Page 175 - Mechatronics with Experiments

MECHANISMS FOR MOTION TRANSMISSION 161
This example matches the ideal model shown in Figure 3.14. The load inertia is

1 2
J = ⋅ m ⋅ (d∕2) (3.171)
l
2
1 2 2
= ⋅ ⋅ ⋅ (d∕2) ⋅ l ⋅ (d∕2) (3.172)
2
1 4
= ⋅ ⋅ ⋅ l ⋅ (d∕2) (3.173)
2
1 4 2
= ⋅ (0.286∕386) ⋅ ⋅ 2.0 ⋅ (3∕2) lb ⋅ in ⋅ s (3.174)
2
= 0.0118 lb ⋅ in ⋅ s 2 (3.175)
Let us assume that we will pick a motor which has a rotor inertia the same as the load so
2
that there is an ideal load and motor inertia match, J = 0.0118 lb ⋅ in ⋅ s . The acceleration,
m
top speed, and deceleration rates are calculated from the kinematic relationships,
1 1
̇
= ⋅ t = ⋅ (3.176)
a
a
2 4 2
̇
= 2 ⋅ ∕t = 2 ⋅ (1∕4) ⋅ ( ∕2) rad∕(0.05 s) (3.177)
a a
= 80 ∕16 rad∕s = 40∕16 rev∕s = 2400∕16 rev∕min = 150 rev∕min (3.178)
̈
̇
2
= ∕t = (80 ∕16)(1∕0.05) = 1600 ∕16 rad∕s = 100 rad∕s 2 (3.179)
a a a
̈
= 0.0 (3.180)
r
̈
=−100 rad∕s 2 (3.181)
d
̈ dw = 0.0 (3.182)
The required torque to move the load through the desired cyclic motion can be calculated
as follows,
̈
T = (J + J ) ⋅ = (0.0118 + 0.0118) ⋅ (100 ) = 7.414 lb ⋅ in (3.183)
m
l
a
T = 0.0 (3.184)
r
̈
T = (J + J ) ⋅ = (0.0118 + 0.0118) ⋅ (−100 ) =−7.414 lb ⋅ in (3.185)
d m l
T = 0.0 (3.186)
dw
Hence, the peak torque requirement is
T max = 7.414 lb ⋅ in (3.187)
and the RMS torque requirement is
)
( 1 ( ) 1∕2
2
2
2
T rms = T ⋅ t + T ⋅ t + T ⋅ t + T 2 dw ⋅ t dw (3.188)
a
d
r
d
a
r
0.250
( ) 1∕2
1 2 2
= (7.414 ⋅ 0.05 + 0.0 ⋅ 0.05 + (−7.414) ⋅ 0.05 + 0.0 ⋅ 0.1) (3.189)
0.250
= 4.689 lb ⋅ in (3.190)
2
Therefore, a motor which has rotor inertia of about 0.0118 lb ⋅ in ⋅ s , maximum speed
capability of 150 rev∕min or better, peak and RMS torque rating in the range of 8.0lb ⋅ in
and 5.0lb ⋅ in range would be sufficient for the task.
This design may be improved further by the following consideration. The top speed
of the motor is only 150 rpm. Most electric servo motors run in the 1500 rpm to 5000 rpm