Page 121 - Dynamics of Mechanical Systems

0593_C04*_fm Page 102 Monday, May 6, 2002 2:06 PM

102 Dynamics of Mechanical Systems

Similarly, from Eq. (4.9.6), the acceleration of end A may be expressed as:

a = a + αα AB × AB + ωω AB ×(ωω AB × AB) (4.9.30)
B
A
Let α be expressed in the form:
AB
αα AB = α n + α n + α n (4.9.31)
x x y y z z

B
Let a be expressed as:
a = a B n (4.9.32)
B
z
Then, by substituting from Eqs. (4.9.19), (4.9.22), (4.9.23), (4.9.31), and (4.9.32) into (4.9.30),
we have:

x (
x ( 3α
B
a n =−3 n + − 12α z) n +−3α − 4α z) n
z y x y (4.9.33)
+( 12α + 4α y) n − 8ω n + − 6ω z) n + 6ω n
x (8ω
x z y x y y z
Also, from Eq. (4.9.20) we have:

−4α + 12α + 3α = 0 (4.9.34)
x y z

Using Eq. (4.9.29) for values of ω , ω , and ω , Eqs. (4.9.33) and (4.9.34) are equivalent to
z
y
x
B
the following four scalar equations for α , α , α , and a :
z
x
y
0α + 3α − 12α + 0a = .
B
0 632
y
x
z
− 3α + 0α − 4α + 0a = .
B
7 1
x y z
(4.9.35)
12α + 4α + 0α − a = .
B
1 776
z
y
x
− 4α + 12α + 3α + 0a = 0
B
z
y
x
B
Solving for α , α , α , and a , we obtain:
z
y
x
.
α =−2 083rad sec 2 , α =−0 641rad sec 2 , α =−0 213rad sec 2 (4.9.36)
.
.
x y z
and
B
.
a =−29 33m s (4.9.37)