In 1851 Foucault set a pendulum vibrating and observed the earth
rotate out from under it. It was a very long pendulum with a heavy
weight at the end so that it would vibrate for a long time without
stopping^{2} .
This is what allowed him to observe the earth rotate out from under it. Clearly such a
pendulum will take 24 hours for the plane of vibration to appear to make one complete
revolution at the north pole. It is also reasonable to expect that no such observed
rotation would take place on the equator. Is it possible to predict what will take place at
various latitudes?
′ ′ ′ ′
= − gk+ T ∕m− 2ω[(− y cosϕ)i+(x cosϕ+ z sin ϕ)j− (y sinϕ)k ]
where T, the tension in the string of the pendulum, is directed towards the point at
which the pendulum is supported, and m is the mass of the weight at the end of the
pendulum. The pendulum can be thought of as the position vector from
(0,0,l)
to the
surface of the sphere x^{2} + y^{2} +
(z − l)
^{2} = l^{2}. Therefore,
T = − Txi− Tyj+T l−-zk
l l l
and consequently, the differential equations of relative motion are
′′ x-- ′
x = − T ml +2ωy cosϕ
′′ -y- ′ ′
y = − Tml − 2ω (x cosϕ + z sinϕ)
and
z′′ = Tl−-z − g+ 2ωy′sin ϕ.
ml
If the vibrations of the pendulum are small so that for practical purposes, z^{′′} = z = 0, the
last equation may be solved for T to get
All terms of the form xy^{′} or y^{′}y can be neglected because it is assumed x and y remain
small. Also, the pendulum is assumed to be long with a heavy weight so that x^{′} and y^{′}
are also small. With these simplifying assumptions, the equations of motion
become
x
x ′′ + gl-= 2ωy′cosϕ
and
y′′ + gy-= − 2ωx ′cosϕ.
l
These equations are of the form
x′′ + a2x = by′,y′′ + a2y = − bx′ (17.22)
(17.22)
where a^{2} =
gl
and b = 2ω cosϕ. There are systematic ways to solve the above linear
system of ordinary differential equations, but for the purposes here, it is fairly tedious but
routine to verify that for each constant c,
( bt) ( √b2 +-4a2 ) ( bt) ( √b2 +-4a2 )
x = c sin -- sin --------t ,y = ccos -- sin --------t (17.23)
2 2 2 2
(17.23)
yields a solution to 17.22 along with the initial conditions,
It is clear from experiments with the pendulum that the earth does indeed rotate out
from under it causing the plane of vibration of the pendulum to appear to rotate. The
purpose of this discussion is not to establish this obvious fact but to predict how long it
takes for the plane of vibration to make one revolution. There will be some instant
in time at which the pendulum will be vibrating in a plane determined by k
and j. (Recall k points away from the center of the earth and j points East. )
At this instant in time, defined as t = 0, the conditions of 17.24 will hold for
some value of c and so the solution to 17.22 having these initial conditions will
be those of 17.23. (Some interesting mathematical details are being ignored
here. Such initial value problems as 17.23 and 17.24 have only one solution
so if you have found one, then you have found the solution. This is a general
fact shown in differential equations courses. However, for the above system of
equations see Problem 13 on Page 1091 found below.) Writing these solutions
differently,
( ) ( ( ) ) (√ ------- )
x (t) sin (b2t) --b2 +-4a2
y(t) = c cos b2t sin 2 t
This is very interesting! The vector, c
( sin(bt))
cos(2bt)
2
always has magnitude equal to
|c|
but its direction changes very slowly because b is very small. The plane of vibration is
determined by this vector and the vector k. The term sin
( √b2+4a2-)
2 t
changes relatively
fast and takes values between −1 and 1. This is what describes the actual observed
vibrations of the pendulum. Thus the plane of vibration will have made one complete
revolution when t = T for
bT
2-≡ 2π.
Therefore, the time it takes for the earth to turn out from under the pendulum
is
4π 2π
T = 2ωcosϕ-= -ω-secϕ.
Since ω is the angular speed of the rotating earth, it follows ω =
2π-
24
=
-π
12
in radians per
hour. Therefore, the above formula implies
T = 24secϕ.
I think this is really amazing. You could determine latitude, not by taking readings with
instruments using the North star but by doing an experiment with a big pendulum. You
would set it vibrating, observe T in hours, and then solve the above equation for ϕ. Also
note the pendulum would not appear to change its plane of vibration at the equator
because lim_{ϕ→π∕2} secϕ = ∞.