Example (Marion)

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 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. 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)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)gωcos(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. 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. 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”!

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 verylow 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  0 = Fp + mg0 - mω (ω r)

15.  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!