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Page 1: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational
Page 2: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

Example (Marion)• Find the horizontal deflection

from the plumb line caused by the

Coriolis force acting on a particle

falling freely in Earth’s gravitational

field from height h above Earth’s

surface. (N. hemisphere):

• Acceleration in the rotating

frame given by “Newton’s 2nd Law”, no external forces S: ar = (Feff /m) = g - 2(ω vr)

g = Effective g already discussed. Along vertical & in same direction as plumb line.

• Local z axis: Vertically upward along -g (fig). (Unit vector ez) ex: South, ey : East. N. hemisphere.

Page 3: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

• Neglect variation of g with altitude.• From figure, at latitude λ:

ωx = - ω cos(λ); ωz = ω sin(λ);

ωy = 0, g = -gez

vr = zez + xex +yey

ar = g - 2(ω vr)

• Component by component:

z = -g - 2(ω vr)z = -g -2ωxy (1)

x = - 2(ω vr)x = 2ωzy (2)

y = - 2(ω vr)y = 2(ωxz- ωzx) (3)

Page 4: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

• Eqtns of motion:

z = -g - 2(ω vr)z = -g -2 ωxy (1)

x = - 2(ω vr)x = 2 ωzy (2)

y = - 2(ω vr)y = 2(ωxz- ωzx) (3)

First approximation, |g| >> All other terms

First approximation:

z -g ; x 0; y 0

z = -gt; x 0; y 0 ;

Put in (1), (2), (3); get next approx:

z -g ; x 0 ;

y 2ωxz = 2gtω cos(λ)

Page 5: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

z -g ; x 0 ; y 2 ωxz = 2gtω cos(λ)

• Integrating y eqtn gives:

y (⅓)gωt3 cos(λ) (A)• Integrating z eqtn gives (standard):

z h – (½)gt2 (B)• Time of fall from (B):

t [(2h)/g]½ (C)• Put (C) into (A) & get (Eastward) Coriolis force

induced deflection distance of particle dropped from height h at latitude λ: d (⅓)gω cos(λ) [(8h3)/g]½

For h = 100 m at λ = 45, d 1.55 cm!• This neglects air resistance, which can be a greater effect!

Page 6: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

Another Example (Marion)• The effect of the Coriolis force on the motion of a

pendulum produces a precession, or a rotation with time of the plane of oscillation. Describe the motion of this system, called a Foucault pendulum. See figure.

Page 7: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

• Acceleration in rotating frame

given by “Newton’s 2nd Law”,

external force T = Tension in

the string: ar = (Feff /m)

ar = g + (T/m) - 2(ω vr)

g = Effective g along the local vertical.• Approximation: Pendulum (length ) moves in

small angles θ Small amplitude

Precession motion in x-y plane; Can neglect z motion in comparison with x-y motion:

|z| << |x|, |y| |z| << |x|, |y|

Page 8: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

• Relevant (approx.) eqtns (fig):

Tx = - T(x/) ; Ty = - T(y/); Tz T

• As before, gx = 0; gy = 0; gz = -g

ωx= - ωcos(λ); ωz= ωsin(λ); ωy = 0

(vr)x = x; (vr)x = y ; (vr)z = z 0

(ω vr)x -y ω sin(λ), (ω vr)y x ω sin(λ)

(ω vr)z -y ω cos(λ)

• Eqtns of interest are x & z components of ar:

(ar)x = x - (T x)/(m) + 2 y ω sin(λ)

(ar)y = y - (T y)/(m) - 2 x ω sin(λ)• Small amplitude approximation: T mg ; Define

α2 T/(m) g/; α2 Square of pendulum natural frequency.

Page 9: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

• Approx. (coupled) eqtns of motion:

x + α2x 2yωsin(λ) = 2yωz (1)

y + α2y -2xωsin(λ) = -2xωz (2)

• One method of solving coupled

eqtns like this is to use complex

variables. Define: q x + i y• Using this & combining (1), (2):

q +2i ωz q + α2 q 0 (3)

• (3) is mathematically identical to a damped harmonic oscillator eqtn, but with a pure imaginary “damping factor”!

Page 10: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

• Define: γ2 α2 + (ωz)2

q(t) exp(-iωzt) [A eiγt +B e-iγt] (I)

A,B depend on the initial conditions

• If the rotation of the Earth is neglected, ωz 0, γ α q + α2 q 0 & (I) is: q(t) [A eiαt +B e-iαt]

(Define with in what follows are functions which ignore

Earth rotation) Ordinary, oscillatory pendulum motion! (Frequency α2 g/ )

• Note that the rotation frequency of Earth, ω, is small ωz = ω sin(λ) is small, even on equator (λ = 0). For

any reasonable α2 g/, it’s always true that α >> ωz.

(I) becomes: q(t) exp(-iωzt) [A eiαt +B e-iαt] (II)

Page 11: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

• Solution is: q(t) exp(-iωzt) [A eiαt +B e-iαt] (II)

or q(t) exp(-iωzt) q(t) where q(t) = solution for

the pendulum with the Earth rotation effects ignored.

• Physics: (II): Ordinary (small angle) pendulum oscillations are modulated (superimposed) with very low frequency (ωz) precession (circular) oscillations in the x-y plane.

• Can see this more clearly by separating q(t) & q(t) into real & imaginary parts (q x + i y; q x + i y) & solving for x(t), y(t) in terms of x(t), y(t). See p 401, Marion, where the author does this explicitly!

• Observation of this precession is a clear demonstration that the Earth rotates! If calibrated, it gives an excellent time standard!

Page 12: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

• Figures (from a mechanics book by Arya) showing precession.

• Precession frequency = ωz= ω sin(λ), λ = Latitude angle Period = Tp = (2π)/[ωsin(λ)]: λ = 45, Tp 34 h;

λ = 90 (N pole); Tp24 h; λ = 0 (Equator); Tp !

Page 13: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

Arya Example• Bucket of fluid spins with angular

velocity ω about a vertical axis.

Determine the shape of fluid surface:• From coordinate system rotating

with bucket, problem is static equilib.!

Free body diagram in rotating frame:

Page 14: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

• Free body diagram in

rotating frame:• Small mass m on

fluid surface. “2nd Law” in

the rotating frame:

r = distance of m from the axis.

Effective force on m:

Feff = Fp + mg0 - (mω)(ω r) - 2m(ω vr)

• Fp = normal force on m at the surface ( Fcont in fig)

Feff = 0 (static), vr = 0

0 = Fp + mg0 - mω (ω r)

Page 15: Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force acting on a particle falling freely in Earth’s gravitational

0 = Fp + mg0 - mω (ω r)

• Horizontal (H) & vertical (V) components, from diagram: (H) 0 = mω2r - Fp sinθ

(V) 0 = Fp cosθ - mg0

Algebra gives: tanθ = (ω2r)/g0

From the diagram: tanθ = (dz/dr)

z = (ω2/2g0)r2 A circular paraboloid!