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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.

Average power divides the work over a nonzero interval by the elapsed time. For the same $2\mathrm{kg}$ point mass driven from rest through $4\mathrm{m}$ by a net $3\mathrm{N}$ force, the displacement is $x(t)=a{t}^2/2$. ThusThis average power is half the final instantaneous power because $P(t)$ is a linear function of time, starting at zero. Its graph of a function bounds a triangle of area $P(t_f)t_f/2=12\mathrm{J}$; the horizontal line at average power encloses a rectangle with the same area. In general, an endpoint value of instantaneous power does not determine average power. ^[ref]

Consider a stipulated ideal cart modeled as a point mass of mass $2\mathrm{kg}$ on a horizontal track in a laboratory inertial frame. Its initial position is $x(0)=0$ and its initial velocity is $v_x(0)=1.5m/s$ to the right. Independent devices exert horizontal forces $+6\mathrm{N}$ and $-6\mathrm{N}$ on this same cart. Its upward support force balances its downward gravitational weight; no other forces contribute. Rotation and deformation are excluded from this model.After $4\mathrm{s}$, the cart is at $6\mathrm{m}$, still moving right at $1.5m/s$. Starting the same model at rest instead would leave it at rest. Force balance determines the zero acceleration, while the initial velocity determines whether the cart moves.

For the plotted model, both horizontal devices are switched off simultaneously at $t=2\mathrm{s}$, with no impulse. The support and gravitational weight remain balanced. The individual horizontal forces change, but their sum stays zero on both sides of the switch, so the velocity stays continuous and unchanged. Switching off just one device would be a different model with a nonzero net external force.

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.

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.

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]

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]

For a cold equatorial axisymmetric magnetohydrodynamic wind with $B_p=C{R}^{-2}$ in the gravity of a point mass $M$, smooth crossing of the Alfvén surface gives ${\rho }_a={\mu }_0{k}^2$ and $\ell =\omega R_a^2$. Put $x=R/R_a$, $y=\rho /{\rho }_a$, $\alpha ={\omega }^2{R}_a^3/(GM)$ and $\beta =C^2/({\mu }_0{\rho }_{a}GM{R}_a^3)$. The corotating energy, scaled by $GM/R_a$, isThe azimuthal velocity relative to field-line rotation is $\omega R_a(x-x^{-1})/(y-1)$ and the poloidal speed is $C/(\sqrt{{\mu }_0{\rho }_a}{R}_a^2{x}^{2}y)$. The apparent singularity at $x=y=1$ is interpreted through a smooth limiting slope. A nondegenerate magnetosonic crossing on a constant-$f$ curve requires $f_x=f_y=0$ and a real admissible tangent slope.

A conical pendulum is a pendulum whose bob executes uniform horizontal circular motion beneath a fixed pivot, with its string at a constant angle $\theta$ to the downward vertical. The string sweeps a right circular conical surface; the bob follows its circular rim, without any physical cone supporting it. The ideal model has a point mass $m>0$, a taut, massless, inextensible string of length $\ell >0$, uniform gravitational acceleration of magnitude $g>0$, and negligible drag force and pivot friction. A nonzero circular motion requires $0<\theta <\pi /2$. This geometry and the corresponding forces are described in Richard Fitzpatrick, The conical pendulum, equations 262–267.

The right triangle formed by the pivot, the orbit centre, and the bob gives $r=\ell \sin \theta$ and $h=\ell \cos \theta$. Here $r$ is the orbit radius and $h$ is the vertical drop from the pivot to the bob's horizontal plane, not the bob's height above a table. Write $T$ for tension, $P$ for the oscillation period of one revolution, and $\omega =2\pi /P$ for the angular speed; then the bob's speed is $v=\omega r$. A side view compresses the horizontal circle to a line segment; the top view reveals the circle and its tangent velocity.

The two actual forces on the bob are tension toward the pivot and gravitational weight vertically downward. The vertical acceleration is zero, while the horizontal centripetal acceleration is $v^2/r={\omega }^{2}r$. Thus Newton's second law givesThe inward net force is the horizontal component of tension, rather than an extra interaction to add to these two forces. Constant speed still requires acceleration because the velocity changes direction. ^[ref]

Dividing the two component equations and using $r=\ell \sin \theta$ gives the useful collection of predictionsIf $\varphi$ is the azimuthal angle, then $\dot{\varphi }=\pm \omega$; either direction of rotation satisfies ${\dot{\varphi }}^2\cos \theta =g/\ell$. The mass cancels from the angular speed and oscillation period, but not from the tension. The motion is a relative equilibrium: in a rotating frame following the bob at angular speed $\omega$, its position is fixed. This is a concrete way to see how rotational symmetry can turn an apparently moving solution into a stationary one.

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 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.

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.

A double pendulum consists of two pendula coupled because the second pivot moves with the first mass. It has two degrees of freedom: even when the upper pivot is fixed, moving the upper arm carries the lower pivot along. This makes the double pendulum a compact example of how mechanical constraints couple motion, and how Lagrangian mechanics converts geometry into dynamics.

Use two equal point masses $m>0$ and two massless rigid rods of equal length $\ell >0$, with joints that allow free rotation in one vertical plane. The upper pivot is fixed; gravity is uniform, with gravitational acceleration $g>0$. Neglect joint friction and drag force. Both angles ${\theta }_1,{\theta }_2$ are measured from the downward vertical, positive toward the right, in radians. In particular, ${\theta }_2$ is an absolute angle, not the bend between the rods: that bend is ${\theta }_2-{\theta }_1$.

The double pendulum Lagrangian below gives the exact dynamics of this ideal model. Near the downward stable equilibrium, normal modes of a double pendulum replace the coupled motion with two independent harmonic oscillators. Their fixed amplitude ratios explain which releases make the rods swing together and which make them swing in opposite directions.

The University of Toronto's institutional demonstration “Chaotic Behaviour (Double Pendulum)”, activity Me-020, uses a metal stand with four leveling feet and a pendulum arm carrying a white dot. Preparation includes leveling the stand and correcting any wobble. The instructions call for holding the arm vertically upward and releasing it; the readout is the visible motion of the dot. A further comparison inserts a metal dowel in a groove on the release arm and repeats the release at the same location. Keeping the stand steady and repeating the release position provide practical controls. The page supplies no numerical angle trace or calibrated timing, and does not explain the dowel's mechanical effect. It therefore documents a qualitative release demonstration, without quantitatively testing the equal-point mass model. Its extended arms also have moments of inertia absent from massless rods. Irregular visible motion alone does not establish sensitive dependence on initial conditions.^[ref]

A separate comparison with normal modes of a double pendulum would require small downward releases and measurement of both angles over time. The actual arm masses, centres of mass and moments of inertia would first have to enter the Lagrangian. The upward release in this demonstration explores a different regime from small oscillations about the downward stable equilibrium.

A normal impulse $j$ on a free planar rigid body of mass $M$ and moment of inertia $I$, at perpendicular lever arm $b$, changes the contact point's normal velocity by $j(1/M+b^2/I)$. Its effective mass in that direction is therefore $(1/M+b^2/I{)}^{-1}$. This follows by adding the translational change $j/M$ to the rotational contribution $b(jb/I)$.

For an incoming point mass $m$, the inverse reduced contact mass is $1/m+1/M+b^2/I$. A smooth elastic collision with initial relative approach speed $u$ has impulse $2u/(1/m+1/M+b^2/I)$, since the relative normal velocity reverses. This separates effective inertia at a contact from the body's total mass and explains the lever-arm dependence in an equal-speed elastic impact on a free rod.

A point mass $m$ strikes a free rod of equal mass, initially at rest, perpendicularly at distance $h$ from its center of mass. Assume a smooth elastic collision and rod moment of inertia $I$ about its centre. The final translational speeds are equal precisely when $h^2=2I/m$. Indeed, momentum conservation then fixes both final velocities to $v_0/2$; angular impulse gives $I\omega =m{v}_{0}h/2$, and conservation of kinetic energy gives $I{\omega }^2=m{v}_0^2/2$. Eliminating $\omega$ proves the formula. A uniform rod has $I=m{l}^2/12$, hence $|h|=l/\sqrt{6}$, which lies within the rod.

The escape velocity is the least initial speed that permits a freely coasting body to move arbitrarily far from an isolated attracting body. Despite its conventional name, it is a speed, not a unique velocity vector. For a fixed point mass $M$, or outside a body with spherical symmetry, set Newtonian gravitational potential energy zero at infinity. An exterior test particle of constant mass $m\ll M$ then hasHere $\mu =GM$ is the standard gravitational parameter, and $r$ is distance from the centre, not height above the surface. Set $r=R$ for surface escape velocity. Evaluate velocity in the body's rest inertial frame. Neglect drag, later rocket thrust, other bodies' gravity, and relativistic effects.^[ref]

Conservation of energy gives the threshold by requiring $E\ge 0$: at infinity the potential energy tends to zero and the kinetic energy cannot be negative. If $E=0$, the outward motion loses speed continuously toward zero. If $E>0$, the asymptotic speed is $v_{\infty }=\sqrt{2E/m}=\sqrt{v_0^2-v_{\mathrm{esc}}(r_0{)}^2}$. If $E<0$, infinity is inaccessible. For a radial outward launch, its outer turning point is $r_{\max }=-GMm/E$. For nonradial motion use angular momentum and the escape energy threshold to distinguish zero radial velocity from zero total speed.

Direction still determines the orbit and whether the body hits an obstacle. An initially inward velocity can cause collision even when $E\ge 0$. For an isolated spherical body, an exterior outward launch, or a tangential launch at or above escape velocity, has an outward escape branch without intersecting the body. A parabolic trajectory is the nonradial threshold path; a hyperbolic Kepler orbit has positive energy. The formula specifies their required local speed, not their orientation.^[ref]

Stipulate a fixed source point mass $M=10^{12}\mathrm{kg}$, two equal test particles of mass $m=1\mathrm{kg}$, and $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. This is a chosen model, not a measurement of a particular object. Place the test particles on the outward radial axis at $R-L/2$ and $R+L/2$, with midpoint distance $R=100\mathrm{m}$ and separation $L=2\mathrm{m}$. Neglect their mutual attraction, other sources and all supporting forces.

Each signed radial gravitational acceleration is inward:Their separation has outward relative accelerationThe exact value is $2.6702540\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-2}$. The midpoint tidal tensor predicts $2\mu L/R^3=2.6697200\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-2}$, an underestimate of $0.019999\%$ relative to the exact value. For these equal $1\mathrm{kg}$ masses, the difference of external forces has the same numerical value in the SI unit of force, the newton. It is not an additional mutual force between the test particles.

The exact formula requires $L<2R$ and describes the instantaneous relative acceleration. Because the two sample positions are symmetric about the midpoint, the Taylor expansion error relative to the linear result starts at order $(L/R{)}^2$; a one-sided comparison with a reference point generally has relative error of order $L/R$. If either test particle enters an extended source, use that source's interior Newtonian gravitational field instead of extrapolating the singular point mass model.

A homogeneous spherical bob rigidly fixed to a massless support has centre-of-mass distance $L$ from its pivot and intrinsic moment of inertia $2M{a}^2/5$. The parallel axis theorem and linear gravitational restoring torque give the title period for small oscillations. Relative to a point mass at the same centre, the leading fractional period increase is $a^2/(5L^2)$. If $L$ is instead measured to the sphere's surface, the centre distance must first be changed to $L+a$.

A smooth central collision leaves a cylinder's angular velocity unchanged if the floor supplies negligible tangential impulse. Imposing rolling without slipping during the impact instead requires a floor impulse and changes the cylinder's effective translational mass to $m+I/r^2$. Consequently a pre-impact rolling condition alone does not determine the outgoing velocities: the impulsive contact law must also be specified. For a uniform solid cylinder striking a stationary point mass $2m$ in an elastic collision, the target acquires $2u/3$ with negligible floor impulse, and $6u/7$ with an enforced rolling constraint, where $u$ is the incoming centre speed.