Page 159 - Dynamics of Mechanical Systems

0593_C05_fm Page 140 Monday, May 6, 2002 2:15 PM

140 Dynamics of Mechanical Systems

Because B has general plane motion, the velocity of Q may be expressed as:
2
v = v + ωω × PQ = 10 n + ω n ×( )λλ = 10 n + 3ω νν
Q
P
0
3.
2 y 2 z 2 x 2 2
/
= 10 n + 3ω ( [ −1 2 n +( 3 2 n y ) ] (5.5.15)
/
x 2 x
)
+
/
/
=[ 10 −(3 2 )ω n ] x ( 3 2 ω n
2 2 y
where ω is the angular speed of B . Note that Q is fixed in both B and B .
2
3
2
2
Because B has pure rotation with center R, Q moves in a circle about R. Hence, v may
Q
3
be expressed as:
v = ωω × RQ = ω n × − ( 495ω ) νν
Q
.
3 3 z 3 3
( [ y] (5.5.16)
+
=−495ω. 2/ n ) 2 x ( 2/ n ) 2 =−3 5ω. n − 3 5ω n
.
3 2 x 3 y
where ω is the angular speed of B .
3
3
Comparing Eqs. (5.5.15) and (5.5.16) we have the scalar equations:
−
10 15ω = − 35ω (5.5.17)
.
.
2 3
and
( 33 2)ω 2 =− 3 5ω 3 (5.5.18)
/
.
Solving for ω and ω we obtain:
2
3
ω = 244 rad sec and ω = − 1 81 rad sec (5.5.19)
.
.
2 3
Hence, v becomes:
Q
Q
.
.
v = 634 n + 6 34 n m sec (5.5.20)
x y
Observe that in calculating the angular speeds of B and B we could also use an analysis
3
2
of the instant centers as discussed in Section 5.4. Because the velocities of P and Q are
perpendicular to, respectively, B (OP) and B (QR), we can construct the diagram shown
3
1
Q
in Figure 5.5.7 to obtain ωω ωω , ωω ωω , and v . By extending OP and RQ until they intersect, we
2
3
obtain the instant center of zero velocity of B . Then, IP and IQ are perpendicular to,
2
respectively, v and v . Triangle IOR forms a 45° right triangle. Hence, the distance between
Q
P
P
I and P is (6.098 – 2.0) m, or 4.098 m. Because v  is 10 m/sec, ω is:
2
P
ω = v / IP = 10 4 098 = 2 44 rad sec (5.5.21)
.
.
2