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Angular momentum measures how a point mass's momentum passes around a chosen origin. Motion straight toward or away from that origin contributes zero; transverse momentum contributes more when it passes farther away. The picture shows the sign convention before the calculation.

Fix an origin $O$ and nonrotating axes in an inertial frame. For a nonrelativistic point mass of constant mass $m>0$, let $\mathbf{r}$ be its position from $O$ and $\mathbf{p}=m\mathbf{v}$ its mechanical momentum in that same reference frame. Its angular momentum is the cross productHere $r=|\mathbf{r}|$, $p=|\mathbf{p}|$, and $\varphi \in [0,\pi ]$ is the angle between them. The right-hand rule gives the direction: curl from $\mathbf{r}$ toward $\mathbf{p}$, and the thumb points along ${\mathbf{L}}_O$. Thus angular momentum is an axial vector, perpendicular to both position and momentum. Its International System of Units unit is $\mathrm{kg}{\mathrm{m}}^2/s$. A point mass need not follow a circle to have angular momentum.^[ref]

With $x$ rightward, $y$ upward, and $+z$ out of the page, the cross product gives $L_z=x{p}_y-y{p}_x$. TakeThen ${\mathbf{L}}_O=(0,0,6)\mathrm{kg}{\mathrm{m}}^2/s$. Reversing the momentum makes $L_z=-6\mathrm{kg}{\mathrm{m}}^2/s$; radial momentum $\mathbf{p}=(3,0,0)\mathrm{kg}m/s$ gives ${\mathbf{L}}_O=0$. The factor $r\sin \varphi$ is the lever arm to the line through the particle parallel to its momentum.

For motion in the $xy$ plane, plane polar coordinates give $\mathbf{v}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta }\hat{\mathbf{\theta }}$, with $\hat{\mathbf{r}}\times \hat{\mathbf{\theta }}=\hat{\mathbf{z}}$. ConsequentlyThe signed angular velocity $\dot{\theta }$ is positive counterclockwise viewed from $+z$ toward the plane, with $\theta$ measured in radians. Radial velocity does not enter the cross product. Under conservation of angular momentum, this relation predicts faster angular motion at smaller $r$ and constant areal velocity $\dot{A}=L_z/(2m)$ for consistently oriented swept area. That is the geometric content of Kepler's second law.

For several point masses, add their angular momenta about the same origin: ${\mathbf{L}}_O={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$. The origin matters: angular momentum under an origin shift explains why the same moving particle can have different values about two fixed points. The central axis of a momentum system and angular momentum decomposition about the centre of mass make this dependence useful.

The Cavendish experiment makes the attraction predicted by Newton's law of universal gravitation measurable between laboratory masses. Two small test masses hang at the ends of a horizontal rod in a Cavendish torsion balance; nearby larger source masses exert a tiny torque. A thin suspension fiber supplies an opposing torque as it twists. Measuring its twist and calibrating its torsion constant lets an experimenter infer the gravitational force and, in a modern analysis, the gravitational constant. ^[ref]

The beautiful feature is that the two horizontal forces can cancel as vectors while their torques add. The instrument converts weak attraction into a rotation, with sensitivity set by the fiber's torsion constant. Earth's much larger downward gravitational weight is supported by the suspension rather than supplying the intended torque about its vertical axis. A laboratory measurement can therefore probe the same Newtonian gravitational field that governs the two-body problem.

Henry Cavendish's original paper was Experiments to determine the Density of the Earth, published in Philosophical Transactions, volume 88 (1798), pp. 469–526, and read on 21 June 1798. On p. 469 he credits John Michell with the apparatus idea and explains that Michell died before using it; the instrument reached Cavendish through Francis John Hyde Wollaston. Cavendish's stated aim was Earth's mean mass density. Calling the experiment a determination of the gravitational constant is a modern interpretation, not Cavendish's own formulation of his result. ^[ref]

The connection between Earth's mean mass density and the gravitational constant follows in a spherical, nonrotating model of Earth. With Earth mass $M_E$, radius $R_E$, and surface gravitational acceleration $g$, the spherical shell theorem and Newton's law of universal gravitation giveThus a local comparison of gravitational attractions, together with knowledge of Earth's size and surface gravitational acceleration, reaches a property of the entire planet. These are modern equations explaining the connection, not equations quoted from Cavendish's paper.

For a worked example, use an explicitly simplified two-near-pair model. Each test point mass is $m$, each source point mass is $M$, and each lever arm is $L$, so the rod's full length is $2L$. Neglect the rod's mass. At zero twist put the test masses at $(-L,0)$ and $(L,0)$, and the sources at $(-L,-r)$ and $(L,r)$. Here $r$ is the near-pair center separation. Ignore each source's attraction on the test mass at the opposite end. Evaluate the two near-pair forces at zero twist and treat their torque as constant for a small twist, requiring $L|\theta |\ll r$. This is a teaching model, not a precision model of Cavendish's actual apparatus.

By Newton's law of universal gravitation, each retained force has magnitude $F=GmM/r^2$. The left force points down and the right force points up. Using torque about the suspension axis, both contributions are counterclockwise when viewed from above:The net retained horizontal force is zero, yet the net torque is positive. Moving the sources to $(-L,r)$ and $(L,-r)$ reverses both forces and hence the torque. In a real apparatus, the opposite source also attracts each test mass; that correction is explicitly included in the university derivation. ^[ref]

For illustrative parameters, take $m=0.010\mathrm{kg}$, $M=1.0\mathrm{kg}$, $r=0.050\mathrm{m}$, $L=0.10\mathrm{m}$, and torsion constant $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$. Use $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, the central value in NIST's 2022 CODATA table; that table gives standard uncertainty $0.00015\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. This is an input for the example, not a measurement made with these assumed parameters. ^[ref]This sub-nanonewton force produces a measurable twist in the Cavendish torsion balance. The displayed digits make the toy calculation reproducible; they do not describe experimental precision.

The PHYWE guide Magnetic moment in the magnetic field makes the magnetic dipole moment measurable. A circular test current loop sits between a pair of Helmholtz coils, with separate supplies and meters for the test electric current $I$ and field-producing electric current $I_H$. The test current loop is mounted on a torsion dynamometer. The guide varies $I_H$, the number of turns, the test electric current, diameter, and the angle between the magnetic dipole moment and magnetic field. Its angular settings are spaced by $15°$. The leads hang loosely and are twisted together to suppress extra torque; the instrument zero is checked frequently because moving leads can shift it. Its reported results include approximately linear dependence on each electric current and on winding count, and approximately quadratic dependence on diameter.^[ref]

The readout is a balance of torques. According to the manufacturer's Torsion dynamometer 0.01 N operating instructions, a metallic band supplies the restoring torque. First set the indication to zero with the applied force absent and center the lever using the zero knob. Apply the load, then turn the indicating knob until the lever returns to its original zero position. The scale then reads a force equivalent, not a torque directly. The quoted scale division is $0.1\mathrm{mN}$. For a force applied perpendicular to the calibration lever arm at distance $\ell$, the conversion is $\tau =F\ell$; a current loop applies a couple, so its torque must be compared with the calibrated restoring torque.^[ref]

For a quantitative repeat, independently check that conversion with known forces at a measured perpendicular lever arm, and check the imposed magnetic field at the test current loop. These are proposed calibration checks, not additional measurements reported here. The ideal Helmholtz coil model predictswhere $N_H$ and $R_H$ are the winding count and radius of each field-producing Helmholtz coil, distinct from $N$ and $R$ of the test current loop. Derive $B_0$ by adding the on-axis magnetic field of a circular current loop at distances $R_H/2$ from both coils. Independently calibrating $B_0$ and the restoring torque lets a fitted torque amplitude determine $m$ as amplitude divided by $B_0$, to compare with $NI\pi R^2$.

Useful additional controls are to reverse only $I$, reverse only $I_H$, and reverse both: the signed torque should reverse in either single reversal and be unchanged in the double reversal. Record baselines with each electric current separately zero, and subtract a nearby zero reading to reduce torsion balance zero drift. Repeat angular settings in both directions to reveal lead-dependent offsets. These are suggested controls. Limits include finite magnetic field uniformity, uncertainty in effective winding area and angular zero, meter calibration, Joule heating, and residual lead torque. An apparently sinusoidal trace alone does not establish the absolute magnetic dipole moment. No raw laboratory data or new measured values are claimed here.

A galvanometer radial field is approximately radial about the rotation axis within each narrow working gap. Its direction follows the moving sides of the coil over the useful sweep. Here the magnetic field points from the left pole toward the core and from the core toward the right pole; it is not outward everywhere. This geometry keeps the magnetic force on a current-carrying wire tangential, with nearly constant lever arm. A spatially uniform magnetic field in a fixed direction would instead give an angle-dependent torque as the coil turns.

Choose a fixed support point $O$ and nonrotating axes in an inertial frame. All angular momenta, torques, and moments of inertia below refer to $O$. Let $\hat{\mathbf{e}}$ point along the axle from $O$ toward the centre of mass, and let $\hat{\mathbf{z}}$ point vertically upward. Write $\theta$ for their angle. For a rapidly spinning symmetric top, the dominant angular momentum is ${\mathbf{L}}_O\simeq L_s\hat{\mathbf{e}}$, where $L_s=I_3{\omega }_3$ is signed and $I_3$ is the axial principal moment of inertia. The sign of the axial angular velocity ${\omega }_3$ follows the right-hand rule.

With mass $M$, centre of mass distance $d$, and constant gravitational acceleration $g$, the gravitational force and its torque areAn ideal support applies a force at $O$, so its lever arm and torque about $O$ vanish. At the instant the axle lies in the $xz$ plane with a positive $x$ component, the gravitational torque points along $+y$. It adds $d{\mathbf{L}}_O={\mathbf{\tau }}_{O}dt$ toward $+y$, rather than along the downward force. The dot product identity $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$ shows why perpendicular torque changes direction without instantly changing magnitude. This vector balance and the slow-turning relation ${\mathbf{\tau }}_O=\mathbf{\Omega }\times {\mathbf{L}}_O$ are developed in Chapter 20: Rotation in space (sections 20-2 and 20-3, equations 20.13–20.15).

For steady precession about the vertical, $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$ and $d\hat{\mathbf{e}}/dt=\Omega \hat{\mathbf{z}}\times \hat{\mathbf{e}}$. Equating the two cross products givesThe angular momentum tip traces a circle of radius $|L_s|\sin \theta$: its sideways speed is $|\Omega L_s|\sin \theta$. Both that radius and the gravitational lever arm contain $\sin \theta$, explaining its cancellation for $0<\theta <\pi$. Exactly vertical alignment has zero gravitational torque and no distinguishable azimuthal precession to infer by this cancellation. Positive $L_s$ gives positive precession, counterclockwise viewed from above; reversing the spin reverses the slow precession. ^[ref]

A body turning about a fulcrum. In static balance, effort and load supply equal and opposite torques: $E{d}_E=L{d}_L$, where the distances are perpendicular lever arms. Its ideal mechanical advantage is $L/E=d_E/d_L$. Measuring distances along the body instead of perpendicular to the force lines can give the wrong answer.

The lever arm of a nonzero force about an origin is the shortest distance from that origin to the line through the application point parallel to the force. Its length is $\ell =|\mathbf{r}\times \hat{\mathbf{F}}|$, with unit vector $\hat{\mathbf{F}}=\mathbf{F}/|\mathbf{F}|$. Thus torque magnitude is $F\ell$, maximal at a right angle for fixed $r$ and $F$, and zero if the line passes through the origin. Sliding the application point along that line leaves torque unchanged, since $(\mathbf{r}+s\hat{\mathbf{F}})\times \mathbf{F}=\mathbf{r}\times \mathbf{F}$. The same perpendicular-distance construction gives angular momentum magnitude $p\ell$ for a line parallel to nonzero momentum through the particle.

For a rectangular winding with $N$ turns, active side length $\ell$, side separation $w$, and per-turn area $A=w\ell$, the magnetic force on a current-carrying wire gives equal and opposite side forces of magnitude $N|I|\ell B$. Their force couple has lever arm $w$. Choosing the positive rotation to match positive electric current in the illustration gives the signed torque and spring balanceHere $B$ is the working-gap magnetic flux density, $k>0$ is the torsional spring constant in $\mathrm{N}m/rad$, and $\theta$ is the deflection angle in radians from the no-current zero. The balance holds after motion has settled, within the radial-field and linear elasticity ranges. Do not identify $\theta$ with the angle between a magnetic dipole moment and a fixed uniform magnetic field: that different geometry has torque magnitude $NIAB|\sin \alpha |$. ^[ref]

For an original illustrative design, choose $N=120$, $A=3.00\times 10^{-4}{\mathrm{m}}^2$, $B=0.250\mathrm{T}$, and $k=3.00\times 10^{-6}\mathrm{N}m/rad$. Then $NAB=9.00\times 10^{-3}\mathrm{N}m/A$. At $I=+40.0\mu \mathrm{A}$, the magnetic torque is $+3.60\times 10^{-7}\mathrm{N}\mathrm{m}$ and $\theta =+0.120\mathrm{rad}\simeq +6.88^{\circ }$. Reversing the electric current reverses both the torque and the deflection, giving $-6.88^{\circ }$ without changing the magnetic field.

A lever is a body free to rotate about a fulcrum. Its usefulness comes from a balance of torques, not from producing energy. For a stationary ideal lever, let $d_E$ and $d_L$ be the perpendicular lever arms of the effort $E$ and load $L$. ThenThe mechanical advantage uses perpendicular distances to the lines of action, which need not equal distances measured along a bent or obliquely loaded arm. In an ideal slow motion, the effort moves through the larger distance when its force is the smaller: work is conserved.

For the three familiar straight arrangements, the fulcrum lies between effort and load in a first-class lever, the load lies between fulcrum and effort in a second-class lever, and the effort lies between fulcrum and load in a third-class lever. The second class has ideal mechanical advantage greater than one; the third has it less than one. The first can have either, depending on the lever arms.

The forearm bends by elbow flexion. The biceps brachii and brachialis pull through short tendons close to the elbow, while a load in the hand acts farther away. This is predominantly a third-class arrangement: a large muscle force buys a comparatively large and rapid hand displacement. The biceps brachii also contributes to forearm supination. Elbow extension is produced chiefly by the triceps brachii, whose tendon acts on the olecranon behind the elbow. The extensor insertion and forearm load lie on opposite sides of the effective pivot, giving the corresponding first-class analogy. Actual anatomical joint motion has changing muscle moment arms, and the joint reaction force also enters the full force balance; the body is not a set of frictionless pins. ^[ref]

Let an upward force $P$ act along $x=0$ and a downward parallel force $Q$ along $x=d>0$. An equivalent resultant force must preserve both total force and torque. Its signed vertical component is $R=P-Q$, and taking torques about the first line gives $R{x}_R=-Qd$. ThusIts magnitude is $|P-Q|$ and its direction agrees with the larger force. Its line lies outside the two original lines on the side of that larger force; for example, $P>Q$ gives $x_R<0$. The products of the original forces and their perpendicular distances from this line are equal.

If $P=Q\ne 0$ and $d\ne 0$, the total force vanishes but the torque is $-Pd$. No single force is equivalent to this system: it is a force couple of magnitude $Pd$. Its torque is independent of the reference point because translating that point changes the torque by the cross product of the shift and the zero resultant force. Equal opposite forces on the same line instead cancel completely.

The eye is rotated by six extraocular muscles. The rectus muscles approach it from behind; the oblique muscles have oblique lines of pull, with the superior oblique muscle tendon redirected by an orbital pulley. Attachments on the sclera give lever arms about the rotation centre. In the straight-ahead position their actions are as follows; vertical and torsional contributions vary with gaze direction.^[ref]

Coordinated contraction selects the desired rotation and balances unwanted components. This illustrates torque control; the full ocular geometry is not simply the parallel-force pair of the first calculation.

Directions. Choose $+x$ along the horizontal spindle from the pivot to the disk, and $+z$ vertically upward. With lever arm $\mathbf{r}=l\hat{\mathbf{x}}$ and weight ${\mathbf{F}}_g=-mg\hat{\mathbf{z}}$, the torque about the pivot isClockwise precession viewed from above has $\mathbf{\Omega }=-\Omega \hat{\mathbf{z}}$. The relation $\mathbf{\tau }=\mathbf{\Omega }\times \mathbf{L}$ therefore requires the spin angular momentum to point along $-x$, from the disk toward the pivot. The disk spins clockwise when viewed from its outer face looking along the spindle toward the pivot. Looking from the pivot reverses that apparent rotation. The right-hand rule fixes these signs unambiguously.

The drawing shows an instantaneous top view; the disk's angular momentum and the horizontal spindle turn clockwise together. The pivot force has zero torque about the pivot.

Rate of precession. For steady horizontal motion the spin contribution has magnitude $L=I\omega$ and turns through $d\varphi =\Omega dt$ in time $dt$. Its change is perpendicular to itself and has magnitude $dL=Ld\varphi$. HenceThis is the usual slow steady gyroscope precession relation. Any constant vertical angular momentum associated with steady precession has zero time derivative; the horizontal spin contribution supplies the changing component here. Transient nutation is excluded by the prescribed horizontal steady motion.

Clock design. Convert both rotation rates to angular frequencies:For the moment of inertia of a uniform disc, $I=m{R}^2/2$, so the mass cancels:

Spin friction. At fixed geometry, reducing $\omega$ increases $\Omega \propto 1/\omega$. The precession period, $T_p=2\pi /\Omega$, therefore decreases. This trend holds while slow steady precession remains a suitable description; it cannot be extrapolated to a stopped disk.

Counterweight. To oppose the disk's gravitational torque, place the counterweight on the other side of the pivot, at $-l_c\hat{\mathbf{x}}$. The net torque isA horizontal axle can remain fixed when this vanishes, givingFor the fixed spin direction already found, counterbalanced gyroscope precession has signed rateMoving the counterweight outward from the pivot first slows the clockwise motion, eliminates its driving torque at balance, and then reverses the steady motion to counterclockwise. A counterweight on the disk's side would add to the gravitational torque. These statements neglect axial spin moment of inertia of the counterweight and assume the stated horizontal motion, without a separately imposed precession at balance.

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.

A torsion balance measures a weak torque by suspending a body from a thin fiber and reading its twist angle. The fiber supports the body's gravitational weight while permitting rotation about the suspension axis. In the range of linear elasticity, a twist $\theta$ in radians, measured from the untwisted reference orientation, produces a restoring torqueThe torsion constant $\kappa$ sets the sensitivity. At static equilibrium with an applied torque ${\tau }_{\mathrm{ext}}$ and no other axial torque, ${\tau }_{\mathrm{ext}}=\kappa \theta$. A perpendicular force $F$ at lever arm $a$ therefore gives $\theta =aF/\kappa$. The torsion balance measures a turning effect: a force whose line of action passes through the suspension axis gives no axial torque. The Cavendish torsion balance is the gravitational application; Coulomb's law supplies an electrical one. ^[ref]

The original schematic separates two ways of reading the same torque. In deflection operation the top attachment stays fixed and the body rotates. In null operation the experimenter rotates the top attachment by $\alpha$ while returning the body's laboratory angle $\beta$ to its original value. If the initially untwisted configuration has $\alpha =\beta =0$, the fiber twist is $\theta =\beta -\alpha$, not $\beta$ alone. ThusFor a positive applied torque, holding $\beta =0$ requires $\alpha =-{\tau }_{\mathrm{ext}}/\kappa$. Keeping the body at the same angle keeps the lever arm and source geometry fixed, making comparison of different applied forces easier. Either readout still requires a calibrated torsion constant and a stable reference orientation.

An illustrative calculation shows the amplification: with $a=0.050\mathrm{m}$, $F=2.0\times 10^{-9}\mathrm{N}$, and $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$, the applied torque is $1.0\times 10^{-10}\mathrm{N}\mathrm{m}$ and the equilibrium twist is $0.010\mathrm{rad}\simeq 0.573^{\circ }$. These are stipulated values, not measured data. Reducing the torsion constant tenfold increases this twist tenfold, provided the fiber remains in its calibrated linear elasticity range. The cost is a slower response, explained by the oscillation period in torsion constant.

A concrete laboratory experiment is PASCO's Coulomb balance: level its counterbalanced rod, discharge both conducting spheres, and set the unloaded zero. Charge the spheres at maximum separation; switch off the charging supply. Move the second sphere to a known center separation $R$, then turn the suspension dial until the rod returns to its index mark. Record the dial change and $R$, recharge, and repeat at several separations. The manual recommends a draft-free room, clean insulating supports, and recharging before each reading: leakage, especially in humid conditions, otherwise changes the electric charges during the sequence. At close separations the conducting spheres redistribute charge, so the point charge approximation needs correction. ^[ref]

For constant electric charges and a fixed perpendicular lever arm, the Coulomb's law prediction is $|\theta |\propto R^{-2}$. Plotting the logarithm of twist magnitude against the logarithm of separation gives $\log |\theta |=\text{constant}-2\log R$, so the predicted slope is $-2$. Absolute force measurement additionally needs the torsion constant; comparing ratios can eliminate it if it remains stable. Repeating a reference separation between other settings distinguishes a separation effect from a changing signal. A declining reading at the reference separation would invalidate the assumption of unchanged electric charges, even if a distance plot looks smooth.

The historical electrical example is Coulomb's first memoir of 1785. His apparatus enclosed a suspended needle with a small ball and a second ball; an adjustable head twisted the suspension. After charging the balls by contact, he reported angular separations $36^{\circ }$, $18^{\circ }$, and $8.5^{\circ }$ with head settings $0^{\circ }$, $126^{\circ }$, and $567^{\circ }$. The total twists were approximately $36^{\circ }$, $144^{\circ }$, and $576^{\circ }$: fourfold restoring torque accompanied approximately halved separation. These are his reported observations, not data from our diagrams. ^[ref] A precise reconstruction must use center distances and the changing lever arm, rather than identify angular separation exactly with distance; Alberto A. Martínez's replication explicitly analyzes that geometry. ^[ref]

Sensitivity also exposes changes of the apparatus itself. Fleischer and colleagues' cryogenic torsion balance used a temperature-controlled enclosure to reduce vacuum-vessel top-plate temperature fluctuations to approximately $\pm 0.01\mathrm{K}$, limiting changes in suspension geometry from thermal expansion. They used a mild bake-out near $40^{\circ }\mathrm{C}$ to speed initial fiber relaxation and reduce slow zero motion. Their optical readout compared the pendulum mirror with an internal reference mirror to reject shared optical motion. Room-temperature thermal radiation entering through the optical opening heated the pendulum; a cold filter reduced that load. The authors could not directly measure the pendulum's temperature without disturbing it, and explicitly distinguished the cold shield's temperature from the body's. Cooling therefore requires checking the actual noise and thermal environment, rather than assuming that the fiber and body share the shield's temperature. ^[ref]

A useful control strategy is to record unloaded readings and temperature before and after each signal sequence, recalibrate the torsion constant under the same conditions, and compare forward and reverse settings. The algebra of torsion balance zero drift shows precisely which offsets such comparisons can cancel. A fine angular scale alone does not establish a small measurement error.

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.