bölüm 7 1 lagrange mechanics

1

ME 302 DYNAMICS OF ME 302 DYNAMICS OF MACHINERYMACHINERY

Lagrangian Mechanics

Dr. Sadettin KAPUCU

© 2007 Sadettin Kapucu

2Gaziantep University

Introduction to BalancingIntroduction to Balancing

Lagrange-Euler Formulation

– Lagrange function is defined

K: Total kinetic energy of the system

P: Total potential energy of the system

: Joint variable of i-th joint

: first time derivative of

: Generalized force (torque) at i-th joint

iq

PKL

iii q

L

q

L

dt

d

)(

iq iq

i

3Gaziantep University

ExampleExample

Derive the force-acceleration relationship for the one degree of freedom system shown in figure using both Lagrangian mechanics, as well as Newtonian mechanics. Assume that the wheels have negligible inertia.

22

2

1

2

1xmmvK

km f(t)

x

iii q

L

q

L

dt

d

• Kinetic energy of the cart

• Potential energy of the cart2

2

1kxP

xmx

L

22

2

1

2

1kxxmPKL

kxx

L

tfkxxm

xmx

L

dt

d

tfkxxm

4Gaziantep University

ExampleExample

Derive the force-acceleration relationship for the one degree of freedom system shown in figure using both Lagrangian mechanics, as well as Newtonian mechanics. Assume that the wheels have negligible inertia.

amF

km f(t)

x

• Freebody diagram of the cart kxFs

tfkxxm

m f(t)

AN BN

mgW

xx maF xmkxtf )(

5Gaziantep University

ExampleExample

Derive the force-acceleration relationship for the two degree of freedom system shown in figure using both Lagrangian mechanics, as well as Newtonian mechanics. Assume that the wheels have negligible inertia.

22

21 2

1

2

1pencartpendulumcart VmVmKKK

k

x

In this problem, there are two degrees of freedom, two coordinates x and , and there will be two equation of motion: one for linear motion of the system and one for the rotation of the pendulum.

F1m

2m

xVcart ?penV

cartpencartpen VVV

ixVcart

jlilVcart

pen

sincos

x

l

6Gaziantep University

ExampleExample

pendulumspring PPP

k

x

F1m

2m

cos22

1

2

1 222

221 xllmxmmK

2

2

1kxPspring

ghmPpendulum 2 cos1lh h

cos12 glmPpendulum

cos12

12

2 glmkxP

7Gaziantep University

ExampleExample

22

21 2

1

2

1pencartpendulumcart VmVmKKK

k

x

F1m

2m

xVcart ?penV

cartpencartpen VVV

ixVcart

jlilVcart

pen

sincos

x

l

222 sincos llxVpen

jlilxVpen

sincos

22 xVcart

22

22

1 sincos2

1

2

1 llxmxmK

cos22

1

2

1 222

221 xllmxmmK

8Gaziantep University

ExampleExample

k

x

F1m

2m

cos22

1

2

1 222

221 xllmxmmK

cos12

12

2 glmkxP

PKL

cos12

1cos2

2

1

2

12

2222

221 glmkxxllmxmmL

9Gaziantep University

ExampleExample k

x

F1m

2m

cos12

1cos2

2

1

2

12

2222

221 glmkxxllmxmmL

cos221

lmxmm

x

L

sincos 22221

lmlmxmmx

L

dt

d

iii q

L

q

L

dt

d

)(

kxx

L

kxlmlmxmmF sincos 22221

10Gaziantep University

ExampleExample k

x

F1m

2m

cos12

1cos2

2

1

2

12

2222

221 glmkxxllmxmmL

cos22

2 xlmlmL

sincos 222

2

xlmxlmlmL

dt

d

iii q

L

q

L

dt

d

)(

sinsin 22 xlmglmL

sincos0 222

2 glmxlmlm

11Gaziantep University

ExampleExamplek

x

F1m

2m

iii q

L

q

L

dt

d

)(

sincos0 222

2 glmxlmlm kxlmlmxmmF sincos 2

2221

sin00

sin0

cos

cos

0 22

22

222

221

glm

kxxlmx

lmlm

lmmmF

12Gaziantep University

ExampleExample

Derive the force-acceleration relationship for the two degree of freedom system shown in figure using both Lagrangian mechanics, as well as Newtonian mechanics. Assume that the wheels have negligible inertia.

AN

k

x

F1m

2m

• Freebody diagram of the cart kxFs m F

BN

gm1

xF

yF

y

x

13Gaziantep University

ExampleExample

Derive the force-acceleration relationship for the two degree of freedom system shown in figure using both Lagrangian mechanics, as well as Newtonian mechanics. Assume that the wheels have negligible inertia.

k

x

F1m

2m

F

2m

• Freebody diagram of the pendulum

gm2

xF

yF AN

• Freebody diagram of the cart kxFs m F

BN

gm1

xF

yF

y

x

n

t

14Gaziantep University

ExampleExample

k

x

F1m

2m

?penVcart

pencartpen VVV

ixVcart

jlilVcart

pen

sincos

• Velocity analysis

jlilixVpen

sincos

15Gaziantep University

ExampleExample

k

x

F1m

2m

?pena

t

cartpen

n

cartpencart

cartpencartpen aaaaaa

txnxixacart

cossin

jllill

tlnlacart

pen

cossinsincos 22

2

• Acceleration analysis

16Gaziantep University

ExampleExample

k

x

F1m

2m

F

AN

• Equations for the cart using x-y

kxFs F

BN

gm1

xF

yF

y

x

ixacart

xmkxFFaxisx x 1:

1mO

0: 1 BAy NNFgmaxisy

BABBAA ddwheredNdNmoment 0:

17Gaziantep University

ExampleExample k

x

F1m

2m

2m

gm2

xF

yF

n

t

• Equations for the pendulum using n-t

txlnxlapen

cossin2

222 sinsincoscos: lxmFFgmaxisn xy

lxmFFgmaxist xy coscossinsin: 22

0sincos: lFlFmoment yx

18Gaziantep University

ExampleExample k

x

F1m

2m

222 sinsincoscos: lxmFFgmaxisn xy

lxmFFgmaxist xy coscossinsin: 22

0sincos: lFlFmoment yx

xmkxFFaxisx x 1:

0: 1 BAy NNFgmaxisy

BABBAA ddwheredNdNmoment 0:

sincos yx FF

0sincos 222 gmlmxm lxmgm cossin 22

19Gaziantep University

ExampleExample k

x

F1m

2m

222 sinsincoscos: lxmFFgmaxisn xy

lxmFFgmaxist xy coscossinsin: 22

0sincos: lFlFmoment yx

xmkxFFaxisx x 1:

0: 1 BAy NNFgmaxisy

BABBAA ddwheredNdNmoment 0:

sin

cossincos x

yyx

FFFF

22

2

2 sinsinsin

coscos

lxmF

Fgm x

x

20Gaziantep University

ExampleExample k

x

F1m

2m

222 sinsincoscos: lxmFFgmaxisn xy

lxmFFgmaxist xy coscossinsin: 22

0sincos: lFlFmoment yx

xmkxFFaxisx x 1:

0: 1 BAy NNFgmaxisy

BABBAA ddwheredNdNmoment 0:

22

2

2 sinsinsin

coscos

lxmF

Fgm x

x

sinsinsinsin

coscossin 2

2

2

2 lxmF

Fgm x

x

21Gaziantep University

ExampleExample k

x

F1m

2m

222 sinsincoscos: lxmFFgmaxisn xy

lxmFFgmaxist xy coscossinsin: 22

0sincos: lFlFmoment yx

xmkxFFaxisx x 1:

0: 1 BAy NNFgmaxisy

BABBAA ddwheredNdNmoment 0:

sinsinsinsin

coscossin 2

2

2

2 lxmF

Fgm x

x

sinsinsincossincos 22

22

222

lmxmFFgm xx

sinsinsincossincos 22

22

222

lmxmFgm x

22Gaziantep University

ExampleExample k

x

F1m

2m

222 sinsincoscos: lxmFFgmaxisn xy

lxmFFgmaxist xy coscossinsin: 22

0sincos: lFlFmoment yx

xmkxFFaxisx x 1:

0: 1 BAy NNFgmaxisy

BABBAA ddwheredNdNmoment 0:

sinsinsincossincos 22

22

222

lmxmFgm x

sinsinsincos 22

222

lmxmFgm x

FkxxmFx 1

23Gaziantep University

ExampleExample k

x

F1m

2m

222 sinsincoscos: lxmFFgmaxisn xy

lxmFFgmaxist xy coscossinsin: 22

0sincos: lFlFmoment yx

xmkxFFaxisx x 1:

0: 1 BAy NNFgmaxisy

BABBAA ddwheredNdNmoment 0:

sinsinsincossincos 22

22

222

lmxmFgm x

sinsinsincos 22

2212

lmxmFkxxmgm

Fgmlmkxxmxm sincossinsin 22

22

21

24Gaziantep University

ExampleExample

ExampleExample k

x

F1m

2mFgmlmkxxmxm sincossinsin 2

22

221

0sincos 222 gmlmxm

sincos0 222

2 glmxlmlm

kxlmlmxmmF sincos 22221

25Gaziantep University

ExampleExample

Derive the Equation Of Motion (EOM) for the two degree of freedom system shown in figure using Lagrangian mechanics.

222

21121 2

1

2

1VmVmKKK

122

12

1122

12

121

21

21 sincos llyxV

1m1

1l

2m

2l

x

y

2

Position of the first mass

111 sinlx

111 cosly Velocity of the first mass

1111 cos lx

1111 sin ly }2

12

12

1 lV

26Gaziantep University

ExampleExample

222

21121 2

1

2

1VmVmKKK

1m1

121121212

1212122

21

22

21

21

22

22

22

22 SSCCllll

yxV

1l

2m

2l

x

y

2

Position of the second mass

12211212112 sinsin SlSlllx 12211212112 coscos ClCllly

Velocity of the second mass

122121112 ClClx

122121112 SlSly

21

21

21 lV

212

12212122

21

22

21

21

22 22 CllllV

27Gaziantep University

ExampleExample

222

21121 2

1

2

1VmVmKKK

1m1

1l

2m

2l

x

y

2

212

12212122

21

22

21

212

21

211

222

12

1

Cllllm

lmK

21 PPP

1111 CglmP

12221122 CglmCglmP

12221121 CglmCglmmP

28Gaziantep University

ExampleExample

1m1

1l

2m

2l

x

y

2

1222112121

212212

2122

21

222

21

2121 2

2

1

2

1

CglmCglmmCllm

lmlmmL

PKL

2221212212212221

2121

1

2

CllmCllmlmlmmL

22221222212

2122121221221

2221

2121

1

22

SllmCllm

SllmCllmlmlmmL

dt

d

122211211

SglmSglmmL

29Gaziantep University

ExampleExample

2221212212212221

2121

1

2

CllmCllmlmlmmL

22221222212

212212

12212212221

2121

1

2

2

SllmCllmSllm

CllmlmlmmL

dt

d

122211211

SglmSglmmL

iii q

L

q

L

dt

d

)(

12222121

222212

2122122221222212212

222

2121 220

SglmSglmmSllm

SllmCllmlmCllmlmlmm

1m1

1l

2m

2l

x

y

2

30Gaziantep University

ExampleExample

1222112121

212212

2122

21

222

21

2121 2

2

1

2

1

CglmCglmmCllm

lmlmmL

PKL

1221221222

2

CllmlmL

2122121221221222

2

SllmCllmlmL

dt

d

1222212

122122

SglmSllmL

1m1

1l

2m

2l

x

y

2

31Gaziantep University

ExampleExample

iii q

L

q

L

dt

d

)(

12222

12212222212212

2220 SglmSllmlmCllmlm

1221221222

2

CllmlmL

2122121221221222

1

SllmCllmlmL

dt

d

1222212

122121

SglmSllmL

1m1

1l

2m

2l

x

y

2

32Gaziantep University

ExampleExample

12222

12212222212212

2220 SglmSllmlmCllmlm

12222121

222212212212

2221222212212

222

2121

2

20

SglmSglmmSllmSllm

CllmlmCllmlmlmm

1222

12221121

12

2122122212

22

21

2212

2212

2

12222212

222

22122222212

222

2121

000

0

2

0

0

Sglm

SglmSglmm

SllmSllm

Sllm

Sllm

lmCllmlm

CllmlmCllmlmlmm

1m1

1l

2m

2l

x

y

2

33Gaziantep University

ExampleExample

34Gaziantep University

ExampleExample

35Gaziantep University

ExampleExample