Sect. 4.10: Coriolis & Centrifugal Forces (Motion Relative to Earth, mainly from Marion) Summary: For motion in an accelerating frame (r), both translating & rotating with respect to a fixed (f, inertial) frame: Velocities: vf = V + vr + ω  r Accelerations: ar = Af + ar + ω  r + ω  (ω  r) + 2(ω  vr)  Newton’s 2nd Law (inertial frame): F = maf = mAf + mar + m(ω  r) + m[ω  (ω  r)] + 2m(ω  vr)  “2nd Law” equation in the moving frame: mar  Feff  F - mAf - m(ω  r) - m[ω  (ω  r)] - 2m(ω  vr)

Motion Relative to Earth  “2nd Law” in accelerating frame: Feff  mar  F - mAf - m(ω  r) - m[ω  (ω  r)] - 2m(ω  vr)  Transformation gave: Feff  F - (non-inertial terms) Interpretations: - mAf : From translational acceleration of moving frame. - m(ω  r): From angular acceleration of moving frame. - m[ω  (ω  r)]:  “Centrifugal Force”. If ω  r: Has magnitude mω2r. Outwardly directed from center of rotation. - 2m(ω  vr):  “Coriolis Force”. From motion of particle in moving system (= 0 if vr = 0) More discussion of last two now!  ≈ 0 for motion near Earth

ω2Re = 3.38 cm/s2 = Centripetal acceleration at equator Motion of Earth relative to inertial frame: Rotation on axis causes small effects! However, this dominates over other (much smaller!) effects: ω = 7.292  10-5 s-1 ; ω2Re = 3.38 cm/s2 = Centripetal acceleration at equator 2ωvr  1.5  10-4 v = max Coriolis acceleration ( 15 cm/s2 = 0.015g for v = 105 cm/s) Even Smaller effects! Revolution about Sun Motion of Solar System in Galaxy Motion of Galaxy in Universe Also, ω = (dω/dt) ≈ 0

Coordinate systems (figure): z direction = local vertical Fixed: (x,y,z) At Earth center Moving: (x,y,z) On Earth surface

Mass m at r in moving system. Physical forces in inertial system: F  S + mg0 S  Sum of non-gravitational forces mg0  Gravitational force on m g0  Gravitational field vector, vertical (towards Earth center; along R in fig). From Newton’s Gravitation Law: g0 = -[(GME)eR]/(R2) G  Gravitational constant, R  Earth radius ME  Earth mass, eR  Unit vector in R direction Assumes isotropic, spherical Earth Neglects gravitational variations due to oblateness; non-uniformity; ...

ez  unit vector along z Effective force on m, measured in moving system is thus: Feff  S + mg0 - mAf - m(ω  r) - m[ω  (ω  r)] - 2m(ω  vr) Earth’s angular velocity ω is in z direction in inertial system (North): ω  ωez ez  unit vector along z Earth rotation period T = 1 day ω = (2π)/T = 7.3  10-5 rad/s (Note: ω  365 ωes) ω  constant  ω  0  Neglect m(ω  r) Consider mAf term in Feff & use again formalism of last time (rotation instead of translation): Af = (ω  Vf ) = [ω  (ω  R)]

Effective force on m is:  Feff  S+mg0 - (mω)  [ω  (r + R)] - 2m(ω  vr) Rewrite as: Feff  S + mg - 2m(ω  vr) Where, mg  Effective Weight g  Effective gravitational field (= measured gravitational acceleration, g on Earth surface!) g  g0 - ω  [ω  (r + R)] Considering motion of mass m, at point r near Earth surface. R = |R| = Earth radius.  |r| << |R|  ω  [ω  (r + R)]  ω  (ω  R)  Effective g near Earth surface: g  g0 - ω  (ω  R)

g = g0 - ω  [ω  (r + R)] Centrifugal force: If m is at point r far from Earth surface, must consider both R & r terms. Effective g for any r: g = g0 - ω  [ω  (r + R)] Second term = Centrifugal force per unit mass (Centrifugal acceleration). Centrifugal force: Causes Earth oblateness (g0 neglects). Goldstein discussion, p 176 Earth  Solid sphere. Earth  Viscous fluid with solid crust. Rotation  “fluid” deforms,  Requator - Rpole  21.4 km gpole - gequator  0.052 m/s2 Surface of calm ocean water is  g instead of g0. Deviation of g from local vertical direction!

Summary: Effective force: Feff = S + mg - 2m(ω  vr) (1) Where, g = g0 - ω  [ω  (r + R)] (2) Often, g  g0 - ω  (ω  R) (3) These are all we need for motion near the Earth!

Direction of g Consider: g = g0 - (ω)  [ω  (r + R)] (2) Effective g = Eqtn (2). Consider experiments. Magnitude of g: Determined by measuring the period of a pendulum (small θ). DIRECTION of g: Determined by the direction of a “plumb bob” in equilibrium. Magnitude of 2nd term in (2): ω2R  0.034 m/s2  (ω2R)/(g0)  0.35% Direction of 2nd term in (2): Outward from the axis of the rotating Earth. Direction of g = Direction of plumb bob = Direction of the vector sum in (2). Slightly different from the “true” vertical  line to the Earth’s center. (Figure next page!)

Direction of plumb bob = Direction of g = g0 - (ω)  [ω  (r + R)] (2) Figure: (r in figure = r in previous figures!) Deviation of g from g0 direction is exaggerated! r = R + z where z = altitude

Coriolis Effects Effective force on m near Earth: Feff = S + mg - 2m(ω  vr) - 2m(ω  vr) = Coriolis force. Obviously, = 0 unless m moves in the rotating frame (moving with respect to Earth’s surface) with velocity vr. Figure again:

- 2m(ω  vr) = Coriolis force. Northern Hemisphere: Earth’s angular velocity ω is in z direction in inertial system (North) ω  ωez ez  unit vector along z (Figures):  In general, ω has components along x, y, z axes of the rotating system. All can have effects, depending on the direction of vr. Most dominant is ω component which is locally vertical in rotating system, that is ωz  Component along local vertical.

- 2m(ω  vr) = Coriolis force, Northern hemisphere. Consider ωz only for now. Particle moving in locally horizontal plane (at Earth surface): vr has no vertical component.  Coriolis force has horizontal component only, magnitude = 2mωzvr & direction to right of particle motion (figure).  Particle is deflected to right of the original direction:

Magnitude of (locally) horizontal component of Coriolis force  ωz = (locally) vertical component of ω  (Local) vertical component of ω depends on latitude! Easily shown: ωz = ω sin(λ), λ = latitude angle (figure).  ωz = 0, λ =0 (equator); ωz = ω, λ = 90 (N. pole)  Horizontal component of Coriolis force, magnitude = 2m ωzvr depends on latitude! 2mωzvr = 2mωvrsin(λ) All of this the in N. hemisphere! S. Hemisphere: Vertical component ωz is directed inward along the local vertical.  Coriolis force & direction of deflections are opposite of N. hemisphere (left of the direction of velocity vr )

Coriolis Deflections: Noticeable effects on: Flowing water (whirlpools) Air masses  Weather. Air flows from high pressure (HP) to low pressure (LP) regions. Coriolis force deflects it. Produces cyclonic motion. N. Hemisphere: Right deflection: Air rotates with HP on right, LP on left. HP prevents (weak) Coriolis force from deflecting air further to right.  Counterclockwise air flow! S. Hemisphere: Left deflection. (Falkland Islands story) Bathtub drains!

More Coriolis Effects on the Weather: Temperate regions: Airflow is not along pressure isobars due to the Coriolis force (+ the centrifugal force due to rotating air mass). Equatorial regions: Sun heating the Earth causes hot surface air to rise (vr has a vertical component).  In Coriolis force need to account ALSO for (local) horizontal components of ω Northern hemisphere: Results in cooler air moving South towards equator, giving vr a horizontal component . Then, horizontal component of Coriolis force deflects South moving air to right (West) Trade winds in N. hemisphere are Southwesterly. Southern hemisphere: The opposite! No trade winds at equator because Coriolis force = 0 there All is idealization, of course, but qualitatively correct!