Page 443 - Dynamics of Mechanical Systems

0593_C12_fm Page 424 Monday, May 6, 2002 3:11 PM

424 Dynamics of Mechanical Systems

By substituting from Eqs. (12.3.1) into (12.3.2) we have:

d  ∂ K  ∂ K ( n)


dt ∂   − q ∂ r = F r r = …,1 , (12.3.3)
q ˙
r
Next, if the system S is such that a potential energy function P exists, the generalized
active forces F may be expressed as (see Eq. (11.11.1)):
r

F =− ∂P ( r =…, n) (12.3.4)
1,
1 q ∂
r
By substituting into Eq. (12.3.3), we have:

d  ∂ L  ∂ L ( n)


dt ∂   − q ∂ r = 0 r = …,1 , (12.3.5)
q ˙
r
where L is defined as:

D
L = K − P (12.3.6)

and is called the Lagrangian.
Equations (12.3.3) and (12.3.5) are the common forms of Lagrange’s equations. The
principle advantage of these equations is that vector accelerations need not be computed;
that is, the inertia forces are developed independently by differentiation of the kinetic
energy function. Another advantage of Lagrange’s equations is that, like Kane’s equations,
nonworking forces are automatically eliminated from the equations through use of gen-
eralized active forces. As such, nonworking forces may be neglected at the onset of an
analysis. Finally, as with Kane’s equations, Lagrange’s equations produce exactly the same
number of equations as the degrees of freedom.
Disadvantages of Lagrange’s equations are that they are not readily applicable with
nonholonomic systems and the scalar differentiations in Eqs. (12.3.3) and (12.3.5) can be
tedious and burdensome for large systems. (References 12.7 and 12.8 discuss extending
Lagrange’s equations to nonholonomic systems.)
To illustrate the use of Lagrange’s equations we can again consider the examples of the
foregoing sections.

Example 12.3.1: The Simple Pendulum
Consider first the simple pendulum of Figure 12.3.1. We considered this system in several
previous sections including Sections 11.5, 11.10. 11.12, and 12.2. The system has one degree
of freedom represented by the angle θ. In Section 11.12, we found the kinetic energy K of
the pendulum to be (see Eq. (11.12.7)):

2 ˙
K = 12 ml θ 2 (12.3.7)