Page 342 - Dynamics of Mechanical Systems

0593_C10_fm Page 323 Monday, May 6, 2002 2:57 PM

Introduction to Energy Methods 323

where n , n , and n are unit vectors parallel to the coordinate axes as shown. We can
y
x
z
determine the work done by F on P as follows: knowing F, we need an expression for ds,
the differential arc length along C. Because the direction of ds is tangent to C, ds may be
expressed as:
ds =ττds (10.2.5)

where ττ ττ is a unit vector tangent to C at the position of P and ds is the differential arc length.
We can obtain expressions for both ττ ττ and ds from Eq. (10.2.3), the defining equations of
C. Recall that the velocity of a particle is tangent to its path (or curve) of travel. Hence,
by considering the parameter t in Eq. (10.2.3) as time, the velocity of P is:

r
r
v = ˙ x n + ˙ y n + ˙ z n = − sint n = cost n + n (10.2.6)
x y z x y z
Then, ττ ττ becomes:
+ ) / 12
2
n /
/
r
ττ= vv = − ( r sint n x + cost n y + ) (1 r (10.2.7)
z
The differential arc length is:
2
2
˙ z
r
ds = (x ˙ 2 + ˙ y 2 + ) / 12 dt = (1 + ) / 12 dt (10.2.8)
By substituting from Eqs. (10.2.4), (10.2.7), and (10.2.8) into (10.2.2), we find the work of
F to be:
t * t *
∫ ⋅ = ( − + 2 t t ) / 12
∫
+
3
W = Fds rtsin t rt cos dt (10.2.9)
0 0
*
where t is the value of t locating the ending position of P. For example, if t is π, W becomes:
*
W =− [ { r sin t tcos t] + r tcos2 t +( t − ) ] t 4 } π
[
−
2
sin
t + /4
2
0 (10.2.10)
=−3π r
By comparing Eqs. (10.2.5) and (10.2.7), we see that ds may be written as:
ds = vds v (10.2.11)
/
By recognizing v as ds/dt, ds becomes:

ds = vdt (10.2.12)

Hence, the definition of work of Eq. (10.2.2) takes the form: