Sect. 4.10: Coriolis & Centrifugal Forces (Motion Relative to Earth, mainly from Marion)

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 inthe 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. Also, ω = (dω/dt) ≈ 0 • 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  10-4v = max Coriolis acceleration ( 15 cm/s2 = 0.015g for v = 105cm/s) Even Smaller effects! • Revolution about Sun • Motion of Solar System in Galaxy • Motion of Galaxy in Universe

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. 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. 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!

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  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!)

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 g0direction 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,ωhas components along x, y, z axes of the rotating system. All can have effects, depending on the direction of vr. • Most dominant is ωcomponent which is locally vertical in rotating system, that is ωzComponent alonglocal vertical.

15. - 2m(ω vr) = Coriolis force, Northern hemisphere. • Consider ωzonly 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 ωdepends on latitude! Easily shown: ωz = ωsin(λ), λ = latitude angle (figure).  ωz = 0, λ =0(equator); ωz = ω, λ = 90(N. pole) Horizontal component of Coriolis force, magnitude = 2m ωzvrdepends on latitude!2mωzvr= 2mωvrsin(λ) • All of this the in N. hemisphere! S. Hemisphere: Vertical component ωz is directed inwardalong the local vertical.  Coriolis force & direction of deflections are opposite of N. hemisphere (leftof the direction of velocity vr)

17. Bathtub drains! • 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)

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 (vrhas 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 vra 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!