Classical mechanics

A mechanism that trades the magnitude or direction of an applied force against the displacement through which it acts. An ideal simple machine conserves work; it does not multiply energy.

Lever

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.

Third-class lever

Second-class lever

First-class lever

Fulcrum

The support or pivot about which a lever turns.

Mechanical advantage ( $MA = F_{load} / F_{effort}$ )

The ratio of the useful load force to the applied effort force. For an ideal simple machine in slow motion, the ratio of effort displacement to load displacement gives the same number by conservation of work.

Mechanical system

A collection of bodies and specified interactions whose configuration, velocities and evolution are studied in mechanics. Generalized coordinates describe its allowed configurations, while constraints restrict them. A model must specify which bodies and surroundings are included before applying conservation of momentum or conservation of energy.

Pulley

A wheel guiding a rope or cable so that it can redirect a tension force. An ideal fixed pulley changes the force direction without multiplying its magnitude. A freely rotating pulley with negligible moment of inertia and friction has equal rope tensions on its two sides; a finite moment of inertia instead relates their difference to its angular acceleration through the net torque.

Bead on a uniformly rotating wire ( $(1 + f^{'2}) \overset{r}{¨} = ω^{2} r - g f^{'} - f^{'} f^{''} \overset{r}{˙}^{2}$ )

A smooth wire has instantaneous shape

z = f (r)

in cylindrical coordinates and rotates at constant angular velocity

\dot{θ} = ω

. The tangent vector at fixed time is

e_{r} + f^{'} (r) e_{z}

. Its dot product with the constraint force vanishes. Projecting Newton's second law along this tangent vector gives the equation in the title.

Multiplying that equation by

\overset{r}{˙}

proves conservation of

H = \frac{1}{2} (1 + f^{'} (r)^{2}) \overset{r}{˙}^{2} + U (r), U (r) = g f (r) - \frac{1}{2} ω^{2} r^{2} .

This is a conserved quantity in the rotating description; the motor can exchange ordinary mechanical energy with the bead. A relative equilibrium at

a > 0

satisfies

U^{'} (a) = g f^{'} (a) - ω^{2} a = 0

. Its linearized equation is

(1 + f^{'} (a)^{2}) \overset{ϵ}{¨} = - U^{''} (a) ϵ

. A strict local minimum with

U^{''} (a) > 0

gives Lyapunov stability, while

U^{''} (a) < 0

gives an unstable equilibrium. If

U^{''} (a) = 0

, higher orders are needed.

Deflection angle ( $δ$ )

For a trajectory with incoming and outgoing straight asymptotes, the deflection angle is the angle between its limiting incoming and outgoing velocity directions. For an attractive hyperbolic orbit of eccentricity

e

and impact parameter

d

sin (δ /2) = 1/ e

and

tan (δ /2) = GM / (d v^{2})

, where

v

is the incoming speed at infinity. The incoming velocity points opposite to the incoming radial direction; ignoring that reversal gives the wrong angle.

Kinematics

The description of motion through position, velocity and acceleration, before specifying the forces responsible. For constant acceleration these are related by integrating time derivatives. Projectile motion under an approximately uniform gravitational acceleration is an elementary example.

Relative motion

The position and velocity of one body relative to another are the differences of their corresponding vectors in the same reference frame. Constant relative velocity gives an affine function of time, even when neither body is at rest.

Closest approach of two uniformly moving particles ( $t_{*} = - d_{0} \cdot v /∣ v ∣^{2}$ )

If the relative position is

d (t) = d_{0} + t v

with

v \neq = 0

, completing the square gives

∣ d (t) ∣^{2} = ∣ v ∣^{2} (t - t_{*})^{2} + ∣ d_{0} ∣^{2} - \frac{( d _{0} \cdot v ) ^{2}}{∣ v ∣ ^{2}}, t_{*} = - \frac{d _{0} \cdot v}{∣ v ∣ ^{2}} .

This proves the minimum distance over all real times. If time is restricted to an interval, project

t_{*}

onto that interval; if

v = 0

, the distance is constant. Both particles are evaluated at the same time, unlike the two independent parameters in the distance between skew lines.

Motion

Change of position with time relative to a specified reference system.

Near-surface recoil height with constant downward thrust

With upward initial speed

v

and a constant downward force

F

on remaining mass

M

, the near-surface excursion is

H = v^{2} / [2 (g + F / M)]

. A spherical body's surface gravity can be inferred from its escape speed:

g = v_{esc}^{2} / (2 R)

. The approximation needs

H ≪ R

and no additional tether constraint.

Contact loss on a smooth sphere

A particle released from an infinitesimal displacement near the top of a smooth sphere has

v^{2} = 2 g R (1 - cos θ)

. The normal reaction vanishes when

m g cos θ = m v^{2} / R

, giving

cos θ = 2/3

. The travelled arc is

R arccos (2/3)

, whereas the vertical drop is

R /3

Coulomb orbit

A classical two-body trajectory in an inverse-distance Coulomb potential. With an effectively fixed centre and coupling

C

, a circular orbit has

v^{2} = C / (m r)

and mechanical energy

- C / (2 r)

Coulomb head-on closest approach with target recoil ( $r_{m i n} = k_{e} q_{1} q_{2} (m_{1} + m_{2}) / (m_{2} K_{lab})$ )

A projectile of mass

m_{1}

, electric charge

q_{1}

, and nonrelativistic laboratory kinetic energy

K_{lab}

approaches an initially stationary free target of mass

m_{2}

and like-sign electric charge

q_{2}

. The relative-motion kinetic energy is

K_{*} = m_{2} K_{lab} / (m_{1} + m_{2})

. At a head-on turning point it is converted to the repulsive Coulomb potential

k_{e} q_{1} q_{2} / r_{m i n}

, proving the displayed separation. The remaining laboratory kinetic energy is the translation of the centre of mass; it cannot be converted into the pair's separation energy. An externally fixed target is the infinite-target-mass limit.

Classical radiation collapse of a Coulomb orbit

An accelerating classical charge radiates according to the Larmor formula. For a slowly evolving nonrelativistic circular electron orbit, energy balance gives

d (r^{3}) / d t = - 4 c r_{0}^{2}

, with

r_{0}

the classical electron radius. The finite formal collapse time exposes classical atomic instability; it is not a model of a stable quantum atom.

Projectile motion

Motion after launch under a specified gravitational field, with negligible drag force and no thrust. In a constant downward gravitational acceleration

g

, horizontal velocity is constant and vertical position obeys

y (t) = y_{0} + v_{y 0} t - g t^{2} /2

. Eliminating time gives a parabola when horizontal velocity is nonzero. Over larger distances the changing direction and magnitude of gravity require an orbital model instead.

Projectile

An object launched into motion whose subsequent trajectory is determined by external forces. The ideal elementary projectile motion model retains uniform gravity and neglects air resistance.

Projectile range ( $R$ )

The horizontal displacement between launch and landing in a projectile motion model. With equal launch and landing heights, uniform gravitational acceleration

g

, no air resistance, launch speed

u

and angle

θ

, it is

R = u^{2} sin (2 θ) / g

. Different landing heights or forces change the result.

Maximum projectile range from an elevated launch ( $X_{m a x} = u u^{2} + 2 g h / g$ )

Under uniform gravity without drag, a projectile launched at speed

u

from height

h

reaches a maximum horizontal projectile range

u u^{2} + 2 g h / g

. For horizontal position

X

, maximize

y = h + X z - g X^{2} (1 + z^{2}) / (2 u^{2})

over

z = tan α

. The maximizer is

z = u^{2} / (g X)

and its envelope meets the ground at the displayed distance. The optimum angle obeys

tan α = u / u^{2} + 2 g h

Projectile range under a ceiling ( $R_{m a x} = 8 H (u^{2} / g - 2 H) if H \leq u^{2} / (4 g)$ )

A projectile launched from and returning to the floor beneath a horizontal ceiling of height

H

has maximum projectile range

8 H (u^{2} / g - 2 H)

when

0 \leq H \leq u^{2} / (4 g)

. For higher ceilings the maximum is

u^{2} / g

. This assumes uniform gravitational acceleration, no air resistance and fixed launch speed

u

Indeed, with

z = sin^{2} θ

, the apex condition is

z \leq 2 g H / u^{2}

and

R^{2} = 4 u^{4} z (1 - z) / g^{2}

. This concave quadratic function is maximized at the admissible value nearest

1/2

. Thus the lower-ceiling expression must not be extrapolated past its regime. If touching the ceiling is excluded, the limiting value is a supremum instead of an attained maximum.

Apex-target condition for a ballistic jet ( $u_{y} = 2 g h, u_{x} = w g / (2 h)$ )

For a jet arriving horizontally at a target displaced horizontally by w and upward by h, the target lies at the projectile apex. Vertical kinematics gives

u_{y}^{2} = 2 g h

and flight time

u_{y} / g

; the horizontal travel then fixes the displayed

u_{x}

. The launch speed is

u_{x}^{2} + u_{y}^{2}

and

tan θ = 2 h / w

. The ideal ballistic approximation neglects air drag, breakup and internal jet stresses after launch.

Jet-area adjustment during ballistic slowing ( $A_{hit} v_{hit} = A_{0} u_{0}$ )

Steady incompressible mass flow rate stays constant along a jet, while speed and cross-sectional area can change under gravity. At a horizontal apex the arriving speed is the initial horizontal component. The force on a target that stops this horizontal momentum is

\overset{m}{˙} v_{hit}

, using the conserved nozzle mass flux. Substituting the nozzle area into

ρ A_{hit} v_{hit}^{2}

without adjusting the area gives the wrong result.

Ballistic orbit insertion with constant gravity ( $h_{b} = h_{o} - v_{o}^{2} tan^{2} θ / (2 g)$ )

In the flat-surface, constant-gravitational acceleration model, a coasting projectile conserves horizontal velocity. If it arrives at the top of its trajectory with horizontal speed

v_{o}

, its earlier velocity at angle

θ

above the horizontal has vertical component

v_{o} tan θ

. Its remaining height gain is therefore

v_{o}^{2} tan^{2} θ / (2 g)

. The displayed burnout height follows. This is a local projectile motion approximation, not an exact Kepler orbit. Computing

v_{o} = g (R_{E} + h_{o})

assumes constant gravitational magnitude also at the orbit; computing

v_{o} = g_{0} R_{E}^{2} / (R_{E} + h_{o})

instead assumes an inverse-square law with surface value

g_{0}

. Those assumptions must be distinguished.

Mechanical impedance ( $Z = F / v$ )

Mechanical impedance is the ratio of a driving force complex amplitude to the resulting velocity complex amplitude at a specified point and in a specified direction. Its units are

N s m^{- 1}

. The direction of the force, the time convention and whether one is describing a travelling component or a total field must be stated. A positive real travelling-wave mechanical impedance transfers mean power

⟨ P ⟩ = Z ∣ v ∣^{2} /2

when

∣ v ∣

is the peak velocity wave amplitude.

Displacement coefficients at a string junction ( $r = (c_{2} - c_{1}) / (c_{2} + c_{1}), t = 2 c_{2} / (c_{1} + c_{2})$ )

For transverse waves on two strings at common tension, displacement continuity gives

1 + r = t

, while transverse-force continuity gives

(1 - r) / c_{1} = t / c_{2}

. Solving gives the displayed displacement-amplitude coefficients. A denser transmitted string has lower speed and a negative reflected coefficient. The energy fractions are

r^{2}

and

(c_{1} / c_{2}) t^{2}

, summing to1. These are displacement coefficients; force or power-normalized wave amplitudes use different conventions.

String-pulse width mapping at a junction ( $W_{t} / W_{i} = c_{2} / c_{1}$ )

At an ideal nondispersive string junction, every temporal feature is scattered with the same constant coefficients. The reflected spatial width equals the incident width, while the transmitted width changes in proportion to wave speed. A reflected asymmetric spatial profile is reversed left-to-right as well as receiving the reflection sign. For a denser second string the transmitted amplitude is smaller but its spatial slopes can be steeper, because its width is compressed.

Trajectory ( $r (t)$ )

Vibration isolation

Vibration isolation reduces the transmission of unwanted mechanical oscillations between an apparatus and its surroundings. Compliant supports and damping can reduce transmitted motion, but their effectiveness depends on the disturbance frequency and the support's resonance.

Inclined plane

An inclined plane is a flat surface tilted through an angle

θ

above horizontal. A ramp exchanges a smaller lifting force for a longer travel distance; for a released body, it reduces the downhill component of gravitational weight and makes the resulting acceleration easier to measure. Its simplest dynamical model is a nonrotating block on a stationary, rigid surface, with constant mass

m > 0

, uniform gravitational acceleration of magnitude

g > 0

, and

0 \leq θ < π /2

. Assume continued contact, negligible drag force, no tipping, and no applied pull. The only forces are gravitational weight, the normal force, and any tangential friction.

The central distinction is between a block resting on the inclined plane and one already sliding. Static friction supplies the force needed to maintain rest, up to its bound; kinetic friction has a prescribed magnitude in the ideal constant-coefficient model. These laws are empirical approximations whose range depends on the contacting surfaces.^[ref]

Gravity components on an inclined plane ( $W_{∥} = m g sin θ, N = m g cos θ$ )

Choose a unit vector

e_{d}

pointing downhill and a perpendicular unit vector

e_{n}

pointing outward from the inclined plane. The gravitational weight points vertically downward. Its scalar projections onto these directions give

W = m g sin θ e_{d} - m g cos θ e_{n} .

The angle between the downward vertical and the inward normal is

θ

: the inward component uses cosine, and the downhill component uses sine. These are two components of one force. They replace its vertical arrow in a component calculation; adding them to that same arrow would count gravitational weight twice.

Figure 1.
Actual contact forces and a separate decomposition of gravitational weight on an inclined plane
. Original geometry for a block sliding downhill at $3 0^{\circ}$ . Left: its free-body diagram shows gravitational weight, normal force, and uphill kinetic friction. Right: the gravitational weight is resolved into downhill and inward-normal components; the dashed arrow is their sum. Arrow lengths are illustrative. **Details**.

Continued contact with a stationary straight inclined plane gives zero perpendicular acceleration. The perpendicular component of Newton's second law therefore gives

N - m g cos θ = 0, N = m g cos θ .

This is why the normal force generally differs from

m g

: it balances the inward component, rather than the whole gravitational weight. An extra lifting force, a curved surface, or motion of the support would require a new perpendicular balance.^[ref]

The same geometry explains the ramp's usefulness. A frictionless block pulled uphill at constant velocity needs a parallel force

P = m g sin θ

. Travelling a distance

L

raises it through height

h = L sin θ

, so the pulling work is

P L = m g h

, its gain in gravitational potential energy. The smaller force costs a larger displacement; the work-energy theorem keeps the accounting consistent.

Sliding acceleration on an inclined plane ( $a = g (sin θ - μ_{k} cos θ)$ )

First test rest. With no other tangential forces, the required uphill static friction is

f_{s} = m g sin θ

. Its maximum is

μ_{s} N

, where

μ_{s}

is the coefficient of static friction. Thus

m g sin θ \leq μ_{s} m g cos θ ⟺ tan θ \leq μ_{s}, θ_{c} = arctan μ_{s} .

The tangent function is increasing on this angle range, and its inverse tangent determines the last permitted resting angle

θ_{c}

. At equality, the ideal model still permits limiting rest; a larger angle excludes it. Below the threshold, setting

f_{s} = μ_{s} N

would incorrectly impose the maximum static friction even when less is required. Static friction on an inclined plane and inclined-plane measurement of static friction develop this onset criterion further.

For a block already sliding downhill, uphill kinetic friction has magnitude

f_{k} = μ_{k} N

, where

μ_{k}

is the coefficient of kinetic friction. Taking downhill as positive, Newton's second law becomes

ma = m g sin θ - μ_{k} m g cos θ, a = g (sin θ - μ_{k} cos θ) .

Every term in the force balance scales with mass; division by

m > 0

removes it. Doubling mass doubles both gravitational weight and the contact forces, while leaving this acceleration unchanged if the coefficients remain unchanged. A mass-independent result is a prediction of these assumptions, rather than a universal property of every ramp.

For an original numerical model, stipulate

m = 3 kg

g = 9.8 m/ s^{2}

μ_{s} = 0.45

and

μ_{k} = 0.30

. The critical angle is

θ_{c} ≃ 24.2 3^{\circ}

. At

2 0^{\circ}

, the normal force is

27.627 N

; rest requires

10.055 N

of uphill static friction, below its

12.432 N

bound. A resting block can stay there. If it is already sliding downhill, however, the kinetic friction is only

8.288 N

, giving downhill acceleration

a ≃ 0.5891 m/ s^{2}

. Specifying the initial state matters.

3 0^{\circ}

, the downhill component of gravitational weight is

14.700 N

, while

N ≃ 25.461 N

and the static friction bound is only

11.458 N

. Rest is impossible. During downhill sliding,

f_{k} ≃ 7.638 N, F_{net} ≃ 7.062 N, a ≃ 2.3539 m/ s^{2} .

Released from rest on a sufficiently long ramp, this ideal block covers its first

1.00 m

t = 2 L / a ≃ 0.9218 s

, reaching speed

v = 2 a L ≃ 2.1697 m/s

. As an independent energy calculation, the lost gravitational potential energy is

m gL sin θ = 14.700 J

and kinetic friction removes

f_{k} L ≃ 7.638 J

of mechanical energy. The remaining

7.062 J

equals

m v^{2} /2

by the work-energy theorem.

Figure 1.
Rest threshold and downhill sliding acceleration for the stipulated inclined-plane model
. Original model predictions with $g = 9.8 m/ s^{2}$ , $μ_{s} = 0.45$ and $μ_{k} = 0.30$ ; no experimental measurements. The shaded range permits rest. The blue curve applies only while downhill contact slip persists, including inside that range. The orange curve gives the frictionless acceleration. Markers show the $2 0^{\circ}$ and $3 0^{\circ}$ examples. **Details**.

θ = arctan μ_{k}

, downhill sliding has zero acceleration and therefore constant velocity. Below that angle, a downhill-moving block slows; when it stops, test the static friction bound again. For uphill sliding, kinetic friction points downhill, so the signed downhill acceleration instead is

g (sin θ + μ_{k} cos θ)

. The sign follows contact slip, and the downhill formula cannot be used unchanged after a reversal.

Photogate measurement of acceleration down an incline ( $a = d v / d t$ )

A modern teaching experiment slows gravity-driven motion enough to resolve it with a photogate. PASCO scientific's Inclined Plane Dynamics Set, ME-6966, Instruction Manual 012-10874A describes “Acceleration Down an Incline” on printed pages 15–16. Its apparatus is a GOcar on a PAStrack with an adjustable incline support, a bracket-mounted photogate, a cart-mounted picket fence, and an interface with recording software. For the interface method, the fence's

1 cm

pattern is aligned to interrupt the beam. Recording starts before release from rest; the cart is caught at the bottom bumper. The procedure specifies three runs per angle, starting at

1 4^{\circ}

and reducing by

2^{\circ}

, with the release position fixed. The readout is velocity against time; its slope estimates acceleration. Mean accelerations are plotted against

sin θ

, testing the ideal slope

g

. An added-mass comparison is also requested. These are protocol settings, not observed results.^[ref]

The cart undergoes rolling without slipping at its wheels, whereas the worked block undergoes contact slip. Even ideal wheel moment of inertia changes the cart prediction. For total cart mass

M

, wheel radii

r_{i}

and axial moments of inertia

I_{i}

, rigid wheels with fixed axles and no losses give

a = \frac{g s i n θ}{1 + i \sum I _{i} / ( M r _{i}^{2} )} .

To derive it, combine translational kinetic energy

M v^{2} /2

with wheel rotational kinetic energy

\sum_{i} I_{i} (v / r_{i})^{2} /2

, then equate their rate of change to the release of gravitational potential energy

M g sin θ v

. This is the multiwheel counterpart of rolling acceleration with rotational inertia. Neglecting wheel moment of inertia yields

a ≃ g sin θ

; treating a cart as a solid rolling cylinder would give a different correction.

Useful additional controls are repeated releases without a push, checking the fence alignment and spacing, and checking that the recorded velocity trace is straight over the fitted interval. A gently moving cart on a level track can expose losses or unintended tilt. These checks help distinguish an angle or timing error from neglected wheel moment of inertia, bearing friction, or drag force. The three-run spread measures repeatability but cannot by itself reveal a shared calibration error. The inspected protocol makes the gravity component measurable; it does not measure the sliding block's coefficient of kinetic friction.

Variable-mass system ( $m \dot{v} = F + (u - v) \overset{m}{˙}$ )

A body exchanging mass with material of inertial velocity

u

satisfies the displayed momentum conservation law for one stream, with signed

\overset{m}{˙}

positive for entry and negative for exit. The force

F

excludes the incoming or outgoing momentum flux. A short time interval gives

d (m v) / d t = F + u \overset{m}{˙}

, proving the formula. For stationary incoming matter,

d (m v) / d t = F

; for matter leaving at the body's own velocity,

m \dot{v} = F

. The rocket equation and accreting-drop motion are applications.

Constant-area capture of stationary dust ( $M^{2} (t) = M_{0}^{2} + 2 ρ A M_{0} v_{0} t$ )

A body of initial mass

M_{0}

and speed

v_{0}

collects all stationary dust swept out by constant area

A

in a uniform cloud of mass density

ρ

. Then

M = M_{0} + ρ A x

and

M v = M_{0} v_{0}

by conservation of momentum. Integration gives the displayed mass law,

v = M_{0} v_{0} / M

and

x = (M - M_{0}) / (ρ A)

. Sticking dissipates kinetic energy at rate

\dot{M} v^{2} /2

. The model assumes no external longitudinal force and negligible relativistic corrections.

Mass-loading power at fixed velocity

A body moving at constant velocity

V

incorporates material at rate

\dot{M}

with zero initial momentum along its motion. The required external force is

F = \dot{M} V

, so the supplied power is

P = \dot{M} V^{2}

. Only

\dot{M} V^{2} /2

becomes the added material's translational kinetic energy; the other half is dissipated during inelastic incorporation. This conclusion concerns the component along the body's motion; an incoming transverse velocity can add further dissipation.

Snow-collecting sled

A sled intercepting snow with zero horizontal velocity in the ground frame is a simple momentum flux example. At arrival rate

σ

, immediate removal with outgoing ground-frame horizontal velocity

v_{out}

gives

M \overset{v}{˙} = - σ v_{out}

for constant moving mass

M

. Removal sideways in the sled frame has

v_{out} = v

and exponential deceleration; removal sideways in the ground frame has

v_{out} = 0

and constant sled velocity. Retained snow instead gives

(M + σ t) v = M v_{0}

. The outgoing reference frame must be specified.

Leaking bucket under constant tension ( $K = T x ln (L / x)$ )

Let a massless bucket approach a wall under constant tension

T > 0

, with remaining distance

x

and remaining sand mass

m = M x / L

. Assume zero-relative-velocity mass shedding horizontally. Then

m \overset{x}{¨} = - T

and, from rest at

x = L

, the speed satisfies

u^{2} = (2 T L / M) ln (L / x)

. Its kinetic energy is

K = T x ln (L / x)

and its momentum magnitude is

p = x (2 MT / L) ln (L / x)

. Differentiating gives

x_{K} = L / e

and

x_{p} = L / e

: the kinetic energy peaks later, closer to the wall, because its dependence on the increasing speed is stronger. Both tend to zero as the remaining mass vanishes, despite divergent ideal speed.

Tangential-exhaust rotor with a fixed dry inertia ( $Δ ω = (v_{e} / a) lo g (I_{i} / I_{f})$ )

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

aω - v_{e}

. Conserving angular momentum of rotor plus newly emitted fuel during

d M_{f} < 0

gives

I d ω + ω d I - d M_{f} a (aω - v_{e}) = 0, I d ω = - a v_{e} d M_{f} .

Since

d I = a^{2} d M_{f}

, integration yields

ω_{f} - ω_{i} = (v_{e} / a) lo g (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} = (4 m + 3 m_{f}) a^{2}

and

I_{f} = 4 m a^{2}

Surface-fed growth of an accreting sphere ( $r (t) = r_{0} + (β / ρ) t$ )

For a sphere with constant mass density

ρ

collecting mass at rate

β

per unit surface area,

\overset{m}{˙} = 4 π r^{2} β

and

m = 4 π ρ r^{3} /3

give

\overset{r}{˙} = β / ρ

. If it falls through stationary material under constant gravitational acceleration

g

without drag, its velocity satisfies

\overset{v}{˙} + 3 \overset{r}{˙} v / r = g

. With

v (0) = 0

, an integrating factor

r^{3}

gives

v = g [r - r_{0}^{4} / r^{3}] / (4 \overset{r}{˙})

Quadratic drag on a sphere with constant radial growth ( $v (t) \sim 8 ρ gμ t / (3 C_{D} ρ_{a})$ )

For surface-fed growth of an accreting sphere with

μ = \overset{r}{˙} > 0

, stationary incoming material, and quadratic drag

F_{D} = C_{D} ρ_{a} π r^{2} v^{2} /2

, the variable-mass system obeys

\overset{v}{˙} + 3 μv / r = g - 3 C_{D} ρ_{a} v^{2} / (8 ρ r)

. For constant positive parameters,

r \sim μ t

and the late balance gives the displayed speed. The omitted acceleration and accretion terms are both

O (t^{- 1/2})

, consistent with the balance. Thus this growing sphere has unbounded speed with vanishing acceleration, rather than a finite terminal velocity. This model has a different growth law from the existing accreting raindrop with quadratic drag, whose growth rate is proportional to speed.

Accreting raindrop with quadratic drag ( $\overset{a}{˙} = c v /4, \overset{v}{˙} = g - 3 (c + k) v^{2} / (4 a)$ )

For a spherical drop of constant mass density

ρ

, radius

a > 0

and downward velocity

v > 0

, suppose its volume grows at rate

c π a^{2} v

by collecting stationary mist and its upward quadratic drag is

k ρ π a^{2} v^{2}

, with

c, k > 0

. The drop mass is

4 π ρ a^{3} /3

. Its incoming mass accretion rate and the variable-mass system momentum law give the displayed equations. The accreted matter must be accelerated from rest, producing the additional

c

term alongside drag.

Speed-radius attractor of an accreting raindrop ( $v^{2} / a ⟶ 8 g / (7 c + 6 k), \overset{v}{˙} ⟶ c g / (7 c + 6 k)$ )

For the accreting raindrop with quadratic drag, set

q = v^{2} / a

and

B = 7 c /8 + 3 k /4

. Its exact equation is

\overset{q}{˙} = 2 (v / a) (g - Bq)

. Dividing by

\overset{a}{˙} = c v /4

gives

q (a) = g / B + (q (a_{0}) - g / B) (a_{0} / a)^{7 + 6 k / c}

. Thus

q

stays between its positive initial value and positive equilibrium. Since

d a / d t = c q /8

, the radius grows without finite-time divergence and tends to infinity. The power correction then vanishes, proving both limits. Positive drag makes the limiting acceleration strictly less than

g /7

Inertia

Inertia is a body's persistence in its state of rest or constant velocity when the net external force vanishes in an inertial frame. It is not an extra force that pushes a moving body forward. Newton's first law describes this persistence; for a point mass, Newton's second law quantifies how a nonzero net external force changes its momentum. At constant mass,

F_{net} = m a

, so a larger mass gives a smaller acceleration for the same net force.

A moving cart does not need a forward net force to maintain its velocity. Ordinary carts slow because friction and drag force produce an opposing net external force; reducing those interactions reveals Newton's first law more clearly. The air-track test of inertial motion separates this persistence from the initial push. In overdamped particle dynamics, the inertial acceleration term is small compared with drag force over the time scales considered; that approximation does not mean that inertia has disappeared.

Mechanical configuration space ( $Q$ )

The configuration space of a mechanical system is the space of its allowed positions or generalized coordinates. Its tangent bundle describes configurations and velocities, and its cotangent bundle describes configurations and conjugate momenta. For smooth holonomic constraints it is a manifold on which geometric Lagrangian and Hamiltonian descriptions can be built. Singular constraints or symmetry quotients can require strata rather than one global smooth manifold.

Mechanical configuration ( $q \in Q$ )

A mechanical configuration specifies the positions and orientations of all parts of a system, subject to its constraints. It is a point of the mechanical configuration space

Q

; generalized coordinates label such points locally. A configuration does not by itself specify velocities, so it generally does not determine the system's full dynamical state. For a bead on a fixed circular hoop, its angle specifies its configuration, while its angular velocity is additional data.

Relational shape space

The relational shape space of a particle configuration is obtained by removing collective translations and rotations; here scale is retained. For generic collision-free, noncollinear configurations of

N \geq 3

labelled particles in three dimensions, the resulting configuration dimension is

3 N - 6

. Collinear configurations have nontrivial rotational stabilizers and require singular strata. Removing position and orientation does not automatically remove conserved angular momentum or its centrifugal effect; a dynamically faithful reduction must retain the appropriate momentum-level data.

Simple pendulum ( $\ddot{θ} + (g / ℓ) sin θ = 0$ )

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{θ} + (g / ℓ) sin θ = 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

- m g ℓ sin θ

. Dividing by its moment of inertia

m ℓ^{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

θ_{0}

, the linearized solution is

θ (t) ≃ θ_{0} cos (g / ℓ t)

, with oscillation period

T_{0} = 2 π ℓ / g

. For

g = 9.81 m/ s^{2}

, lengths one and four metres give

T_{0} \approx 2.0061 s

and

4.0121 s

. This same scaling follows from dimensional analysis; the dynamical calculation supplies the coefficient

2 π

Finite angular amplitude changes the exact ideal period. Conservation of energy gives

\dot{θ}^{2} = (2 g / ℓ) (cos θ - cos θ_{0})

for release from rest. Integrating a quarter oscillation and substituting

sin (θ /2) = sin (θ_{0} /2) sin ϕ

gives

T (θ_{0}) = 4 \frac{ℓ}{g} K (sin \frac{θ _{0}}{2}), \frac{T ( θ _{0} )}{T _{0}} = \frac{2}{π} \int_{0}^{π /2} \frac{d ϕ}{1 - s i n ^{2} ( θ _{0} /2 ) s i n ^{2} ϕ}, 0 < θ_{0} < π .

Here

K

is the complete elliptic integral of the first kind. The small-angle approximation gives a ratio tending to one; at

1 0^{\circ}, 6 0^{\circ}, 9 0^{\circ}

, the ratios are approximately

1.00191, 1.07318, 1.18034

. Thus almost constant period at small amplitude has a quantitative limit. As

θ_{0} \to π

, the period diverges, connecting to the pendulum separatrix approach takes infinite time.

Figure 1.
Length scaling and finite-amplitude period corrections for an ideal simple pendulum
. **Details**.

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.

Spherical pendulum ( $L = \frac{1}{2} m r^{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) + m g r cos θ$ )

A point mass constrained to a fixed distance

r

from a pivot and free to move in three dimensions under uniform gravitational acceleration. With the vertical axis downward and polar angle

θ

measured from it, the Lagrangian is the expression in the title. The azimuth

ϕ

is a cyclic coordinate, so

m r^{2} sin^{2} θ \dot{ϕ}

is conserved angular momentum about the vertical axis. A rigid massless rod enforces the constraint without the loss-of-tension limitation of a string.

Small nutation about a conical pendulum ( $ω_{θ}^{2} = \frac{g}{l} \frac{1 + 3 c o s ^{2} θ _{0}}{c o s θ _{0}}$ )

For a spherical pendulum with nonzero conserved vertical angular momentum

h

, eliminate the azimuthal angular velocity using

h = m l^{2} sin^{2} θ \dot{ϕ}

. The polar effective potential is

V_{eff} = h^{2} / (2 m l^{2} sin^{2} θ) - m g l cos θ

. A conical pendulum at

0 < θ_{0} < π /2

is a minimum satisfying

h^{2} = m^{2} g l^{3} sin^{4} θ_{0} / cos θ_{0}

. Linearization at fixed

h

yields polar angular frequency

ω_{θ}^{2} = V_{eff}^{''} (θ_{0}) / (m l^{2}) = Ω^{2} (1 + 3 cos^{2} θ_{0})

, where

Ω^{2} = g / (l cos θ_{0})

is the conical revolution frequency squared. Holding

\dot{ϕ}

fixed during the perturbation would describe a different, externally driven constraint. The polar perturbation is a nutation; the exactly planar

h = 0

limit requires separate treatment.

Exact Lagrangian of two spring-coupled pendulums ( $L = \frac{1}{2} m l^{2} (\dot{θ}_{1}^{2} + \dot{θ}_{2}^{2}) + m g l (cos θ_{1} + cos θ_{2}) - \frac{1}{2} k (d - r)^{2}$ )

Two identical planar simple pendulums of length

l

and bob mass

m

, with pivots separated horizontally by

r > 0

, are coupled by a straight spring of natural length

r

and spring constant

k

. In the ideal model without contact restrictions, their configuration space is the torus

S^{1} \times S^{1}

. Measure both angles from the downward vertical towards the same horizontal direction. Their bob separation is

d = (r + l (sin θ_{2} - sin θ_{1}))^{2} + l^{2} (cos θ_{2} - cos θ_{1})^{2} .

The exact Lagrangian is the expression in the title. To derive it, use bob positions

(l sin θ_{1}, - l cos θ_{1})

and

(r + l sin θ_{2}, - l cos θ_{2})

. Kinetic energy from Cartesian position vectors gives

T = m l^{2} (\dot{θ}_{1}^{2} + \dot{θ}_{2}^{2}) /2

. Gravitational potential energy is

- m g l (cos θ_{1} + cos θ_{2})

and spring potential energy is

k (d - r)^{2} /2

; subtracting these from

T

proves the formula. Near the downward equilibrium,

d - r = l (θ_{2} - θ_{1}) + O (∣ θ ∣^{3})

for fixed

r

, giving the familiar normal modes of two spring-coupled equal pendulums.

Pendulum with a prescribed translating support ( $l \ddot{θ} + \overset{a}{¨}_{x} cos θ + (g + \overset{a}{¨}_{y}) sin θ = 0$ )

For support displacement

a (t)

and angle measured from the downward vertical, the mass position is

(a_{x} + l sin θ, a_{y} - l cos θ)

. Its Euler-Lagrange equation is

l \ddot{θ} + \overset{a}{¨}_{x} cos θ + (g + \overset{a}{¨}_{y}) sin θ = 0

. In the support frame the effective acceleration is

g - \ddot{a}

; taking its torque about the hinge gives

l^{2} \ddot{θ} \hat{z} = l \times (g - \ddot{a})

. The prescribed support can do work, so pendulum energy need not be conserved.

Harmonic forcing of a horizontally driven pendulum ( $\ddot{θ} + (g / l) θ = (A ω^{2} / l) cos ω t$ )

For small angles about the downward vertical and prescribed pivot position

x = A cos ω t

, a simple pendulum obeys the displayed forced-oscillator equation. Off resonance, its harmonic particular solution is

[A ω^{2} / (l (ω_{0}^{2} - ω^{2}))] cos ω t

, with

ω_{0}^{2} = g / l

. At exact resonance, a pendulum initially vertical with zero angular velocity has

θ = (A ω_{0} / (2 l)) t sin ω_{0} t

in the undamped linear model. The growing envelope eventually violates the small-angle assumption; damping and nonlinear frequency shifts matter in a real pendulum.

Horizontal-vibration averaged pendulum potential ( $V_{eff} = - m g l cos Θ + m a^{2} ω^{2} cos^{2} Θ/4$ )

For a small, rapid horizontal displacement

a cos ω t

, the leading fast angular excursion is

ξ = - (a / l) cos Θ cos ω t

. Correlating that excursion with the driving torque gives the averaged force

\ddot{Θ} = - (g / l) sin Θ + (a^{2} ω^{2} /2 l^{2}) sin Θ cos Θ

. Its effective potential is

- m g l cos Θ + m a^{2} ω^{2} cos^{2} Θ/4

. The downward state loses linear stability above

aω = 2 g l

and two tilted minima appear. Small

a / l

and slow dynamics compared with

ω

are required; horizontal and vertical support vibration produce different effective potentials.

Pendulum clock below a uniform-density surface ( $t_{read} / t = 1 - h / R$ )

Inside a spherical body of uniform density, the shell theorem gives gravitational acceleration

g (r) = g (R) r / R

. A simple pendulum at depth

h

therefore has period

P_{h} = P_{0} / 1 - h / R

. If its clock is calibrated at the surface, its indicated elapsed time is

t 1 - h / R

and it loses

t [1 - 1 - h / R] ≃ t h / (2 R)

. This is a mechanical clock-rate change caused by weaker local gravity, not gravitational time dilation. The uniform-density assumption is essential; Earth's actual density profile gives a different interior field.

Conical pendulum

A conical pendulum is a pendulum whose bob executes uniform horizontal circular motion beneath a fixed pivot, with its string at a constant angle

θ

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

ℓ > 0

, uniform gravitational acceleration of magnitude

g > 0

, and negligible drag force and pivot friction. A nonzero circular motion requires

0 < θ < π /2

. This geometry and the corresponding forces are described in Richard Fitzpatrick, The conical pendulum, equations 262–267.

Figure 1.
Conical pendulum geometry, horizontal orbit, and the two forces on the bob
. **Details**.

The right triangle formed by the pivot, the orbit centre, and the bob gives

r = ℓ sin θ

and

h = ℓ cos θ

. 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

ω = 2 π / P

for the angular speed; then the bob's speed is

v = ω 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 = ω^{2} r

. Thus Newton's second law gives

T cos θ = m g, T sin θ = m ω^{2} r = \frac{m v ^{2}}{r} .

The 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 = ℓ sin θ

gives the useful collection of predictions

ω^{2} cos θ = \frac{g}{ℓ}, T = \frac{m g}{c o s θ}, v^{2} = g ℓ \frac{s i n ^{2} θ}{c o s θ}, P = 2 π \frac{ℓ c o s θ}{g} = 2 π \frac{h}{g} .

ϕ

is the azimuthal angle, then

\dot{ϕ} = \pm ω

; either direction of rotation satisfies

\dot{ϕ}^{2} cos θ = g / ℓ

. 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

ω

, its position is fixed. This is a concrete way to see how rotational symmetry can turn an apparently moving solution into a stationary one.

Magnetically charged conical pendulum ( $Q = CB ω r^{2} /2$ )

A conducting conical pendulum connected to a grounded pivot develops a motional electric potential difference

B ω r^{2} /2

between the rotation axis and a bob at radius

r

. If the bob dominates the capacitance, its electric charge is

CB ω r^{2} /2

. An additional uniform vertical electric field changes both the force on that charge and, while the grounding connection remains, the background electric potential entering the charging relation. These effects must be treated consistently when calculating a first-order mechanical response.

Rotating bob experiment for a conical pendulum

The motivating observation is a bob going around at almost constant height while its string traces a cone. The University of Iowa's demonstration places a camera above a short pendulum to reveal this motion. ^[ref] The following proposed measurement procedure develops that observation into a test of the conical pendulum model; it is not a report of collected data.

Suspend a compact bob from a rigid, fixed support using a light string, and measure

ℓ

from the pivot to the bob's centre. Put a stationary marker beside the orbit as a timing reference. Displace the bob sideways and give it a horizontal push tangent to the intended circle, then let go. For a chosen starting angle

θ

, the predicted initial speed is

v = g ℓ sin^{2} θ / cos θ

. Adjust the launch between trials until a short observation interval has nearly constant orbit radius and height. Keep the pivot fixed while recording: circling the support by hand changes the problem, and a maintained drive would introduce another force.

Record from above with the camera looking approximately along the vertical axis. Check that the bob's centre follows a circle and that successive returns to the marker take nearly equal times. A separate side recording checks whether the bob's height stays constant. A markedly elliptical orbit or visibly rising and falling bob is not the uniform circular motion assumed by the conical pendulum equations. Measure the orbit diameter with a scale at the orbit's height; in a level side view, use the two extreme horizontal positions, not one arbitrary frame. Halve the diameter to obtain

r

, or measure the vertical drop

h

directly in a calibrated side view. An oblique camera view and a scale at a different depth distort apparent lengths, so arrange the geometry before interpreting the images.

Time

N

complete returns to the same marker in the same direction and estimate

P = Δ t / N

. Counting ten revolutions, when the orbit stays sufficiently steady, spreads an endpoint timing error over ten periods. Repeat the timing and compare the first and last few revolutions to detect drift; using a longer interval helps only if the motion remains close to uniform. From independently measured geometry, compute

sin θ = r / ℓ

cos θ = h / ℓ

, and check

r^{2} + h^{2} ≃ ℓ^{2}

if both were measured. Then compare the measured oscillation period with

P^{2} = 4 π^{2} h / g

. MIT's 8.01X experimental assignment explicitly asks students to compare the conical pendulum's period, length, and orbit radius with theory. ^[ref]

Repeat at several launch radii with the same

ℓ

. Larger

r

means smaller

h = ℓ^{2} - r^{2}

, so the ideal oscillation period decreases. A plot of measured

P^{2}

against independently measured

h

should follow a straight line through the origin with slope

4 π^{2} / g

; conversely,

g = 4 π^{2} h / P^{2}

estimates gravitational acceleration. Do not infer

h

from the timing equation and then use that same equation as a test. Compare discrepancies with the spread of repeated timings and the resolution of the length measurements. String stretch, a moving pivot, and appreciable bob size challenge the stated geometry; drag force and pivot friction challenge the assumption of undriven uniform motion. Observe a short interval and record any shrinking or drifting orbit rather than silently treating it as an exact conical pendulum.

For an illustrative calculation, assume

ℓ = 1.00 m

m = 0.050 kg

g = 9.81 m/ s^{2}

, and

θ = 4 0^{\circ}

. The model gives

r \approx 0.643 m

h \approx 0.766 m

P \approx 1.756 s

v \approx 2.300 m/s

, and

T \approx 0.640 N

. Its vertical tension component is

0.4905 N

, equal to the gravitational weight; its inward component is approximately

0.412 N

. These are predictions from chosen parameters, not measured data or measurements of local

g

. The next figure uses the same stipulated

ℓ

and

g

, including illustrative angles

1 5^{\circ}

4 0^{\circ}

, and

6 0^{\circ}

Period and small-angle limit of a conical pendulum ( $P / P_{0} = cos θ, P_{0} = 2 π ℓ / g$ )

Figure 1.
Ideal conical pendulum period decreases with cone angle and its square is proportional to the vertical drop
. **Details**.

For fixed

ℓ

and

g

, the exact conical pendulum oscillation period satisfies

P / P_{0} = cos θ

. In the small-angle approximation, with

θ

in radians,

\frac{P}{P _{0}} = 1 - \frac{θ ^{2}}{4} + O (θ^{4}), P_{0} = 2 π \frac{ℓ}{g} .

Thus the limiting revolution period equals the small-amplitude oscillation period of a planar simple pendulum. This equality has a geometric explanation: to first order,

x (t) = r cos (ω t)

and

y (t) = r sin (ω t)

are perpendicular simple harmonic motions of equal amplitude, a quarter period apart, with

ω \to g / ℓ

. At finite angle, however, the conical pendulum's oscillation period decreases with

θ

, whereas a planar simple pendulum's oscillation period increases with its angular amplitude.

The endpoint

θ = 0

itself is a stationary bob with

r = 0

and no observable revolution;

P_{0}

is a limit of nonzero circular motions, not a measured revolution period at rest. For fixed

ℓ

, nonzero conical pendulum motion also requires

ω > g / ℓ

, because

cos θ = g / (ℓ ω^{2}) < 1

. At the other endpoint,

θ \to π /2

gives

h \to 0

P \to 0

, and unbounded tension and speed. These mathematical limits explain why the ideal string cannot sustain a horizontal orbit at finite tension; a real experiment ceases to satisfy the ideal assumptions before that limit.

Damped pendulum ( $\ddot{θ} + c \dot{θ} + sin θ = 0$ )

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

ℓ > 0

, uniform gravitational acceleration

g > 0

, and resisting torque

- b d θ / d t

with constant

b > 0

. The angle

θ

is measured from the downward vertical in radians. An ideal viscous rotary dashpot realizes this torque law; tangential linear drag

- αv

on the bob gives the same equation with

b = α ℓ^{2}

. The moment of inertia is

m ℓ^{2}

, so

m ℓ^{2} \frac{d ^{2} θ}{d t ^{2}} + b \frac{d θ}{d t} + m g ℓ sin θ = 0, ω_{0} = \frac{g}{ℓ}, γ = \frac{b}{m ℓ ^{2}}, τ = ω_{0} t, c = \frac{γ}{ω _{0}} > 0.

Here

t

is physical time,

ω_{0}

is the undamped small-amplitude angular frequency, and

γ

has units of inverse time. In the normalized equation

\ddot{θ} + c \dot{θ} + sin θ = 0

, dots mean differentiation with respect to

τ

, not

t

. The parameter

c

is dimensionless. This separates the chosen damping law from the small-angle approximation: retaining

sin θ

allows finite angles even when the resisting torque is linear in angular velocity.

Figure 1.
Proposed viscous-damper experiment and illustrative small-angle decay: angle versus time and successive peak ratios, not measured data
. **Details**.

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

ℓ

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 ℓ^{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} θ / d t^{2} + b d θ / d t + M g d sin θ = 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

θ = atan2 (x, y)

; record the frame timestamps rather than assuming a frame rate. Release from rest at about

1 0^{\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} = ∣ θ (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 < γ < 2 ω_{0}

and release from rest at

θ_{0}

, its exact linear-model solution is

θ_{lin} (t) = θ_{0} e^{- γ t /2} (cos ω_{d} t + \frac{γ}{2 ω _{d}} sin ω_{d} t), ω_{d} = ω_{0}^{2} - \frac{γ ^{2}}{4}, T_{d} = \frac{2 π}{ω _{d}} .

The decaying sinusoidal envelope has coefficient

∣ θ_{0} ∣ ω_{0} / ω_{d}

; because release is from rest, its successive same-side extrema occur at

t_{n} = n T_{d}

and have

A_{n} = ∣ θ_{0} ∣ e^{- γn 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

- γ /2

, giving

γ = \frac{2}{t _{n + 1} - t _{n}} ln \frac{A _{n}}{A _{n + 1}}, b = m ℓ^{2} γ, ω_{d}^{2} + \frac{γ ^{2}}{4} = \frac{g}{ℓ} .

Use several peaks and repeated trials rather than a single ratio, and compare the frequency identity with independently measured

ℓ

and

g

. For the illustrative choices

ℓ = 1.00 m

g = 9.81 m/ s^{2}

c = 0.20

, and

θ_{0} = 1 0^{\circ}

, the linear model gives

γ \approx 0.6264 s^{- 1}

T_{d} \approx 2.0162 s

, and

A_{n + 1} / A_{n} \approx 0.5318

. The first two positive peaks after release are therefore approximately

5.31 8^{\circ}

and

2.82 8^{\circ}

. The undamped small-amplitude oscillation period is about

2.0061 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 θ

θ

is about

θ^{2} /6

for small angles: about

0.51%

1 0^{\circ}

, but about

17.3%

6 0^{\circ}

when compared with

θ

. 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

(θ, \dot{θ}) = (kπ, 0)

. Even multiples of

π

are attracting: their linearization has eigenvalues

(- c \pm 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

π

have eigenvalues

(- c \pm 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.

Energy dissipation of a damped pendulum ( $\dot{E} = - c \dot{θ}^{2}$ )

For the normalized damped pendulum, the sum of kinetic energy and gravitational potential energy, divided by

m g ℓ

, is

E (τ) = \frac{1}{2} \dot{θ}^{2} + 1 - cos θ, \dot{E} = \dot{θ} (\ddot{θ} + sin θ) = - c \dot{θ}^{2} \leq 0.

The chain rule and the full nonlinear equation give this identity exactly, without a small-angle approximation. Here

\dot{θ} = d θ / d τ

as in damped pendulum. In physical units the mechanical energy and the power of the resisting torque are

E (t) = \frac{1}{2} m ℓ^{2} (\frac{d θ}{d t})^{2} + m g ℓ (1 - cos θ), \frac{d E}{d t} = - b (\frac{d θ}{d t})^{2} .

The pendulum's lost mechanical energy is transferred to the damper and surroundings; it does not violate conservation of energy. In an ideal viscous dashpot this loss is frictional heating. The exact cumulative balance is particularly useful for checking a numerical solution:

Q (τ) = c \int_{0}^{τ} \dot{θ} (s)^{2} d s, E (τ) + Q (τ) = E (0) .

Figure 1.
Illustrative nonlinear and linear damped pendulum solutions at 10 and 60 degrees, with exact mechanical energy transfer and loss; no measured data
. **Details**.

The figure uses

c = 0.20

and release from rest. Over

0 \leq τ \leq 25

, the largest angular discrepancy between the nonlinear solution and the damped harmonic oscillator solution is approximately

0.000702 rad

(

0.040 2^{\circ}

) for a

1 0^{\circ}

release, but

0.14995 rad

(

8.59 2^{\circ}

) for a

6 0^{\circ}

release. Their nonlinear initial energies are

1 - cos 1 0^{\circ} \approx 0.015192

and

1 - cos 6 0^{\circ} = 0.5

, respectively. At

τ = 25

the remaining fractions are approximately

0.0064080

and

0.0063757

. These are illustrative calculations from specified initial conditions, not observations. The generator uses a Runge-Kutta method and integrates

Q

alongside the motion; halving its step and independently integrating

c \dot{θ}^{2}

checks the result.

For

ℓ = 1 m

and

g = 9.81 m/ s^{2}

, the first three nonlinear same-side swing intervals after a

6 0^{\circ}

release are approximately

2.1008

2.0378

, and

2.0221 s

. They approach the linear-model oscillation period

2.0162 s

as the amplitude shrinks. After a

1 0^{\circ}

release the corresponding intervals are already about

2.0185

2.0168

, and

2.0164 s

. This makes the experiment's period drift a quantitative test of the small-angle approximation, rather than an automatic reason to change the damping coefficient.

At a turning point,

\dot{θ} = 0

and

\dot{E} = 0

even though the swing is losing mechanical energy over the cycle. Its potential energy there is

E_{n} = 1 - cos A_{n}

, so measured turning angles give successive energies without estimating angular velocity. In the proposed decay experiment, multiply these values by

m g ℓ

to obtain energies in joules, then subtract adjacent same-side energies to estimate the loss over each full swing. Only for small amplitudes is

E_{n} ≃ A_{n}^{2} /2

. In the linear underdamped model, same-side peak energies consequently satisfy

E_{n + 1} / E_{n} = e^{- γ T_{d}}

, the square of the angular amplitude ratio. For the illustrative parameters above this energy ratio is approximately

0.2828

. Between turning points, kinetic energy and potential energy continue to exchange while their sum decreases; the energy slope vanishes briefly at each extremum and is negative whenever the bob moves.

It is therefore inaccurate to replace the exact identity by

\dot{E} = - c E

. Even for the linear model,

E_{lin} = (\dot{θ}^{2} + θ^{2}) /2

has

\dot{E}_{lin} = - c \dot{θ}^{2}

. For weak damping and small angles, cycle averaging gives

⟨ \dot{θ}^{2} ⟩ ≃ ⟨ E_{lin} ⟩

, so the averaged energy approximately follows

e^{- c τ}

while angular amplitude follows

e^{- c τ /2}

. This exponential decay is an approximation to the energy trace; the same-side peak ratio is exact within the linear model. The figure distinguishes the actual energy curves from this averaging approximation.

The energy identity also gives a global conclusion without solving the nonlinear equation. If

E

were constant on a positive-length interval, then

\dot{θ}

would vanish throughout it; the equation would force

sin θ = 0

, making the entire trajectory an equilibrium. Every nonstationary trajectory therefore loses mechanical energy over every positive-length interval, even though its instantaneous derivative can vanish at a turning point. A nonconstant periodic motion would have to return to its initial mechanical energy, so it is impossible for this undriven damped pendulum. Oscillations near a stable focus are successive, shrinking swings, not a repeating cycle. Moreover, if

E (0) < 2

, the angle cannot reach an upright equilibrium, whose potential energy is

2

; such motion stays within its original potential well. This is the usefulness of the energy description: one scalar balance controls both the observed decay and which nonlinear motions remain accessible.

Foucault pendulum ( $\dot{ψ} = - Ω sin φ$ )

A Foucault pendulum is a long simple pendulum whose suspension allows the bob to swing in both independent horizontal directions. Its slowly changing swing direction relative to the building reveals Earth's rotation. A hinge restricting the bob to one vertical plane would prevent this measurement: the support would impose the very direction we want the pendulum to select freely. Léon Foucault demonstrated the effect in 1851.^[ref]

The bob's back-and-forth oscillation and the swing direction's precession are different motions. The bob traverses its arc in seconds, while successive arcs change direction over hours. In a conical pendulum the bob itself circulates around the vertical; a nearly planar Foucault pendulum instead reverses along an almost straight horizontal trace, whose orientation changes slowly. The measured quantity is that orientation against a fixed floor or wall reference, rather than the bob's instantaneous azimuth.

UNSW Foucault pendulum observation

The University of New South Wales documents an undriven, visitor-started Foucault pendulum. A fixed vertical wire and wall line define its initial north–south swing plane. The documented comparison one hour later shows reduced amplitude and about nine degrees of anticlockwise precession. The apparatus swings for several hours without an electromagnetic drive.^[ref]

Figure 1.
Functional Foucault pendulum apparatus, fixed sighting reference, and predicted latitude dependence; schematic with no historical dimensions
. **Details**.

To turn that demonstration into a quantitative experiment, the following is a proposed measurement protocol. Establish a rigid support with no preferred horizontal swing direction, measure the wire length to the bob's centre, and mark two perpendicular horizontal axes fixed to the building. Start with a small displacement along the sighting plane, then release with as little transverse velocity as possible. Record the actual launch method; burning a fine restraining thread is one possible proposed release, not a claim about the documented UNSW launch. Keep people clear of the bob and avoid touching the suspension during the observation.

Record the horizontal position from above with a stationary, calibrated camera, or sight the bob against the wire and wall line at recorded times. Fit the long axis of each short group of swings to estimate its orientation

ψ

; use the extrema on both sides rather than following the bob's rapidly reversing velocity. A swing line has orientation modulo

π

, so unwrap successive small direction changes continuously. Rotate no camera or reference between measurements. Record the fast oscillation period separately, and compare the slope of

ψ (t)

with the Latitude law of Foucault precession. Repeat launches along different directions and monitor ellipses, pivot motion, and air currents before attributing the slow change to Earth's rotation.

Latitude law of Foucault precession ( $\dot{ψ} = - Ω sin φ, T_{p} = T_{sid} /∣ sin φ ∣$ )

Use local east, north, and up unit vectors, with

x

eastward,

y

northward, and positive orientation

ψ

anticlockwise as viewed from above. Let

φ

be signed geographical latitude, positive north, and

Ω > 0

the magnitude of Earth's angular velocity. In the spherical-Earth approximation,

Ω = Ω cos φ e_{N} + Ω sin φ e_{U}, α = Ω sin φ, ω_{0}^{2} = \frac{g}{ℓ} .

Here

g

is the local effective gravity in a rotating frame, including the steady terrestrial centrifugal acceleration. For small horizontal displacements compared with

ℓ

, the horizontal restoring acceleration from tension is

- ω_{0}^{2} (x, y)

. Vertical displacement is second order in the swing amplitude, so vertical velocity can be omitted at this order. The horizontal Coriolis acceleration in local Earth coordinates then gives the coupled linear differential equations

\overset{x}{¨} + ω_{0}^{2} x = 2 α \overset{y}{˙}, \overset{y}{¨} + ω_{0}^{2} y = - 2 α \overset{x}{˙} .

These equations neglect damping, finite-amplitude corrections, support anisotropy, and spatial changes of the centrifugal force of order

Ω^{2}

; they retain the leading terrestrial Coriolis force. A related derivation uses south/east axes instead.^[ref]

Introduce the complex number

z = x + i y

. Multiplying the second equation by

i

and adding gives

\overset{z}{¨} + 2 i α \overset{z}{˙} + ω_{0}^{2} z = 0.

Set

z = e^{- i α t} u

. Direct differentiation cancels the first derivative of

u

and yields

\overset{u}{¨} + (ω_{0}^{2} + α^{2}) u = 0, β = ω_{0}^{2} + α^{2}, z = e^{- i α t} (C cos βt + D sin βt) .

(4)

The

α^{2}

β

comes from solving these reduced equations exactly; it is not a complete prediction of terrestrial corrections of order

Ω^{2}

. Thus the horizontal trajectory is an ellipse in coordinates rotating at angular velocity

- α

relative to the building. When

∣ α ∣ ≪ ω_{0}

, each rapid swing samples an almost fixed long axis; its slow precession is

\dot{ψ} = - Ω sin φ, T_{p} = \frac{2 π}{Ω∣ s i n φ ∣} = \frac{T _{sid}}{∣ s i n φ ∣} .

(5)

The sidereal day

T_{sid} = 2 π /Ω

is a rotation relative to distant directions.

T_{p}

denotes a full

2 π

turn of a continuously followed orientation; an unlabelled swing line repeats after

T_{p} /2

. At the equator there is no finite precession period in this model. The Coriolis parameter is

f = 2 α

, so the plane's rate is

- f /2

, not

- f

The limiting cases check the sign and scale. At the north pole

α = Ω

: the swing direction turns clockwise once per sidereal day. At the equator

α = 0

: both horizontal equations reduce to independent simple harmonic motions and a line launch stays on its original line. At the south pole

α = - Ω

: the direction turns anticlockwise once per sidereal day. Vanishing horizontal precession at the equator does not mean that every component of Coriolis acceleration vanishes there.

For a stipulated

Ω = 7.292115 \times 1 0^{- 5} s^{- 1}

, the sidereal day is approximately

23.9345 h

. At

4 5^{\circ}

north the predicted rate is

- 10.635 6^{\circ} / h

and

T_{p} ≃ 33.8485 h

; at

4 5^{\circ}

south the rate changes sign. At the illustrative geographical latitude

3 4^{\circ}

south, the prediction is

+ 8.410 9^{\circ} / h

. These calculations are model predictions, not new observations of the UNSW apparatus.

A swing plane fixed among the stars is generally only a polar picture. A north–south swing at the equator turns with Earth in an inertial frame, while having no precession against the building.^[ref]

Exact linear Foucault pendulum trajectory ( $z = A e^{- i α t} (cos βt + i α sin βt / β)$ )

For release from rest relative to the building, take

x (0) = A > 0

y (0) = \overset{x}{˙} (0) = \overset{y}{˙} (0) = 0

. The exact solution of the stated coupled linear differential equations is

z (t) = A e^{- i α t} (cos βt + i \frac{α}{β} sin βt), β = ω_{0}^{2} + α^{2} .

Indeed

u (0) = A

and

\overset{u}{˙} (0) = i α A

, so

\overset{z}{˙} (0) = 0

. In the slowly rotating coordinates, the ellipse has semiaxes

A

and

A ∣ α ∣/ β

. A bob released from rest therefore does not describe a perfectly straight line even in this ideal linear model. Replacing the bracket by

cos ω_{0} t

gives the familiar leading approximation but introduces a spurious initial transverse velocity

- i α A

. The exact expression keeps the initial condition and separates that tiny orbit width from the slow rotation of its long axis.

Figure 1.
Exact trajectories of the linear Foucault pendulum equations at the actual Earth rotation rate, showing one swing near zero, three, and six hours at northern, equatorial, and southern latitudes
. **Details**.

For the illustrative

ℓ = 20 m

and

g = 9.81 m/ s^{2}

, the fast period

2 π / β

is approximately

8.9714 s

. At

4 5^{\circ}

north,

∣ α ∣/ β ≃ 7.3624 \times 1 0^{- 5}

: the transverse semiaxis is too small to distinguish at the figure's scale. Nevertheless the long-axis orientation changes by about

- 63.813 8^{\circ}

in six hours. Changing the sign of geographical latitude conjugates the entire solution, reflecting the trajectory across the eastward axis; setting

α = 0

gives

z = A cos ω_{0} t

exactly. These are stronger checks than merely drawing a rotating line.

The coupled equations give an additional independent constraint. Multiplying them by

\overset{x}{˙}

and

\overset{y}{˙}

and adding cancels the Coriolis force's contributions:

\frac{d}{d t} [\frac{1}{2} (\overset{x}{˙}^{2} + \overset{y}{˙}^{2}) + \frac{1}{2} ω_{0}^{2} (x^{2} + y^{2})] = 0.

This is mechanical energy per unit mass in the reduced model; precession does not require a loss of mechanical energy. Also, direct differentiation gives

\frac{d}{d t} [x \overset{y}{˙} - y \overset{x}{˙} + α (x^{2} + y^{2})] = 0.

For this release the constant is

α A^{2}

. The exact trajectory satisfies both identities. They test the coupled dynamics separately from the visible slow rotation and express the fact that the Coriolis force does no work.

Controls and limitations of a Foucault pendulum

An unwanted ellipse is an experimental warning: a finite-amplitude pendulum can develop additional orbit-axis precession even without Earth's rotation. A symmetrical, firm support and careful launch reduce this effect.^[ref] Distinguish that nonlinear effect from the tiny ellipse already present in the Exact linear Foucault pendulum trajectory. The linear calculation establishes a controlled approximation, not the complete motion of an arbitrary large swing.

Support anisotropy creates a second competing mechanism. As an illustrative modification, let the two horizontal restoring angular frequencies be

ω_{x}

and

ω_{y}

. Even with

α = 0

, a diagonal release from rest gives

x = A_{x} cos ω_{x} t, y = A_{y} cos ω_{y} t .

When

ω_{x} \neq = ω_{y}

, the relative phase changes on the scale

1/∣ ω_{x} - ω_{y} ∣

: the trace changes shape without any terrestrial Coriolis force. It need not be uniform precession of a Foucault pendulum. To cleanly resolve the terrestrial effect over times of order

1/∣ α ∣

, a useful design target is

∣ ω_{x} - ω_{y} ∣ ≪ ∣ α ∣

. Near the equator that target becomes particularly demanding. Compare fast periods along perpendicular launch directions to look for this frequency splitting; repeat at several launch orientations to detect preferred suspension axes.

Other controls address distinct alternatives: shelter the apparatus from air currents, track the support against the building, measure both orbit axes, and compare early and late fitted directions as the amplitude decays. A sustaining drive, if used in a different apparatus, must supply mechanical energy without imposing an azimuthal bias; an undriven run provides a useful comparison. A change in swing direction alone cannot exclude these effects. Agreement of the signed rate with independently known geographical latitude and Earth's rotation, together with stable repeat launches and small competing effects, is the inference the experiment is designed to support.

Eisenhart-Duval lift ( $g = d x^{2} + 2 d t (d u - 2 A \cdot d x - U d t)$ )

A mechanical Lagrangian quadratic in velocities can be represented by geodesics of a Lorentzian metric with an extra null coordinate. For

L = \frac{1}{2} ∣ \dot{x} ∣^{2} - 2 A \cdot \dot{x} - U

, use

g = d x^{2} + 2 d t (d u - 2 A \cdot d x - U d t)

. The null Killing vector field

\partial_{u}

gives the conserved momentum

p_{u} = t'

. Fixing

p_{u} = 1

projects the geodesic equation to the mechanical Euler-Lagrange equations. This is useful for relating mechanical forces to spacetime geometry.

Gauge transformations of an Eisenhart-Duval lift ( $A \mapsto A + \nabla χ, u \mapsto u + 2 χ$ )

A time-independent gauge transformation

A \mapsto A + \nabla χ

in an Eisenhart-Duval lift is the coordinate change

u \mapsto u + 2 χ

. It preserves the metric one-form

d u - 2 A \cdot d x - U d t

and adds only

- 2 d χ / d t

to the reduced Lagrangian.

Null Kaluza-Klein reduction of a stationary force ( $p_{u} = t' \neq = 0$ )

For an Eisenhart-Duval lift, reduction along the null isometry fixes a nonzero conserved

p_{u} = t'

and projects geodesics to

(t, x)

. One never divides by

g_{uu} = 0

. When

E = - \nabla U

and

\nabla \times A = B

, the reduced equation is

\ddot{x} = E + 2 B \times \dot{x}

. Every trajectory admits a null lift by choosing

\overset{u}{˙} = U + 2 A \cdot \dot{x} - \frac{1}{2} ∣ \dot{x} ∣^{2}

Newtonian mechanics ( $F = m a$ )

Newtonian mechanics describes motion using forces, masses and acceleration in inertial frames with absolute time. It is the low-speed limit of special relativity; for constant proper acceleration with

∣ a t ∣ ≪ c

, relativistic hyperbolic motion reduces to

x ≃ a t^{2} /2

Galilean spacetime ( $(M, t_{a}, h^{ab}, \nabla)$ )

A geometric background for Newtonian mechanics consisting of a four-dimensional affine space, an absolute time covector

t_{a}

, a rank-three positive spatial inverse metric tensor

h^{ab}

with

t_{a} h^{ab} = 0

, and a flat torsion-free affine connection preserving both. A trajectory parametrized by absolute time has tangent

u

with

t_{a} u^{a} = 1

. Its absolute acceleration is

\nabla_{u} u

; an unforced particle obeys

\nabla_{u} u = 0

. The structure distinguishes inertial motion without selecting an absolute state of uniform rest. Additional force laws, including a Newtonian gravitational force when wanted, specify a dynamical theory on this background.

Small-slope particle motion on a height graph ( $\ddot{r} ≃ - g \nabla h$ )

A particle constrained to the graph

z = h (x, y)

has kinetic energy

m [∣ \dot{r} ∣^{2} + (\nabla h \cdot \dot{r})^{2}] /2

and potential energy

m g h

. For a gently varying graph and velocities for which its geometric kinetic correction is higher order, the leading Euler-Lagrange equations are the displayed planar gradient-force equations. Small slope alone at arbitrarily large speed or curvature does not justify discarding the normal inertial acceleration. For a radial height, the leading planar angular momentum is conserved, allowing an effective potential reduction.

Radial turning points for an inverse-square height well ( $\overset{r}{˙}^{2} = 2 E + 2 g / r - (L^{2} + 2 g) / r^{2}$ )

For the height profile

h (r) = r^{- 2} - r^{- 1}

, set

A = L^{2} + 2 g

and use the planar small-slope model. Its effective potential per unit mass is

V_{eff} = A / (2 r^{2}) - g / r

. Release with negligible radial speed at

r_{0}

gives

E = A / (2 r_{0}^{2}) - g / r_{0}

and the displayed radial-speed law. A bounded trajectory requires

E < 0

, equivalently

r_{0} > A / (2 g)

. The two positive turning radii are

r_{0}

and

A r_{0} / (2 g r_{0} - A)

; they coincide at the circular solution

r_{0} = A / g

. Zero energy is marginal escape, not bounded motion.

Tangentially driven motion on an axisymmetric surface ( $\dot{ϕ}^{2} = \frac{g h ^{'} ( r )}{r + 2 r _{ϕ}^{2} / r - r _{ϕϕ}}$ )

Neglecting vertical acceleration and the vertical driving-force component, the radial normal reaction is

- M g h^{'} (r)

. A drive parallel to the horizontal velocity cancels when the radial acceleration equation is combined with

r_{ϕ} / r

times the azimuthal equation. For a prescribed path

r (ϕ)

this gives the displayed speed constraint. It describes the projected approximation, not arbitrary trajectories with substantial vertical motion.

// Target: classical-mechanics.bigb

Friction

Friction is a contact force tangent to the surfaces. Kinetic friction opposes contact slip; static friction maintains contact without contact slip, including rolling without slipping. The perpendicular contact component is the normal force. These two components answer different questions: how strongly the surfaces press together, and how strongly they resist sliding past one another.

The ideal Coulomb dry-contact model combines a bound on static friction with a sliding law for kinetic friction. Its parameters are the coefficient of static friction and the coefficient of kinetic friction; constant coefficients are an empirical approximation, rather than a consequence of Newton's laws of motion. Surface condition matters, and the static coefficient need not universally exceed the kinetic coefficient. ^[ref]

The model's historical setting was practical machinery: Coulomb's Théorie des machines simples investigated friction and the stiffness of ropes. The Académie's bibliography records its 1781 prize and its appearance in volume 10, pages 161–332, of the Recueil des savants étrangers. ^[ref]

Frictional sliding

Relative tangential motion along a contact or fault resisted by friction.

Frictional strength

The shear traction required for frictional failure or sliding under specified effective normal load and material conditions.

Byerlee's law

An empirical relationship summarizing the high friction measured for many rock surfaces as a function of effective normal load. Weak fault gouge or high pore pressure can produce different effective fault strengths. Named for James Byerlee.

Coulomb friction

A friction law relating limiting shear traction to effective normal load, commonly with a cohesion term in geological applications.

Coefficient of friction ( $μ$ )

The dimensionless ratio between an appropriate tangential friction force and normal load in a specified contact model. Static limiting and sliding coefficients differ, and nanoscale adhesion can invalidate a simple load-independent ratio.

Lubricant

Kinetic friction ( $∣ F_{fr} ∣ = μ N$ )

Kinetic friction is friction during nonzero contact slip. For one-dimensional motion along a surface, choose right as positive and write

v

for the body's contact-point velocity and

U

for the surface's contact-point velocity, both measured in the same inertial frame. The signed relative velocity is

w = v - U

. The ideal Coulomb law for the tangential force on the body is

f_{k} = - μ_{k} N sgn (v - U), v - U \neq = 0.

, coefficient of static friction

Here

N \geq 0

is the magnitude of the normal force,

μ_{k} \geq 0

is the coefficient of kinetic friction, and

sgn

is the sign function. Thus kinetic friction opposes contact slip, which can differ from the body's direction of motion. In several tangential directions the same law reads

f_{k} = - μ_{k} N w /∣ w ∣

for a nonzero tangential relative velocity

w

. The magnitude law is documented in Ling, Sanny and Moebs, OpenStax University Physics 1, §6.2: static and kinetic friction.

For a stipulated toy model, take a nonrotating box with mass

m = 1 kg

, normal force

N = 10 N

μ_{s} = 0.5

, and coefficient of kinetic friction

μ_{k} = 0.3

. These are assumed model parameters, rather than measured properties of any particular pair of materials; the prescribed normal force is consistent with horizontal support under

g = 10 m/ s^{2}

and no other vertical forces. On stationary ground, a box already sliding right has

f_{k} = - 3 N

. With an applied horizontal force

F = 6 N

to the right, Newton's second law gives

a = (F + f_{k}) / m = 3 m/ s^{2}

, for as long as rightward contact slip persists.

Direction is clearer by comparing two instantaneous states of this same model. If the box has

v = 1 m/s

on stationary ground,

U = 0

and

w = + 1 m/s

, so kinetic friction acts left with magnitude

3 N

. If the box still has

v = 1 m/s

but a belt beneath it has

U = 2 m/s

, then

w = - 1 m/s

: kinetic friction acts right with magnitude

3 N

, even while the box moves right.

Figure 1.
The same rightward box velocity gives opposite friction directions on stationary ground and a faster moving belt
. **Details**.

In the belt case, the power delivered to the box is

P_{box} = f_{k} v = + 3 W

. With no other horizontal force, its kinetic energy therefore increases by the work-energy theorem. By Newton's third law, the box exerts

- f_{k}

on the belt, so the contact power delivered to the belt is

P_{belt} = - f_{k} U = - 6 W

. Their sum is

- 3 W

: frictional heating produces thermal internal energy at

3 W

. An ideal motor maintaining the prescribed belt velocity supplies

+ 6 W

against this contact load, neglecting other losses, so its input divides into

3 W

of box kinetic energy and

3 W

of frictional heating. These are instantaneous rates, since the box's velocity subsequently changes. This balance follows directly from power as force times velocity and conservation of energy; the power law is given in Ling, Sanny and Moebs, OpenStax University Physics 1, §7.4: power.

When contact slip vanishes, including a transition to rolling without slipping, the nonzero-slip law no longer applies. Determine the required static friction from the no-slip condition and the equations of motion, then test its bound. Substituting

sgn (0) = 0

into the sliding law would incorrectly exclude a nonzero static friction.

Pressure-driven cork ejection ( $\overset{x}{¨} = P_{g} x / (ρ l^{2})$ )

A cylindrical plug of radius

a

, length

l

and density

ρ

is pushed outward by gauge pressure

P_{g}

. If its remaining contact length is

l - x

and its constant radial contact pressure is

P_{c}

, the resisting kinetic friction is

2 πa (l - x) μ P_{c}

. At the marginal initial condition

π a^{2} P_{g} = 2 πa l μ P_{c}

, the net outward force becomes

π a^{2} P_{g} x / l

. With mass

ρ π a^{2} l

\overset{x}{¨} = b^{2} x, b = \frac{1}{l} \frac{P _{g}}{ρ}, x (t) = \frac{v _{0}}{b} sinh (b t), \overset{x}{˙}^{2} = v_{0}^{2} + b^{2} x^{2} .

Thus the exit speed at

x = l

tends to

P_{g} / ρ

v_{0} \to 0^{+}

. Exactly

v_{0} = 0

leaves the marginal configuration stationary, and the exit time

b^{- 1} arsinh (b l / v_{0})

diverges in that limit. An attached extra mass changes the acceleration but not the pressure-minus-friction work, if the contact geometry is unchanged.

Uniform-pressure friction torque on a circular contact ( $τ_{f} = 2 μ N a /3$ )

A flat circular contact of radius

a

carrying total normal force

N

has uniform pressure

p = N / (π a^{2})

. In steady relative rotation, a ring of radius

s

and thickness

d s

experiences tangential kinetic friction

d F = μ p 2 π s d s

. Its resisting torque is

d τ = s d F

. Integration from zero to

a

gives

τ_{f} = 2 μ N a /3

. A different pressure distribution changes the result; for an axisymmetric distribution use

τ_{f} = 2 π μ \int_{0}^{a} p (s) s^{2} d s

Coefficient of kinetic friction ( $μ_{k}$ )

The coefficient of kinetic friction is the dimensionless quantity

μ_{k} = ∣ f_{k} ∣/ N

for sliding contact with positive normal force

N

in the ideal kinetic friction model. It characterizes the pair of contacting surfaces and their condition. A constant

μ_{k}

means the model neglects dependence on contact slip speed; unlike the coefficient of static friction, it specifies a sliding force magnitude rather than a no-slip upper bound.

The ratio compares the tangential kinetic friction magnitude with the perpendicular normal force:

μ_{k} = \frac{∣ f _{k} ∣}{N}, N > 0, ∣ w ∣ > 0.

Here

w

is the tangential relative velocity defining contact slip. Both numerator and denominator have the same physical dimension, so the coefficient of kinetic friction has no units. It belongs to a contact pair under specified conditions, rather than to one material alone. The coefficient of static friction instead describes the threshold

∣ f_{s} ∣_{m a x} / N

before contact slip; a pull required to start motion cannot be substituted for a sliding measurement.^[ref]

As an original model example, stipulate

μ_{k} = 0.30

and gravitational acceleration

g = 9.8 m/ s^{2}

. An already sliding, nonrotating block of total mass

m = 2 kg

is pulled horizontally across a stationary horizontal surface at constant velocity. With no other vertical force, Newton's second law gives

N = m g = 19.6 N

. Zero horizontal acceleration then requires a pulling force

F = ∣ f_{k} ∣ = μ_{k} m g = 5.88 N

. Doubling the total mass to

4 kg

gives

N = 39.2 N

and

F = 11.76 N

. The force doubles while the coefficient of kinetic friction stays

0.30

by assumption. These numbers are model predictions, not observations or a calibration for a particular material pair.

Figure 1.
Original horizontal sliding model: a 2 kg block has balanced forces at constant speed with a stipulated kinetic friction coefficient of 0.30
. **Details**.

A constant coefficient of kinetic friction is useful because one ratio can summarize many loads, as tested by load dependence of kinetic friction. It is an empirical approximation over a stated range of normal forces, speeds and surface conditions. It implies no universal independence from speed or apparent contact area: a change of face or sliding regime can change the result. Temperature, wear and lubrication are further conditions to report. Newton's second law still applies when the constant-coefficient approximation fails.^[ref]

Constant-speed measurement of kinetic friction ( $μ_{k} = \overline{F} / (m g)$ )

A force gauge reads the pulling force. For horizontal straight-line sliding at constant velocity, its reading equals the opposing kinetic friction magnitude, provided these are the only horizontal forces. For total supported mass

m

, including added loads,

N = m g

, so a mean sliding reading

\overline{F}

gives

μ_{k} = \overline{F} / (m g)

. The important readout is the sustained sliding force after the starting peak; the peak belongs to the static friction threshold.

The University of Mississippi's Experiment 6 uses a wooden block on a wooden plane, a triple-beam balance, added masses, and a dual-range 50 N force probe connected through LabPro to Logger Pro. Its procedure calibrates the probe with a

0.5 kg

load on a hanger, entering the corresponding force in newtons, and measures the block's mass separately. With the plane horizontal, it starts with

0.5 kg

on the block and pulls horizontally at constant speed. The recorded force trace has a starting peak; the procedure selects the subsequent constant-force interval and takes its mean. It repeats with

0.5 kg

increments up to

3 kg

added, then plots sliding force against normal force to estimate the coefficient of kinetic friction from the slope.^[ref]

For a replication, check the gauge's zero with the string slack and its response to known gravitational weights; zeroing alone does not check its scale. Keep the plane level, the pull parallel to it, and the block face and track region consistent. Specify the contacting materials and whether the surfaces are dry, cleaned or lubricated; repeat runs to see whether debris or wear changes the sliding force. Verify constant velocity independently from position versus time, rather than assuming that a flat force trace proves zero acceleration. These are proposed measurement controls, not additional observations from the laboratory writeup. Record the selected interval, variation between runs and gauge resolution as sources of measurement error.

Load dependence of kinetic friction ( $∣ f_{k} ∣ = μ_{k} N$ )

Changing the added mass while retaining the same contact pair and sliding speed tests whether a constant coefficient of kinetic friction is a useful model. Compute

N = (m_{block} + m_{added}) g

for each level horizontal run, and plot the mean sliding force vertically against normal force horizontally. The constant-coefficient prediction is a straight line through the origin; its slope is

μ_{k}

. This extracts a ratio from several loads rather than relying on one reading.

Figure 1.
Horizontal wooden-block force-probe apparatus and original load model: 19.6 N and 39.2 N normal loads predict 5.88 N and 11.76 N sliding forces
. **Details**.

The apparatus sketch identifies the contact pair and horizontal pull used by the cited laboratory procedure. The plotted line and points are the original stipulated

μ_{k} = 0.30

model from coefficient of kinetic friction, not laboratory data. The two model points give

\frac{Δ∣ f _{k} ∣}{Δ N} = \frac{11.76 - 5.88}{39.2 - 19.6} = 0.30.

In actual measurements, a linear regression

\overline{F} = b + μ_{k} N

can expose an unexplained intercept

b

. A nonzero intercept calls for checking the gauge zero, other resisting forces and the contact model; forcing the line through zero can hide those problems. Conversely, constant ratios within measurement error support the approximation only over the loads, speeds and surface conditions tested. A load sweep alone cannot establish independence from contact area or sliding speed; those require separate controlled variations.

Corrections to a pulling measurement of kinetic friction ( $μ_{k} = (F cos α - ma) / (m g - F sin α)$ )

For a block sliding right on a stationary horizontal plane, take rightward acceleration

a

as positive. A gauge reading

F

for a pull at angle

α

above horizontal has horizontal component

F cos α

and upward component

F sin α

. With no vertical acceleration or other forces, Newton's second law gives

N = m g - F sin α, F cos α - ∣ f_{k} ∣ = ma,

and therefore

μ_{k} = \frac{F c o s α - ma}{m g - F s i n α} .

This inference requires continued contact with

N > 0

and rightward contact slip, so its numerator must be nonnegative. An upward pull reduces normal force; a downward pull, represented by

α < 0

, increases it. The horizontal constant-velocity formula is the special case

α = 0

a = 0

For the original

2 kg

μ_{k} = 0.30

model, an upward pull at

α = 3 0^{\circ}

maintaining constant velocity needs

F = \frac{μ _{k} m g}{c o s α + μ _{k} s i n α} ≃ 5.7873 N, N ≃ 16.7064 N .

Dividing this gauge reading by

m g

would give about

0.2953

, rather than

0.30

: the gauge reads the full inclined force, and the pull has changed the normal force.

Even with a horizontal pull, an accelerating measurement needs the

ma

correction:

μ_{k} = (F - ma) / (m g)

. In the same model, a horizontal

7.88 N

pull produces

a = 1 m/ s^{2}

while the block slides right; subtracting

ma = 2 N

recovers the

5.88 N

kinetic friction. Dividing the uncorrected reading by

19.6 N

instead gives about

0.4020

. Measure acceleration in an inertial frame, along the sliding direction, over the same interval as the gauge reading. These corrections follow from the force balance, independently of whether the resulting coefficient of kinetic friction is constant across runs.

// Target: classical-mechanics.bigb

Two-stage stopping under linear drag and dry friction ( $t = \frac{M}{λ} lo g \frac{U}{u _{}} + \frac{u _{}}{μg}, d = \frac{M}{λ} (U - u_{}) + \frac{u _{}^{2}}{2 μg}$ )

For a body of mass

M

initially at speed

U

, suppose the total resistance is linear drag

λ u

until speed

u_{*} > 0

, followed by kinetic friction

μ M g

until rest. The first stage solves

M \overset{u}{˙} = - λ u

, giving duration

(M / λ) lo g (U / u_{*})

and distance

(M / λ) (U - u_{*})

. Constant deceleration in the second stage gives duration

u_{*} / (μg)

and distance

u_{*}^{2} / (2 μg)

. The stopping distance is finite because the switch removes the infinite exponential stopping time of pure linear drag.

Contact slip ( $w$ )

Contact slip is the tangential relative velocity between the two material points instantaneously in contact. For a nonrotating box on a belt it is

w = v - U

; for a rigid body, use the velocity of its contact point, including the contribution from angular velocity. Zero contact slip defines rolling without slipping at a rolling contact, even when the center of mass moves.

Frictional heating ( $\dot{E}_{thermal} = μ_{k} N ∣ w ∣$ )

In the ideal kinetic friction model, the loss of combined mechanical energy of the two contacting bodies becomes thermal internal energy at rate

\dot{E}_{thermal} = - f_{k} (v - U) = μ_{k} N ∣ v - U ∣ \geq 0.

This follows by adding the contact powers

f_{k} v

and

- f_{k} U

from Newton's third law. It depends on contact slip, whereas the contact power received by either body separately depends on its velocity in the chosen inertial frame. Thus positive frictional heating is compatible with increasing kinetic energy of one body, as the moving-belt example in kinetic friction shows. Heating need not cause a temperature rise if thermal energy escapes by heat transfer.

Newton–Cartan theory

Newton–Cartan theory describes Newtonian mechanics geometrically using absolute time, a degenerate spatial inverse metric, and a compatible torsion-free affine connection. It is not a theory with a nondegenerate Lorentzian metric. Gravitationally accelerated particles are geodesics of that connection, whose curvature measures tidal gravity.

Newton–Cartan gravitational connection ( $Γ^{i}_{00} = \partial_{i} Φ$ )

In Cartesian spatial coordinates and absolute time

t

, the gravitational potential

Φ (t, x)

determines an affine connection by the displayed coefficients, all others zero. The geodesic equation with affine time is

d^{2} x^{i} / d t^{2} = - \partial_{i} Φ

. It preserves

d t

and the spatial inverse metric

h^{ij} = δ^{ij}

, reproducing freely falling Newtonian mechanics in a curved spacetime connection.

Newtonian tidal curvature ( $R^{i}_{0 j 0} = \partial_{i} \partial_{j} Φ, R_{00} = \nabla^{2} Φ$ )

Using

[\nabla_{c}, \nabla_{d}] V^{a} = R^{a}_{b c d} V^{b}

, the Newton–Cartan gravitational connection has only the displayed Hessian components and their antisymmetric partners in the last two indices. The geodesic deviation equation gives relative acceleration

- \partial_{i} \partial_{j} Φ δ x^{j}

. The Poisson equation for Newtonian gravity is therefore

R_{ab} = 4 π Gρ (d t)_{a} (d t)_{b}

. A spatially uniform gravitational acceleration has zero tidal curvature even if the connection coefficients do not vanish in the original coordinates.

Equation of motion

An equation of motion determines how a physical state evolves. A variational action gives Euler-Lagrange equations; a phase-space action instead gives canonical Hamiltonian mechanics equations together with any imposed constraints.

Time reversal in classical mechanics ( $x (t) \mapsto x (- t)$ )