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A point mass constrained to a fixed-length massless arm under uniform gravitational acceleration is a simple pendulum. With no drag its angular equation is $\ddot{\theta }+(g/\ell )\sin \theta =0$. The small-angle approximation yields simple harmonic motion, whereas the full restoring force is nonlinear. Adding drag gives a damped pendulum.

For an ideal point mass $m$ swinging in a vertical plane, the torque about the pivot is $-mg\ell \sin \theta$. Dividing by its moment of inertia $m{\ell }^2$ gives the displayed equation. Both the inertia and the gravitational torque scale with $m$, explaining why the point mass cancels. This derivation and the small-amplitude period are given in Ling, Sanny and Moebs, University Physics, pendulums.

Released from rest at a small angular amplitude ${\theta }_0$, the linearized solution is $\theta (t)\simeq {\theta }_0\cos (\sqrt{g/\ell }t)$, with oscillation period $T_0=2\pi \sqrt{\ell /g}$. For $g=9.81m/{\mathrm{s}}^2$, lengths one and four metres give $T_0\approx 2.0061\mathrm{s}$ and $4.0121\mathrm{s}$. This same scaling follows from dimensional analysis; the dynamical calculation supplies the coefficient $2\pi$.

Finite angular amplitude changes the exact ideal period. Conservation of energy gives ${\dot{\theta }}^2=(2g/\ell )(\cos \theta -\cos {\theta }_0)$ for release from rest. Integrating a quarter oscillation and substituting $\sin (\theta /2)=\sin ({\theta }_0/2)\sin \varphi$ givesHere $K$ is the complete elliptic integral of the first kind. The small-angle approximation gives a ratio tending to one; at $10^{\circ },60^{\circ },90^{\circ }$, the ratios are approximately $1.00191,1.07318,1.18034$. Thus almost constant period at small amplitude has a quantitative limit. As ${\theta }_0\to \pi$, the period diverges, connecting to the pendulum separatrix approach takes infinite time.

These predictions assume the specified ideal system. Drag changes the mechanical energy; a massive support changes the moment of inertia. The distinction is useful in experiments: a change in period can test an approximation rather than merely indicate an arithmetic mistake.

A magnetic dipole of fixed magnitude $m_B$, free to rotate about an axis with moment of inertia $J$, has potential energy $-m_{B}B\cos \theta$ in a uniform magnetic field. Near stable alignment, $J\ddot{\theta }=-m_{B}B\theta$, so its small-angle period is $2\pi \sqrt{J/(m_{B}B)}$. This neglects damping, bearing friction and any other restoring torque.

For rolling without slipping inside a stationary circular cylinder, take the inner body's axial moment of inertia to be $I=\gamma m{r}^2$. Its center of mass follows a circle of radius $d=R-r$, and the reduced Lagrangian is $L=m(1+\gamma )d^2{\dot{\theta }}^2/2-mgd(1-\cos \theta )$. The Euler-Lagrange equation gives $(1+\gamma )d\ddot{\theta }+g\sin \theta =0$. Linearizing near the lowest position yields the displayed small oscillation frequency. A uniform solid cylinder has $\gamma =1/2$. This reduction requires sustained contact and enough static friction to prevent slipping.

Purdue's Bicycle Wheel Gyroscope demonstration uses a bicycle wheel on a long axle, with a ball at one end seated in a freely rotating socket on a heavy stand. Spin the wheel by hand or with the supplied motor, seat the ball, and release the nearly horizontal axle. The readout is the axle sweeping around the vertical while the wheel continues spinning. Compare with the stopped wheel, which falls when one end is released; reverse the spin with the same support and the precession reverses. These are the page's documented apparatus, presentation, and reversal checks. ^[ref]

An alternative in Physics Demonstrations by Julien Clinton Sprott (section 1.3, Materials and Procedure) suspends one axle end by stainless steel wire and a fishing-line swivel. The swivel allows precession without winding up the suspension. Sprott describes the axle bobbing as nutation and precession becoming faster as spin decreases. A suspended support can move, whereas the calculation here assumes a fixed $O$; use the socket version for the closest realization. ^[ref]

A proposed quantitative extension is to record an overhead view of an axle marker against a fixed angle scale, together with a side view for $\theta (t)$. Time successive overhead passages to estimate $\Omega =2\pi /T$; count passages of a rim mark to estimate $|{\omega }_3|=2\pi f$. Record the spin sense too. Measure $M$ and the centre of mass offset $d$ of the whole moving assembly, and determine its axial moment of inertia $I_3$ independently. Wheel, axle, support, release angle, and observation interval should stay the same between spin trials. Measuring both $\Omega$ and $\theta (t)$ distinguishes azimuthal precession from nutation; an overhead movie alone can miss the latter. These camera readouts and parameter checks are proposed here, not reported measurements from the demonstration pages.

For slow precession, compare $\Omega I_3{\omega }_3$ with $Mgd$, and test whether reversing ${\omega }_3$ reverses $\Omega$ with nearly unchanged magnitude. A brief interval minimizes spin loss from friction; repeat at several starting spin rates rather than interpreting an entire slowing run as one steady state. A suspension restoring torque governed by its torsion constant, socket friction, support motion, a misplaced centre of mass, and a push at release can all supply extra torque or change the initial conditions. The swivel reduces the twist that activates the suspension's torsion constant; it does not demonstrate zero torque.

Steady precession keeps $\theta$ constant; nutation changes $\theta$ while the axle may also precess. Sudden release generally excites both. A small initial drop is consistent with conservation of energy and conserved vertical angular momentum: some gravitational potential energy supplies the motion around the vertical, and the axial contribution to vertical angular momentum adjusts. Bearing friction may damp the bobbing, but an ideal undamped heavy symmetric top need not settle into steady precession. Chapter 20: Rotation in space explicitly distinguishes uniform precession from release transients and nutation (section 20-3, paragraphs following equation 20.15).^[ref]

Attach a spring of spring constant $k$ to the axle of a round body of mass $m$, radius $R$ and axial moment of inertia $I$. Assume rolling without slipping, a horizontal drag force $-b\dot{x}$ applied through the axle, and no additional resisting torque. The translational force equation is $m\ddot{x}=-kx-b\dot{x}+f$, where $f$ is signed static friction. The rotational equation and rolling constraint give $f=-I\ddot{x}/R^2$, yielding the displayed equation. Equivalently, the kinetic energy is $(m+I/R^2){\dot{x}}^2/2$. This added effective inertia lowers the undamped angular frequency to $\sqrt{k/(m+I/R^2)}$. A distributed viscous torque would add a separate damping term and cannot be inferred from the drag force alone.

A rigid body's axial moment of inertia $I$ and a constant resisting torque ${\tau }_f>0$ give constant angular acceleration $-{\tau }_f/I$. Hence $t_{\mathrm{stop}}=I{\omega }_0/{\tau }_f$ and the angular displacement is the displayed expression. Equivalently, the lost rotational kinetic energy $I{\omega }_0^2/2$ equals the resisting work ${\tau }_f\Delta \varphi$. The total number of revolutions is $\Delta \varphi /(2\pi )$; the number of whole revolutions completed is its floor function.

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

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

Let fuel of remaining mass $M_f$ be carried at radius $a$ on a rotor whose dry moment of inertia is $I_d$. Then $I=I_d+a^2{M}_f$. Fuel leaves tangentially backwards with speed $v_e$ relative to the rim, so its inertial tangential velocity is $a\omega -v_e$. Conserving angular momentum of rotor plus newly emitted fuel during $d{M}_f<0$ givesSince $dI=a^{2}d{M}_f$, integration yields ${\omega }_f-{\omega }_i=(v_e/a)\log (I_i/I_f)$. Including the fuel's co-rotation is essential: the rotor alone is an open system and its angular momentum is not conserved. For three radial rods of mass $m$ and length $a$, each carrying a point rocket of dry mass $m$ and fuel mass $m_f$, $I_i=(4m+3m_f)a^2$ and $I_f=4m{a}^2$.

An ideal circular ring whose mass lies at a single radius $R$ from its symmetry axis. About that axis, the moment of inertia is $\int R^{2}dm=M{R}^2$. The result assumes negligible radial and axial thickness; it is different from the $M{R}^2/2$ of a uniform solid disk.

A thin uniform circular rod of radius $r$ and length $\ell$ undergoing a small end twist of complex amplitude $\Theta$ has shear-strain amplitude $\gamma (\rho )=\rho \Theta /\ell$. A complex viscoelastic modulus $G^{\ast }(\omega )$ gives shear-stress amplitude $G^{\ast }\rho \Theta /\ell$. Integrating the restoring torque givesThe area integral is the polar second moment of the cross-section, not a mass moment of inertia. For a disc of moment of inertia $I$ and negligible rod inertia, harmonic angular motion satisfies $[-I{\omega }^2+K^{\ast }(\omega )]\Theta =0$ in a free-mode calculation, or equals the external torque amplitude in a driven measurement. The complex stiffness is a frequency-domain constitutive representation; a general real-time response has memory and cannot be described by putting an arbitrary frequency-dependent complex number into an instantaneous real equation.

For a slender homogeneous circular shaft in small-amplitude linear elasticity, torque is $G{J}_p{\theta }_x$ and axial angular inertia per length is $\rho J_p$. Their common geometric factor cancels in the angular equation of motion, leaving the displayed wave equation. Here $J_p=\int r^{2}dA$ is the polar second moment of area, rather than the mass moment of inertia itself.

For small torsional displacement $\varphi (x,t)$, the transmitted torque is $C=G{J}_p{\varphi }_x$. A slice of length $dx$ receives net torque $G{J}_p{\varphi }_{xx}dx$ and has axial moment of inertia $\rho J_{p}dx$. Its angular equation therefore gives the displayed wave equation, with wave speed $\sqrt{G/\rho }$. The same geometric factor cancels from inertia and stiffness, making this ideal torsional speed independent of rod radius. This long-wave circular-rod model neglects additional elastic modes and dissipation.

The torsion constant, or torsional spring constant, is the positive restoring torque per unit twist angle of a complete suspension in its linear elasticity range:Its SI unit is $\mathrm{N}m/rad$. The minus sign describes restoration toward the untwisted orientation. The derivative specifies a local stiffness if the response is not globally linear; using one constant over a finite range additionally requires a straight calibration curve. Return to the same zero after unloading tests recoverable elastic deformation; repeat loading in both directions to check whether the response depends on its previous loading. Staying below permanent deformation is necessary, but does not alone prove proportionality between torque and twist.

For an ideal uniform circular wire of length $\ell$, radius $r$, and shear modulus $G_s$, the filament torsional stiffness is $C=G_s{J}_p$, with polar second moment of area $J_p=\pi r^4/2$. Uniform twist gives $\theta /\ell$ as the twist per unit length, soHere $U$ is stored elastic energy. Doubling the wire's radius multiplies the ideal torsion constant by $16$; doubling its length halves it. This explains why a long thin fiber is useful for weak torque measurement. The formula assumes linear elasticity with uniform material and circular section; it does not calibrate arbitrary ribbons, clamps, or a compound suspension. Within this ideal model, the equipartition theorem predicts an equilibrium thermal mean-square angular displacement $⟨{\theta }^2⟩=k_B\mathcal{T}/\kappa$, where $k_B$ is the Boltzmann constant and $\mathcal{T}$ the uniform absolute temperature. A softer suspension magnifies both the signal and its thermal fluctuations.

Static calibration supplies known torques without assuming the unknown interaction being measured. Apply a known perpendicular force $F$ at lever arm $a$, record the change from the unloaded twist, and useSeveral positive and negative settings test proportionality; fit the slope of applied torque against twist and retain a possible intercept. A nonzero intercept must not be silently absorbed into the slope. Convert degrees to radians before reporting $\mathrm{N}m/rad$. PASCO's procedure turns the Coulomb balance on its side and adds small known masses to generate a gravitational weight at the sphere's lever arm, returning the rod to its index mark by turning the dial. Its tabulated coefficient is force per degree: multiply by the lever arm and $180/\pi$ to obtain our torsion constant. ^[ref]

The figure's ideal static calibration uses $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$: an applied torque of $2.0\times 10^{-10}\mathrm{N}\mathrm{m}$ gives $0.020\mathrm{rad}$. Both signs lie on the same straight line. The plotted points are calculated examples, not experimental observations; they cannot demonstrate a real fiber's linearity.

Dynamic calibration instead measures the oscillation period $T_0$ with known total suspended moment of inertia $I$. For negligible damping, small free oscillations about a fixed reference obey simple harmonic motion:Include the rod, mirrors, and any attachments in $I$ unless their contributions are demonstrably negligible. Keep the applied torque absent or independent of angle. If an external torque has derivative $g_{\tau }=d{\tau }_{\mathrm{ext}}/d\theta$ at equilibrium, linearization gives effective stiffness $\kappa -g_{\tau }$: timing alone then measures that combination, and stability requires $\kappa -g_{\tau }>0$. A constant applied torque merely shifts equilibrium. Non-negligible damping changes the measured period; damped torsional oscillation gives a correction under a specified damping model. Agreement between static and dynamic calibrations is a useful check on the assumed moment of inertia, reference stability, and restoring law, rather than a reason to ignore their assumptions.

A harmonic oscillator in which a restoring torque is proportional to angular displacement; its moment of inertia and torsion constant set the natural angular frequency.

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

A uniform circular disc of radius $a$ and mass $m$, pivoted on a perpendicular axis through its rim, has moment of inertia $I_A=3m{a}^2/2$ by the parallel axis theorem. Its hanging small oscillations have period $2\pi \sqrt{3a/(2g)}$. Measured from the upright position, the falling equation is $\ddot{\theta }=2g\sin \theta /(3a)$. Exact upright rest remains an unstable equilibrium; the energy level of that state describes a limiting separatrix after perturbation, with ${\dot{\theta }}^2=4g(1-\cos \theta )/(3a)$. The center acceleration and pivot reaction follow by resolving radial and tangential components.