Page 870 - Mechanical Engineers' Handbook (Volume 2)

9 Electromechanical Modeling Example 861

• Electrical equations:

di in
V L Ri K emf motor (91)

in
in
dt
¨
I motor motor (92)
in
1
¨
spring I (93)
aa
2
¨
where 1 (1/N) 2 and 2 spring I a . Substituting into (92) yields
a
1
¨
¨
( I ) I
in spring aa motor motor
N (94)
1
¨
¨
m
a
N
The arm can then be modeled with an effective spring constant:
spring
1
¨
¨
I a
I (95)
in m motor m
N N N
Then spring can be derived as
K (96)
spring effective a
LF KL 2 a (see Fig. 52)
(97)
K effective
Substituting (95)–(97) into (94) yields
KL 2 I
¨
¨
arm a I motor motor (98)
in
m
N 2 N 2
Inertial forces Spring force reflected to motor axis
I a ¨ KL 2

motor
in
N 2 I m N 2 m (99)
Arm inertia Motor inertia
reflected to
motor axis
L
– L a
a
F = Kx = KL a Figure 52 Free-body diagram of the arm.