Page 120 - Dynamics of Mechanical Systems

0593_C04*_fm Page 101 Monday, May 6, 2002 2:06 PM

Kinematics of a Rigid Body 101

Solution: From Eq. (4.9.4), the velocity of end B may be expressed as:

V = V + ωω AB × AB (4.9.21)
B
A
For the rod position of Figure 4.9.4, AB is:

AB =−4 n + 12 n + 3 n m (4.9.22)
x y z
AB
Let ωω ωω be expressed in the form:

ωω AB = ω n + ω n + ω n (4.9.23)
x x y y z z
B
Let V be expressed as:

V = v B n (4.9.24)
B
z
Then, by substituting from Eqs. (4.9.19), (4.9.22), (4.9.23), and (4.9.24) we have:

x (
x (
B
v n = 6 n + 3ω − 12ω z) n +−3ω − 4ω z) n
z y x y (4.9.25)
(
+ 12ω x + 4ω y) n z
Also, from Eq. (4.9.20) we have:

−4ω + 12ω + 3ω = 0 (4.9.26)
x y z

Equations (4.9.25) and (4.9.26) are equivalent to the following four scalar equations for ω ,
x
B
ω , ω , and v :
z
y
0ω + 3ω − 12ω + 0v = − 6
B
x y z
− 3ω + 0ω − 4ω + 0v = 0
B
x y z
(4.9.27)
12ω + 4ω + 0ω − v = 0
B
x y z
− 4ω + 12ω + 3ω + 0v = 0
B
x y z
B
Solving for ω , ω , ω , and v , we obtain:
x
y
z
.
.
.
ω =−0 568rad sec , ω =−0 296rad sec , ω =−0 426rad sec (4.9.28)
x y 3
and
B
v =−8m s (4.9.29)