Page 344 - Dynamics of Mechanical Systems

0593_C10_fm Page 325 Monday, May 6, 2002 2:57 PM

Introduction to Energy Methods 325

P 1
k
P(m)

h C θ
ds
dh
τ
ˆ
C
P
2
FIGURE 10.2.5 FIGURE 10.2.6
ˆ
A particle P moving from P 1 to P 2 under the action A differential arc of an arbitrary curve C.
of gravity.

It happens that Eq. (10.2.18) is a valid expression for the work of the gravity force
regardless of the path of descent of the particle P. In the pendulum example, the particle
ˆ
path was a circular arc. However, if P had moved on some other path (say, , as in Figure
C
10.2.5), the work done by gravity would still be mgh, where h is the change in elevation
ˆ
ˆ
C
C
of P as it descends from P to P along . To see this, consider a differential arc of as
2
1
in Figure 10.2.6 where ττ ττ is a unit vector tangent to C ˆ and k is a vertically downward
directed unit vector. Then, the integrand of Eq. (10.2.2) becomes:
τ
τ
⋅
wds = mg k⋅ ds = mgcosθ ds = mgdh (10.2.19)
ˆ
C
where θ is the angle between and the vertical as in Figure 10.2.6 and dh is the differential
elevation change as shown. Hence, integrating the work done by gravity is:
W = ∫ mgdh = mgh (10.2.20)
For a third example, consider the work done by a force in stretching or compressing a
linear spring. Specifically, consider a spring as depicted in Figure 10.2.7. Let F be the
magnitude of the stretching force F. Let the natural length of the spring be , and let its
stretched length, under the action of F, be + δ. Then, from Eq. (10.2.2), the work is:

δ δ
W = ∫ Fdx = ∫ kxdx = ( ) kδ 2 (10.2.21)
12
0 0
where k is the spring modulus, and x is the end displacement along the axis of the spring.
Similarly, if the spring is compressed, the work done by the compressing force is (1/2)kδ ,
2
where now δ is a measure of the shortening of the spring.

F
FIGURE 10.2.7
A linear spring stretched by a force F.