lect2physicalsystemmodels_10feb14

8/21/2019 Lect2PhysicalSystemModels_10Feb14

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(a) Discrete Dynamical Systems:1. Inverted Double Pendulum: Consider the double

pendulum shown:

1 2,

( )

k k linear torsional springl length of 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


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Inverted Double Pendulum

Equation of motion

Here,

- are the generalized coordinates,

T

andV

- are the respective kineticand potentialEnergies for the system,

-- the generalized forcesdue to non-

conservative effects.

nc

i

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 Lagranges equations:


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Equations of motion

2

2 2 2

1 2 G1 G1 1 G2 G2 2

2 2

1 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 2

2 2 1 2 1 2 2 1T (ml / 12) / 2 ml [ / 4 cos( ) / 2] / 2

For the double pendulum- kinetic energy:


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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 2

1 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:


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Equations of motion:

Equation for 1:

2

1 1 2 2 1

1

2 1 2 2 1

2

2 1 2 1

1

1 1 2 1 2 2

1

2 2

1 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 Pl sin


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Equation for 2:

2

2 1 2 1

2

1 1 2 2 1

2

2 1 2 1

2

2 2 1 2 2

1

2 2

2 1 2 1 1 2 1

2 1 2 2 2 2

d T( ) ml [ / 3 cos( ) / 4

dt

( )sin( ) / 4]

Tml sin( ) / 4

V

mglsin / 2 k k

ml [ / 3 cos( ) / 4 sin( ) / 4]

k k mglsin / 2 Plsin


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We now consider a simplified version with k1= k2=kLet

Then, the equations are:

Equation for 1:

Equation for 2:

2

2 1 2 1 1 2 1

1 2 2

[ / 3 cos( ) / 4 sin( ) / 4]

k k (P M)sin

2

1 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


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Discrete dynamical systems.

2. Inverted Double Pendulum with Follower Force:

Consider the same system as in last example,except that the force P changes directiondepending on the orientation of the body on which

it acts.

The force P now acts atan angle to the rod AB andalways maintains thisdirection relative to the rod

regardless of the position inspace of the system duringits oscillations.

1

2

l

l

k2

k1

P

g

O

A

y

x

B


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Inverted Double Pendulum with Follower.

Equations of motion

The work done by the external force Pin a virtual

displacement from straight vertical position is:

1 2 1 1 2

2 2

1 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:


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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 2

1 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:


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Inverted Double Pendulum with Follower

Equations of motion

In the reduced case, with equal springs etc., theequations are:

2 2

2 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 P sin( ) 3M sin

2

2 1 2 1 1 2 1

1 2 2

[ / 3 cos( ) / 4 sin( ) / 4]

k k PSin Msin

and


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Discrete dynamical systems.

3. Dynamics of a Bouncing Ball: Consider aball 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)

ball

gmg

ormy mg y g ( ) sin( )X t A t


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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 ) 20 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

fi

re

mpac

stit

ssumet,

ution.ground

table

Y(t)

ball

gmg

( ) sin( )X t A t


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Thus, we have the relation:

These relations provide a

complete description of

motion of the ball. Note

that, given an initialcondition, 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)

ball

gmg


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Let us now proceed in a systematic manner. For

nonlinear analysis, it is always advisable to non-

dimensionalizeequations: 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

A

X g


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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

2

0 0 0 0

( ) 2( ),

( ) ( ) ( )

Z Z

Z Z Z

ground

table

Y(t)

ball

gmg

Integrating,

Now,


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Thus, we have

Ball motion:

Table motion:

Initially: ball starts at time 0

when it is in contact with the

table, and just about toleave

X( ) sin(2 ) (3)

2

0 0

2

0 0 0 0

2; ( ) 2( ), (1)

( ) ( ) ( ) (2)

Z Z Z

Z Z Z

0 0 0 0X( ) Z( ) Z sin(2 ) (4)

2ground

table

Y(t)ball

gmg


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Ball velocity at =0:

Next collision at time 1 > 0when

Now,

Using (3)-(5)

Z( ) X( ) W( ) 0

0 0 00

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)ball

gmg


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Then, the time instant 1 is defined by

(this is a relation in W0, 1and 0, and it depends on )

0

2

0 0 0 0

W( ) [sin(2 ) sin(2 )]

2[W cos(2 )]( ) ( )

1 0 1

0 0 1 0

2

1 0

W( ) [sin(2 ) sin(2 )]2

[W cos(2 )]( )

( ) 0 (7)

ground

table

Y(t)ball

gmg


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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)

ball

gmg

When just about to contact at this time instant 1the

relative velocity is :

On impact, the ball relative velocity changes:


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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

2

i i i 1 i i 1 i

W ( ) [sin(2 ) sin(2 )]2

[W cos(2 )]( ) ( ) 0 (A)

and

ground

table

Y(t)

ball

gmg

Knowing (I,Wi), equations

(A) and (B) can be used tocompute (I+1,Wi+1), thus

generating the trajectory.


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(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 inbuoyancy 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.

heating

cooling

R

r


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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 isnegligible flow in the

radial direction. Anotherapproach is to averagethe velocity andtemperature over the

tube radius. Then, theequations for fluid

motion are:

heating

cooling

R

r


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Rotating thermosyphon: equations

Continuity:Here, Vaverage flow velocity at any section ;

- density of the fluid and (V) is independent of .Momentum:

Here, pfluidpressure at a section,

w- shear stress at the

wall

1 (V)0 (1)

t R

heating

cooling

R

r 2

2

(V) 1 (V) 1cos

2cos sin (2)w

pg

t R R

Rr

Rotating thermosyphon:


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Equations

Energy:

where Cp- specific heat of the fluid, Tmean fluidtemperature at a section,

kthermal conductivity,

q applied heat sourceper unit length.

Remark:Here viscousdissipation term isneglected

2

2 22 2

( )( ) (3)t

p

T V T k T r C r qR R

heating

cooling

R

r



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Equations

Simplification and nondimensionalization:

Integrating the momentum eqn. (2) along theloop

Shear stress:

Friction factor:

heating

cooling

R

r

2 2 2 22

0 0 0 0

2 2

2

0 0

(V) 1 (V) 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



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Equations

Simplification:

Introduce the variation of density with temperature

in buoyancy and centrifugal terms, and use

periodicity of variables (eqn. (4))

2

2

0

22

0

V 32( )cos

(2 ) 2

( )cos sin (6)2

r

r

V gT T d

t r

R

T T d

where -kinematic viscosity, Tr-reference temperature

- coefficient of thermal expansion



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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 T U V

r R T

R q Ra r T Qg r C T R

H er e C k

r q CRa

k

andtl number

M odif ied Rayleigh number



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Equations

Non-dimensional equations:

The resulting momentum and energy equations are:

Note: - combination of geometric parameterand the Prandtl number Pr. Pr > 1 for ordinary

fluids (air, water etc.) and


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Solution approach: The solutions to (7) and (8) can

be expressed as:

1

( , ) [ ( )sin( ) ( )cos( )] (9)

n n

n

B n C n

1

( ) [ sin( ) cos( )] (10)

n n

n

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), andcollecting the appropriate Fourier coefficients gives an infiniteset of ordinary differential equations In the unknowns U(),Bn(), and C

n().


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Solution approach:It turns out that only five of

these equations are independent-masterequations.The remaining equations are linear equations for the

remaining variable (called slave variables and slave

equations). Equation (8) gives:

2

2

0 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 nU A n D n


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Solution approach:Collecting terms of different ns 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 (cos 3 ): (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 Dn p B p B pUC A p

C p C

(cos ) (11)

3,4,5,6,...........

p p ppUB D p

p



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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


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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


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Buckling of elastic..

V(s)vertical displacement of the centroidal axis,

Xdistance 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


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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 hplate thickness. The equations of

motion for the plate, formoderately large

displacementsvon Karman

equations.

In here, we give a short

review of the derivation of

these equations.x, u

y, v

z, w

a


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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


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The constitutive equations for a linearly elastic and

isotropic material are:

In these expressions, N, Niforces, M - moments

2 1 2

2

2 1 2

2

(1 ) [ / 2

/ 2] (4)

(1 ) [ / 2

/ 2] (5)

x x x

i

y y x

y y y

i

x x y

N Eh u w

v w N

N Eh v w

u w N

xy

zNx

Ny

Nxy

Nxy

Ny

Nx

F

[ ] (6)

( ) (7)

i

xy y x x y xy

x xx yy

N Gh u v w w N

M D w w

Thi l l


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The constitutive equations for a linearly elastic and

isotropic material are:

Also u,v,wdisplacements

Substituting the forcedisplacement 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

Thi t l l t


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The dynamic equations for a plate made of linearly

elastic and isotropic material are then simplified byintroducing 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 xy N N N

x yy x

( ) ( ) 0 (11) xyy yxx y x xy

u v u v

(b) C ti D i l S t


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4. Flow between concentric rotating cylinders:

Consider two concentric cylinders with radii a, b;

Let 1, 2angular

velocities of inner and outer

cylinders; let (ur,u,uz)velocity components in a

cylindrical coordinate system;

ppressure at a point;

We now define the equations

of motion for the system.

a

b 12


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Flow between concentric rotating.

Equations of motion for the system:In cylindrical

coordinate system the NS equations are

a

b 12

2

r rr 2 2

r r

2 2

zz

r z

u uDu u1 p 2( u ),

Dt r r r r

Du u u u u1 p 2( uDt r r r r

Du 1 p( u ), (1)

Dt zwhere

Du u u

Dt t r r z

Fl b t t i t ti


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Equations of motion for the system:and

There is also the equation for mass

conservation:

The basic flowis defined by:

r r zuu u u1

0. (2)r r r z

a

b 12

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

Fl b t t i t ti


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The basic flow: Equations (3) have solution of

the form

Let us consider perturbations to

the basic flow:We can then obtain linearizedequationsabout the basic flow.

a

b 12

2

22

1 1 12 2

2 1 1 2

(r) A B / r

1where A , B R

1 1and / , R / R (4)

r zu (u ,V u ,u ), p P p

Fl b t t i t ti


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Linearized equationsabout the basic flow:

r r rr 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 zuu u u1

and 0r r r z

lect2physicalsystemmodels_10feb14

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