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0593_C04*_fm Page 89 Monday, May 6, 2002 2:06 PM

Kinematics of a Rigid Body 89

V 0
R N
N
k
k
O V 0
O
j R
j
i
i
S
A A S
(a) (b)
FIGURE 4.6.2
Vertical launch of a rocket from the surface of the Earth. (a) Launch at the North Pole; (b) launch at the Equator.
Example 4.6.1: Effect of Earth Rotation
Equation (4.6.6) is useful for developing expressions for velocity and acceleration of
particles moving relative to rotating bodies. To illustrate this, consider the case of a rocket
R launched vertically from the Earth’s surface. Let the vertical speed of R relative to the
Earth (designated as E) be V . Suppose we want to determine the velocity of R in an
0
astronomical reference frame A in which E rotates for the cases when R is launched from
(a) the North Pole, and (b) the equator.
Solution: Consider a representation of R, E, and A as in Figure 4.6.2a and b, where i,
j, and k are mutually perpendicular unit vectors fixed in E, with k being along the
north/south axis, the axis of rotation of E. In both cases the velocity of R in A may be
expressed as:

V = dOR dt (4.6.7)
A R A

where O is the center of the Earth which is also fixed in A.

a. For launch at the North Pole, the position vector OR is simply (r + h)k, where
r is the radius of E (approximately 3960 miles) and h is the height of R above
R
A
the surface of E. Thus, from Eq. (4.6.6), V is:
dr h k ) ]
+
A
A V = [ ( + dt+ ωω E ×( r h k ) (4.6.8)
R
E
Observe that angular velocity of E in A is along k with a speed of one revolution
per day. That is,

×
A E 2 π −5
ωω= ωk = k = 727 10 krad sec (4.6.9)
.
24
( )(3600 )
Hence, by substituting into Eq. (4.6.8), we have:
V = h = V k (4.6.10)
A R ˙
k
o