Mathematics of Physics and Engineering

Uniform Rotation of Frames 53

2u> x ri, and the centripetal acceleration o,c — u} x [u) x fi). Note that o,c
is orthogonal to both u> and wxri.
Assume that the fixed frame O is inertial, and let F be the force acting
on the point mass m in O. By Newton's Second Law, we have mro = F in
the inertial frame O and, by (2.1.23),

mfi = F - 2mw xn-mwxfwxri) (2.1.24)

in the rotating frame 0. Similar to (2.1.16), inertial forces appear as
corrections to Newton's Second Law in the non-inertial frame 0. There
are two such forces in (2.1.24): the Coriolis force Fcor = -2m« x r\ and
the centrifugal force Fc = — mw x (a; x n).
As an example illustrating the relation (2.1.23), consider a point mass
m moving on the surface of the Earth along a meridian (great circle through
the poles) with constant angular speed 7; the axis of rotation goes through
the North and South Poles. We place the origins O and 0\ of the fixed and
rotating frames at the center of the Earth and assume that the frame 0\
is rotating with the Earth. (Figure 2.1.5).

Fig. 2.1.5 Motion Along a Meridian

Denote by P the current position of the point, and consider the plane
(NOP) Relative to the Earth, that is, in the frame Ox, the plane (NOP)
is fixed, and the motion of m is a simple circular rotation in this plane
with constant angular speed 7 so that 6(t) = jt. Relative to the fixed
frame O, the plane (NOP) is rotating, and the rotation vector is u>. We
will determine the three components of the acceleration of the point in the
frame O according to (2.1.23).
Introduce the polar coordinate vectors r, 6 in the plane (NOP). By