Download ME375 Dynamic System Modeling and Control

ME375
System Modeling and Analysis
Rotational Mechanical Systems
Rotational Mechanical Systems
•
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Variables
Basic Modeling Elements
Interconnection Laws
Derive Equation of Motion (EOM)
Variables
 (t)
J
q
K
B
• q : angular displacement [rad]
• w : angular velocity [rad/sec]
• a : angular acceleration [rad/sec2]
•  : torque [Nm]
• p : power [Nm/sec]
• w : work ( energy ) [Nm]
1 [Nm] = 1 [J] (Joule)
d
q q w
dt
d  d  d2
w   q   2 q q a
dt  dt  dt
d
p    w   q  w
dt
t1
w(t1 )  w(t0 )   p(t ) dt
t0
t1
 w(t0 )   (  q ) dt
t0
Basic Rotational Modeling Elements
• Damper
• Spring
– Friction Element
– Stiffness Element
D
w1
S
w2
D
B
S
q1
q2
b
 S  K q 2  q1
g
– Analogous to Translational
Spring.
– Stores Potential Energy.
– E.g. shafts
d
i b
 D  B q 2  q 1  B w 2  w 1
– Analogous to Translational
Damper.
– Dissipate Energy.
– E.g. bearings, bushings, ...
g
Basic Rotational Modeling Elements
• Moment of Inertia
• Parallel Axis Theorem
– Inertia Element
J  JO  M r 2
d
Ex:
J  JO  M r2

J q    i
i
– Analogous to Mass in
Translational Motion.
– Stores Kinetic Energy.
1
MR 2  Md 2
2
J  JO  M r2
2
1
 L
2

ML  M  
12
2
1
 ML2
3
Interconnection Laws
• Newton’s Second Law
• Lever
(Motion Transformer Element)
af
d
J w  J q    EXTi


dt 
i
Angular
Momentum
• Newton’s Third Law
– Action & Reaction Torque
J q     f1d1  f 2d 2
Massless or small
acceleration
 Jq 0
x1  d1q
x2  d 2q
f 2 x1 

f1 x 2
f1  x1  f 2  x2
f1d1  f 2 d 2
f 2 d1
1



f1 d 2 N
x1 d1
1


x2 d 2 N
Motion Transformer Elements
• Rack and Pinion
• Gears
1

q
f12
I ocm , r
Bg
F gr
1
f 21
2
Mr
Br
Ideal case
Gear 1
0  J1 q1     1  f12 r1
Gear 2
0  J 2 q2     2  f 21r2
 1  f12 r1
1
f12 r1
r1
1





2
f 21r2
r2
N
 2  f 21r2
r1w1  r2w2 
x
w2
r
1
 1 
w1
r2
N
 1  w1   2  w2
  Fr  0
x  rq

F
x
q
 r
 r
 q  F  x
Example
Derive a model (EOM) for the
following system:
q
 (t)
J
q
K
B
FBD: (Use Right-Hand Rule
to determine direction)
Translational

Equivalent

0
s
s
s
J
K
q
D
D
J q       s   D
 s  K ( 0  q )   Kq
 D  B ( 0  q )   Bq
J q    Kq  Bq
J q  Bq  Kq  
B
0
D
Energy Distribution
• EOM of a simple Mass-Spring-Damper System
Jq

0
Contribution
of Inertia
Bq
1
Contribution
of the Damper
Kq

 (t )

1
Contribution
of the Spring
3
Total
Applied Torque
We want to look at the energy distribution of the system. How should we start ?
• Multiply the above equation by angular velocity term w :
 What have we done ?
bg
J q  q  Bq  q  Kq  q   t  w
• Integrate the second equation w.r.t. time:

t1
t0
J q  q dt
KE

change of
kinetic energy


t1
t0
Bw  w dt
t
2
t01 Bw dt  0

energy dissipated
by damper


t1
t0
 What are we doing now ?
K q  q dt
PE

change of
potential energy

  ( t )  w dt
t1
t0
W
Total work done by the
applied torque  ( t ) from
time t0 to t1
Energy Method
• Change of energy in system equals to net work done by external
inputs
KE 
change of
kinetic energy
•
PE

change of
potential energy
work done by
external inputs
(non-conservative inputs)
Change of energy in system per unit time equals to net power exerted by
external inputs
d
( KE  PE ) 
dt
change of
total energy
per unit time
•
W
P
power exerted by
external inputs
(non-conservative inputs)
EOM of a simple Mass-Spring-Damper System
 (t)
q
J
B


d 1
1
2
2 
Jq 
Kq     q  Bq  q
dt 
2
2
P


KE
PE


1
1
J  2q  q 
K  2q  q    q  Bq  q
2
2
J q  Kq    Bq
J q  Bq  Kq  
K
Example
• Rolling without slipping
Elemental Laws:
Mx   F  f  f s  f D  f f
Coefficient
of friction m
x
q
R
f(t)
K
Jq  f f R
B
f s  Kx
f D  Bx
J, M
Kinematic Constraints: (no slipping)
FBD:
x  Rq  x  Rq
x
q
f(t)
fs
fd
Mx  f  Kx  Bx  f f
ff
J
Mg
N
x
 f fR
R
Example (cont.)
I/O Model (Input: f, Output: x):
Q: How would you decide whether or not
the disk will slip?
Mx  f  Kx  Bx  f f
x
J  f fR f
R
1
Jx
f 
R2
If
ff 
1
Jx  ms Mg
2
R
Slipping happens.
1
Mx  f  Kx  Bx  2 Jx
R
1 

 M  2 J  x  Bx  Kx  f
R 

M eff
Q: How will the model be different if the
disk rolls and slips ?
x  Rq
does not hold any more
f f  mk Mg
Example (Energy Method)
•
Rolling without slipping
Coefficient
of friction m
x
q
R
P   Bx  x  f  x
K
f
B
f(t)
Power exerted by non-conservative inputs
J, M
D
Energy equation
d
( KE  PE )  P
dt
Kinetic Energy
1
1
KE  Mx 2  J q 2
2
2
Potential Energy
1
PE  Kx 2
2
 Mx  x  J q  q  Kx  x  f  x  Bx  x
Kinematic Constraints: (no slipping)
x  Rq  x  Rq
EOM
Mx  x  J
x x

 Kx  x  f  x  Bx  x
R R
J 

 M  2  x  Bx  Kx  f
R 

M eff
Example (coupled translation-rotation)
q
Bg

I ocm , r
F gr
Mx  Fgr  Br x
I o  cmq    Fgr r  Bgq
x
x  rq
Mr
Br
Fgr 

r

Bg
I o  cm
x

x
r2
r2
Real world
Bg  
I ocm  

 M  2  x   Br  2  x 
R  
R 
r

M eff
Think about performance of lever and
gear train in real life
Beff