Page 418 - Dynamics of Mechanical Systems

0593_C11_fm Page 399 Monday, May 6, 2002 2:59 PM

Generalized Dynamics: Kinematics and Kinetics 399

Using Eq. (11.12.5), the generalized inertia forces are then:

F =− mx ˙˙ + m(l + x )θ 2 ˙ (11.12.18)
*
x 1 1 1
F =− mx ˙˙ + ( l x 2 ˙ (11.12.19)
m 2 + )θ
*
x 2 2 2
m 3 + )θ
F =− mx ˙˙ + ( l x 2 ˙ (11.12.20)
*
x 3 3 3
[
2
2
θ
*
x
m
x
3
x
F =− (l + ) +( l2 + ) +( l + ) 2 ] ˙˙
θ
1
2
3
+ ( [ + ) ]
− m l2 x ˙ ) x +( l2 + x ˙ ) x +( l3 x ˙ x θ ˙ (11.12.21)
1 1 2 2 3 3
− ( 2 θ ) 3 ˙˙
ML
These results are the same as those of Eqs. (11.10.30) to (11.10.33).
Example 11.12.5: Rolling Circular Disk
For an example where Eq. (11.12.5) cannot be used to determine generalized inertia forces,
consider again the nonholonomic system consisting of the rolling circular disk on the flat
horizontal surface as shown in Figure 11.12.7. (Recall that Eq. (11.12.5) was developed for
holonomic systems [that is, systems with integrable or nonkinematic constraint equations]
and, as such, it is not applicable for nonholonomic systems [that is, systems with kinematic
or nonintegrable constraint equations].)
Due to the rolling constraint, the disk has three degrees of freedom instead of six, as would
be the case if the disk were unrestrained. Recall from Eqs. (11.10.34), (11.10.35), and (11.10.36)
that the condition of rolling (zero contact point velocity) produces the constraint equations:
˙ x = ( [ ˙ + ˙ sin ) θ cos +φ θ ˙ cos sinφ ] (11.12.22)
θ
r ψφ
˙ y = ( [ ˙ + ˙ sin ) θ sin −φ θ ˙ cos cosφ ] (11.12.23)
θ
r ψφ

˙ z =− θ ˙ sinθ (11.12.24)
r

FIGURE 11.12.7
Circular disk rolling on a horizontal surface S.