Bölüm 7 1 lagrange mechanics
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1
ME 302 DYNAMICS OF ME 302 DYNAMICS OF MACHINERYMACHINERY
Lagrangian Mechanics
Dr. Sadettin KAPUCU
© 2007 Sadettin Kapucu
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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
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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
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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 )(
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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
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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
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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
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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
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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
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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
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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
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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
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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
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14Gaziantep University
ExampleExample
k
x
F1m
2m
?penVcart
pencartpen VVV
ixVcart
jlilVcart
pen
sincos
• Velocity analysis
jlilixVpen
sincos
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15Gaziantep University
ExampleExample
k
x
F1m
2m
?pena
t
cartpen
n
cartpencart
cartpencartpen aaaaaa
txnxixacart
cossin
jllill
tlnlacart
pen
cossinsincos 22
2
• Acceleration analysis
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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:
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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
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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
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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
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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
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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
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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
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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
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24Gaziantep University
ExampleExample
ExampleExample k
x
F1m
2mFgmlmkxxmxm sincossinsin 2
22
221
0sincos 222 gmlmxm
sincos0 222
2 glmxlmlm
kxlmlmxmmF sincos 22221
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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
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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
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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
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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
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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
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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
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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
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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
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33Gaziantep University
ExampleExample
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34Gaziantep University
ExampleExample
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35Gaziantep University
ExampleExample