EE/ME/AE324:
Dynamical Systems
Chapter 2: Modeling Translational
Mechanical Systems
Common Variables Used
• Assumes 1 DoF per mass, i.e., all motion scalar
• Displacement: x(t ) [m]
• Velocity: v(t )  dx(t ) [m/s]
dt
2
dv
(
t
)
d
x(t )
2
• Acceleration: a(t ) 

[m/s
]
2
dt
dt
• Force:
d ( M (t )v(t ))
dv(t )
kg  m
f (t ) 
M
 Ma(t ) [N or
]
2
dt
dt
s
t
• Energy (work): w(t )  w(t0 )  p( )d   J or N  m 

t0
• Power: p(t )  dw(t )  f (t )v(t ) [W or J/s]
dt
Variable Conventions
• Position, velocity and force:
• Supplied power (p(t )  0) :
Element Laws: Mass
• Mass: M [kg]
– Assumes constant, non-relativistic motion with 1 DoF
 f  Ma
1
• Kinetic Energy: wk  Mv 2
2
• Potential Energy: wp  Mgh
– g=9.81 [m/s2] @ surface of the earth
– h is height above(below) a specified reference point
Element Laws: Viscous Friction
• Viscous friction:
f  B v  B(v2  v1 )
Element Laws: Viscous Friction
Element Laws: Stiffness (Spring)
• Stiffness: f  K x  K ( x2  x1 )
– f is a tensile (stretching) force rather than a compressive force
when
x2  x1  0 (as drawn)
– d0 is the natural length of the spring, e.g., when x=0
• Potential Energy stored in spring: w  1 K ( x) 2
p
2
Interconnection Laws
• D’Alembert’s Law (restatement of Newton’s 2nd Law):
( f
i
)  Ma   fi  0
ext i
i
• Law of Reaction Forces (Newton’s 3rd Law):
Interconnection Laws
• Law of Displacements:
(
i
x)i  0, around any closed path
Free-body Diagrams
• Free-body diagrams are used as an intermediate step to
obtaining system equations of motion (EoM)
• Assume all elements at equilibrium (EQ) when position
and velocity references equal to zero
Equilibrium  Net force on body = 0,
with all inputs constant (zero)
• Suggested order applying forces to a free-body diagram
–
–
–
–
–
Applied forces, i.e., specified inputs
Inertial forces, i.e., opposite position reference
Spring forces
Viscous friction (damping) forces
All others, e.g., gears, pulleys, levers, etc.
Simple Free-body Example
• The MSD systems below have equivalent dynamics
– Assume elements at equilibrium (EQ) when
– Spring “stretched” when x  0, as drawn
x x0
Equations of Motion
• To obtain EoM from free-body diagram is straight forward
 left (up) pointing forces =  right (down) pointing forces
• Given the prior free-body diagram
• We have the following EoM
Mx  Bx  Kx  f a (t )
Example 2.2
• Equations of Motion:
M1 x1  K1 x  B( x2  x1 )  K 2 ( x2  x1 )
M 2 x2  B( x2  x1 )  K 2 ( x2  x1 )  f a (t )
Ex. 2.4: Relative Displacements
• Assume springs in EQ when x  z  0
• As drawn: Elongation of spring K2 is x  z
Inertial forces proportional to absolute
(not relative) accelerations, e.g., x  z
Relative Displacements
• As drawn x,z>0, which implies K2 “stretched”, K1 “compressed”
Relative Displacements
• EoM:
M1 x  B3 x  B1 x  K 1 x  B2 z
M 2 ( x  z )  B2 z  K 2 ( x  z )  f a (t )
Problem w/Absolute References
• As drawn x2 > x1 >0, which implies K2 “stretched,” K1
“compressed”
Problem w/Absolute References
• Free-body:
Ex. 2.5: Vertical Motion
• Assume x is position of the system at EQ
• Gravitational force on mass is downward Fg  Mg
Vertical Motion
• EoM: Mx  Bx  Kx  f a (t )  Mg
• At EQ with zero applied force, the constant
displacement caused by gravity can be calculated
Mg
Kx0  Mg  x0 
K
• When the mass is moving and f a (t ) is non-zero
– we can redefine x  x0  z
– substituting into the EoM yields
Mz  Bz  K  x0  z   f a (t )  Mg
 Mz  Bz  Kz  f a (t )
Ex. 2.6: Vertical Motion
• Assume springs relaxed when x1  x2  0
• Determine the static-EQ positions due to gravity
• Note: x2 is relative to M1
– total displacement of M2 is x1 +x 2
• As drawn: x2 , x1  0
– springs K1, K2 are stretched
– spring K3 is compressed
– friction B can be thought of as a
damper acting parallel to spring K2
with forces in the same direction
Ex. 2.6: Vertical Motion
Ex. 2.6: Vertical Motion
• EoM: M1 x1  K1 x1  f a (t )  Bx2  ( K 2  K3 ) x2  M 1 g
M 2 ( x1  x2 )  Bx2  ( K 2  K3 ) x2  M 2 g
• To find displacement due
to gravity set applied
force and all derivatives
to zero to obtain:
K1 x10  ( K 2  K 3 ) x20  M 1 g
( K 2  K 3 ) x20  M 2 g
M2g
 x20 
K 2  K3
(M1  M 2 ) g
x10 
K1
Ex. 2.7: Ideal Pulley
• An ideal pulley is a mass and friction-less element that
changes direction of motion, e.g., horizontal to vertical,
without cable slippage or stretch (assume in tension)
Ex. 2.7: Ideal Pulley
• The pulley system is the same as the one below, except
that is includes the gravitational force M2g
Ex. 2.8: Parallel Combinations
• For springs with a common natural length, the variable x
can be used to describe their displacement; otherwise
use d(t) as shown in the free-body below:
Ex. 2.8: Parallel Combinations
The resulting EoM:
Mx  Bx  ( K1  K 2 ) x  Mx  Bx  K eq x  f a (t )
where
Keq  K1  K 2
Ex. 2.8: Series Combinations
Ex. 2.8: Series Combinations
K1 x1
EoM at A: K1 ( x1  x2 )  K 2 x2  x2 
( K1  K 2 )
EoM at Mass:
K1 x1
Mx1  Bx1  K1 ( x1  x2 )  Mx1  Bx1  K1 ( x1 
)
( K1  K 2 )
 K1 K 2 
Mx1  Bx1  
 x1  Mx1  Bx1  K eq x1  f a (t )
 K1  K 2 
K1 K 2
where K eq 
K1  K 2
Ex. 2.11: Parallel-Series Combinations
B1 ( B2  B3 )
Beq 
B1  B2  B3
B1 B2  B1 B3

B1  B2  B3
( B2 B3 )
Beq  B1 
B2  B3
B1 B2  B1 B3  B2 B3

B2  B3
P2.14
Displacements x1 , x2 are relative to x3
Assume system at EQ when all xi  0; therefore, as drawn:
K1 compressed by amount x1  x2
K 2 , B2 , and B1 stretched by amount x2 , x2 and x1 , respectively
An absolute reference to M1 is y1  x3  x1  y10
An absolute reference to M 2 is y2  x3  x2  y20
P2.14
M 1 y1 
M 1 ( x3  x1 )
K1 ( x1  x2 )
M1
B1 x1
M 2 y2 
M 2 ( x3  x2 )
B1 x1
K 2 x2
M 3 x3
B2 x2
M3
f a (t )
M2
K 2 x2
B2 x2
P2.14
M 1 y1 
M 1 ( x3  x1 )
K1 ( x1  x2 )
M1
B1 x1
M 2 y2 
M 2 ( x3  x2 )
B1 x1
K 2 x2
M 3 x3
B2 x2
M3
f a (t )
EoM:
M 1 ( x3  x1 )  B1 x1  K1 ( x1  x2 )  0
M 2 ( x3  x2 )  K1 ( x1  x2 )  K 2 x2  B2 x2
M 3 x3  B2 x2  K 2 x2  B1 x1  f a (t )
M2
K 2 x2
B2 x2
P2.21
Assume system at EQ when all xi  0,
and x1  x2  0; then, as drawn:
K1 compressed by amount x1
K 2 stretched by amount x1  x2
K 3 stretched by amount x2
B1 stretched by amount x1  x2
B2 stretched by amount x2
P2.21
M 2 x2
K 3 x2
B2 x2
M2
B1 ( x1  x2 )
f a (t )
M2g
M 1 x1
K 2 ( x1  x2 )
M1
fb (t )
K1 x1
M1 g
P2.21
M 2 x2
K 3 x2
B2 x2
B1 ( x1  x2 )
M 2 g K 2 ( x1  x2 )
K 2 ( x1  x2 )
M1
M2
B1 ( x1  x2 ) f a (t )
M 1 x1
fb (t )
K1 x1
M1 g
EOM:
M 2 x2  B2 x2  K3 x2  B1 ( x1  x2 )  K 2 ( x1  x2 )  f a (t )  M 2 g
M1 x1  B1 ( x1  x2 )  K 2 ( x1  x2 )  K1 x1  f b (t )  M 1 g
Modified P2.21
Assume system at EQ when all xi  0,
and x1 , x2  0; then, as drawn:
K1 stretched by amount x1
K 2 compressed by amount x1  x2
K 3 stretched by amount x2
B1 compressed by amount x1  x2
B2 stretched by amount x2
Modified P2.21
Assume system at EQ when all xi  0,
and x0  x1  x2  0; then, as drawn:
x0
M0
P2.24
Assume system at EQ when all xi  0,
and x1  x2  0; then, as drawn:
K1 stretched by amount x1  x2
K 2 stretched by amount x2
B1 stretched by amount x1
B2 stretched by amount x2
P2.24
M 2 x2 B2 x2 K 2 x2
M2
K1 ( x1  x2 )
M 1 x1
f a (t )
M1
EOM:
M 2 x2  B2 x2  K 2 x2  K1 ( x1  x2 )  M 2 g
M1 x1  B1 x1  K1 ( x1  x2 )  f a (t )
M2g
K1 ( x1  x2 )
B1 x1
Questions?