Page 158 - The Mechatronics Handbook

0066-frame-C09 Page 32 Friday, January 18, 2002 11:00 AM

z a
A

Ω
y a
^
k
z o a
^
j
^ a
i a
x a
O y o

x o
FIGURE 9.29 Often it is necessary to find the time derivative of vector relative to a axes, x o , y o , z o , given its value
A
in the translating-rotating system x a , y a , z a .
need to find its value in x o , y o , z o . The vector A is expressed in the axes x a , y a , z a using the unit vectors
shown as

A = A x i a + A y j a + A z k a
ˆ
ˆ
ˆ
To find the time rate of change, we identify that in the moving reference the time derivative of A is

 dA  = --------- i a + -------- j a + dA z ˆ
dA y
ˆ
dA x
ˆ
dt 
 ------- a dt dt -------- k a
dt
ˆ
ˆ ˆ
Relative to the x o , y o , z o axes, the direction of the unit vectors i a, j a, and change only due to rotation
k a
Ω, so,
ˆ
ˆ
ˆ
dA  dA  di a dj a dk a
------- = ------- + A x ------ + A y ------ + A z --------
dt  dt  dt dt dt
ˆ ˆ ˆ
ˆ
ˆ
------ = Ω × ˆ i a, ------ = Ω × j a, -------- = Ω × k a
di a
dj a
dk a
dt dt dt
then,
dA  dA 
------- = ------- + Ω × A (9.9)
dt  dt  a
This relationship is very useful not only for calculating derivatives, as derived here, but also for
formulating basic bond graph models. This is shown in the section titled “Rigid Body Dynamics.”
Motion of a Body Relative to a Coordinate System
Translating Coordinate Axes
The origin of a set of axes x a , y a , z a is fixed in a rigid body at A as shown in Fig. 9.30(a), and translates
without rotation relative to the axes x o , y o , z o with known velocity and acceleration. The rigid body is
ω
subjected to angular velocity and angular acceleration α in three dimensions.
Motion of Point B Relative to A. The motion of point B relative to A is the same as motion about a
fixed point, so v B/A = ω × v B/A, and a B/A = α × r B/A + ω × (ω × r B/A).
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