There is such a thing as centrifugal

force, but it's a

misnomer;
the

phenomenon isn't a force at all. Rather, it's a

mathematical

fudge factor introduced by

physicists when they want to use

Newton's laws of

motion in a rotating, non-inertial

reference
frame. Newtonian

mechanics is only valid in frames of reference that
are either

stationary or moving with constant

velocity (according
to

relativity theory, there's no difference between the two anyway).
Imagine a

marble on a rotating platform on a table. In attempting to
describe this situation, one traditionally chooses a coordinate

axis
attached to the table, which is not accellerating, and compares the
marble's

movement-resisting

inertia with the

centripetal force of

friction between the marble and the platform to determine where the
marble will go. Unfortunately, this can be conceptually difficult. The
alternative, often discouraged in introductory

physics courses but
allowed later on, is to choose a coordinate

axis rotating relative to
the table, and attached to the platform. Newton's classic

equations

*will not work* in this reference frame because it is
accellerating, but they can be

*made to work* by treating it as if
it weren't. To do this we have to introducing a

virtual
outward-pointing force, the centrifugal force. A

derivation of the
centrifugal force using

vector calculus follows.

F=mA_{i}

Newton's Second Law in an inertial reference frame.
(d/dt)_{i}=(d/dt)_{r}+(w
x r)

Apply this coordinate transformation...

V_{i}=V_{r}+(w x r)

...to the radius vector.

(d/dt)_{i}=(d/dt)_{r}+(w
x r)

And again...

A_{i}=A_{r}+2(w x
V_{r})+(w x (w x r))

...to the velocity vector.

F_{i}-2m(w x V_{r})-m(w x
(w x r))=mA_{r}

Substitute into the initial Second Law equation...

F_{eff}=F_{i}-2m(w x
V_{r})-m(w x (w x r))

...and get the effective force.

The third term on the right, **-m(w x (w x r))** is the
**centrifugal force.**

The second term on the right is the icing on the cake. **-2m(w x
V**_{r}) is the **Coriolis force.**