Mathematics of Physics and Engineering

54 Kinematics and Dynamics of a Point Mass

(1.3.27), page 36, we find ag = 0 and so

rl=ar = -j^2 Rr, (2.1.25)

where R is the radius of the Earth.
Similarly, we use (1.3.28), page 36, to find the velocity rj, of the point
in the frame 0\ ti — "fR9, and then the Coriolis acceleration in O is
acor = 2CJ x r\ — 2^Ru> x 9.
To evaluate the product w x 9, we need some additional constructions.
Note that both u> and 9 are in the plane (NOP). Let b be the unit vector
that lies both in the (NOP) and the equator planes as shown in Figure
(2.1.5). This vector is rotating with the plane (NOP) and therefore changes
in time; by (2.1.18), b'(t) = u> x b(t). With this construction, the vector
LJ x 9 is in the plane of the equator and has the same direction as — b'(t).
By definition, 9 is orthogonal to r. By construction, the vectors CJ and b
are also orthogonal, and so the angle 6 between the vectors b and r is equal
to the angle between the vectors u> and 9. Since ||b'(£)|| — u = ||u>|| and
||0|| = 1, we find UJ x 9 — -sinOb', and therefore

acor(t) = -27-Rsin(7i)b'(i). (2.1.26)

Finally, we use 9 and b to write the centripetal acceleration ac =wx(uxr)

of the point. We have

u>xr = Rsm(n/2-e)b' = RcosQb', u xb'=-uj^2 b. (2.1.27)

Hence,

ac = -(<Jj^2 Rcos'yt)b. (2.1.28)

Combining (2.1.25), (2.1.26), and (2.1.28) in (2.1.23), we find the total

acceleration r(t) of the point mass in the fixed frame O:

r(t) = --y^2 Rr(t) -2^Rsia{jt)b'(t) - cj^2 Rcos(-ft)b(t). (2.1.29)

EXERCISE 2.1.8.c Verify all the equalities in (2.1.27).

We now use (2.1.24) to study the dynamics of the point mass m moving

on (or close to) the surface of the Earth. Let r\ = r\(t) be the trajectory

of the point in the frame 0\. Unlike the discussion leading to (2.1.29), we

will no longer assume that the point moves along a meridian. Suppose that

O is an inertial frame. If F is the force acting on m in the frame O, then