A Foucault pendulum demonstrates the rotation of the earth – but the details are subtle.
It's easy enough to understand the motion of a pendulum suspended at one of the Earth's poles, because in this case the point of suspension is not accelerating. At any other latitude, the point of suspension accelerates and this introduces some more complicated timedependent forces acting on the pendulum. At the poles, the (apparent) precession period is 23.9 hours, whereas elsewhere it is longer by a factor of the reciprocal sine of the latitude. But why? And is the precession uniform? On the way, we'll derive expressions for the fictitious forces mentioned in the introduction.
This page is a physical analysis of the motion. I have kept it as simple as I could (though see footnote), but it does use vector calculus – without this tool, the analysis would be very long and awkward. Before you start on the analysis below, see the Introduction to the Foucault pendulum.
We treat the earth as rotating about its axis with angular velocity Ω relative to an inertial frame. At
a point O on the surface with latitude λ, we choose a Cartesian coordinate system with the x axis
horizontal and South, the y axis horizontal and East and the z axis vertical and up. In the diagram, the i,k plane (the NorthSouth, vertical plane at the position of our pendulum) is shaded.
This coordinate system is not inertial: it is rotating with the Earth, so not only the position, r_{o}, of its origin, but also i, j and k are all varying in time
P is the point on the earth's axis closest to O and r_{o} is the vector from P to O. If the position vector of a moving particle is r relative to O, then its position vector relative to P is r_{p} = r + r_{o}. Further let r be sufficiently small that the gravitational field does not vary significantly between O and r. Now P is in an inertial frame, so the equation of motion for the particle is: mr"_{p} = mG + F (1) where m is the mass of the particle, G is the gravitational field (i.e. force per unit mass) at O, F is the sum of all forces other than gravity, and dashes signify derivatives with respect to time.

Point O rotates with the angular velocity of the earth Ω, so
r"_{o} = – Ω^{2} r_{o} = Ω^{2}r_{o}(sin λ i − cos λ k) (2)
Consider now a pendulum consisting of a particle at the end of a string length a attached to a point
with coordinates in the O frame (0,0,L). Suppose initially that the pendulum were stationary and
vertical in that frame. The tension T_{o} is the only external force and: T_{o} = mgk (4) where g is the (local) apparent gravitational acceleration at O. (Note that g and G are neither parallel nor equal in magnitude because of the rotation of the earth. In Sydney, the centripetal acceleration due to the Earth's rotation (a_{c} ≈ G  g) has a magnitude of 28 mm.s^{1}: small compared to g ≈ 9800 mm.s^{1} and therefore often neglected. In the Southern hemisphere, g points North of the centre of the Earth, as shown. See Why g is not parallel to F_{g} for an explanation.) For this vertical pendulum,

mr" = mG + mgk + mΩ^{2} r_{o}(sin λ i − cos λ k) (5).
The last term in (5) is the first of our fictitious forces, called the centrifugal force. Called fictitious because it doesn't exist: r"_{p} is the acceleration in an inertial frame and, in (1), there is no centrifugal force. r" is the acceleration in a noninertial frame and, in such frames, Newton's first and second laws do not apply.
This 'pendulum', remember, is not moving with respect to the earth: it's a plumb bob and it just hangs vertically. So r, r' and r" are all zero. Thus for this case, (5) gives:
G + Ω^{2}r_{o}(sin λ i − cos λ k) = − gk (6).
In this equation, G is the gravitational field, −gk is the apparent gravitational acceleration, and the difference is the centripetal acceleration at that latitude, r"_{o} = Ω^{2}r_{o}(sin λ i − cos λ k) = − Ω^{2}r_{o}.
Now consider a moving pendulum with tension T, and use Newton’s second law in an inertial frame:
mr"_{p} = T + mG (7).
Now we need to relate the acceleration r"_{p} in the inertial frame to the r", the (apparent) acceleration in the frame on the surface of the earth. We write
r = r_{x}i + r_{y}j + r_{z}k
remembering that the i,j,k coordinates vary: they rotate with Ω = − Ω cos λ i − Ω sin λ k. Now r_{p} = r_{o} + r so
r_{p} = r_{o} + r_{x}i + r_{y}j + r_{z}k
so r'_{p} = r'_{o} + (r_{x})'i + (r_{y})'j + (r_{z})'k + r_{x} i' + r_{y} j'+ r_{z}k'
Now i' is the motion of a point at i relative to O due to the rotation Ω of the frame so i' = Ω X i and similarly j' = Ω X j and k' = Ω X k, so
r' = {(r_{x})' i + (r_{y})' j + (r_{z})' k} + Ω X r
which, by definition, = r' + Ω X r and so
r"_{p} = r"_{o} + r" + 2 Ω X r' + Ω X (Ω X r) (8).
For a pendulum small compared with the earth, r is much less than r_{o} (typically a million times) so the last term in (8) is much less r"_{o} which, as we have seen is tens of mm.s^{−2}. So, neglecting this last term, and noting that r"_{o}= Ω X (Ω X r_{o}):
r"_{p} = r" + Ω X (Ω X r_{o}) + 2 Ω X r' (8').
where the first term is the apparent acceleration in the frame, the second is the centripetal acceleration and the last is the Coriolis acceleration. (6), (7) and (8) together then give the equation of motion of the pendulum in the frame of the earth's surface:
m r" = T − mgk − 2m Ω X r' (9).
Note the significance of the terms: the first is the string tension, the second is the apparent weight in the rotating frame and the third term, which depends on the velocity of the pendulum and on the earth's rotation, is the other fictitious force, the Coriolis force. What has happened to the centrifugal force? Note the approximation made above: for altitudes small compared with the radius of the earth, the centripetal acceleration is not only small (tens of mm.s^{−2}) but always has the same amplitude and direction in the earth frame. It is therefore 'included' in g. One could say that the 'centrifugal force' is the difference between G, the gravitational field, and g, the apparent gravitational acceleration. The Coriolis force depends on r', the velocity in the earth frame and so cannot be thus 'included'. Of course, it is very small, because Ω is only 2π radians per day, or 0.000073 s^{−1}. Which is why the pendulum precesses so slowly.
Resolving (9) in coordinates fixed with respect to the earth gives:
mx" = T_{x} + 2mΩ sin λ y' (9.1),
my" = T_{y} − 2mΩ (sin λ x' + cos λ z') (9.2)
and mz" = T_{z} − mg + 2mΩ cos λ y' (9.3).
For a swing of small angle, z << a, motion in the z (vertical) direction is small and so z' and z" are approximately zero. Equation (9.3) therefore becomes:
T_{z} = mg − 2mΩ cos λ y' (9.3*),
whose last term is very small.
For small oscillations (x, y, z all << a), the components of the tension T in the string are
T_{x} = − T x/a , . . . T_{y} = − T y/a, . . . T_{z} = T(L − z)/a (10)
(where, to first order, T_{z} = T = mg.) Let us define the constant ω^{2} = g/a and substitute (10) and (9.3*) in (9.1) and (9.2), taking advantage of z' approx = 0 :
x" − 2Ω sin λ y' + ω^{2}x = 0 (9.1*)
y" + 2Ω sin λ x' + ω^{2}y = 0 (9.2*).
Now let us define the complex variable ζ = x + iy = ζexp(i φ) (which represents the position of the pendulum in the horizontal plane). Multiplying (9.2*) by i and adding it to (9.1*) gives the differential equation for motion in the horizontal plane:
ζ" + 2iΩ sin λ . ζ' + ω^{2}ζ = 0 (11).
This is a version of the equation of the harmonic oscillator, but note that the coefficient of the the first derivative of ζ is purely imaginary and so it does not contribute damping.
It may be verified by substitution that the solution to (11) is:
ζ = {Aexp(iωt) + Bexp(−iωt)} exp(−iΩt sin λ) (12)
where ω^{2} has been neglected in comparison with Ω^{2} and where the complex constants A and B depend on the initial conditions of position and velocity. (Two simple cases are A = +/B, both real, corresponding to motion passing through the equilibrium point. Note, however, that if the pendulum is released from a point of maximum amplitude, it never passes through ζ = 0.)
Equation (12) is readily interpreted: if the earth were not rotating (Ω = 0), the solution is just that for a simple pendulum with period t = 2π/ω whose elliptical path is given by
Finally, a complication that applies in principle to all pendulums, including those of clocks. Taking x and y components of equation (8) shows an asymmetry in the x and y directions:
x" − 2Ω sin (λ) y'+((g/a)−Ω^{2} sin (λ)^{2}) x = 0, and
y" + 2Ω sin (λ) x'+((g/a)−Ω^{2}) y = 0
Thus, at positions other than the poles, there is a small difference in the pendulum frequencies for the NS and EW directions. Ω/2π is of course about once per day, and for Foucault pendula, (g/a)^{1/2}/2π is of order once every several seconds, so this variation is small. For a grandfather clock, (g/a)^{1/2}/2π is faster and so the variation with orientation is smaller still. It is occasionally said that one might detect the difference in the timekeeping of a good pendulum clock when it is rotated between a NS and an EW swing. However, other effects, such as the angle of the floor, may be larger.
Links to other sites with information on the Foucault pendulum:
© 1996. Modified 9/8/04 J.Wolfe@unsw.edu.au, phone 61
29385 4954 (UT + 10, +11 OctMar). School of Physics, University of New South Wales, Sydney, Australia.
Joe's scientific home page

