User:John R. Brews/Coriolis force

The Coriolis force is a force experienced by a object traversing a path in a rotating framework that is proportional to its speed and also to the sine of the angle between its direction of movement and the axis of rotation. It is one of three such inertial forces that appear in an accelerating frame of reference due to the acceleration of the frame, the other two being the centrifugal force and the Euler force. The mathematical expression for the Coriolis force appeared in an 1835 paper by a French scientist Gaspard-Gustave Coriolis in connection with the theory of water wheels, and also in the tidal equations of Pierre-Simon Laplace in 1778.

Although sometimes referred to as an apparent force, it can have very real effects.

Reference frames

Newton's laws of motion are expressed for observations made in an inertial frame of reference, that is, in any frame of reference that is in straight-line motion at constant speed relative to the "fixed stars", an historical reference taken today to refer to the entire universe. However, everyday experience does not take place in such a reference frame. For example, we live upon planet Earth, which rotates about its axis (an accelerated motion), orbits the Sun (another accelerated motion), and moves with the Milky Way (still another accelerated motion).

The question then arises as to how to connect experiences in accelerating frames with Newton's laws that are not formulated for such situations. The answer lies in the introduction of inertial forces, which are forces observed in the accelerating reference frame, due to its motion, but are not forces recognized in an inertial frame. These inertial forces are included in Newton's laws of motion, and with their inclusion Newton's laws work just as they would in an inertial frame. Coriolis force is one of these inertial forces, the other two being the centrifugal force and the Euler force.

These motions are slight, but the Coriolis force does affect aiming artillery pieces and plotting transoceanic air flights. The way Coriolis forces work is illustrated below by a few examples.

Foucault pendulum

(PD) Image: John R. Brews
While the pendulum P swings in a fixed plane about its hanger at H, the planes of the Earth observer rotate.

The Foucault pendulum, or Foucault's pendulum, demonstrated in Paris in 1851, is named after the French physicist Léon Foucault. It is a device that demonstrates the rotation of the Earth. According to an inertial observer, the pendulum swings in a plane fixed relative to the "fixed stars", while as observed on Earth this plane appears to rotate. For most pendulums, the effect is masked by other complications, so some care must be taken in constructing a successful pendulum, particularly to insure the hanger exerts no torque upon the pendulum and is immune to wear.

The easiest way to understand the operation is to imagine the pendulum at the North pole, shown in the figure. As the pendulum swings, the Earth rotates and it appears to the Earth-bound observer that every 24 hours the pendulum returns to its initial plane of oscillation.

Approximations and idealizations

In the inertial frame of reference, the pendulum swings back in forth in a plane. However, at the top of its swing it has zero velocity and the bob is subject to no centrifugal force, while at the bottom of its swing it moves rapidly, and is subject to considerable centrifugal force. Consequently, the tension in the wire varies appreciably if the swing has large amplitude, and the wire will stretch as the bob descends, meaning the orbit is not a circle. That complicates the analysis without adding to any insight into the experiment.

In the rotating Earth frame of reference, the pendulum does not swing in a plane, because the Earth turns during its swing, making the projection onto a horizontal plane of the pendulum's path describe an arc, not a straight line. That introduces further complications for analysis in the Earth frame of reference.

To simplify the analysis and make clear what the experiment is about, the pendulum is idealized as executing very small amplitude swings, such that the bob's change in elevation is negligible, the bob's velocity is low enough that centrifugal force has negligible effect, and the tension in the wire has always a magnitude given by the force of gravity, mg, where g is the acceleration due to gravity and m is the mass of the bob.

Under these circumstances, if we introduce coordinate axes with z the vertical and x and y in the horizontal plane, the movement of the pendulum is analyzed entirely in the xy-plane. The problem in the inertial frame becomes one of watching the bob travel back and forth along the same direction parallel to the ground at the North pole, while from the Earth frame, one watches the same thing from a rotating carousel.

Inertial frame

In the inertial frame the path of the bob is a straight line in the xy-plane. If the origin is taken at the center of the swing, the path is described as:

$\mathbf {r} (t)=A\sin(\omega t)\mathbf {u_{r}} \ ,$

where u_r is a unit magnitude vector pointing along the path of the bob, and r is the radius vector from the origin to the bob. The frequency ω is related to the period T of the swing as:

$\omega =\pi /T\ .$

The velocity of the bob is then:

$\mathbf {\dot {r}} (t)=A\omega \cos \omega t\ \mathbf {u_{r}} \ ,$

The net force on the bob is therefore:

$\mathbf {F_{Intl}} =m\mathbf {\ddot {r}} (t)=-mA\omega ^{2}\sin \omega t\ \mathbf {u_{r}} =-m\omega ^{2}\mathbf {r} (t)\ .$

At time t =T/2, the radial coordinate is r = Au_r and the bob is at its furthest extreme. The tension in the string and gravity combine in a radially inward component, accelerating the bob toward the origin with acceleration A ω². At time t=0, there is zero net force on the bob, and it travels at maximum speed through the origin.

This net force upon the bob (within the idealizations outlined above) is the resultant of the tension in the wire and the force due to gravity, both of which are agreed upon by the Earth observer and the inertial observer as identical forces acting upon the bob.

Earth frame

The Earth coordinate axes rotate with the angular rate Ω of the Earth about the pole. Thus, the unit vector u_r is described from Earth at any given time t as:

$\mathbf {u_{r}} =\cos \Omega t\ \mathbf {i} -\sin \Omega t\ \mathbf {j} \ ,$

with i, j unit vectors in the x- and y-directions. The time derivative of this unit vector is:

${\frac {d}{dt}}\mathbf {u_{r}} =-\Omega \left(\sin \Omega t\ \mathbf {i} +\cos \Omega t\ \mathbf {j} \right)=-\Omega \mathbf {u_{t}} \ ,$

where u_t is a unit magnitude vector orthogonal to the radial vector u_r in the xy-plane. Likewise:

${\frac {d}{dt}}\mathbf {u_{t}} ={\frac {d}{dt}}\left(\sin \Omega t\ \mathbf {i} +\cos \Omega t\ \mathbf {j} \right)=\Omega \mathbf {u_{r}} \ .$

At t=0 the plane of the pendulum is along the x-axis, and at t =T/2 the Earth has rotated counterclockwise by 90°, placing the pendulum along the negative y-direction. Consequently, the path of the bob in the Earth frame is:

$\mathbf {r} (t)=A\sin(\omega t)\left(\cos \Omega t\ \mathbf {i} -\sin \Omega t\ \mathbf {j} \right)\ ,$

The velocity of the bob is:

$\mathbf {\dot {r}} (t)=A\omega \cos \omega t\ \mathbf {u_{r}} +A\sin \omega t\ {\dot {\mathbf {u_{r}} }}$

$=A\omega \cos \omega t\ \mathbf {u_{r}} -A\Omega \sin \omega t\ \mathbf {u_{t}} \ ,$

The acceleration of the bob is:

$\mathbf {\ddot {r}} (t)=A\omega {\frac {d}{dt}}\left(\cos \omega t\ \mathbf {u_{r}} \right)-A\Omega {\frac {d}{dt}}\left(\sin \omega t\ \mathbf {u_{t}} \right)\ ,$

$=-A\omega ^{2}\sin \omega t\ \mathbf {u_{r}} -A\omega \Omega \cos \omega t\ \mathbf {u_{t}} -A\Omega \omega \cos \omega t\ \mathbf {u_{t}} -A\Omega ^{2}\sin \omega t\ \mathbf {u_{r}}$

$=-A(\omega ^{2}+\Omega ^{2})\sin \omega t\ \mathbf {u_{r}} -A2\omega \Omega \cos \omega t\ \mathbf {u_{t}} \ .$

Unlike the inertial frame acceleration, which always is in the radial direction, the Earth observer sees a perpendicular component that causes the plane of the pendulum to rotate clockwise (opposite to the Earth's rotation).

This result can be compared with the inertial frame acceleration:

$\mathbf {\ddot {r}} (t)=-A\omega ^{2}\sin \omega t\ \mathbf {u_{r}} \ .$

Evidently the Earth observer sees the acceleration of the inertial observer with two additional terms.

$\mathbf {F_{Earth}} =\mathbf {F_{Intl}} -mA\Omega ^{2}\sin \omega t\ \mathbf {u_{r}} -mA2\omega \Omega \cos \omega t\ \mathbf {u_{t}} \ .$

The force on the bob due to wire tension and gravity F_intl must be the same for both observers. But the forces upon the bob seen by the Earth observer supplement the forces from tension and gravity with the centrifugal force observed due to rotation:

$\mathbf {F_{Cfgl}} =mA\Omega ^{2}\sin \omega t\ \mathbf {u_{r}} =m\Omega ^{2}\mathbf {r} (t)\ ,$

and the observed Coriolis force:

$\mathbf {F_{Cor}} =-2mA\Omega ^{2}\mathbf {r} (t)-mA2\omega \Omega \cos \omega t\ \mathbf {u_{t}} \ ,$

which accounts (for the Earth observer) for the daily precession of the pendulum's orbit. Introducing the vector representing the rotation of the Earth along the z-axes as:

$\mathbf {\Omega } =\omega \mathbf {u_{z}} \ ,$

and noting that:

$\mathbf {u_{z}\times u_{r}} =\mathbf {u_{z}\times } \left(\cos \Omega t\ \mathbf {i} -\sin \Omega t\ \mathbf {j} \right)=\mathbf {u_{t}} \ ,$

the Coriolis force can be represented as:

$\mathbf {F_{Cor}} =-2mA\Omega ^{2}\mathbf {r} (t)-2mA\ \mathbf {\Omega \times {\dot {r}}} (t)\ .$

If the pendulum is "frozen" at some position r, so the bob does not swing and simply is stationary in the inertial frame, the Earth observer sees the bob execute a circular orbit of radius r and the Earth observer sees a net force on the bob of:

$\mathbf {F_{Earth}} =\mathbf {F_{Cfgl}} +\mathbf {F_{Cor}} =-mA\Omega ^{2}\mathbf {r} \ ,$

which is to say, the Earth observer sees the centripetal force that is demanded by Newton's laws for the apparent circular motion seen from Earth. The centrifugal force, always outward directed, is overcome by the Coriolis force to provide the centripetal force. Thus, the body seen as stationary by the inertial observer and so subject to no force whatsoever, also is seen as requiring no outside agency of centripetal force by the Earth observer because the Coriolis and centrifugal forces look after the matter by themselves.

Notes

http://books.google.com/books?id=-3H5V0LGBOgC&pg=PA122&dq=Foucault+pendulum&hl=en&ei=CV1gTenZOpGisQO9ruzNCA&sa=X&oi=book_result&ct=result&resnum=9&ved=0CGQQ6AEwCA#v=onepage&q=Foucault%20pendulum&f=false

http://books.google.com/books?id=NAo7yv7Jmq0C&pg=PA22&dq=Foucault+pendulum&hl=en&ei=VopgTdmBCoK4sQP6p-ncCA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCoQ6AEwADgK#v=onepage&q=Foucault%20pendulum&f=false

http://books.google.com/books?id=NAo7yv7Jmq0C&pg=PA22&dq=Foucault+pendulum&hl=en&ei=VYVgTcujHoOusAO0vOXNCA&sa=X&oi=book_result&ct=result&resnum=1&ved=0CCoQ6AEwADgK#v=onepage&q=Foucault%20pendulum&f=false

http://books.google.com/books?id=hrBe52GPHrYC&pg=PA351&dq=Foucault+pendulum&hl=en&ei=VYVgTcujHoOusAO0vOXNCA&sa=X&oi=book_result&ct=result&resnum=9&ved=0CFQQ6AEwCDgK#v=onepage&q=Foucault%20pendulum&f=false

http://books.google.com/books?id=GfCil84YTm4C&pg=PA116&dq=Foucault+pendulum&hl=en&ei=E4ZgTeaDGIa-sQP1zvDHCA&sa=X&oi=book_result&ct=result&resnum=8&ved=0CE0Q6AEwBzgU#v=onepage&q=Foucault%20pendulum&f=false

http://books.google.com/books?id=mms6MXH9CuoC&pg=PA22&dq=Foucault+pendulum&hl=en&ei=34ZgTeukEIuesQPMvPHYCA&sa=X&oi=book_result&ct=result&resnum=2&ved=0CCsQ6AEwATge#v=onepage&q=Foucault%20pendulum&f=false

http://books.google.com/books?id=imrm2aOs9_8C&pg=PA90&dq=Foucault+pendulum&hl=en&ei=34ZgTeukEIuesQPMvPHYCA&sa=X&oi=book_result&ct=result&resnum=7&ved=0CEQQ6AEwBjge#v=onepage&q=Foucault%20pendulum&f=false

Maxwell

http://books.google.com/books?id=wr2QOBqOBakC&pg=PA184&dq=Foucault+pendulum&hl=en&ei=U4hgTd-9Foa6sQP3lt3ACA&sa=X&oi=book_result&ct=result&resnum=2&ved=0CDAQ6AEwATgy#v=onepage&q=Foucault%20pendulum&f=false

Practical matters

http://books.google.com/books?id=sSPLspTUYEEC&pg=PA73&dq=Foucault+pendulum&hl=en&ei=U4hgTd-9Foa6sQP3lt3ACA&sa=X&oi=book_result&ct=result&resnum=4&ved=0CDsQ6AEwAzgy#v=onepage&q=Foucault%20pendulum&f=false

http://books.google.com/books?id=1J2hzvX2Xh8C&pg=PA272&dq=Foucault+pendulum&hl=en&ei=D4lgTdrSGIi6sQPrsunYCA&sa=X&oi=book_result&ct=result&resnum=3&ved=0CDIQ6AEwAjg8#v=onepage&q=Foucault%20pendulum&f=false