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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.
• 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)
• 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)
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!
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.
• 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|
• 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.
• 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”!
• 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)
• 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!
• 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 !
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:
• 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)
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!