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Newton's law in a rotating frame, the centrifugal force $-m\vec{\Omega} \times (\vec{\Omega} \times \vec{r})$ and the Coriolis force $-2m\vec{\Omega} \times \vec{v}$, effective gravity, free fall on Earth and the Foucault pendulum.
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By the end of this lesson you will be able to write Newton's second law in a rotating frame and use the centrifugal and Coriolis forces to predict motion on the Earth and aboard spinning spacecraft.
The first lesson of this course found the centripetal and Coriolis terms of the acceleration in polar coordinates, measured in an inertial frame. You know the angular velocity vector and cross products, and that the Earth turns once a day. This lesson writes Newton's second law in a frame that itself rotates, which is the frame every experiment on Earth is actually done in.
| Term | What it means |
|---|---|
| Rotating frame | A frame turning at angular velocity $\vec{\Omega}$ relative to an inertial one. |
| Inertial force | A term added to $\vec{F}$ so that $m\vec{a} = \vec{F}_{\text{eff}}$ holds in a noninertial frame; not caused by any body. |
| Centrifugal force | $-m\vec{\Omega} \times (\vec{\Omega} \times \vec{r})$, outward from the rotation axis with size $m\Omega^2\rho$. |
| Coriolis force | $-2m\vec{\Omega} \times \vec{v}$, perpendicular to the velocity in the rotating frame. |
| Effective gravity | $\vec{g}_{\text{eff}} = \vec{g} - \vec{\Omega} \times (\vec{\Omega} \times \vec{r})$, what a plumb line points along. |
| Foucault pendulum | A long pendulum whose swing plane turns at $\Omega\sin\lambda$, showing the Earth rotates. |
| Sidereal day | $23.93$ hours, the Earth's rotation period relative to the stars. |
For any vector $\vec{Q}$, its rate of change seen from an inertial frame and from a frame rotating at $\vec{\Omega}$ differ by the rotation carrying it round:
$$\left(\frac{d\vec{Q}}{dt}\right)_{\text{inertial}} = \left(\frac{d\vec{Q}}{dt}\right)_{\text{rot}} + \vec{\Omega} \times \vec{Q}.$$
Applying this twice to the position vector and substituting into $m\ddot{\vec{r}}_{\text{inertial}} = \vec{F}$ gives, for steady rotation,
$$m\ddot{\vec{r}} = \vec{F} - 2m\vec{\Omega} \times \dot{\vec{r}} - m\vec{\Omega} \times (\vec{\Omega} \times \vec{r}),$$
with every derivative now taken in the rotating frame. The two extra terms are inertial forces: the Coriolis force $-2m\vec{\Omega} \times \vec{v}$, which acts only on moving bodies and is always perpendicular to their velocity, and the centrifugal force, which points straight away from the axis with size $m\Omega^2\rho$, where $\rho$ is the distance from the axis. Neither comes from any other body; both are the price of describing motion from a turning platform, and both are exactly the acceleration terms of the first lesson moved to the force side.
Another way: picture
Sit on a spinning merry-go-round and roll a ball straight toward the center. To someone on the ground the ball goes straight; to you it curves off to the side, as if pushed. That apparent push is the Coriolis force. Meanwhile you feel yourself pressed outward against the rail, though nothing pushes you out: the centrifugal force, which is really your body trying to go straight.
Another way: steps
A body at rest on the Earth's surface feels no Coriolis force, but it does feel the centrifugal force $m\Omega^2R\cos\lambda$, pointing away from the Earth's axis. Adding it to true gravity gives the effective gravity $\vec{g}_{\text{eff}}$, which is what a scale weighs and what a plumb line hangs along.
At the equator the centrifugal acceleration is $R\Omega^2 = 0.034$ m/s², about a third of a percent of $g$, and it points straight up, so things weigh slightly less. At other latitudes it has a horizontal part, $R\Omega^2\sin\lambda\cos\lambda$, which tilts the plumb line by up to about $0.1°$ at $45°$. The Earth itself has responded to the same effective gravity over its history: its equatorial radius exceeds its polar radius by $21$ km, the shape a rotating fluid takes.
On a body moving over the Earth's surface, only the vertical component of $\vec{\Omega}$, $\Omega\sin\lambda$, gives a horizontal Coriolis force: $2m\Omega v\sin\lambda$, directed to the right of the motion in the Northern Hemisphere and to the left in the Southern. It vanishes at the equator and is largest at the poles.
The effect is small, a few thousandths of a meter per second squared for a car on a highway, but it acts on weather systems for days. Air flowing toward a low-pressure center is turned to the right, so it circulates counterclockwise around the low: that is why hurricanes approaching Florida and the Gulf Coast spin counterclockwise, and those near Australia spin clockwise. Long-range artillery and ballistic missiles must be corrected for it too.
A ball dropped from rest in the rotating frame starts moving downward, and the Coriolis force on that downward velocity points east. The deflection is small but systematic. Integrating to first order in $\Omega$, a ball falling a height $h$ in time $t = \sqrt{2h/g}$ lands a distance $x = \tfrac{1}{3}\Omega gt^3\cos\lambda$ east of the point directly below.
From $100$ m at the equator the deflection is about $2.2$ cm. The inertial explanation is simple: the top of the tower is farther from the Earth's axis than its base and so moves east slightly faster; the ball keeps that extra eastward speed as it falls. Experiments dropping balls down mine shafts in the nineteenth century confirmed the effect, one of the first direct demonstrations that the Earth turns.
In 1851 Léon Foucault hung a $67$ m pendulum in the Panthéon in Paris and showed that its plane of swing slowly turned. In the rotating frame of the floor, the Coriolis force $-2m\vec{\Omega} \times \vec{v}$ on the swinging bob has a horizontal part $2m\Omega\sin\lambda\,v$ that nudges the bob sideways on every swing. The result is that the swing plane turns at the rate $\Omega\sin\lambda$, clockwise seen from above in the Northern Hemisphere.
At the North Pole the plane turns once per sidereal day: the Earth simply turns under a pendulum that keeps its plane fixed in space. At the equator it does not turn at all. In between, a full turn takes $23.93/\sin\lambda$ hours, about $38$ hours in Washington, D.C. Many American science museums keep a Foucault pendulum knocking over pegs arranged in a circle, one every few minutes.
A ring-shaped space station spinning at angular velocity $\omega$ provides artificial gravity $R\omega^2$ at its rim. For Earth-like gravity with $R = 100$ m, $\omega = \sqrt{g/R} = 0.31$ rad/s, about three revolutions per minute. In the station's frame, astronauts stand on the rim with the centrifugal force playing the role of weight.
The Coriolis force is the catch. An astronaut climbing toward the hub at $1$ m/s feels a sideways acceleration $2\omega v = 0.63$ m/s², six percent of the artificial gravity, and turning the head quickly stimulates the inner ear in confusing ways. Studies of rotating rooms by NASA suggest people adapt to up to about $2$ to $4$ revolutions per minute, which means rings with radii of at least $50$ to $200$ m for comfortable gravity.
Checking an answer. Set $\Omega = 0$ and the answer must reduce to the inertial one. The Coriolis force must be perpendicular to the velocity and so do no work. In the Northern Hemisphere, horizontal deflections must be to the right. And the same physics must be explainable in the inertial frame, which is the best check of all.
Is the centrifugal force real? In the rotating frame it has every property of a force: it presses the astronaut into the floor, flattens the Earth and pulls water up the sides of a spinning bucket. But it has no source and no reaction partner, and it vanishes in an inertial frame, where the same effects are explained by bodies trying to move in straight lines.
The rule is to use one frame consistently. In an inertial frame, never add a centrifugal force: the centripetal acceleration is already on the other side of the equation. In a rotating frame, always add both inertial forces. Most sign errors in problems with rotation come from mixing the two, for example drawing a centrifugal force on a free-body diagram and then also writing $ma = mv^2/r$.
A hurricane forms around a region of low pressure over warm ocean. Air flows in toward the low, and the Coriolis force, $2\Omega v\sin\lambda$, turns it to the right in the Northern Hemisphere. The inflowing air therefore circles the center counterclockwise, spinning up into the storm that satellites photograph approaching the Gulf Coast and the Carolinas.
The same term explains why hurricanes almost never form within about $5°$ of the equator, where $\sin\lambda$ is too small for the Coriolis force to organize the inflow into rotation, and why they tend to curve northward and eastward as they travel. The National Hurricane Center's forecast models, run by NOAA in Miami, solve the equations of fluid motion in the Earth's rotating frame, with the Coriolis term included at every grid point.
Foucault pendulums hang in science museums across the United States: at the Smithsonian in Washington, at Griffith Observatory in Los Angeles, at the Franklin Institute in Philadelphia and elsewhere. A heavy bob on a long wire, often $10$ to $30$ m, swings slowly, and a ring of small pegs stands around it. Over the day, the bob knocks the pegs over one by one as its plane turns.
In Washington, at $38.9°$ north, the plane takes $23.93/\sin 38.9° = 38$ hours to turn once, about $9.5°$ per hour. Visitors see the Earth's rotation measured in a single room, with no reference to the stars. The long wire is not needed for the physics, which is independent of length, but it makes the swing slow and lets the bob keep swinging for hours with only a small electromagnetic push at the top of each swing.
It is often said that sinks and toilets drain counterclockwise in the Northern Hemisphere because of the Coriolis force. At the scale of a sink, draining in seconds, the Coriolis acceleration is about $10^{-4}$ m/s², far weaker than the swirl left by filling the basin or the shape of the drain. Carefully controlled experiments, with water left to settle for a day in a large symmetric tank, can show the effect; your bathroom cannot.
A second error is to add a centrifugal force in an inertial frame. In the ground's frame, a car rounding a curve has no outward force on it; the inward friction supplies its centripetal acceleration.
Find the centrifugal acceleration at the equator.
$R\Omega^2 = 6.371 \times 10^6 \times (7.292 \times 10^{-5})^2 = 0.0339\ \text{m/s}^2$
$\Omega = 2\pi/(23.93\ \text{h})$.
Find its size at latitude $45°$.
$R\Omega^2\cos 45° = 0.0339 \times 0.707 = 0.0240\ \text{m/s}^2$
The distance to the axis is $R\cos\lambda$.
Find its horizontal part there.
$R\Omega^2\cos\lambda\sin\lambda = 0.0240 \times 0.707 = 0.0170\ \text{m/s}^2$
The centrifugal force points away from the axis, not straight up.
Find the plumb line's tilt from true vertical.
$\alpha \approx \dfrac{0.0170}{9.8} = 1.73 \times 10^{-3}\ \text{rad}$
Horizontal part over gravity, for a small angle.
Convert to degrees.
$\alpha = 0.099°$
Tilted toward the equator by about a tenth of a degree.
A station of radius $100$ m must give $9.8$ m/s² at its rim. Write the condition.
$R\omega^2 = g$
Centrifugal acceleration equals Earth's gravity.
Solve for the spin rate.
$\omega = \sqrt{\dfrac{9.8}{100}} = 0.313\ \text{rad/s}$
Take the root.
Convert to revolutions per minute.
$\dfrac{0.313 \times 60}{2\pi} = 2.99\ \text{rpm}$
About three turns a minute.
Find the Coriolis acceleration on an astronaut climbing at $1.0$ m/s.
$2\omega v = 2 \times 0.313 \times 1.0 = 0.63\ \text{m/s}^2$
Perpendicular to the climb.
Compare it with the artificial gravity.
$\dfrac{0.63}{9.8} = 0.064$
A noticeable sideways push.
A ball is dropped $100$ m at the equator. Find the fall time.
$t = \sqrt{\dfrac{2 \times 100}{9.8}} = 4.52\ \text{s}$
Ignoring air resistance.
Write the eastward Coriolis acceleration during the fall.
$\ddot{x} = 2\Omega v_{\text{down}}\cos\lambda = 2\Omega gt\cos\lambda$
The downward velocity grows as $gt$.
Integrate twice from rest.
$x = \tfrac{1}{3}\Omega gt^3\cos\lambda$
$\int\int 2\Omega gt\,dt\,dt = \Omega gt^3/3$.
Substitute the values at the equator.
$x = \tfrac{1}{3} \times 7.29 \times 10^{-5} \times 9.8 \times 4.52^3$
$\cos 0° = 1$.
Evaluate the deflection.
$x = 0.022\ \text{m} = 2.2\ \text{cm}$
Eastward.
Check it from the inertial frame.
$\Delta v_{\text{east}} = \Omega h = 7.3 \times 10^{-3}\ \text{m/s} \text{ at the top}$
The top moves east faster than the base; the ball keeps that speed, which gives the same order of drift.
Write the turning period.
$T_F = \dfrac{23.93}{\sin\lambda}\ \text{h}$
The sidereal day over the sine of the latitude.
Substitute the sine.
$T_F = \dfrac{23.93}{0.5}$
$\sin 30° = 0.5$.
Evaluate the period.
A Foucault pendulum $16$ m long is set swinging exactly on the equator. How does its plane of swing turn relative to the floor?
Complete the worked solution: a Foucault pendulum hangs where $\sin\lambda = 0.6$, and the Earth turns $15°$ per hour. Find the rate at which its plane turns in degrees per hour, the time for a full turn in hours, and the angle it turns in $2$ hours.
Multiply the Earth's rate by the sine of the latitude.
$\text{rate} = 15\sin\lambda =$ r
The vertical component of the rotation.
Divide a full turn by the rate.
$T = \dfrac{360°}{\text{rate}} =$ p
Hours for the plane to come back to its start.
Multiply the rate by the elapsed time.
$\text{angle} = \text{rate} \times \text{hours} =$ a
It turns steadily.
Check the direction of turning.
$\text{clockwise seen from above, in the Northern Hemisphere}$
The Coriolis force pushes the bob to the right of its motion.
Match each inertial force of a noninertial frame to its expression.
| $-m\vec{\Omega} \times (\vec{\Omega} \times \vec{r})$ | $-2m\vec{\Omega} \times \vec{v}$ | $-m\vec{A}$ | $-m\dot{\vec{\Omega}} \times \vec{r}$ | |
|---|---|---|---|---|
| centrifugal | ||||
| Coriolis | ||||
| accelerating frame | ||||
| changing rotation |
A ring-shaped space station of radius $200$ m spins at $0.2$ rad/s. An astronaut climbs a ladder toward the hub at $1$ m/s. Fill in the artificial gravity at the rim in m/s², the Coriolis acceleration on the astronaut in m/s², and their ratio.
| value | |
|---|---|
| artificial gravity (m/s²) | |
| Coriolis acceleration (m/s²) | |
| Coriolis over gravity |
A planet turns $21°$ per hour. Write the rate, in degrees per hour, at which a Foucault pendulum's plane of swing turns relative to the ground, as a formula in the latitude $t$.
Answer:
In Chicago, at latitude $41.9°$ north, a vehicle moves horizontally at $250$ m/s. What is the horizontal Coriolis acceleration on it, in mm/s²? Use $\Omega = 7.292 \times 10^{-5}$ rad/s.
Answer: mm/s² of Coriolis acceleration
A Foucault pendulum swings at the Science Museum of Minnesota in St. Paul, at latitude $44.95°$ north. The Earth turns once in a sidereal day of $23.93$ hours. How many hours does the pendulum's plane take to turn a full circle?
Answer: hours for a full turn
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A planet turns $23°$ per hour. Write the rate, in degrees per hour, at which a Foucault pendulum's plane of swing turns relative to the ground, as a formula in the latitude $t$.
Answer:
You can use rotating frames. Explain to someone why a Foucault pendulum in Houston turns more slowly than one in St. Paul.
18. Your turn: find how long a Foucault pendulum takes to turn once at latitude $30°$., step 3
$T_F = 47.9\ \text{h}$
About two days.