Page 543 - Dynamics of Mechanical Systems

0593_C15_fm Page 524 Tuesday, May 7, 2002 7:05 AM

524 Dynamics of Mechanical Systems

Then, by comparing Eqs. (15.5.16) and (15.5.17) we have:

2
4
cosφ =−( )( ) l 2 sin θ −( )( ) l 4 sin θ −… (15.5.18)
1 8 r
1
1 2 r
By substituting into Eq. (15.5.13), the displacement x has the approximate form:
2
2
l
+ −( )l
x ≈ cosθ l 12 r ( ) sin θ (15.5.19)
r
2
Using the identity sin θ ≡ (1/2) – (1/2)cos2θ, we then have:
2
2
l
l
x ≈ cosθ l r )( ) +( )( ) cos θ (15.5.20)
l
r
4
2
r
+ −(14
Let us assume that the crank rotates at a constant angular rate ω. Then, by differentiating
in Eq. (15.5.20), the acceleration of the mass C is given by the approximate expression:
θ
˙˙ x ≈− rω 2 [ cos +( ) l cos θ2 ] (15.5.21)
r
Hence, the inertia force F θ * on C is approximately:
C [
θ
F ≈ mrω 2 cos + r ( ) l cos θ n x ] (15.5.22)
*
2
C
The first term of Eq. (15.5.22) is called the primary unbalancing force and the second smaller
term is called the secondary unbalancing force.
We can counteract the primary unbalancing force by adding a balancing mass ˆ m at a
point Q on the crank, at a distance h away from A, on the opposite side of B, as in Figure
*
15.5.5. Then, the resultant F of the inertia forces on the system due to the masses ˆ m and
m become:
C
F* =− ( [ ˆ + m r ω 2 θ 2 cos θ ]
)
( )rωl
mh
C cos + m r 2 n x
C
(15.5.23)
+− [ ˆ mhω 2 sin n ] θ
y
Then, by adjusting the magnitude of ˆ mh , we can eliminate the primary unbalancing
2
force (m rω cosθn ). In so doing, however, we introduce an unbalance in the n direction.
C
x
y
2
Also, the secondary unbalancing force (m (r/ )rω cos2θn ) remains. Therefore, as a com-
C
x
promise, mh is usually selected as a fraction (say, 1/2 to 2/3) of m r [15.7]. In the following
C
section, we will explore means for counteracting the secondary unbalancing forces.
n y
B
r
C(m )
C
h
FIGURE 15.5.5 n x
Balancing mass Q ( ) on the rotating crank. Q(m)
ˆ m