University Physics — Year 1 · Bachelor Year 1

18Non-Inertial Frames; Dynamics on Earth

Standing in a bus that turns left, you lean right, pushed by a force nobody exerts. A bucket of water spun on a turntable hollows its surface into a bowl. Under the dome of a museum a pendulum swings for hours, and the line of its swing creeps slowly round, a few degrees an hour, as if an invisible hand turned it. None of these is a new force: they are what Newton’s laws look like when the frame in which one describes the motion is itself accelerating or rotating. This chapter shows how velocities and accelerations transform between frames, how the laws of dynamics are repaired by adding inertial forces, and what those forces do on the most important rotating frame of all — the Earth.

The Foucault pendulum of the Panthéon in Paris, where it was first hung in 1851: 67\, m of wire, a 28\, kg bob, and a plane of swing that turns with respect to the floor. Photograph: Olga Khomitsevich, CC BY 2.0. — The Foucault pendulum of the Panthéon in Paris, where it was first hung in 1851: $67\,\mathrm{m}$ of wire, a $28\,\mathrm{kg}$ bob, and a plane of swing that turns with respect to the floor. Photograph: Olga Khomitsevich, CC BY 2.0.

18.1 Changing frame: kinematics

Definition 18.1 (Relative motion; frame in translation, in rotation)

Let $\mathcal R$ be a frame (origin $O$ ) and $\mathcal R'$ another (origin $O'$ ) moving with respect to it. $\mathcal R'$ is in translation if its axes keep fixed directions in $\mathcal R$ (every point of $\mathcal R'$ has the velocity of $O'$ ); it is in rotation about a fixed axis $\Delta$ if $\Delta$ is fixed in both frames and $\mathcal R'$ turns about it at the angular velocity $\Omega(t)$ . The velocity and acceleration of a point $M$ relative to $\mathcal R'$ , $\vect v'$ and $\vect a'$ , are those a stationary observer of $\mathcal R'$ measures.

Theorem 18.2 (Composition of velocities and accelerations)

$\mathcal R'$ in translation with velocity $\vect v_{O'}$ and acceleration $\vect a_{O'}$ :

$\vect v = \vect v' + \vect v_{O'}, \qquad \vect a = \vect a' + \vect a_{O'} .$
$\mathcal R'$ rotating about the fixed axis $\Delta$ at $\Omega$ (vector $\vect\Omega = \Omega\vect e_z$ along the axis); $H$ the projection of $M$ on the axis:

$\vect v = \vect v' + \vect\Omega\wedge\vect{HM}, \qquad \vect a = \vect a' + \underbrace{\big(-\Omega^2\vect{HM} + \dot\Omega\,\vect e_z\wedge\vect{HM}\big)}_{\vect a_e} + \underbrace{2\vect\Omega\wedge\vect v'}_{\vect a_C} .$

$\vect a_e$ is the entrainment acceleration (the acceleration of the point of $\mathcal R'$ that coincides with $M$ : centripetal toward the axis for uniform rotation), $\vect a_C = 2\vect\Omega\wedge\vect v'$ the Coriolis acceleration, present only when $M$ moves relative to the rotating frame.

Proof. Translation: $\vect{OM} = \vect{OO'} + \vect{O'M}$ with the components of $\vect{O'M}$ in the common basis; differentiate twice. Rotation: use cylindrical coordinates about $\Delta$ in $\mathcal R$ , $(r, \theta, z)$ , and in $\mathcal R'$ , $(r, \theta', z)$ with $\theta = \theta' + \int\Omega\,\dd t$ . Then Theorem 11.7 in $\mathcal R$ with $\dot\theta = \dot\theta' + \Omega$ : $\vect v = \dot r\vect e_r + r(\dot\theta' + \Omega)\vect e_\theta + \dot z\vect e_z = \vect v' + r\Omega\vect e_\theta$ , and $r\Omega\vect e_\theta = \vect\Omega\wedge\vect{HM}$ . For the acceleration, $\ddot\theta = \ddot\theta' + \dot\Omega$ : $\vect a = [\ddot r - r(\dot\theta' + \Omega)^2]\vect e_r + [r(\ddot\theta' + \dot\Omega) + 2\dot r(\dot\theta' + \Omega)]\vect e_\theta + \ddot z\vect e_z$ ; expanding and grouping, $\vect a = \vect a' - r\Omega^2\vect e_r + r\dot\Omega\vect e_\theta + 2\Omega(-r\dot\theta'\vect e_r + \dot r\vect e_\theta)$ , and the last bracket is $2\vect\Omega\wedge\vect v'$ since $\vect e_z\wedge\vect e_r = \vect e_\theta$ , $\vect e_z\wedge\vect e_\theta = -\vect e_r$ . ∎

Example 18.3 (Walking on a turntable)

Exercise 11.12 computed the acceleration of a person walking radially at $u$ on a turntable rotating at $\Omega$ : $-ut\Omega^2\vect e_r + 2u\Omega\vect e_\theta$ . In the new language: relative acceleration zero, entrainment $-r\Omega^2\vect e_r$ , Coriolis $2\vect\Omega\wedge u\vect e_r = 2u\Omega\vect e_\theta$ .

18.2 Dynamics in a non-inertial frame: inertial forces

Theorem 18.4 (Newton’s law in a non-inertial frame)

In a frame $\mathcal R'$ that is not inertial, the motion of a particle obeys

m\vect a' = \sum\vect F + \vect F_e + \vect F_C, \qquad \vect F_e = -m\vect a_e, \qquad \vect F_C = -m\vect a_C = -2m\vect\Omega\wedge\vect v' :

Newton’s second law holds provided one adds to the real forces the entrainment force (for a uniformly rotating frame, the centrifugal force $+m\Omega^2\vect{HM}$ , directed away from the axis; for a frame in translation, $-m\vect a_{O'}$ ) and the Coriolis force. These inertial forces are not interactions: no body exerts them, they have no reaction, and they vanish in an inertial frame.

Proof. $m\vect a = \sum\vect F$ in the inertial frame $\mathcal R$ ; substitute $\vect a = \vect a' + \vect a_e + \vect a_C$ and move the two terms to the right. ∎

Example 18.5 (The leaning passenger, the plumb bob)

In a bus braking at $a = 3.0\,\mathrm{m}/\mathrm{s}^{2}$ , a hanging strap feels, in the bus frame, gravity plus a forward inertial force $ma$ : it hangs along the apparent gravity $\vect g' = \vect g - \vect a_{O'}$ , tilted forward by $\arctan(a/g) = 17^\circ$ . A passenger who wants to stand still must lean into that apparent vertical. In an elevator falling freely, $\vect g' = \vect 0$ : weightlessness, seen from inside.

Inertial forces. Left: in a braking bus the plumb bob hangs along the apparent gravity g - a_O', tilted forward. Right: in a frame rotating at , a particle feels a centrifugal force away from the axis and, if it moves, a Coriolis force perpendicular to its relative velocity. — Inertial forces. Left: in a braking bus the plumb bob hangs along the apparent gravity $\vect g - \vect a_{O'}$ , tilted forward. Right: in a frame rotating at $\Omega$ , a particle feels a centrifugal force away from the axis and, if it moves, a Coriolis force perpendicular to its relative velocity.

Example 18.6 (The spinning bucket)

Water in a bucket rotating at $\Omega$ about its vertical axis is at rest in the rotating frame under gravity and the centrifugal force: its free surface is perpendicular to the apparent gravity $\vect g' = -g\vect e_z + \Omega^2r\vect e_r$ (the surface of a fluid at rest is normal to the field, Chapter 21). Its slope is $\dd z/\dd r = \Omega^2r/g$ , so $z = \Omega^2r^2/2g$ : a paraboloid. At $10\,\mathrm{rad}/\mathrm{s}$ the rim of a $10\,\mathrm{cm}$ bucket stands $5\,\mathrm{cm}$ above the center — the principle of the liquid-mirror telescope, whose spinning mercury takes exactly the parabolic shape a mirror needs.

Method 18.7 (Working in a non-inertial frame)

Choose the frame in which the question is simplest (the bus, the turntable, the Earth); list the real forces; add $-m\vect a_e$ and, if the body moves in the frame, $-2m\vect\Omega\wedge\vect v'$ ; then proceed as in Method 12.10. Check: inertial forces never appear in the energy balance as “work done by someone” — the Coriolis force, perpendicular to $\vect v'$ , does no work; the centrifugal force derives from the potential energy $-\tfrac12 m\Omega^2r^2$ in a uniformly rotating frame.

18.3 The terrestrial frame

Definition 18.8 (Geocentric and terrestrial frames)

The geocentric frame has its origin at the Earth’s center and axes pointing to fixed stars; it is inertial to an excellent approximation for motions of a day or less (its own acceleration around the Sun is only $6 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$ , and nearly uniform over the Earth). The terrestrial frame, attached to the ground, rotates with respect to it about the polar axis at

\Omega = \frac{2\pi}{86\,164\,\mathrm{s}} = 7.29 \times 10^{-5}\,\mathrm{rad}/\mathrm{s} .

Proposition 18.9 (Weight, and the variation of $g$ )

In the terrestrial frame a body at rest feels the gravitational attraction $m\vect{\mathcal G}$ and the centrifugal force $m\Omega^2\vect{HM}$ ; their sum is what a balance measures and a plumb line shows: the weight $m\vect g = m\vect{\mathcal G} + m\Omega^2\vect{HM}$ . The centrifugal term is largest at the equator, $\Omega^2R = 0.034\,\mathrm{m}/\mathrm{s}^{2}$ , $0.35\%$ of $g$ ; it reduces $g$ from the poles to the equator and tilts the vertical away from the Earth’s center by up to $0.1^\circ$ at mid-latitudes.

Proof. Apply Theorem 18.4 with $\vect v' = \vect 0$ (no Coriolis force); $\Omega^2R = (7.29\times10^{-5})^2 \times 6.37\times10^6$ . The deviation angle follows from the triangle of $m\vect{\mathcal G}$ and the centrifugal force (Exercise 18.6). ∎

Proposition 18.10 (The Coriolis force on Earth)

At latitude $\lambda$ , a body moving horizontally at speed $v'$ feels a horizontal Coriolis force of magnitude

F_C = 2m\Omega v'\sin\lambda ,

directed to the right of the motion in the northern hemisphere and to the left in the southern one, whatever the direction of travel; a body falling vertically is deflected eastward by $2m\Omega v'\cos\lambda$ .

Proof. Decompose $\vect\Omega = \Omega(\cos\lambda\,\vect e_N + \sin\lambda\,\vect e_z)$ (north and local vertical). For horizontal $\vect v'$ , $-2m\vect\Omega\wedge\vect v'$ has a horizontal part $-2m\Omega\sin\lambda\,\vect e_z\wedge\vect v'$ — rotated $90^\circ$ clockwise from $\vect v'$ seen from above, for $\lambda > 0$ — and a vertical part (negligible against gravity). For vertical $\vect v' = -v'\vect e_z$ : $-2m\Omega\cos\lambda\,\vect e_N\wedge(-v'\vect e_z) = 2m\Omega v'\cos\lambda\, \vect e_E$ , eastward. ∎

Left: at latitude the Earth’s rotation vector has a vertical component and a northward one . Right: seen from above in the northern hemisphere, a body moving horizontally is pushed to its right by the Coriolis force and its path curves clockwise. — Left: at latitude $\lambda$ the Earth’s rotation vector has a vertical component $\Omega\sin\lambda$ and a northward one $\Omega\cos\lambda$ . Right: seen from above in the northern hemisphere, a body moving horizontally is pushed to its right by the Coriolis force and its path curves clockwise.

Example 18.11 (Small forces, big consequences)

A train at $100\,\mathrm{km}/\mathrm{h}$ at latitude $45^\circ$ : $a_C = 2 \times 7.29\times 10^{-5} \times 27.8 \times 0.71 = 2.9 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$ , three thousandths of $g$ — the right-hand rail of a one-way track wears a little faster. Air flowing at $20\,\mathrm{m}/\mathrm{s}$ toward a low-pressure center is turned right all the way in and ends up circling it counterclockwise: every cyclone of the northern hemisphere turns that way, every southern one the other. A stone dropped down a $100\,\mathrm{m}$ shaft at $45^\circ$ lands $1.6\,\mathrm{cm}$ east of the plumb line (Exercise 18.7).

Proposition 18.12 (Foucault’s pendulum)

A pendulum swinging with small amplitude at latitude $\lambda$ keeps its plane of oscillation fixed with respect to the geocentric frame’s local vertical rotation only: seen in the terrestrial frame, the plane turns clockwise (from above, northern hemisphere) at the angular rate $\Omega\sin\lambda$ , i.e. one full turn in

T_F = \frac{23.93\,\mathrm{h}}{\sin\lambda} .

Partial proof. At the pole ( $\lambda = 90^\circ$ ) the argument is geometric: no horizontal force acts on the bob except the tension, whose horizontal component is along the swing, so in the geocentric frame the plane of swing is fixed while the Earth turns beneath it at $\Omega$ ; seen from the ground it turns at $-\Omega$ . At latitude $\lambda$ , the horizontal Coriolis force is $-2m\Omega\sin\lambda\,\vect e_z\wedge\vect v'$ : exactly what a frame rotating at $\Omega\sin\lambda$ about the local vertical would produce. In a frame turning with the plane at $-\Omega\sin\lambda$ , that force disappears to first order and the pendulum is an ordinary plane oscillator; hence the plane precesses at $\Omega\sin\lambda$ in the terrestrial frame. The full calculation, with the neglected second-order terms, is the weekend problem. ∎

Remark 18.13 (Tides, in two lines)

In the frame of the Earth’s center, which falls freely toward the Moon, the Moon’s attraction is not uniform: a little stronger on the near side, weaker on the far side. The difference, of order $2GM_{\text{Moon}}R/d^3 \approx 1 \times 10^{-6}\,\mathrm{m}/\mathrm{s}^{2}$ , stretches the oceans into two bulges — two tides a day — and the Sun’s contribution, $46\%$ of the Moon’s, makes spring and neap tides as the two align or cross (Exercise 18.12).

18.4 Exercises

Exercise 18.1 ★

A plumb bob hangs in a car accelerating at $2.5\,\mathrm{m}/\mathrm{s}^{2}$ : angle with the vertical, and the tension in units of $mg$ . Same car braking at the same rate?

Solution

Solution of Exercise 18.1.

$\tan\alpha = a/g = 0.255$ , $\alpha = 14^\circ$ (the bob trails backward); $T = m\sqrt{g^2 + a^2} = 1.03\,mg$ . Braking: same angle and tension, bob tilted forward.

Exercise 18.2 ★

A child sits $3.0\,\mathrm{m}$ from the axis of a merry-go-round turning once every $4.0\,\mathrm{s}$ . Centrifugal acceleration in the rotating frame, as a fraction of $g$ ; what real force provides it?

Solution

Solution of Exercise 18.2.

$\Omega = 2\pi/4.0 = 1.57\,\mathrm{rad}/\mathrm{s}$ ; $\Omega^2r = 7.4\,\mathrm{m}/\mathrm{s}^{2} = 0.75g$ , directed outward in the rotating frame; the seat (friction and restraint) supplies the real inward force.

Exercise 18.3 ★

Compute the Earth’s angular velocity $\Omega$ , the centrifugal acceleration at the equator, and its fraction of $g$ . At what rotation period would objects at the equator float off?

Solution

Solution of Exercise 18.3.

$\Omega = 2\pi/86164 = 7.29 \times 10^{-5}\,\mathrm{rad}/\mathrm{s}$ ; $\Omega^2R = 0.034\,\mathrm{m}/\mathrm{s}^{2}$ , $0.35\%$ of $g$ . Floating off needs $\Omega^2R = g$ : $T = 2\pi\sqrt{R/g} = 84\,\mathrm{min}$ — seventeen times faster than now.

Exercise 18.4 ★

A train of $400\,\mathrm{t}$ runs at $100\,\mathrm{km}/\mathrm{h}$ at latitude $45^\circ$ . Coriolis acceleration and the sideways force on the rails; which rail is pushed, heading north? heading east?

Solution

Solution of Exercise 18.4.

$a_C = 2\Omega v\sin\lambda = 2 \times 7.29\times10^{-5} \times 27.8 \times 0.707 = 2.9 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$ ; $F = 1.1\,\mathrm{kN}$ . Heading north: pushed east, onto the right-hand (eastern) rail; heading east: pushed south, again the right-hand rail.

Exercise 18.5 ★★

A bucket of radius $10\,\mathrm{cm}$ spins at $2.0\,\mathrm{rad}/\mathrm{s}$ , then $10\,\mathrm{rad}/\mathrm{s}$ . Rise of the water at the rim relative to the center. Why is a spinning mercury bath a good telescope mirror?

Solution

Solution of Exercise 18.5.

$z = \Omega^2r^2/2g$ : $2.0\,\mathrm{mm}$ at $2\,\mathrm{rad}/\mathrm{s}$ , $5.1\,\mathrm{cm}$ at $10\,\mathrm{rad}/\mathrm{s}$ . A paraboloid focuses a parallel beam to a point: spinning mercury is a perfect, cheap, self-polishing parabolic mirror (that cannot be tilted).

Exercise 18.6 ★★

Show that the plumb line at latitude $\lambda$ deviates from the Earth’s center by about $\Omega^2R\sin\lambda\cos\lambda/g$ radians, and compute it at $45^\circ$ in minutes of arc. By how much is $g$ smaller at the equator than at the pole from rotation alone?

Solution

Solution of Exercise 18.6.

The centrifugal acceleration $\Omega^2R\cos\lambda$ points away from the axis; its component perpendicular to the local vertical is $\Omega^2R\cos\lambda \sin\lambda$ , which tilts $\vect g$ by $\delta \approx \Omega^2R\sin\lambda\cos\lambda/g = 0.034 \times 0.5/9.81 = 1.7 \times 10^{-3}\,\mathrm{rad} = 6'$ . Equator: $g$ reduced by $\Omega^2R = 0.034\,\mathrm{m}/\mathrm{s}^{2}$ from rotation (the flattening adds more).

Exercise 18.7 ★★

A stone falls from rest down a $100\,\mathrm{m}$ shaft at latitude $45^\circ$ . Treat the Coriolis acceleration $2\Omega v\cos\lambda$ (eastward) as a small perturbation with $v = gt$ ; integrate twice to find the eastward deviation at the bottom. Why eastward?

Solution

Solution of Exercise 18.7.

$\ddot x_E = 2\Omega gt\cos\lambda$ : $x_E = \tfrac13\Omega g\cos\lambda\,t^3$ ; $t = \sqrt{2h/g} = 4.5\,\mathrm{s}$ : $x_E = \tfrac13 \times 7.29\times10^{-5} \times 9.81 \times 0.707 \times 92 = 1.6\,\mathrm{cm}$ . Eastward: the top of the shaft moves east faster (larger radius) than the bottom; the stone keeps that excess — or, in the rotating frame, the Coriolis force on a downward velocity points east.

Exercise 18.8 ★★

Precession period of a Foucault pendulum in Paris ( $\lambda = 48.9^\circ$ ), at the pole, and at the equator. Through what angle does the plane turn in one hour in Paris?

Solution

Solution of Exercise 18.8.

$T_F = 23.93/\sin\lambda$ : Paris $31.8\,\mathrm{h}$ ; pole $23.9\,\mathrm{h}$ ; equator infinite (no precession). Paris: $360^\circ/31.8 = 11.3^\circ$ per hour.

Exercise 18.9 ★★

Air flows at $20\,\mathrm{m}/\mathrm{s}$ toward a low-pressure center at $45^\circ$ north. Coriolis acceleration; in which sense does the air end up circling the center? Why do small eddies (a sink, a tornado) not obey this rule?

Solution

Solution of Exercise 18.9.

$a_C = 2\Omega v\sin\lambda = 2.1 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$ , to the right. Air converging from all sides is deflected right, so it spirals in counterclockwise (seen from above): the northern cyclone. A sink drains in seconds and a tornado forms in minutes: the Coriolis deflection over such times is negligible against any initial swirl.

Exercise 18.10 ★★★

A helium balloon on a string floats inside a car. When the car accelerates forward, which way does the balloon lean? Explain with the apparent gravity and buoyancy.

Solution

Solution of Exercise 18.10.

In the car’s frame the apparent gravity $\vect g - \vect a$ tilts backward. The air, heavier, “falls” backward; the balloon, lighter than the air it displaces, is buoyed up the apparent vertical — it leans forward, opposite to a plumb bob.

Exercise 18.11 ★★★

A space station is a ring of radius $100\,\mathrm{m}$ spun to give $1g$ of artificial gravity at the rim. Rotation period? An astronaut walks along the rim at $1.5\,\mathrm{m}/\mathrm{s}$ : Coriolis acceleration, and its effect on his apparent weight when walking with and against the rotation. He drops a ball from $2.0\,\mathrm{m}$ : roughly how far from his feet does it land?

Solution

Solution of Exercise 18.11.

$\Omega^2R = g$ : $\Omega = 0.31\,\mathrm{rad}/\mathrm{s}$ , $T = 20\,\mathrm{s}$ . $a_C = 2\Omega v = 0.94\,\mathrm{m}/\mathrm{s}^{2}$ , about $0.1g$ : walking with the rotation he feels heavier by $10\%$ (the Coriolis force points outward), against it lighter. A dropped ball: in the inertial frame it moves straight while the floor curves up; over the fall time $\sqrt{2h/g} = 0.64\,\mathrm{s}$ the deviation is of order $\tfrac13\Omega\,g\,t^3 \cdot 2 \approx 2 \times 0.31 \times 9.81 \times 0.26/3 \approx 0.5\,\mathrm{m}$ — a few decimeters, against the rotation.

Exercise 18.12 ★★★

Show that the difference between the Moon’s gravitational acceleration at the Earth’s center and at the point of the surface nearest the Moon is about $2GM_Mr/d^3$ ( $r$ the Earth’s radius, $d$ the Earth–Moon distance); compute it, and the same for the Sun; deduce the ratio of solar to lunar tides.

Solution

Solution of Exercise 18.12.

$GM/(d - r)^2 - GM/d^2 \approx 2GMr/d^3$ (first order in $r/d$ ). Moon: $2 \times 6.67\times10^{-11} \times 7.35\times10^{22} \times 6.37\times10^6/ (3.84\times10^8)^3 = 1.1 \times 10^{-6}\,\mathrm{m}/\mathrm{s}^{2}$ ; Sun: $2 \times 1.33\times10^{20} \times 6.37\times10^6/(1.50\times10^{11})^3 = 5.0 \times 10^{-7}\,\mathrm{m}/\mathrm{s}^{2}$ ; ratio $0.46$ : the Sun’s tides are almost half the Moon’s, though its attraction is $180$ times larger — tides go as $M/d^3$ .

18.5 Problem: Foucault’s pendulum

Problem 18.1

Weekend problem — a 28-kilogram bob on a 67-meter wire under a dome: why its plane of swing turns, how fast, what it proves, and how small the force is that does it

In 1851 a pendulum of length $\ell = 67\,\mathrm{m}$ and mass $m = 28\,\mathrm{kg}$ swung under the dome of the Panthéon in Paris, latitude $\lambda = 48.85^\circ$ , with an amplitude of about $x_0 = 3.0\,\mathrm{m}$ . $\Omega = 7.29 \times 10^{-5}\,\mathrm{rad}/\mathrm{s}$ , $g = 9.81\,\mathrm{m}/\mathrm{s}^{2}$ .

Part I — The pendulum as such.

Period of small oscillations; check that the amplitude is small ( $x_0/\ell$ ).
Maximal speed of the bob, and its maximal height above the lowest point.
Tension at the lowest point, in units of $mg$ .
Estimate the air drag on the bob at maximal speed (quadratic law, $C_xS \approx 0.05\,\mathrm{m}^{2}$ , $\rho = 1.2\,\mathrm{kg}/\mathrm{m}^{3}$ ) and compare with the maximal restoring force $m\omega_0^2x_0$ .
Why a long wire and a heavy bob? (Two reasons, one about the period, one about damping.)

Part II — At the pole. Imagine the same pendulum at the north pole.

In the geocentric frame, list the forces on the bob and show that nothing can turn its plane of swing.
Deduce what an observer standing on the ice sees the plane do, and in how long it returns to its initial direction.
Describe the path of the bob on the ground (a rosette): how many petals in one day?

Part III — At latitude $\lambda$ : the Coriolis term. Use local axes: $x$ east, $y$ north, $z$ up; small oscillations, so the motion is nearly horizontal, $\vect v' \approx (\dot x, \dot y, 0)$ ; the restoring force is $-m\omega_0^2(x, y, 0)$ with $\omega_0 = \sqrt{g/\ell}$ .

Write $\vect\Omega$ in these axes.
Compute $-2m\vect\Omega\wedge\vect v'$ and keep its horizontal components; show they involve only $\Omega\sin\lambda$ .
Write the two equations of motion for $x$ and $y$ ; set $\omega_F = \Omega\sin\lambda$ and compare $\omega_F$ with $\omega_0$ .
Introduce $u = x + jy$ and show that $\ddot u + 2j\omega_F\dot u + \omega_0^2u = 0$ .
Try $u = \eu^{j\alpha t}$ : find the two values of $\alpha$ , and show that for $\omega_F \ll \omega_0$ they are $\alpha \approx \pm\omega_0 - \omega_F$ .
Deduce that $u \approx \eu^{-j\omega_Ft}(A\eu^{j\omega_0t} + B\eu^{-j\omega_0t})$ : interpret the factor $\eu^{-j\omega_Ft}$ — in which sense, at what rate, does the plane of swing turn?
Compute the precession period in Paris and the angle turned in one hour.
Why does the plane not turn at the equator?

Part IV — How small the force is.

Compute the maximal Coriolis force on the bob and compare it with the weight and with the maximal restoring force.
By how much is the swing’s end point displaced sideways during one half-period? (The plane turns by $\omega_FT/2$ .)
In 1851 the bob carried a stylus that knocked over small pegs placed in a ring at the amplitude: about how long before the next peg fell?
The pendulum was launched by burning a thread that held the bob aside. Why not push it by hand?
Air drag stops a pendulum after a few hours. Modern museum pendulums use a small electromagnet at the bottom to restore energy without touching the plane of swing: why must that kick be strictly radial?

Part V — What it proves, and what else it does.

Foucault’s demonstration convinced the public of the Earth’s rotation. What exactly does the precession rate measure, and could the experiment determine the latitude?
At which latitude would the plane turn once in exactly $48\,\mathrm{h}$ ?
A gyroscope (a fast-spinning wheel in gimbals) keeps its axis fixed in the geocentric frame. Explain in one sentence why it too reveals the rotation, and name an application.
Summarize: the one inertial force responsible, its dependence on latitude, and the period it gives in Paris.

Solution

Solution of Problem 18.1.

1. $T = 2\pi\sqrt{67/9.81} = 16.4\,\mathrm{s}$ ; $x_0/\ell = 0.045$ ( $2.6^\circ$ ): small.

2. $v_{\max} = x_0\omega_0 = 3.0 \times 0.383 = 1.15\,\mathrm{m}/\mathrm{s}$ ; height $x_0^2/2\ell = 6.7\,\mathrm{cm}$ .

3. $T = mg + mv_{\max}^2/\ell = mg(1 + 0.002)$ : $1.002\,mg$ .

4. $\tfrac12\rho C_xSv_{\max}^2 = 0.5 \times 1.2 \times 0.05 \times 1.3 = 0.04\,\mathrm{N}$ , against $m\omega_0^2x_0 = 12\,\mathrm{N}$ : three hundred times smaller, yet enough to stop the pendulum in a few hours.

5. A long wire gives a long period, hence a slow bob and a small relative drag, and makes the precession per swing visible; a heavy bob stores much energy for the drag to eat — hours of swinging.

6. Weight (vertical) and tension (in the plane of the wire and the vertical, i.e. in the plane of swing): no force has a component perpendicular to that plane.

7. The plane is fixed among the stars; the ice turns beneath it eastward at $\Omega$ , so the observer sees the plane turn clockwise, back to its start after one sidereal day, $23.93\,\mathrm{h}$ .

8. Each half-swing traces a slightly rotated chord: a rosette; $86164/8.2 \approx 10500$ half-swings, each $0.034^\circ$ further round.

9. $\vect\Omega = \Omega(0, \cos\lambda, \sin\lambda)$ .

10. With $\vect v' = (\dot x, \dot y, 0)$ : $\vect\Omega\wedge\vect v' = \Omega(-\sin\lambda\,\dot y,\ \sin\lambda\,\dot x,\ -\cos\lambda\,\dot x)$ , so $-2m\vect\Omega\wedge\vect v'$ has the horizontal components $2m\Omega\sin\lambda\,(\dot y, -\dot x)$ : only $\Omega\sin\lambda$ enters.

11. $\ddot x = -\omega_0^2x + 2\omega_F\dot y$ , $\ddot y = -\omega_0^2y - 2\omega_F\dot x$ ; $\omega_F = 7.29\times10^{-5} \times 0.753 = 5.5 \times 10^{-5}\,\mathrm{rad}/\mathrm{s} \ll \omega_0 = 0.38\,\mathrm{rad}/\mathrm{s}$ .

12. $\ddot u = -\omega_0^2u + 2\omega_F(\dot y - j\dot x) = -\omega_0^2u - 2j\omega_F\dot u$ .

13. $-\alpha^2 - 2\omega_F\alpha + \omega_0^2 = 0$ : $\alpha = -\omega_F \pm \sqrt{\omega_F^2 + \omega_0^2} \approx -\omega_F \pm \omega_0$ .

14. The bracket is a plane oscillation (a fixed line through the origin for real $A = B$ ); the factor $\eu^{-j\omega_Ft}$ rotates the whole pattern clockwise at $\omega_F$ : the plane of swing precesses clockwise at $\Omega\sin\lambda$ .

15. $T_F = 2\pi/\omega_F = 1.14 \times 10^{5}\,\mathrm{s} = 31.8\,\mathrm{h}$ ; $11.3^\circ$ per hour.

16. $\sin\lambda = 0$ : the local vertical component of $\vect\Omega$ vanishes; the horizontal Coriolis force is zero for horizontal motion.

17. $2m\omega_Fv_{\max} = 2 \times 28 \times 5.5\times10^{-5} \times 1.15 = 3.5\,\mathrm{mN}$ ; weight $275\,\mathrm{N}$ ( $10^{-5}$ of it), restoring force $m\omega_0^2x_0 = 12\,\mathrm{N}$ : three thousand times smaller.

18. $\omega_FT/2 = 5.5\times10^{-5} \times 8.2 = 4.5 \times 10^{-4}\,\mathrm{rad}$ : $x_0 \times 4.5\times10^{-4} = 1.4\,\mathrm{mm}$ per half-swing.

19. Pegs a centimeter apart on the ring fall every $\sim 7$ half-swings, about a minute.

20. A hand push gives a sideways velocity: the bob then describes an ellipse, whose own slow precession (an effect of the finite amplitude) masks Foucault’s.

21. A radial kick adds energy along the swing without adding transverse velocity, so the plane is left untouched; a sideways component would drive the pendulum into an ellipse and fake or hide the precession.

22. The precession rate measures $\Omega\sin\lambda$ , the vertical component of the Earth’s rotation; knowing $\Omega$ from astronomy, the rate gives the latitude (and vice versa).

23. $\sin\lambda = 23.93/48 = 0.499$ : $\lambda = 30^\circ$ .

24. Its angular momentum is conserved, so its axis stays fixed among the stars while the Earth turns under it: the gyrocompass, which finds true north without a magnet.

25. The Coriolis force $-2m\vect\Omega\wedge\vect v'$ ; its horizontal part grows as $\sin\lambda$ ; one turn in $23.93\,\text{h}/\sin\lambda$ , $31.8\,\mathrm{h}$ in Paris.

Terms defined in this chapter

See all 393 terms in the glossary