Where is the space elevator falling? And the cable holding it? I guess the concept of a space elevator is pretty well known. The
idea, first published by Konstantin Tsiolkovsky in 1895, and
popularized (among others) by Arthur C. Clarke in The Fountains of
Paradise, is to have a geosynchronous satellite with a very strong and
light cable hanging down to the ground at some anchoring point on the
equator circle, and a counterweight extending outward in space to keep
the center of mass at the rigth geosynchronous altitude (this variant is
due to Yuri N. Artsutanov). Whether it
can be done for Earth remains doubtful, as we do not (yet?) know of
materials that can sustain the strain (afaik), but the concept is
interesting and has been reused in several science-fiction stories,
for Earth and other planets.
In at least two of the novels I read that use the concept, the
elevator cable breaks, or is broken, below the center of mass and the
cable falls back to the ground (with or without the car).  But these
two novels do not seem to agree on how it falls.
In one novel, the cable falls in a heap on its ground anchoring
place. In another, it falls on a big circle (the equator), making
a "straight" line across the planet surface, though I do not recall
whether it is forward (ahead of its anchor, with respect to the planet
rotation) or backward (behind its anchor, with respect to the planet
rotation). The planet may not be Earth.
My knowledge of mechanics no longer being was it may once have been, I
am not sure I can analyze the problem correctly. I seriously doubt the
cable would fall as a heap onto its anchor (or that the car would do
that if it were to get loose from the cable, as suggested in one
novel).
So my question is what are the machanical laws of the phenomenon. and
where do the car and the cable fall and how.
They could fall ahead of the anchoring point, or behind the anchoring
point (with respect to earth rotation). The cable could be taut on the
equator, or zigzaging because it fell too fast with respect to
earth rotation. It could start falling in one direction (forward or
backward) and later reverse the other way for the remaining span.
I just have no real idea of what might happen, and I wonder how it is
to be analyzed.
I tried to get a first understanding by considering the car alone
getting loose from the cable, and I am putting it in a first tentative
partial answer, so that this question does not get too long. My
conclusion for the car felt counter-intuitive at first, becoming obvious in retrospect. But what of the cable?
 A: You could use the Coriolis force to analyze this, or just a little common sense. 
Assuming the cable is anchored at the equator, the linear velocity of a point "high up" will be greater than the velocity of the anchor point. When each point of the cable is in free fall, that velocity will carry the higher parts of the cable "ahead of" the anchor point. It will land to the East, and not in a heap (although elastic forces may further complicate things, it is unlikely to completely reverse this).
You can reach the same conclusion by thinking about conservation of angular momentum.
A: The car alone falls ahead of the anchoring point
This is a partial answer to illustrate the question in a simpler case.
It is separated from the question to keep it short enough, and because
it is more an answer than a question.
If one considers the car alone, getting suddenly loose, its angular
speed would be to low to keep it in a circular orbit, at its current
altitude, since the whole elevator is rotating at planet-synchronous
speed, which corresponds to a satellite above the car altitude. Thus
the car would dive into an elliptical orbit that might, or might not
intersect the ground level (I am ignoring the atmosphere for
simplification, though the elevator is not really needed when there is
none).
Car touching ground at perigee
The orbital period is actually determined by the length $a$ of the
semi-major axis of the orbit, according to the formula $T =
2\pi\sqrt{a^3/GM}$, where $M$ is the mass of the planet. That is $a$ is
the initial height of the car above the planet center. If it touches
ground just at perigee, it will have taken half a period of that
elliptical orbit, a time shorter than half a period of
planet-synchronous orbit (which has a longer semi-major axis).  Hence
the car will reach the ground in less than half of a planet rotation
period, at a point that would be half the equator from the anchor, if
Earth were not rotating.
Thus the car will fall ahead of the anchoring point. This reasonning
can be done without the exact formula, using only Kepler third law.
The general case for the car alone
I have not done the precise calculation for all cases, and would not
be very good at it, but it seems that qualitative reasonning is enough
to show that, if the car hits the ground (i.e. its orbits intersect
the planet surface), then it is always ahead of the anchoring point.
A first remark is that, if the car crashes on the ground, it does so
somewhere along its first half orbit from apogee (when it gets loose)
to perigee (when it is closest to the planet center, if it
can). According to Kepler second law of equal areas (using a
differential form of it) its angular speed increases continuously from
apogee to perigee, as its distance to the planet decreases.
The car's angular speed at apogee is the same as the planet's, i.e., when
the car gets loose. From then on, when it crashes, its angular speed
will only increase until it crashes. Hence it will continuously get
further ahead of the anchoring points as its follows its orbit to the
ground, and will necessarily crash on the equator ahead of the
elevator anchoring point.
The cable case
Thus I would tend to believe that a ruptured cable will fall similarly
ahead of the anchoring point. But I have no idea how forces
propagating along the cable might affect its motion, and whether it
will be taut on the ground.
Examining the above reasonning, it is clear that the car alone falls on the
first half of the equator great circle, ahead of the anchoring point.
Now, it we consider a space elevator for Earth, if the cable is broken
right under the geosynchronous satellite, its length is about 35,786
km, while half of the equator great circle is only about 20,000 km.
Hence there is no way the cable can be taut along the equator while
falling only on its first half, which is shorter.
Could it be that some extra energy is propagated to the end (highest
part) of the cable allowing it to stay "in orbit" beyond the half great
circle limit?
