2. Example (Marion) Acceleration in the rotating
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.

3. 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)

4. z = -g - 2(ω  vr)z = -g -2 ωxy (1) x = - 2(ω  vr)x = 2 ωzy (2)
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)

5.  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)gt (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!

6. 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.

7. 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|

8. Tx = - T(x/) ; Ty = - T(y/); Tz  T
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.

9. y + α2y  -2xωsin(λ) = -2xωz (2) One method of solving coupled
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  (3)
(3) is mathematically identical to a damped harmonic oscillator eqtn, but with a pure imaginary “damping factor”!

10. 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)

11. 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!

12. 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); Tp24 h; λ = 0 (Equator); Tp   !

13. 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:

14. 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
 = Fp + mg0 - mω  (ω  r)

15.  = 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!