Page 167 - Dynamics of Mechanical Systems

0593_C05_fm Page 148 Monday, May 6, 2002 2:15 PM

148 Dynamics of Mechanical Systems

Let n be a unit vector normal to the X–Y plane generated by n × n . Then, ω may be
y
x
z
expressed as:
ωω= ωn (5.7.5)
z
where ω is positive when B rotates counterclockwise, as viewed in Figure 5.7.1.
P
Using Eqs. (5.7.1) to (5.7.5), v may be expressed as:
ωω
p
v = ˙ x n + ˙ ˙ n = − × r
y
P x P y
n n n
x y z
=− 0 0 ω (5.7.6)
rcosθ rsinθ 0
= r sinω θ n − r cosω θ n
x y

By comparing components, we obtain:

rsinθ = x ˙ ω and rcosθ = − y ˙ ω (5.7.7)
P P
We can readily locate C using these results: from Eqs. (5.7.1), (5.7.2), and (5.7.7), we have:

p = p + = x n + y n = ( x + cosθ n ) x ( y + sin n ) θ (5.7.8)
+
r
r
r
C P C x C y P P y
Therefore, by comparing components, we have:
x = x − ˙ y ω and y = y + ˙ x ω (5.7.9)
C P P C P P
Equation (5.7.9) shows that if we know the location of a typical point P of B, the velocity
of P, and the angular speed of B, we can locate the center C of zero velocity of B.
Let Q be a second typical point of B (distinct from P). Then, from Eq. (5.7.9) we have:

x = x − ˙ y ω and y = y + ˙ x ω (5.7.10)
C Q Q C Q Q
By comparing the terms of Eqs. (5.7.9) and (5.7.10), we have:

x − ˙ y ω = x − ˙ y ω and y + ˙ x ω = y + ˙ x ω (5.7.11)
P P Q Q P P Q Q

Solving for ω we obtain:

˙ x − x ˙ y − ˙ y
ω = Q P and ω = P Q (5.7.12)
y − y x − x
P Q P Q