the normalized equation $\ddot{\theta }+c\dot{\theta }+\sin \theta =0$, dots mean differentiation with respect to $\tau$, not $t$. The parameter $c$ is dimensionless. This separates the chosen damping law from the small-angle approximation: retaining $\sin \theta$ allows finite angles even when the resisting torque is linear in angular velocity.

To motivate the model, release a pendulum without a push and record how its swing shrinks. A published example by Gintautas and Hubler uses a rotary sensor built from a computer mouse to measure the frequency and damping coefficient of a physical pendulum. ^[ref] The following is a proposed experiment with a chosen viscous rotary dashpot; the plotted observations are predictions, not a reproduction of their measurements.

Mount a compact bob on a light rigid rod with a fixed pivot, couple the pivot to an adjustable viscous rotary dashpot, and measure $\ell$ from the pivot to the bob's centre. First record decay with the dashpot disconnected, then repeat with it connected at several settings; this establishes how much decay is already due to other losses. Keep the bob, rod, camera, and release angle unchanged between these trials. The simple model requires rod and damper-rotor moments of inertia to be negligible compared with $m{\ell }^2$. Otherwise measure the total moment of inertia $I$ and the distance $d$ to the assembly's centre of mass: the physical pendulum equation becomes $I{d}^2\theta /d{t}^2+bd\theta /dt+Mgd\sin \theta =0$, where $M$ is the total moving mass. A damper that also exerts an elastic restoring torque would require another term.

Place a camera perpendicular to the swing plane, include the pivot and bob centre in every frame, and determine their coordinates relative to the resting vertical. For horizontal and downward separations $x,y$, use $\theta =atan2(x,y)$; record the frame timestamps rather than assuming a frame rate. Release from rest at about $10^{\circ }$, record through several complete swings, and repeat the release. For every same-side turning point, tabulate its time $t_n$ and angular amplitude $A_n=|\theta (t_n)|$. Measure the oscillation period from differences $t_{n+1}-t_n$; comparing several periods exposes drift. Interpolating around each observed extremum reduces frame discretization error. Check that late peaks still exceed the angular resolution; taking the natural logarithm of tiny, uncertain peaks amplifies their relative errors.

In the small-angle approximation, the damped pendulum becomes a damped harmonic oscillator. For $0<\gamma <2{\omega }_0$ and release from rest at ${\theta }_0$, its exact linear-model solution isThe decaying sinusoidal envelope has coefficient $|{\theta }_0|{\omega }_0/{\omega }_d$; because release is from rest, its successive same-side extrema occur at $t_n=n{T}_d$ and have $A_n=|{\theta }_0|e^{-\gamma n{T}_d/2}$. Thus the envelope and the sequence of actual peak heights are different descriptions of the same trace. The linear damping regimes and this solution are derived in Richard Fitzpatrick, Damped Oscillatory Motion, equations 82–93.

The experiment tests two predictions independently: nearly equal same-side peak spacing and a straight line when plotting $\ln (A_n/A_0)$ against $t_n-t_0$. Its predicted slope is $-\gamma /2$, givingUse several peaks and repeated trials rather than a single ratio, and compare the frequency identity with independently measured $\ell$ and $g$. For the illustrative choices $\ell =1.00\mathrm{m}$, $g=9.81m/{\mathrm{s}}^2$, $c=0.20$, and ${\theta }_0=10^{\circ }$, the linear model gives $\gamma \approx 0.6264{\mathrm{s}}^{-1}$, $T_d\approx 2.0162\mathrm{s}$, and $A_{n+1}/A_n\approx 0.5318$. The first two positive peaks after release are therefore approximately $5.318^{\circ }$ and $2.828^{\circ }$. The undamped small-amplitude oscillation period is about $2.0061\mathrm{s}$: substantial loss per swing need not imply a large change in period.

Repeat at a larger release angle to distinguish two limitations. The relative error in replacing $\sin \theta$ by $\theta$ is about ${\theta }^2/6$ for small angles: about $0.51\%$ at $10^{\circ }$, but about $17.3\%$ at $60^{\circ }$ when compared with $\theta$. Finite-amplitude swing spacing can therefore change as the swing shrinks even if $b$ is constant. Separately, friction at the pivot or quadratic drag can invalidate the assumed linear resisting torque; exponential peak decay should be checked, not imposed on all recorded motion. Peters discusses why real pendulum damping can require nonlinear terms. ^[ref] Repeatability, residual pivot motion, changes in damper temperature, and finite camera resolution are useful checks on this proposed measurement.

The normalized equilibrium points are $(\theta ,\dot{\theta })=(k\pi ,0)$. Even multiples of $\pi$ are attracting: their linearization has eigenvalues $(-c\pm \sqrt{c^2-4})/2$, giving a stable focus for $0<c<2$, repeated negative eigenvalues at $c=2$, and a stable node for $c>2$. These are the underdamped, critically damped, and overdamped small-displacement regimes. Odd multiples of $\pi$ have eigenvalues $(-c\pm \sqrt{c^2+4})/2$ of opposite signs and are saddle equilibria. The energy dissipation of a damped pendulum explains why trajectories in the phase portrait spiral inward near a stable focus instead of closing into the recurring orbit of an undamped simple pendulum.

A damped torsional oscillation is rotation about a fiber's fixed reference orientation with a restoring torque and dissipative torque. In the ideal viscous model, damping contributes $-b\dot{\theta }$ with constant $b\ge 0$. This is a rotational damped harmonic oscillator. For known constant moment of inertia $I$ and torsion constant $\kappa$, define $\gamma =b/(2I)$. In the oscillatory regime ${\gamma }^2<\kappa /I$,$A$ and $\phi$ are set by the initial state. Successive positive peaks measured from the same equilibrium have magnitudes $p_n$ and $p_{n+1}$; their ratio gives $\delta =\ln (p_n/p_{n+1})=\gamma T_d$. Consequently a measured damped oscillation period and decay together recoverUsing only $4{\pi }^{2}I/T_d^2$ underestimates the torsion constant in this model. The toy curve in torsion constant sets $I=2.0\times 10^{-6}\mathrm{kg}{\mathrm{m}}^2$, $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$, and $\gamma =0.020{\mathrm{s}}^{-1}$. It gives $T_0\simeq 88.86\mathrm{s}$, $T_d\simeq 92.64\mathrm{s}$, and $\delta \simeq 1.853$. Ignoring the decay recovers $9.2\times 10^{-9}\mathrm{N}m/rad$, an $8\%$ underestimate; including it recovers the stipulated torsion constant.

These equations assume a single rotational mode, a stationary zero, linear elasticity, and viscous damping. Internal fiber losses need not behave as a constant viscous coefficient: Fleischer and colleagues model their precision apparatus using a loss angle. Their response is evidence for specifying a damping model, not for applying the viscous correction universally. ^[ref]

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

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

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