Page 411 - Dynamics of Mechanical Systems

0593_C11_fm Page 392 Monday, May 6, 2002 2:59 PM

392 Dynamics of Mechanical Systems

of freedom represented by the coordinates q (r = 1,…, n). Then, from Eq. (11.6.7), we
r
ˆ
F
recall that the contribution of the spring force to the generalized force on S, for the
r
coordinate q , is:
r
F =− () x ∂
ˆ
r f x q ∂ (11.11.14)
r
where f(x) is the magnitude of the spring force due to a spring extension, or compression,
ˆ
F
x. For a linear spring f(x) is simply kx; hence, for a linear spring, is:
r
ˆ
F =− kx x ∂ (11.11.15)
r q ∂
r
ˆ
From Eq. (11.11.10), if we let the potential energy function P be (1/2)kx , we obtain from
2
F
r
Eq. (11.11.1) as:
ˆ
F =−∂ ∂ q =− ∂ ( [ 12 kx ) ] =− kx x ∂ (11.11.16)
2
P
r r q ∂ q ∂
r r
which is consistent with Eq. (11.11.15).
Alternatively, for a nonlinear spring, we may let the potential energy function have the
form:
x
P = ∫ f ()dξξ (11.11.17)
0
ˆ
Then, from Eq. (11.11.1), becomes
F
r
d  =− ()
ˆ
F =−∂ P q∂ = ∂   ∫ x f () ξξ  f x x ∂ (11.11.18)
r r q ∂   q ∂
r  r
 0
which is consistent with Eq. (11.11.14).
The definition of Eq. (11.11.1) and these simple examples show that if a potential energy
function is known we can readily obtain the generalized applied (or active) forces. Indeed,
the examples demonstrate the utility of potential energy for finding generalized forces.
Moreover, for gravity and spring forces, Eqs. (11.11.5), (11.11.10), and (11.11.17) provide
expressions for potential energy functions. This, however, raises a question about other
forces; that is, what are the potential energy functions for forces other than gravitational
or spring forces? The answer is found by considering again Eq. (11.11.1): If

∫
∂x
=−F then P =− Fdq r (11.11.19)
r
r
∂q
r
Unfortunately, it is not always possible to perform the integration indicated in Eq.
(11.11.19). Indeed, the integral represents an anti-partial differentiation with respect to