16.61 Aerospace Dynamics Spring 2003
Lecture #9
Virtual Work
And the
Derivation of Lagrange?s Equations
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 1
16.61 Aerospace Dynamics Spring 2003
Derivation of Lagrangian Equations
Basic Concept: Virtual Work
Consider system of N particles located at
( )
123 3
,,,
N
xxx x…
)
with
3 forces per particle
(
123 3
,,,
N
F FF F… , each in the positive
direction.
F
3
F
2
F
1
(x
1
, x
2
, x
3
)
F
5
F
6
F
4
(x
4
, x
5
, x
6
)
x
i
x
j
x
k
F
N
F
N-1
F
N-2
(x
N-2
, x
N-1
, x
N
)
Assume system given small, arbitrary displacements in all
directions.
Called virtual displacements
- No passage of time
- Applied forces remain constant
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 2
16.61 Aerospace Dynamics Spring 2003
The work done by the forces is termed Virtual Work.
3
1
N
jj
j
WFxδ δ
=
=
∑
Note use of δx and not dx.
Note:
? There is no passage of time
? The forces remain constant.
In vector form:
3
1
ii
i
Wδ δ
=
= ?
∑
Fr
Virtual displacements MUST satisfy all constraint relationships,
! Constraint forces do no work.
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 3
16.61 Aerospace Dynamics Spring 2003
Example: Two masses connected by a rod
l
1
R
2
R
?
r
e
Constraint forces:
12
?
r
Re
2
= ?=?RR
Now assume virtual displacements
1
δr , and
2
δr - but the
displacement components along the rigid rod must be equal, so
there is a constraint equation of the form
21
rere
rr
δδ ?=?
Virtual Work:
()
11 2 2
212
22 1
??
?
0
RrRr
rr
r
rr
r
W
Re Re
RRe
=?+?
=? ? + ?
=? ?
=
2
δ δδ
δ δ
δ
So the virtual work done of the constraint forces is zero
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 4
16.61 Aerospace Dynamics Spring 2003
This analysis extends to rigid body case
? Rigid body is a collection of masses
? Masses held at a fixed distance.
! Virtual work for the internal constraints of a rigid body
displacement is zero.
Example: Body sliding on rigid surface without friction
Since the surface is rigid and fixed, 0, 0
s
rWδ δ= →=
For the body,
1
W
1
δ δ=?R
W
r, but the direction of the virtual
displacement that satisfies the constraints is perpendicular to the
constraint force. Thus 0δ = .
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 5
16.61 Aerospace Dynamics Spring 2003
Principle of Virtual Work
i
m = mass of particle i
i
R = Constraint forces acting on the particle
i
F = External forces acting on the particle
! For static equilibrium (if all particles of the system are
motionless in the inertial frame and if the vector sum of all
forces acting on each particle is zero)
0
ii
+ =RF
The virtual work for a system in static equilibrium is
()
1
0
N
ii i
i
Wδδ
=
=+?
∑
RF r=
But virtual displacements must be perpendicular to constraint
forces, so
0
ii
δ? =Rr ,
which implies that we have
1
0
N
ii
i
δ
=
? =
∑
Fr
Principle of virtual work:
The necessary and sufficient conditions for the static
equilibrium of an initially motionless scleronomic system
which is subject to workless bilateral constraints is that zero
virtual work be done by the applied forces in moving through
an arbitrary virtual displacement satisfying the constraints.
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 6
16.61 Aerospace Dynamics Spring 2003
Example: System shown consists of 2 masses connected by a
massless bar. Determine the coefficient of friction on the floor
necessary for static equilibrium. (Wall is frictionless.)
Virtual Work:
12
Wmgx Nx
2
δ δμδ= ?
Constraints and force balance:
122
,2xxNmgδ δ= =
Substitution:
( )
12mg x 0μ δ? =
Result:
1
2
μ =
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 7
16.61 Aerospace Dynamics Spring 2003
So far we have approached this as a statics problem, but this is a
dynamics course!!
Recall d?Alembert who made dynamics a special case of statics:
()
()
i
1
1
0
0
RF r r
Frr
N
ii i
i
N
iii i
i
Wm
m
=
=
=+??
? ? =
∑
∑
""
""
δδ
δ
=
?
! So we can apply all of the previous results to the dynamics
problem as well.
Comments:
" Virtual work and virtual displacements play an important
role in analytical dynamics, but fade from the picture in the
application of the methods.
" However, this is why we can ignore the calculation of the
constraint forces.
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 8
16.61 Aerospace Dynamics Spring 2003
Generalized Forces
Since we have defined generalized coordinates, we need
generalized forces to work in the same ?space.?
Consider the 2 particle problem:
(x
1
, x
2
)
(x
3
, x
4
)
x
i
x
j
l θ
()()
1234
22
2
13 24
Coordinates: , , ,
Constraint:
DOF: 4 1 3
xxxx
xx xx l? +? =
?=
? Select n=3 generalized coordinates:
q
xx
q
xx
q
xx
xx
1
13
2
24
3
1
42
31
22
=
+
=
+
=
?
?
?
() ()
tan
()
()
? Can also write the inverse mapping:
( )
123
,,, ,
ii n
xfqqq qt= …
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 9
16.61 Aerospace Dynamics Spring 2003
Virtual Work:
3
1
N
j j
j
WFxδ δ
=
=
∑
Constraint relations:
1
n
j
ji
i i
x
x q
q
=
?
??
δ =δ
??
?
??
∑
Substitution:
3
11
Nn
j
ji
ji i
x
WF
q
==
?
??
qδ =δ
??
?
??
∑∑
Define Generalized Force:
3
1
N
j
ij
j
i
x
q
=
QF
?
??
=
??
?
??
∑
! Work done for unit displacement of q
i
by forces acting on the
system when all other generalized coordinates remain constant.
?=
=
∑
δδWQ
ii
i
n
1
q
n
? If is an angle, is a torque
i
q
i
Q
? If is a length, is a force
i
q
i
Q
? If the ?s are independent, then for static equilibrium must
have:
i
q
0, 1, 2,
i
Qi= = …
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 10
16.61 Aerospace Dynamics Spring 2003
Derivation of Lagrange?s Equation
? Two approaches
(A) Start with energy expressions
Formulation
Lagrange?s Equations (Greenwood, 6-6)
Interpretation
Newton?s Laws
(B) Start with Newton?s Laws
Formulation
Lagrange?s Equations (Wells, Chapters 3&4)
Interpretation
Energy Expressions
(A) Replicated the application of Lagrange?s equations in
solving problems
(B) Provides more insight and feel for the physics
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 11
16.61 Aerospace Dynamics Spring 2003
Our process
1. Start with Newton
2. Apply virtual work
3. Introduce generalized coordinates
4. Eliminate constraints
5. Using definition of derivatives, eliminate explicit use of
acceleration
? Start with a single particle with a single constraint, e.g.
o Marble rolling on a frictionless sphere,
o Conical pendulum
θ
φ
r = L
x
y
z
m
F
Free body
Diagram
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 12
16.61 Aerospace Dynamics Spring 2003
1. Newton: m=Fa
? For the particle: ,,
xyz
F mx F my F mz==="" "" "" where axes x, y, z
describe an inertial frame
? Note that ,,
x yz
F FF are the vector sum of all forces acting on
the particle (applied and constraint forces)
2. Apply Virtual Work:
? Consider δs, which is an arbitrary displacement for the
system, then the virtual work associated with this
displacement is:
xyz
WFxFy Fzδ δδδ= ++
? Note that δs may violate the applied constraints, because F
contains constraint forces
? Combine Newton and Virtual Work
x
y
z
Fx mxx
Fymy
Fz mzz
y
δ δ
δ δ
δ δ
=
=
=
""
""
""
? Add the equations
( )
xyz
mxx y y z z F x F y F zδ δδ δ δ++ = + +
"" "" ""
δ
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 13
16.61 Aerospace Dynamics Spring 2003
? Called D?Alembert?s equation:
()
xyz
mxx y y z z F x F y F zδ δδ δ δ++ = + +"" "" "" δ
? Observations:
o Scalar relationship
o LHS ≈ kinetic energy
o RHS ≈ virtual work term
3. Introduce generalized coordinates
o Assumed motion on a sphere ! 1 stationary constraint
o DOF = 3 - 1 = 2 generalized coordinates
( ) ( ) ( )
112 212 312
,, ,, ,x f qq y f qq z f qq===
? Define virtual displacements in terms of generalized
coordinates:
1
n
j
ji
i i
x
x q
q
=
?
??
δ= δ
??
?
??
∑
!
12
12
12
12
12
11
xx
xq
qq
yy
yq
qq
zz
zq
qq
q
q
q
? ?
=+
??
??
=+
??
??
=+
??
δ δδ
δ δδ
δ δδ
? Note: these virtual displacements conform to the constraints,
because the mapping of the generalized coordinates conforms
to the surface of the sphere.
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 14
16.61 Aerospace Dynamics Spring 2003
? Substitute virtual displacements into D?Alembert?s equation
12
12
12
12
12
11
xx
xq
qq
yy
yq
qq
zz
zq
qq
q
q
q
? ?
=+
??
??
=+
??
??
=+
??
δ δδ
δ δδ
δ δδ
()
xyz
mxx y y z z F x F y F zδ δδ δ δ++ = + +"" "" "" δ
12
111 2 22
111 2 22
xyz x y z
xyz xyz
mx y z q mx y z q
qqq qqq
xyz xyz
F FF qFFF
qqq qqq
???
??? ???
++ + ++
???
??? ???
???
??? ???
=++ +++
??? ???
"" "" "" """" ""δδ
q
?
?
?
δ δ
? Facts:
o Virtual displacements
1
qδ and
2
qδ conform to constraints
o Virtual work Wδ is work that conforms to constraints
o
1
qδ and
2
qδ are independent and can be independently
moved without violating constraints
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 15
16.61 Aerospace Dynamics Spring 2003
? Conclusion:
o Force of the constraint has been eliminated by selecting
generalized coordinates that enforce the constraint
(Reason 2 for Lagrange, pg 24)
o Further, we can split the equation into two equations in two
unknowns due to independence of
1
qδ and
2
qδ .
111 1 1 1
222 2 2
xyz
xyz
xyz x y z
mx y z F F F
qqq q q q
xyz x y
mx y z F F F
qqq q q q
???
??? ? ? ?
++ = + +
???
??? ? ? ?
???
???
??? ? ? ?
++ = + +
???
??? ? ? ?
???
"" "" ""
"" "" ""
2
z
?
?
?
?
?
?
5. Finally, eliminate acceleration terms
? Consider the total derivative
11
dx xdx
xxx
dt q q dt q
?? ?
?? ?
=+
?? ?
?? ?
?? ?
""""
1
?
?
?
? Rearrange
11
xd x dx
xxx
qdt q dtq
?? ?
??
=?
?? ?
?? ?
"" " "
1
?
?
?
?
(1)
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 16
16.61 Aerospace Dynamics Spring 2003
? Recall
() ()
112 112
,,
d
x f qq x f qq
dt
? ?=∴=
? ?
"
? Perform the derivative (chain rule):
1
12
xx
xq
qq
2
q
? ?
=+
??
"" " (2)
? Partial derivative of (2) with respect to q gives
1
"
11
x
qq
x? ?
=
? ?
"
"
(3)
? Since
() (
112 112
1
,, ,
x
xfqq gqq
q
)
?
==
?
is a ftn of both q
1
and q
2
the time derivative of
1
x
q
?
?
gives (chain rule again)
1
111 21
dx x x
q
dt q q q q q
?? ?? ??
??? ??
=+
?? ?? ??
??? ??
?? ?? ??
"
2
" (4)
? Partial derivative of (2) with respect to gives x"
1
q
1
111 12
xx x
q
qqq qq
?? ??
??? ??
=+
?? ??
??? ??
?? ??
"
"" (5)
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 17
16.61 Aerospace Dynamics Spring 2003
? Note RHS of 4 and 5 are the same, thus
!
11
xdx
qdtq
??
? ?
=
??
? ?
??
"
(6)
? Now, insert (3) and (6) into (1):
11
xd x dx
xxx
qdt q dtq
?? ?
??
=?
?? ?
?? ?
"" " "
1
?
?
?
?
(1)
11
x
qq
??
=
??
"
"
x
11
dx
dt q q
??
x? ?
=
??
? ?
??
"
(3 and 6)
? Results in:
11
xd x
xx
qdt q q
??
??
=?
??
??
"
"" " "
"
1
x?"
(7)
? Note that
2
11
2
x
x
x
qq
??
?
??
?
?
=
??
"
"
"
""
?
and
2
11
2
x
x
x
qq
??
?
??
?
?
=
??
"
"
"
?
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 18
16.61 Aerospace Dynamics Spring 2003
? Finally
22
11
2
xx
xd
x
qdt q q
??
?? ??
??
??
?? ??
?
?? ?
=?
??
??
??
""
""
"
1
2
?
?
(8)
? The above process is identical for y and z.
? Recall our virtual work equation for :
1
q
111 1 1
xyz
x
1
y zxy z
mx y z F F F
qqq q q q
???
??? ? ? ?
++ = + +
???
??? ? ? ?
???
"" "" ""
?
?
?
? Insert equations (8) for x, y and z and collect terms to
eliminate acceleration terms. (Reason 3 for Lagrange, pg 24)
222 222
11
111
xyz
dxy zxy z
mm
dt q q
xyz
FFF
qqq
??
??
?++ ?++
?
??
??
??
????
??
???
=++
??
???
??
"" """"
"
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 19
16.61 Aerospace Dynamics Spring 2003
? Observe that:
()
222
11
22
Tmv mxyz== ++
"""
2
which is the kinetic energy of the particle
? Finally:
11 11
xyz
dx
TTFFF
dt q q q q q
?? ?
?? ???
?= + +
?? ?
?? ???
?? ?
"
1
y
?
?
?
? Similarly:
22 22
xyz
dx
2
y z
TTFFF
dt q q q q q
?? ?
?? ???
?= + +
?? ?
?? ???
?? ?
"
?
?
?
? The general form of Lagrange?s equation is thus:
r
q
rr
d
TT
dt q q
??
??
?=
??
??
??
"
Q
r
qx y z
rrr
xyz
QF F F
qqq
? ??
=++
???
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 20
16.61 Aerospace Dynamics Spring 2003
? Some observations:
o One Lagrange equation needed for each DOF
o Easily extendable for a system of particles
o T ? Expression of system kinetic energy
o All inertial forces contained in the LHS
o only contains external forces
q
r
Q
? How to use this ?.
1. Determine number of DOF and constraints
2. Identify generalized coordinates and equations of constraint
a. Iterate on 1 and 2 if needed
3. Write expression for T
a. v inertial velocity that can be written in terms of the
coordinates of any frame
b. Find required derivatives of T
4. Find generalized forces Q
r
q
a. If forces are known in inertial coordinates, transform
them to generalized coordinates
b. Apply generalized force equation for each force
r
qx y z
rr
x
r
y z
QF F F
qq
??
q
? ??
=++
??
???
??
5. Substitute into Lagrange?s equation
6. Solve analytically or numerically
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 21
16.61 Aerospace Dynamics Spring 2003
Example: Projectile Problem:
z
x
y
1, 2, 3. DOF = 3, no constraints
4.
()
222
11
22
Tmv mx
2
y z== ++"""
5. ,,
TTT
mx my mz
xyz
???
===
???
"""
"""
,,
dT d T dT
mx my mz
dt x dt y dt z
?????
?? ??
==
???? ??
?? ??
??
"" "" ""
"" "
=
0
TTT
xyz
???
===
???
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 22
16.61 Aerospace Dynamics Spring 2003
6. Generalized forces:
r
qx y z
rr
x
r
y z
QF F F
qq
??
???
=++
??
???
??
q
?mg z=?F
0,
xy z
qq qz
z
QQ QF mg
z
?
== = =?
?
7. EOMs: 0, 0,mx my mz mg== = =?"" "" ""
8. Solve differential equations.
? Comments:
o Method is overkill for this problem
o Inspection shows agreement with Newton
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 23
16.61 Aerospace Dynamics Spring 2003
Example: Mass moving along a frictionless track.
θ
φ
X
Y
Z
ρ
z
? Track geometry defined such that:
azρ = , and bzφ =?
DOF = 3 ? 2 =1
? Constraint equations: azρ = , and bzφ = ?
? Generalized coordinate: z
? Find
2
1
2
m=T , what is v? v
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 24
16.61 Aerospace Dynamics Spring 2003
? Define rotating coordinate frame such that mass remains in
plane. ? ?x? z
φ
X
Y
Z
ρ
?x
?y
?x
z
ρ
z
?z
r x zz azx zz=+= +ρ
# # # #
and
ω φ= = ?
"
##
z bzz
" "# "#
(
#
)
# #
"# " # "#
r azx zz bzz azx zz
azx abzzy zz
=++?×+
=? +
!
vrr
az abzz z
aabzz
2
22
2222
1
=?
=+ +
=+ +
""
( ")(" ) "
()"
2
2
T
m
aabzz=++
2
1
2222
()"
2
and
2 222
(1 )
T
maabz
z
z
?
=++
?
"
"
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 25
16.61 Aerospace Dynamics Spring 2003
()
2 222 22 2
(1 ) 2
dT
maabzzmabzz
dt z
?
??
=++ +
??
?
??
"" "
"
22 2
()
T
mabzz
z
?
=
?
"
? External force is gravity
r
qx y z
rr
xy
QF F F
qq
??
r
z
q
? ??
=++
??
???
??
?mg z= ?F
0,
xy z
qq qz
z
QQ QF m
z
g
?
== = =?
?
? Equation of Motion:
( )
2 222 22 2
1aabz zabz+++ ="" " g?
? Comments:
o Solution highly nonlinear
o ?Trick? was finding inertial velocity
o Still need to use FARM approach
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 26
16.61 Aerospace Dynamics Spring 2003
Extending Lagrange?s Equation to Systems
with Multiple Particles
? Assume a system of particles and apply Newton?s laws:
111
11
,,
,,
ppp
xyz1
x py pz p
F mx F my F mz
F mx F my F mz
===
===
"" "" ""
$$$
"" "" ""
? As before, the F?s contain both external and constraint forces.
? Multiply both sides of each equation by the appropriate
virtual displacement and add all the equations together.
()
()
11
iii
pp
ii ii ii x i y i z i
mx x y y z z F x F y F zδ δδ δ δ δ
==
++ = + +
∑∑
"" "" ""
? Recall that this is D?Alembert?s equation
? Assume the system has n DOF, np3≤
? Select generalized coordinates, that enforce the constraints:
i
q
( )
()
12
12
12
,,,,
,,,,
,,,,
ii n
ii n
ii n
xfqq qt
ygqq qt
zhqq qt
=
=
=
…
…
…
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 27
16.61 Aerospace Dynamics Spring 2003
? Express virtual displacements in terms of generalized
coordinates:
12
12
12
12
12
12
ii i
in
n
ii i
n
ii i
in
n
xx x
xqq
qq q
yy y
yqq
qq q
zz z
zqq
qq q
?? ?
=+++
?? ?
?? ?
=+++
?? ?
?? ?
=+++
?? ?
…
…
…
q
q
q
δ δδ δ
δ δδ δ
δ δδ δ
? Substitute the relations into D?Alembert?s equation
1
1
iii
p
iii
iiir
i
rrr
p
iii
x yz
i
rrr
xyz
mx y z q
qqq
xyz
r
F FF
qqq
=
=
??
???
++
??
???
??
???
=++
???
∑
∑
"" "" "" δ
qδ
? As before, have used fact that the generalized coordinates
automatically enforce the constraints.
o Sum over the entire system of particles decouples for each
of the generalized coordinates.
o This leaves us n such equations.
? Using the calculus relations (chain rule), one can show that
r
q
rr
d
TT
dt q q
??
??
?=
??
??
??
"
Q
? Once again, one Lagrange equation for each DOF.
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 28
16.61 Aerospace Dynamics Spring 2003
Lagrange?s Equation for Conservative Systems
? Conservative forces and conservative systems
o Forces are such that the work done by the forces in moving
the system from one state to another depends only on the
initial and final coordinates of the particles (path
independence).
? Potential Energy, V
o Work done by a conservative force in a transfer from a
general configuration A to a reference configuration B is
the potential energy of the system at A with respect to B.
o Note: V is defined as work from the general state to the
reference state.
? Examples of conservative forces:
o Springs (linear elastic)
o Elastic bodies
o Gravity force
? Non-conservative forces
o Friction
o Drag of a fluid
o Any force with time or velocity dependence
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 29
16.61 Aerospace Dynamics Spring 2003
? General Expression for V, the potential energy
()
1
iii
A
P
xi yi zi
i
B
VFdxFdF
=
=? + +
∑
∫
d
? Note the ??? sign since the path is from B to A. The sum is
over the P particles in the system.
? For path independence, integrand must be an exact
differential. Thus:
iii
xyz
ii
VV
FFF
x
i
V
y z
??
=? =? =?
?
(C1)
? Observe that:
3
4
2
44 3 3
2
33 4 3
x
y
F
VV
yyx x
F
VV
xxy x
?
??
?? ?
=?=?
??
??? ??
??
?
??
?? ?
=?=?
??
??? ??
??
4
4
y
y
? Thus, in general
i r
x y
ri
F F
yx
? ?
=
? ?
(C2)
? Equation (C1) represents a necessary condition for a force to
be conservative, Equation (C2) is a sufficient condition.
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 30
16.61 Aerospace Dynamics Spring 2003
? Recall expression for generalized forces:
1
iii
p
ii
qr x y z
i
rr
xy
QFFF
qq
=
??
i
r
z
q
? ??
=++
??
???
??
∑
o Separate forces into conservative and non-conservative
1
p
N
iii
qr qr
i
ir ir ir
N
qr
r
xyzVVV
QQ
xq yq yq
V
Q
q
=
??
??????
=? + + +
??
?? ?? ??
??
?
=? +
?
∑
? Lagrange?s Equation:
r
q
rr
d
TT
dt q q
??
??
?=
??
??
??
"
Q
? Substitute in generalized force:
()
N
qr
rr r
N
qr
rr
dV
TT Q
dt q q q
d
TTV
dt q q
??
???
?=?+
??
?? ?
??
??
??
??=
??
?
??
?
"
"
Q?
? Since conservative forces are not functions of
velocities: 0
r
V
q
?
=
?"
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 31
16.61 Aerospace Dynamics Spring 2003
? Thus, can define the Lagrangian LTV= ? to obtain the
final form of Lagrange?s equation:
qr
rr
dL L
F
dt q q
??
??
?=
??
??
??
"
Example: Planar pendulum with an inline spring.
m
θ
y
x
k
r
? DOF = 3 ? 1 = 2
? Constraint equation: z = 0
? Generalized coordinates: r, θ
? Coordinate mapping: cos , sinxr yrθ θ= =
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 32
16.61 Aerospace Dynamics Spring 2003
Massachusetts Institute of Technology ? How, Deyst 2003 (Based on Notes by Blair 2002) 33
? Kinetic energy
()
22
11
22
Tmv mxy== +""
2
? Derivatives of coordinates:
cos sin , sin cosxr r yr rθ θθ θθ=? =+
""
"" "" θ
? Substitute into kinetic energy equation
()
222
1
2
Tmrrθ=+
"
"
? Potential energy
()
21
cos
2
o
Vkrrmgrθ=??
? Lagrangian
()
()
2
222
11
cos
22
o
LTV mr r krr mgrθ θ=?= + ? ? +
"
"
? Derivatives of L (note need to do this for each GC)
()
2
,,
o
LdL L
mr mr mr k r r mg
rdtr r
cosθ θ
???
??
===??+
??
??
"
"""
""
22
,2,
LdL L
mr mr mrr mgr
dt
sinθ θθ
θθ θ
?? ?
??
==+=?
??
??
""""
"
""
θ
? Substitute into Lagrange?s Equation:
( )
2
2
cos
2sin0
o
mr mr k r r mg
mr mrr mgr
?+?=
+? =
"
""
"" "
"
θ θ
θθ θ
