Page 118 - Dynamics of Mechanical Systems

0593_C04*_fm Page 99 Monday, May 6, 2002 2:06 PM

Kinematics of a Rigid Body 99

By differentiating in Eq. (4.9.10), the angular acceleration of W in S is:

π
αα = 2π dn dt + = 2π ωω × n = −( )(22 25 (4.9.11)
S W S S C
2
0
x x y )n
Finally, from Eqs. (4.9.4) and (4.9.5), the desired relative velocity and acceleration expres-
sions are:
S L R S W [ y )
V = ωω × RL = 2π n − n × n
z ) ] ( 3 2
x (22 25
(4.9.12)
= 0 762 n + 5 44. n ft sec
.
x z
and
S W
S
S L R = αα × RL+ ωω W ×( ωω W × RL)
S
a
[
)
×
×
=− (22 25π )n y ( 3 2 n y ) + 2πn x −(22 25 n z] (0 762n. x + 5 44n. z) (4.9.13)
2
.
=−33 5n ft sec 2
y
Example 4.9.2: Absolute Movement of Sports Car Operator’s Hands
To further illustrate the application of Eqs. (4.9.4) to (4.9.6), suppose we are interested in
determining the velocity and acceleration of the sports car operator’s left hand in the
previous example. Specifically, find the velocity and acceleration of L in S.
Solution: Observe that the steering wheel hub O is fixed in both the steering wheel W
and the car C. Observe further that O moves on the 25-ft-radius curve (or circle) at 15
mph (or 22 ft/sec). Therefore, the velocity and acceleration of O are:
S O SO 2
22
.
V = 22 n ft sec and a = −( ) 25 n =−19 36 n y (4.9.14)
y
x
L
S L
O
S O
S
S
By knowing V and a , we can use Eqs. (4.9.4) and (4.9.6) to determine V and a . That is,
V = V + ωω × OL
S L S O S W
and
a + αα
S
L
S a = S O S W × OL+ ωω W ×( ωω W × OL) (4.9.15)
S
where from Figure 4.9.3 the position vector OL is:
OL = ( )( )
.
n + 05
.
05 0866
.
.
y
y . n = 0433 n + 025 n z (4.9.16)
z
S
S L
L
Hence, from Eqs. (4.9.10), (4.9.11), and (4.9.14), V and a are seen to be:
S L n + ( z)
V = 22 x (2π n − 22 25 n z) × 0 433. n + 0 25 n
.
x
y