2. 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)

3. 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

4. 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 =  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

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

6. 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; ...

7. 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)]

8. 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)

9. 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  m/s2
Surface of calm ocean water is  g instead of g0.
Deviation of g from
local vertical direction!

10. 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!

11. 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  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!)

12. 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

13. 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:

14. - 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.

15. - 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:

16. 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 )

17. 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!

18. 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!