discrete dynamical systems : inverted double pendulum : consider the double pendulum shown:
DESCRIPTION
Some analytical models of nonlinear physical systems. P. 2. l. x. A. 1. k 2. l. g. k 1. O. y. Discrete Dynamical Systems : Inverted Double Pendulum : Consider the double pendulum shown:. Inverted Double Pendulum – E quation of motion. - PowerPoint PPT PresentationTRANSCRIPT
1
(a) Discrete Dynamical Systems:1. Inverted Double Pendulum: Consider the double
pendulum shown:
1 2,
( )
k k linear torsional spring
l lengthof each rod
P external conservative load
m mass of each rod
2. Some analytical models of nonlinear physical systems
1
2
l
l
k2
k1
P
g
O
A
y
x
2
Inverted Double Pendulum – Equation of motion
Here,
• - are the generalized coordinates, • T and V- are the respective kinetic and potential
Energies for the system, • -- the generalized forces due to non-
conservative effects.
nci
i i i
d T T V( ) Q , i 1,2,3,......
dt q q q
The equations of motion can be determined by using Lagrange’s equations:
3
Inverted Double Pendulum – Equations of motion
2 2 2 21 2 G1 G1 1 G2 G2 2
2 21 1
G2 A G2/A A 1 OA
OA 1 1 A 1 1 1
G2/A 2 G2/A G2/A 2 2
G2/A 2 2 2
T T T [(mv I ) (mv I )] / 2
T (ml / 3) / 2;
v v v ; v k r
r l(cos i sin j ); v l( sin i cos j );
v k r ; r l(cos i sin j ) / 2;
v l( sin i cos j ) / 2;
2 2 2 2 22 2 1 2 1 2 2 1T (ml / 12) / 2 ml [ / 4 cos( ) / 2] / 2
For the double pendulum- kinetic energy:
4
Inverted Double Pendulum – Equations of motion
The work done by the external force P in a virtual
displacement from straight vertical position is:
1 2 1 1 2
2 21 1 2 2 1
V V V mglcos / 2 mg[lcos lcos / 2]
[k k ( ) ] / 2
1 1 2 2
1 1 1 2 2 2
1 1 2 2
1 1 2 2
B
B
B
W P i r ;
r l(cos i sin j ) l(cos i sin j )
r l( sin i cos j ) l( sin i cos j )
W P[ lsin lsin ]
The are :
Q Plsin ; Q Plsin ;
generalized forces
Potential energy:
5
Inverted Double Pendulum – Equations of motion:
Equation for 1:
21 1 2 2 1
1
2 1 2 2 1
22 1 2 1
1
1 1 2 1 2 21
2 21 2 2 1 2 2 1
1 2 1 2 2
d T( ) ml [ / 3 cos( ) / 4
dt
( )sin( ) / 4]
Tml sin( ) / 4
V3mglsin / 2 (k k ) k
ml [4 / 3 cos( ) / 4 sin( ) / 4]
(k k ) k 3mglsin 1 1/ 2 Plsin
6
Inverted Double Pendulum – Equations of motion:
Equation for 2:
22 1 2 1
2
1 1 2 2 1
22 1 2 1
2
2 2 1 2 21
2 22 1 2 1 1 2 1
2 1 2 2 2 2
d T( ) ml [ / 3 cos( ) / 4
dt
( )sin( ) / 4]
Tml sin( ) / 4
Vmglsin / 2 k k
ml [ / 3 cos( ) / 4 sin( ) / 4]
k k mglsin / 2 Plsin
7
Inverted Double Pendulum –Equations of motion:
We now consider a simplified version with k1= k2=k
Let
Then, the equations are:
Equation for 1:
Equation for 2:
22 1 2 1 1 2 1
1 2 2
[ / 3 cos( ) / 4 sin( ) / 4]
k k (P M)sin
21 2 2 1 2 2 1
1 2 1
[4 / 3 cos( ) / 4 sin( ) / 4]
2k k (P 3M)sin
2 2 2k k /ml , P Pl /ml , M mgl / 2ml
8
Discrete dynamical systems…….
2. Inverted Double Pendulum with Follower Force: Consider the same system as in last example, except that the force P changes direction depending on the orientation of the body on which
it acts.The force P now acts
at an angle to the rod AB and always maintains this
direction relative to the rod
regardless of the position in space of the system during its oscillations.
1
2
l
l
k2
k1
P
g
O
A
y
x
B
9
Inverted Double Pendulum with Follower….– Equations of motion
The work done by the external force P in a virtual
displacement from straight vertical position is:
1 2 1 1 2
2 21 1 2 2 1
V V V mglcos / 2 mg[lcos lcos / 2]
[k k ( ) ] / 2
2 2
1 1 2 2
B
B
W P[cos( ) i sin( ) j ] r ;
r l(cos i sin j ) l(cos i sin j )
Note that the only change is in the effect of the external force P. The potential and kinetic energy expressions remain the same. So,potential energy:
10
Inverted Double Pendulum with Follower– Equations of motion
The resulting equations of motion are:
1 1 1 2 2 2
1 1 2 2
Br l( sin i cos j ) l( sin i cos j )
So,the virtual work done is
W Pl[ sin( ) sin ]
1 1 2 2Q Plsin( ); Q Plsin .
2 21 2 2 1 2 2 1
1 2 1 2 2 1 1 2
ml [4 / 3 cos( ) / 4 sin( ) / 4]
(k k ) k 3mglsin / 2 Plsin( )
Thus, the generalized forces are:
The virtual displacement is:
11
Inverted Double Pendulum with Follower– Equations of motion
In the reduced case, with equal springs etc., the
equations are:
2 22 1 2 1 1 2 1
2 1 2 2 2
ml [ / 3 cos( ) / 4 sin( ) / 4]
k k mglsin / 2 Plsin
21 2 2 1 2 2 1
1 2 1 2 1
[4 / 3 cos( ) / 4 sin( ) / 4]
2k k Psin( ) 3Msin
22 1 2 1 1 2 1
1 2 2
[ / 3 cos( ) / 4 sin( ) / 4]
k k PSin Msin
and
12
Discrete dynamical systems…….
3. Dynamics of a Bouncing Ball: Consider a
ball bouncing above a horizontal table. The table
oscillates vertically in a specified manner (here
we assume harmonic oscillations).
The motion of the ball, during free flight, is governed by
Integrating once gives:
0 0y=y ( )g t t ground
table
Y(t)
ballgmg
ormy mg y g ( ) sin( )X t A t
13
Dynamics of a Bouncing Ball …….
Integrating once gives:
Integrating again, we get:
The position of the ball has to remain above the
table Finally, we have the law
of interaction between the table and the ball:
0 0y=y -g(t-t )2
0 0 0 0
1( ) ( )
2y y y t t g t t
0y(t) X(t), t t
the relative velocities
before and after impact are relate
If w
d
by c
simple law o
oefficient
e a
of
f
i
re
mpac
stit
ssume
t,
ution.ground
table
Y(t)
ballgmg
( ) sin( )X t A t
14
Dynamics of a Bouncing Ball …….
Thus, we have the relation:
These relations provide a complete description of
motion of the ball. Note that, given an initial condition, we have to piece together the
motion in forward time
i i i i
i
i
V(t )-X(t )=e[X(t ) U(t )]
V(t ) - velocity of the ball immediately after impact
and U(t ) - velocity of the ball just before impact
where
ground
table
Y(t)
ballgmg
15
Dynamics of a Bouncing Ball …….
Let us now proceed in a systematic manner. For
nonlinear analysis, it is always advisable to non-
dimensionalize equations: So we define
ground
table
Y(t)
ballgmg
2 2
2 2
2
: / , 2
: 2
: ( 2)
. : ( ) 2
, ( ) (2 ) ( )
sin( ) sin(2 )
( ) sin(2 );2
Time t t T whereT
Acceleration units g
Veloctiy units g T g
Pos units g T g
So X t g X
A t A
AX
g
16
Dynamics of a Bouncing Ball …….
Ball motion: z(t)= (22g/2)Z(),
2 2
2 2
2 2 2 2
2 2 2 2
dz 2 g dZ d 2 g dZ g dZ( ) ( ) ( )
dt d dt d 2 d
d z g d Z d g d Z g d Z( ) ( ) ( )
dt d dt d 2 2 d
2 .z g Z
0 0
20 0 0 0
( ) 2( ),
( ) ( ) ( )
Z Z
Z Z Z
ground
table
Y(t)
ballg
mg
Integrating,
Now,
17
Dynamics of a Bouncing Ball …….
Thus, we have
Ball motion:
Table motion:
Initially: ball starts at time 0 when it is in contact with the table, and just about to
leave
X( ) sin(2 ) (3)
2
0 0
20 0 0 0
2; ( ) 2( ), (1)
( ) ( ) ( ) (2)
Z Z Z
Z Z Z
0 0 0 0X( ) Z( ) Z sin(2 ) (4)2
ground
table
Y(t)
ballgmg
18
Dynamics of a Bouncing Ball …….
Ball velocity at =0:
Next collision at time 1 > 0 when
Now,
Using (3)-(5)
Z( ) X( ) W( ) 0
0 0 0
0
0 0 0
0
dZZ W X ( )
d
Z W cos(2 ) (5)
(H is relative velocity of the balere W 0 an unknl, own)
0 0 0
20
W( ) Z Z ( )
( ) sin(2 ) (6)2
ground
table
Y(t)
ballgmg
19
Dynamics of a Bouncing Ball …….
Then, the time instant 1 is defined by
(this is a relation in W’0, 1
and 0, and it depends on )
0
20 0 0 0
W( ) [sin(2 ) sin(2 )]2
[W cos(2 )]( ) ( )
1 0 1
0 0 1 0
21 0
W( ) [sin(2 ) sin(2 )]2
[W cos(2 )]( )
( ) 0 (7)
ground
table
Y(t)
ballgmg
20
Dynamics of a Bouncing Ball …….
or:
1 1 1 1
1 0 0 1 0
1
W W ( ) Z ( ) X ( )
dW( )cos(2 ) [W cos(2 )] 2( ) (8)
d
1 1
coefficient of restitut
W eW
(e ion)
1 0
1 0 1 0
W e[ {cos(2 )
cos(2 )} W 2( )] (9)ground
table
Y(t)
ballgmg
When just about to contact at this time instant 1 the
relative velocity is :
On impact, the ball relative velocity changes:
21
Dynamics of a Bouncing Ball …….
One can write the equations now in a more compact
form:
i 1 i i 1 i
i 1 i
W e[ {cos(2 ) cos(2 )} W
2( )] (B)
i 1 i i 1
2i i i 1 i i 1 i
W ( ) [sin(2 ) sin(2 )]2
[W cos(2 )]( ) ( ) 0 (A)and
ground
table
Y(t)
ballgmg
Knowing (I,W’i), equations (A) and (B) can be used to compute (I+1,W’i+1), thus generating the trajectory.
22
(b) Continuous Dynamical Systems
1. Rotating Thermosyphon: Consider a closed
circular tube in a vertical plane. The tube is filled with
a liquid of constant properties, except for variation of its density with
temperature in buoyancy and
centrifugal terms, i.e. in body forces. One part
of the loop is heated, and the other cooled.
The tube is spun about the vertical axis.
α
heating
cooling
R
r
23
Rotating Thermosyphon…
There are many ways to develop a model for the system. If the tube radius ‘r’ is much smaller than the torus radius ‘R’, one can assume that there is
negligible flow in the radial direction. Another approach is to average
the velocity and temperature over the
tube radius. Then, the equations for fluid
motion are:
α
heating
cooling
R
r
24
Rotating thermosyphon: equations…
Continuity:Here, V – average flow velocity at any section ; - density of the fluid and (V) is independent of .
Momentum:
Here, p – fluid pressure at a
section,
w- shear stress at the wall
1 (ρV)0 (1)
t R
α
heating
cooling
R
r 2
2
(ρV) 1 (ρV) 1cos
2cos sin (2)w
pg
t R R
Rr
25
Rotating thermosyphon:Equations…
Energy:
where Cp- specific heat of the fluid, T – mean fluid temperature at a section,
k – thermal conductivity,
q’ – applied heat source per unit length.
Remark: Here viscous dissipation term is neglected
22 2
2 2
( )( ) (3)
tp
T V T k Tr C r q
R R
α
heating
cooling
R
r
26
Rotating thermosyphon: Equations…
Simplification and nondimensionalization:Integrating the momentum eqn. (2) along the loop
Shear stress:
Friction factor: α
heating
cooling
R
r
2 2 2 22
0 0 0 0
2 22
0 0
(ρV) 1 (ρV) 1cos
2cos sin (4)w
pd d d g d
t R R
d R dr
2 / 2 (5)w f V
16 / Re, Re 2 /f Vr
27
Rotating thermosyphon: Equations…
Simplification:Introduce the variation of density with temperature in buoyancy and centrifugal terms, and use periodicity of variables (eqn. (4))
2
20
22
0
V 32( )cos
(2 ) 2
( )cos sin (6)2
r
r
V gT T d
t r
RT T d
where -kinematic viscosity, Tr-reference temperature
- coefficient of thermal expansion
28
Rotating thermosyphon: Equations…
Non-dimensional variables:
2
2
2
4 2
3
2
32 (2 ), , ,
(2 ) 32
2048 '( ) * 1, ( ) ( )
(2 ) 8 Pr 8192
Pr /
2 ( '
P
)*
r
2
r
p
p
p
t r T TU V
r R T
R q Ra rT Q
g r C T R
Here C k
r q CRa
k
andtl number
Modified Rayleigh number
29
Rotating thermosyphon: Equations…
Non-dimensional equations:The resulting momentum and energy equations are:
Note: - combination of geometric parameter and the Prandtl number Pr. Pr > 1 for ordinary fluids (air, water etc.) and <1 for liquid metals.
2 2
0 0
2
2
2 2
cos cos sin (7)
( ) (8)
/ , ( ) / 8Pr
UU d d
U Q
rwhere R g
R
2( / )r R
30
Rotating thermosyphon: equations…
Solution approach: The solutions to (7) and (8) can
be expressed as:
1( , ) [ ( )sin( ) ( ) cos( )] (9)
n n
nB n C n
1( ) [ sin( ) cos( )] (10)
n nn
Q A n D n
where as, the externally imposed heat flux can be represented in a Fourier series as:
Substituting (9) and (10) in equations (7) and (8), and collecting the appropriate Fourier coefficients gives an infinite set of ordinary differential equations In the unknowns U(), Bn(), and Cn().
31
Rotating thermosyphon: equations…
Solution approach…: It turns out that only five of these equations are independent-master equations.The remaining equations are linear equations for the remaining variable (called “slave variables” and “slave equations”). Equation (8) gives:
2
20 0
0
1
[ ( )sin( ) ( ) cos( )] [ ( )sin( ) ( ) cos( )]
[ ( )sin( ) ( ) cos( )]
[ sin( ) cos( )]
n n n n
n n
n n
n n
n
n
B n C n B n C n
B n C n
U A n D n
32
Rotating thermosyphon: equations…
Solution approach…: Collecting terms of different n’s give:
0 0
1 1 1 1
1 1 1 1
2 2 2 2
2 2 2 2
3 3 3 3
3 3 3 3
2
2
0 :
1: (sin )
(cos )
2 : 4 2 (sin 2 )
4 2 (cos 2 )
3 : 9 3 (sin 3 )
9 3 (cos3 )
: (sin )
p p p p
p
n C C
n B B UC A
C C UB D
n B B UC A
C C UB D
n B B UC A
C C UB D
n p B p B pUC A p
C p C
(cos ) (11)
3,4,5,6,...........
p p ppUB D p
p
33
Rotating thermosyphon: Equations…
Solution approach…: Now, considering equation (7)we get:
Evaluating the two integral terms on the right-hand side, it is clear that only coefficients of will survive. Thus, we get
Equations (11) and (12) govern the dynamics.
2
0
2
0
1
1
[ ( )sin( ) ( )cos( )]cos
[ ( )sin( ) ( )cos( )]cos sin
n n
n n
n
n
UU B n C n d
B n C n d
cos( ) sin(2 )and
1 2 (12) U U C B
34
(b) Continuous Dynamical Systems
2. Buckling of Elastic Columns: Consider a thin
beam that is initially straight. O xyz is coordinate
system with x-y plane coinciding with undisturbed
neutral axis of the beam. Let EI is the bending stiffness
X
V
35
Buckling of elastic…..
V(s) – vertical displacement of the centroidal axis,
X – distance measured along the centroidal axis from left end.
Define:
Then, the strain energy of the system is:
X
V
2 2 3 3x X /L; u V /L; L P / EI; L K / EI;
2
2
2
2 2
0
0
1 u(u, , ) )dx
2 1 u
1 u )dx
1 1u (0) u ( )
2 2
(
(1
36
(b) Continuous Dynamical Systems
3. Thin rectangular plates: Consider a thin plate
that is initially flat. Oxyz is coordinate system with x-y
plane coinciding with undisturbed middle surface of
the plate. Let h – plate thickness. The equations of motion for the plate,
for moderately large displacements
– von Karman equations.
In here, we give a short review of the
derivation of these equations.
x, u
y, v
z, w
a
37
Thin rectangular plates…..
Consider a differential plate element: The equations
in the three directions are: 2
2
2
2
(1)
(2)
xyx
xy y
NN uh
x y t
N N vh
x y t
x
y
z
Nx
Ny
Nxy
Nxy
Ny
Nx
F
2 22
2 2
2
2
2
( ) ( )
( ) ( ) (3)
y xyx
x y
xy xy
M MM
x yx y
w wN N
x x y y
w w wN N h F
x y y x t
38
Thin rectangular plates…..
The constitutive equations for a linearly elastic and
isotropic material are:
In these expressions, N, Ni – forces, M - moments
2 1 2
2
2 1 2
2
(1 ) [ / 2
/ 2] (4)
(1 ) [ / 2
/ 2] (5)
x x x
iy y x
y y y
ix x y
N Eh u w
v w N
N Eh v w
u w N
x
y
z
Nx
Ny
Nxy
Nxy
Ny
Nx
F
[ ] (6)
( ) (7)
ixy y x x y xy
x xx yy
N Gh u v w w N
M D w w
39
Thin rectangular plates…..
The constitutive equations for a linearly elastic and
isotropic material are:
Also u,v,w – displacements
Substituting the force – displacement
relations in the dynamic equations give:
x
y
z
Nx
Ny
Nxy
Nxy
Ny
Nx
F
3 2
( ) (8)
(1 ) (9)
2 /(1 );
/12(1 )
y yy xx
xy xy
M D w w
M D w
G E
D Eh
40
Thin rectangular plates…..
The dynamic equations for a plate made of linearly
elastic and isotropic material are then simplified by
introducing a stress function such that:
Then, (1) and (2) are automatically satisfied if in-
plane inertia terms are neglected. Furthermore, the
expressions (7)-(9) can be substituted in (3) to get
equation for transverse displacement. Also, a
compatibility condition is (gives an equation for ):
2 2 2
2 2, , (10)
x y xyN N Nx yy x
( ) ( ) 0 (11) xyy yxx y x xyu v u v
41
(b) Continuous Dynamical Systems
4. Flow between concentric rotating cylinders:
Consider two concentric cylinders with radii a, b;
Let 1, 2 – angular velocities of inner
and outer
cylinders; let (ur,uθ,uz) – velocity
components in a cylindrical coordinate system;
p – pressure at a point;
We now define the equations of motion for the system.
a
b 12
42
Flow between concentric rotating….
Equations of motion for the system: In cylindrical
coordinate system the NS equations are
ab 1
2
2r r
r 2 2
r r2 2
zz
r z
u uDu u1 p 2( u ),
Dt r r r r
Du u u u u1 p 2( u ),
Dt r r r r
Du 1 p( u ), (1)
Dt z
where
Du u u
Dt t r r z
43
Flow between concentric rotating….
Equations of motion for the system: and
There is also the equation for mass conservation:
The basic flow is defined by:
r r zuu u u10. (2)
r r r z
ab 1
2
2 2 2
2 2 2 2r rr r z
r z
2
u 0, u 0, u V(r) r (r) and p P(r)
1 dP Vor and DD V 0 (3)
dr r
44
Flow between concentric rotating….
The basic flow…: Equations (3) have solution of the form
Let us consider perturbations to the basic flow:
We can then obtain linearized equations about the basic
flow.
ab 1
2
2
22
1 1 12 2
2 1 1 2
(r) A B / r
1where A , B R
1 1
and / , R /R (4)
r zu (u ,V u ,u ), p P p
45
Flow between concentric rotating….
Linearized equations about the basic flow:
r r r
r 2 2
rr 2 2
z zz
r r z
uu u u1 p 22 u ( u )
t r r r
u u u u1 p 2(D V)u ( u )
t r r r
u u 1 p( u )
t z
uu u u1and 0
r r r z