Page 414 - Dynamics of Mechanical Systems

0593_C11_fm Page 395 Monday, May 6, 2002 2:59 PM

Generalized Dynamics: Kinematics and Kinetics 395

By multiplying Eq. (11.12.1) by the mass m of P, we have:

d  ∂  1   ∂  1 
⋅
mav =  mv 2  − v 2 (11.12.4)
q r ˙ dt ∂ ˙ q  2   q ∂ r  2 

r
or by using Eqs. (11.12.2) and (11.12.3) we have:
d  ∂ K  ∂ K
F =−   + (11.12.5)
*

q r dt ∂ q ˙ r  q ∂ r
Consider next a set of particles P (i = 1,…, N) as parts of a mechanical system S having
i
n degrees of freedom. Then, by superposing (or adding together) equations as Eq. (11.12.5)
for each of the particles, we obtain an expression identical in form to Eq. (11.12.5) and valid
for the set of particles. Finally, if the set of particles is a rigid body, Eq. (11.12.5) also holds.
To illustrate the use of Eq. (11.12.5), consider again the elementary examples of the
foregoing section.

Example 11.12.1: Simple Pendulum
Consider first the simple pendulum as in Figure 11.12.1, where we are using the same
notation as before. Recall that this system has one degree of freedom represented by the
angle and that the velocity and partial velocity of the bob P are (see Eqs. (11.10.1) and
(11.10.4)):

v = lθ ˙ n and v = l n θ (11.12.6)
P
P
θ
˙ θ
The kinetic energy of P is then:
2
22 ˙
K = 1 m( ) = ( ) ml θ (11.12.7)
P
v
12
2
Then, by using Eq. (11.12.5), the generalized inertia force F θ * is:
˙˙
F =− d   ∂  1 ml 22    + ∂  1 ml 22  =−ml 2 θ (11.12.8)
θ
˙
θ
*
θ
dt  ∂ θ  2   ∂ θ  2 
This result is identical with that of Eq. (11.10.5).
R O

n θ
θ

n
r
FIGURE 11.12.1
The simple pendulum. P