mech370: modelling, simulation and analysis of physical

MECH370: Modelling, Simulation and Analysis of

Physical Systems

Chapter 5Rotational Mechanical Systems

VariablesElement LawsInterconnection LawsObtaining the system model

Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Youmin Zhang (CU), Simulation and Analysis of Physical Systems, Youmin Zhang (CU)

Lecture Notes on Lecture Notes on MECH 370MECH 370 –– ModellingModelling, Simulation and Analysis of Physical Systems, Simulation and Analysis of Physical SystemsChapter 5Chapter 5 2

Course OutlineCourse OutlineModellingModelling (Ch. 2,3,4,5,6,9,10,11,12) Simulation (4) Analysis (7(Ch. 2,3,4,5,6,9,10,11,12) Simulation (4) Analysis (7,8),8)

1. Definition and classification of dynamic systems (chapter 1)

2. Translational mechanical systems (chapter 2) 3. Standard forms for system models (chapter 3)4. Block diagrams and computer simulation with

Matlab/Simulink (chapter 4)5. Rotational mechanical systems (chapter 5)6. Electrical systems (chapter 6)7. Analysis and solution techniques for linear systems

(chapters 7 and 8)8. Developing a linear model (chapter 9)9. Electromechanical systems (chapter 10)10.Thermal and fluid systems (chapters 11, 12)


torque/velocity torque/position

Inertia T = J dω/dt T = J dθ2/dt2

Viscous friction

T = B ω T = B dθ/dt

Stiffness T = K ∫ ω dt T = K θ

T, T, θθ

JJ

BB

T, T, θθ

T, T, θθ

KK

Overview of Element Models in Physical SystemsOverview of Element Models in Physical SystemsRotational Models (Review)Rotational Models (Review)


Lever

Gears

TT1 1 , , θθ11

TT2 2 , , θθ22

NN11

NN22

ff11 , x, x11

ff22 , x, x22

LL11

LL22

f1 L2=

f2 L1

x1 L1=

x2 L2

T1 N1=

T2 N2

θ1 N2=

θ2 N1

Overview of Element Models in Physical SystemsOverview of Element Models in Physical SystemsTransformation Models (Review)Transformation Models (Review)


Rotational Mechanical SystemsExample:

The above system is the rotational analog of the mass-spring-damper system.

Rotational variables:

Notation Variable Units Translational AnalogAngular displacement radians (rad) Displacement (x, m)Angular velocity rad/s Velocity (v, m/s)Angular acceleration rad/s2 Acceleration (a, m/s2)Torque N-m Force (F, N)Moment of inertia Kg-m2 Mass (M, kg)

θθω &=

θα &&=τ

J

θ

KK

)(taτ

ω, α

JB


Rotational Mechanical Systems (cont’d)

thatsodirection same in the ,,, Make .conventionsign a need weso vectorsof componentsscalar are ,,,

αωθταωθ

−−

2

2

dtθd

dtdω,α

dtdθω ===

Remarks:

Assumptions:• Rigid body motion• Pure mechanical system• Fixed axis rotation in inertial reference frame

Valid Not Valid

+ω


Element Laws

ω

Moment of Inertia (J)

∫= dMrJ 2

(for a point mass J=Mr2)

∑τ=ω

=ω idtdJJ

dtd )(

systems) tionalin transla ( :energy Potential

systems) tionalin transla 21(

21 :energy Kinetic 22

MghEMghE

MνEJωE

pcp

kk

==

==

)m(kgrotation of axis theabove inertia ofmoment -* 2⋅J

+

hc-height of center of mass

hc

dM

rJ

Newton’s 2nd Law:

dM: differential mass element

∫+=

=⋅=1

01 :k)Energy(wor

)( :Powert

to

p(t)dt)w(t)w(t

fvpωτp



2

121 ΜLJ =

2

21 MRJ =

Special cases:

a) Slender barAxis of rotation passes through the center of mass:

b) Disk

How to Calculate Moment of Inertia (J)?



2220

20

31)

41

121()

2(

121

MLMLLMJJ

MLJ

=+=+=

=

c) Slender bar

d) Slender bar

Parallel-axis theorem: for cases that axis does not pass through the center of mass

rotationofaxis parallel from mass ofcenter of distance -

mass ofcenter about inertia ofmoment - 0

20

aJ

MaJJ +=

c)in 2

e.g. La =

AB

C

Axis of rotation does not pass through the center of mass:

For case c):



22

21

2121

21

21

2

21

1

31

31)

2(

121)

2(

:d) caseFor

0

dMJ

dMdMdMdMJJ

Mdd

dM

Mdd

dM

BCBC

ABABABABABAB

BC

AB

=

=+=+=

+=

+=

)(31][

31 3

23

121

22

21

221

21

1 dddd

MMddd

dMddd

dJJJ BCAB ++

=+

++

=+=

)constant (for :Law JτωJ =&

2

2100 :energy Kinetic Jω)(wωdt

dtdωJ)(wωdtτw κ

t

tκ

t

tκoo

+=+=⋅= ∫∫Mghwp = :energy Potential

AB

C



Example: Develop a set of state-variable equations

ωθ

MgθcosLf(t)θsinLωJ

τ

=

⋅−⋅=

&

4444 34444 21&

2Note: equations are nonlinear and cannot be put into matrix form

FBD:

L

θ

Mg

)t(fJ

θMgcosL2

)t(fθsinL ⋅

Newton’s 2nd Laws:

x

y


Rotational Friction

Torque due to friction is proportional to angular velocity (assumption), and in directions that tend to reduce the relative angular velocity

Friction in response to rotational movement:

(a) (b) (c)Clutch Torsional system

ω

τ B

Rotational viscous friction12 where,ω

ωωΔωBΔτ

−==

relative angular velocity


Rotational FrictionIdealized dashpot, no inertia

)ωΒ(ω)ωΒ(ωτ 2112 −=−−=)ωΒ(ωτ 12 −=

Example:

0)( 11 =−−+∴ ωωΒΒωωJ &

B

τ2ω 1ω

τ

B

τ2ω 1ω τ

1B

τ

1ω

ω JB

+

ω)(ωB −11

Bω ω

ω&J

J1ω

Note: denotes the torque due to the friction, not the applied external torque as stated in the book.

τ

Bω

ωJ &ω)(ωB −11

J

FBD


Rotational Spring/Stiffness

mequilibriu fromnt displacemeangular 21 =,θθ

For a linear torsional spring (or shaft), the torque duo to the stiffness is:

∫ ∫ θ===

=−=Δ=

2

21

21))((

(Nm/rad)constant stiffness where,)(

KΔdtθΔθΔKpdtw

KθθKθKτ

k&&

Analogous to “regular” spring:

Idealized torsional spring (or shaft), no mass


LeverAssumption for an ideal lever:• no mass• no friction• rigid

? smallfor and c) and b) and a)

between relation theisWhat

12

12

12

θffvxx

⎪⎩

⎪⎨

⎧ν

1

2

1

2

11

2222

1

1111

)(

a)

dd

xx

xddθdθsindx

dxθθdθsindx

=∴

=≅=

=⇒≅=

! small is rotation of Angle θ

ff11 , x, x11

ff22 , x, x22

LL11

LL22

x1 L1=

x2 L2


Lever (cont’d)

1

2

1

2

11

22

1

1

22

or

)()( b)

dd

νν

νddν

dtdx

dd

dtdx

=

=⇒=

c) FBD

{ {

2

1

1

21

2

12

11

121

2

or )(

0

dd

fff

ddf

fθcosdfθcosd

==

=⋅−⋅≈≈

Newton’s 2nd Law:

Note: The sum of moments around pivot is zero! Why? – Mass-less!

11 fθcosd ⋅

22 fθcosd ⋅


Gears

• Purpose of gears is to decrease angular velocity and increase torque. Gears also change direction of rotational motion.

• Ideal gear: no inertia, rigid bodies, no friction, perfect meshing

• Most gears have friction and backlash, but these can be represented by additional elements.


Gears (cont’d)Ideal gear:

1

2

2

12211 r

rrr =⇒=∴θθθθ

Differentiating gives

ratioGear 1

2

2

1

=

==

N

Nrr

ωω

Or since the velocities of gear edges are equal, i.e.

1

2

1

2

2

1

2211

nnN

rr

ωω

ωrωr

===∴

=

Perfect meshing implies arc lengths PA and PB are equal

TT1 1 , , θθ11

TT2 2 , , θθ22

NN11

NN22

θ1 N2=

θ2 N1


Gears (cont’d)FBD:

Newton’s 2nd Law:

00

22

11

=+=−

ττ

c

c

frfr Since the gears have no inertia, sum of the torques

on each of the gears must be zero

equal.not are they and directions oppositein bemust torques1

2

1

2

⇒

−=−= Nrr

ττ

rquecontact toforcecontact

2 ==

c

c

frf

1τ

cfr1

cf1r

2τ

cfr2

cf

2r

Eliminating fc yields:

External torque

External torque


Gears (cont’d)Example:

FBD:

Nrr==

1

2

1

2

ττ

(t)τ(t)NτωJ

(t)τ(t)τrr(t)ττωJ

3122

311

23222

−=

−=−=∴

&

&

rigid

(t)τ32J

2τ2τ2θ

1θ(t)τ11r

2r

2τ 2J


Gears (cont’d)

gears. for theFBD a usingnot are wesince on right sign get the toimportant isit that Note 2τ

get to substitute Now1

2

2

1 Nrr

ωω

==

(t)τN

(t)τωNJ

(t)τ(t)NτNωJ

31122

311

2

1)(

)(

−=

−=

&

&

.N

tτN

1by )( torqueload and 1by reduced is inertia ofEffect 32∴

.

(t)τ(t)NτωJ

(t)τ(t)τrr(t)ττωJ

3122

311

23222

−=

−=−=∴

&

&


Interconnection LawsD’Alembert’s Law

θ,ω,αωJ

ωJτn

iiext

toopposite torque,Inertial where,

0)(

=

=−∑&

&

Equilibrium condition: 0=∑n

iiτ

Law of Reaction Torques:

A simple consequence of Newton’s 3rd law.Elements must be connected along the same axis (not applicable to gear meshing)

Torque exerted by one element on another produces an equal and opposite reaction torque.

K Kτ

J

θ Kτ


Law of Angular Displacement

K

Jθ

Jθ θ

Not applicable to gear meshing since the axes are not the same.

Two elements connected along the same axis have the same angular displacement.


Obtaining the System Model

diagramsblock and/or equations,output -input equations variable-state

:forms standard toequations obtained theRepresent equations. aldifferentiget tolaw 2 sNewton'Apply

laws.element using of in terms torquesExpress motion.unknown ith junction wor inertiaeach for FBD a Draw

nd

⋅⋅⋅

••

••

θ,ω

Procedure:


Obtaining the System Model (cont’d)Example 5.1: Find the input-output and state-variable equations for the

Example given in pp. 5.

( ):0 Law sAlembert'D'by or Law 2nd sNewton' =iiτΣ

(t)τKθBωωJBωKθ(t)τωJ

a

a

=++−−=

&

&

or

Very important step towards to obtaining correct system model!

Are the above equations input-output equation format?

(t)τKθBωωJ(t)τKθBωωJ

a

a

=++=+−−−

&

&

or 0

(t)τKθBωωJ(t)τKθBωωJ

a

a

=++=−++

&

&

or 0

+ +


Obtaining the System Model (cont’d): iableoutput varfor equation output -Input θ

⎥⎦

⎤⎢⎣

⎡θω

=

τ⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+⎥

⎦

⎤⎢⎣

⎡⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ −−=⎥⎦

⎤⎢⎣

⎡

=

τ+−−=

1] [0

0

1

01

1

y

)t(Jθω

JK

JB

θω

ωθ

))t(KθBω(J

ω

a

a

&&

&

&

⎟⎠⎞

⎜⎝⎛ ====++ θω,θ

dtdθω(t)τKθθBθJ a

&&&&&&&

[ ] ⎥⎦

⎤⎢⎣

⎡=

τ⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=⎥

⎦

⎤⎢⎣

⎡

τ+−−=

=

ωθ

y

)t(Jω

θ

JB

JK

ωθ

)t(J

ωJBθ

JKω

ωθ

a

a

01

1010

1

&

&

&

&

State-variable equations:

or

or

(t)τKθBωωJBωKθ(t)τωJ

a

a

=++−−=

&

&

or


Obtaining the System Model (cont’d)Block diagram:

( )(t)τKθθBJ

θ a+−−= &&& 1Based on input-output equation:

Simulink diagram:

J1

dt∫ dt∫input−(t)τa θ&& θ& output−θ

B

K

+-

-θJ &&

J1

s1

+dotdot

Theta

B

K

s1

step

−− dotTheta Theta

scope


Obtaining the System Model (cont’d)

J1

dt∫ dt∫(t)τa ω& ω θ

JB

JK

+-

-

Block diagram:Based on state-variable equation:

Simulink diagram:

(t)τJ

ωJBθ

JKω

ωθ

a1

+−−=

=

&

&

J1

s1

JB

JK

+-

-+

step

−−

scopes1



)t(fKxxBxM a=++ &&&

• Can you see the difference between Block Diagram and SimulinkDiagram?

• See Example 5.2 for two shafts

• What are differences between these two Simulink and Block Diagrams, respectively?

• Compare this example with M-S-D system.

Questions:

KM fa(t)

B x, v

(t)τKθθBθJ a=++ &&&


Obtaining the System Model (cont’d)Example 5.2 :

FBDs:



22

11

1222222

122111111

0)(0)(

ωθ

ωθ

(t)τθθKωΒωJθθKθKωΒωJ

a

=

=

=−−++=−−++

&

&

&

&

Rearrange to obtain state-variable equation:

22

22212222

11

1122211111

)]([

][

ωθ

/JtτθKθKωΒωωθ

/JθKθKθKωΒω

a

=

+−+−==

−+−−=

&

&

&

&

22

22222122

11

122121111

)]([

)])([

ωθ

/JtτθKωΒθKωωθ

/JθKθKKωΒω

a

=

+−−==

++−−=

&

&

&

&or

+


Obtaining the System Model (cont’d):formatvector -matrix have then we, defining If 1θy =

{

[ ] 0D 0010

0

100

0100

00001

0

2

2

1

1

C

B

12

2

1

1

A

2

2

2

2

2

2

1

2

1

21

1

1

2

2

1

1

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

+−−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

θωθω

y

(t)τJ

θωθω

JK

JB

JK

JK

JKK

JB

θωθω

a

4434421

444444 344444 21

&&

&&

)?( and or )( and relatingequation output -input about the How] [q as vector state choosingby equation variable-state about the How

system?theoftion representadiagramblock about How

21

T2211

tθtθ?ωθωθ

aa ττ=

Easy for Matlab script implementation in matrix-vector form.

Easy for Simulinkimplementation

22

22222122

11

122121111

)]([

)])([

ωθ

/JtτθKωΒθKωωθ

/JθKθKKωΒω

a

=

+−−==

++−−=

&

&

&

&



(1b) 0)((1a) 0)(

1222222

122111111

=−−++=−−++

(t)τθθKωΒωJθθKθKωΒωJ

a&

&

(2) Eq.by (1) Eq. following rewrite need we),( and relatingequation output -inputan obtain To 2 tτθ a

)()()()()(

)()(

:gives (2a) to ngsubstitutiThen

][1

:for solvecan we(2b), From

21112212221221

221221221(3)21221

)4(221

1

2222222

1

1

tτKKtτBtτJθKKθKBKBKB

θBBKJKJKJθBJBJθJJ

θ

(t)τθKθΒθJK

θ

θ

aaa

a

+++=+++

++++++

−++=

&&&&

&&

&&&

(2b)

(2a) 0

22222212

221211111

(t)τθKθΒθJθK

θK)θK(KθΒθJ

a=+++−

=−+++&&&

&&&

Or a simpler way of using p-operator as we shown in slides for Chapter 3 (pp. 29-31).


Obtaining the System Model (cont’d)Example 5.4: Develop the state-variable equations for the system and

show how to replace the two sections by an equivalent stiffness element.

FBDs

+


Obtaining the System Model (cont’d)Summing the torques on each of the FBDs gives:

.τr find wish to weifonly needed is (1) Eq.(3)0)(

(2)0)((1)0

2

21

1

LL&

LL

LL

=−−++=−−

=−

(t)τθθKBωωJθθKθK

θKτ

aA

AA

Ar

(t)τθKK

KθKBωωJ

θKK

Kθ

θθ

a

A

A

=+

−++

+=

)(

:yields (3) into (4) ngSubstituti

(4)

i.e. other,each toalproportion are and seecan we(2), From

21

22

21

2

&

LL

(t)τθKBωωJ

(t)τθKK

KKBωωJ

aeq

a

=++

=+

++

&

&

or 21

21

21

21 where,KK

KKKeq +=

Equivalent stiffness constant for the series combination of the two shafts.

+



[ ]

⎥⎦

⎤⎢⎣

⎡⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

+=⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡

−−=⎥⎦

⎤⎢⎣

⎡

+−−=

=

ωθ

KKK

θθ

(t)τJω

θ

JB

JK

ωθ

θθ

(t)τBωθKJ

ω

ωθ

ωθ

A

aeq

A

aeq

001

1010

:model variable-state offormat matrix have then weiables,output var theas and Selecting

1

have then we, variablesstate theas and Selecting

21

2

&

&

&

&


Obtaining the System Model (cont’d)Example 5.6: Create a state-variable model for:

State variables: There is one mass and two springs. However, the two springs are dependent, so they act as one spring. Therefore, we have two state variables: x1,v1.


Obtaining the System Model (cont’d)Balancing torques about lever:

(1))())(( 21123421 LLxxKdxtxKd +=−

We also have:

22

13

22

13 or xddx

θdxθdx

⎟⎟⎠

⎞⎜⎜⎝

⎛=

==

Substitution into (1) gives:

( ) (2))()( 21121421 LLθdxKdθ-dtxKd +=

Balancing forces on mass:

(3)0or

0

211111

211111

LL&

&

=+++

=+++

θdKxKBννM

xKxKBννM



(t)xKKddαKx

KdKαdK

ddKKf

xxKx(t)xKffdd

KK

α

(t)xKddαν

x

MB

MαK

νx

(t)xKddαBν)xK(α

Mν

νx

θθ

r

rr

4211

221

21

121

1

221

211342

2

1

2

2

1

41

1

21

11

1

1

411

21111

11

)]([))((or

)(][ :for equation Output

1

1 Where,

010

])([1

:have we(3), into plug and for (2) solve So, . eliminatemust we(3) and (2) Eqs.Between

−++⎥⎦

⎤⎢⎣

⎡−+=

++−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎥⎦

⎤⎢⎣

⎡

⎥⎥⎦

⎤

⎢⎢⎣

⎡−−=⎥

⎦

⎤⎢⎣

⎡⇒

⎪⎩

⎪⎨⎧

++−=

=

&

&

&

&


Obtaining the System Model (cont’d)Example 5.9: Derive the state-variable equations for the gear-driven

disk.

00

1

2

=−=−++

ca

c

fr(t)τfrKθBωωJ &

By D’Alembert’s Law:

Eliminating fc yields:[ ]⎪⎩

⎪⎨⎧

+−−=

=

=

(t)NτBωKθJ

ωωθ

ωθrrN

a1

:have wevariables, state aschosen are and If

where1

2

&

&

(t)Nτ(t)τrrKθBωωJ aa ==++

1

2&

fc=contact force

+


Example 5.11:

Obtaining the System Model (cont’d)Derive the state-variable equations for the system shown below.

FBDs:

c

a

fν,x,(t)τ

:Output :Input

)t(τa

ωB1

)θK(θ A−ωJ &

cRf

)θK(θ A−cf

vB2

vM &

x

y +


Obtaining the System Model (cont’d)Summing the torques in FBDs yields:

(3)0(2)0)(

(1)0)(

2 LL&

LL

LL&

=−+=−−

=−−++

cfνBνMθθKRf

(t)τθθKBωωJ

Ac

aA

In addition,(4)LLxRθA =

State variables:

AR θθθω,ν −= and , :chocie Possible

input. theand offunction algebraican as expressed becannot since variablesstatefour need we thusoutputs, specified theof one is andinterest ofusually is However,

Rω,ν,θx

x

J M KEnergy-storing elements:

)t(τa

ωB1

)θK(θ A−ωJ &

cfvB2

vM &

x

y +

)θK(θ A−



⎟⎠⎞

⎜⎝⎛ −−=

=

⎥⎦⎤

⎢⎣⎡ ++−−=

=

=−++

=−−++

νBxRKθ

RK

Mν

νx

(t)τxRKωBKθ

Jω

ωθ

θRKx

RKνBνM

f

(t)τxRKKθωBωJ

θνx,ω,θ,

a

c

a

A

22

1

22

1

1

1

are equations variable-state The

0

in results (3)in eliminate to(2) and (4) using

0

gives (1)in eliminate to(4) using and : variablesstate of Choice

&

&

&

&

&

&



. variablesstate only three with formulated be alsocan problem this:Note

0010000100

00

10

01000

00010

:model variable-state offormat vector -Matrix

)(

have then we(2), into (4) ngsubstitutiby found becan .variablesstateofpartsare and outputs The

2

22

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

+

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−−

−−=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−=

νxωθ

RK

RKf

vx

(t)τJ

νxωθ

MB

MRK

MRK

JRK

JB

JK

νxωθ

Rxθ

RKf

fνx

c

a

c

c

&

&

&

&


Reading and ExerciseReading and Exercise• Reading

Chapter 5

• Assignment #2Ex. 4.2, 4.3, 5.10, 5.18, 5.24 (to be

handed in)

Due: Wed. 18/7/07 at lecture.

Ex. 4.5, 4.8, 5.7, 5.13, 5.28 (for your practice, no need to hand in)

mech370: modelling, simulation and analysis of physical

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