Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force...
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Transcript of Example (Marion) Find the horizontal deflection from the plumb line caused by the Coriolis force...
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!