Consider a weight of mass m suspended from a fixed point by a string of length l. Usual axes, but with origin at the point of suspension: x-axis points East y-axis points North z-axis points vertically upwards. The position of the bob is x, with coordinates (x, y, z). Note the constraint x2 + y 2 + z 2 = l2 .
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The equation of motion is m¨ x = −2mω × x˙ − T x ˆ + mg where T is the tension in the pendulum string We have ignored (ω × (ω × x)) and ω˙ × x. In components, we have ω = (0, ω cos λ, ω sin λ), g = (0, 0, −g) where λ is the lattitude.
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Thus m¨ x = −T x/l − 2ω(z˙ cos λ − y˙ sin λ)
(i)
m¨ y = −T y/l + 2ω x˙ sin λ
(ii)
m¨ z = −mg − T z/l − 2ω x˙ cos λ
(iii)
Now x/l 1 and y/l 1, so from x2 + y 2 + z 2 = l2 we have
z =− 1− l
x2
+ l2
y2
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=− 1−
x2
+ 2l2
y2
+ ···
(Taylor series) so z/l differs from −1 by second order quantities (x/l is first order); z˙ and z¨ are second order quantites and, as we have already assumed in ignoring the centrifugal forces, ω is first order.
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Thus, ignoring second order quantities, equation (iii) above becomes T = mg. Substituting this into equations (i) and (ii), and again ignoring second order quantities, gives g x ¨ = − x + 2ω y˙ sin λ, l g y¨ = − y − 2ω x˙ sin λ. l
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(∗) (∗∗)
Now let ζ = x + iy. Taking (∗) + i(∗∗) gives g ¨ ζ = − ζ − 2iω ζ˙ sin λ. l Setting ζ = eαt gives the auxiliary equation g α + 2iωα sin λ + = 0, l 2
i.e., r α = −iω sin λ ±
p g −ω 2 sin2 λ − ≈ −iω sin λ ± i g/l. l
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The general solution is therefore p p −iωt sin λ ζ=e A cos( g/l t) + B sin( g/l t) . the with-ω solution is that same as the without-ω solution except that the argand diagram for ζ, which is just the x-y plane, is rotating in the negative sense, i.e. clockwise viewed from above, This rotation is relatively slow: its period is 2π/(ω sin λ), and ω = 2π/(1 day), so at λ = 48◦ 520 N (Paris) the period is around 32 hours.
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