16.61 Aerospace Dynamics
Spring 2003
Lecture #9
Virtual Work
And the
Derivation of Lagrange’s Equations
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16.61 Aerospace Dynamics
Spring 2003
Derivation of Lagrangian Equations
Basic Concept: Virtual Work
Consider system of N particles located at ( x1 , x2 , x3 , … x3 N ) with
3 forces per particle ( F1 , F2 , F3 , … F3 N ) , each in the positive
direction.
xk
F3
F1
F5
F6
F4
F2
(x4, x5, x6)
(x1, x2, x3)
xj
FN
xi
(xN-2, xN-1, xN)
FN-2
FN-1
Assume system given small, arbitrary displacements in all
directions.
Called virtual displacements
- No passage of time
- Applied forces remain constant
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The work done by the forces is termed Virtual Work.
3N
δ W = ∑ F jδ x j
j =1
Note use of δx and not dx.
Note:
• There is no passage of time
• The forces remain constant.
In vector form:
3
δ W = ∑ Fi • δ ri
i =1
Virtual displacements MUST satisfy all constraint relationships,
! Constraint forces do no work.
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16.61 Aerospace Dynamics
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Example: Two masses connected by a rod
l
R1
R2
eˆr
Constraint forces:
R1 = − R 2 = − R2eˆr
Now assume virtual displacements δ r1 , and δ r2 - but the
displacement components along the rigid rod must be equal, so
there is a constraint equation of the form
er • δ r1 = er • δ r2
Virtual Work:
δ W = R1 • δ r1 + R 2 • δ r2
= − R2eˆr • δ r1 + R2eˆr • δ r2
= ( R2 − R2 ) eˆr • δ r1
=0
So the virtual work done of the constraint forces is zero
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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, δ rs = 0, → δ W = 0
For the body, δ W = R1 • δ r1 , but the direction of the virtual
displacement that satisfies the constraints is perpendicular to the
constraint force. Thus δ W = 0 .
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16.61 Aerospace Dynamics
Spring 2003
Principle of Virtual Work
mi = mass of particle i
R i = Constraint forces acting on the particle
Fi = 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)
R i + Fi = 0
The virtual work for a system in static equilibrium is
N
δ W = ∑ ( R i + Fi ) • δ ri = 0
i =1
But virtual displacements must be perpendicular to constraint
forces, so
R i • δ ri = 0 ,
which implies that we have
N
∑F •δ r = 0
i =1
i
i
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.
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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:
δ W = mgδ x1 − µ N 2δ x2
Constraints and force balance:
δ x1 = δ x2 , N 2 = 2mg
Substitution:
Result:
mg (1 − 2 µ ) δ x = 0
µ = 12
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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:
N
δ W = ∑ ( R i + Fi − mi""ri ) • δ ri = 0
i =1
N
⇒ ∑ ( Fi − mi""
ri ) • δ ri = 0
i =1
! 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.
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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:
xj
(x3, x4)
θ
l
(x1, x2)
xi
Coordinates:
x1 , x2 , x3 , x4
Constraint:
( x1 − x3 )
DOF:
4 −1 = 3
2
+ ( x2 − x4 ) = l 2
2
• Select n=3 generalized coordinates:
( x1 + x3 )
( x + x4 )
( x − x2 )
q2 = 2
q3 = tan −1 4
2
2
( x3 − x1 )
• Can also write the inverse mapping:
q1 =
xi = fi ( q1 , q2 , q3 , … qn , t )
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Spring 2003
3N
δ W = ∑ F jδ x j
Virtual Work:
j =1
 ∂x j
δx j = ∑ 
i =1  ∂qi
n
Constraint relations:

δqi

 ∂x j
δW = ∑∑ F j 
j =1 i =1
 ∂qi
3N n
Substitution:

δqi

 ∂x j 
Qi = ∑ Fj 

q
∂
j =1
 i
3N
Define Generalized Force:
! Work done for unit displacement of qi by forces acting on the
system when all other generalized coordinates remain constant.
n
⇒ δW = ∑ Qi δqi
i =1
• If qi is an angle, Qi is a torque
• If qi is a length, Qi is a force
• If the qi ’s are independent, then for static equilibrium must
have:
Qi = 0, i = 1, 2, … n
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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
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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
z
m
r=L
φ
θ
x
y
F
Free body
Diagram
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Spring 2003
1. Newton: F = ma
• For the particle: Fx = mx"", Fy = my"", Fz = mz"" where axes x, y, z
describe an inertial frame
• Note that Fx , Fy , Fz 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:
δ W = Fxδ x + Fyδ y + Fzδ z
• Note that δs may violate the applied constraints, because F
contains constraint forces
• Combine Newton and Virtual Work
Fxδ x = mx""δ x
Fyδ y = my""δ y
Fzδ z = mz""δ z
• Add the equations
m ( ""
xδ x + ""
yδ y + ""
zδ z ) = Fxδ x + Fyδ y + Fzδ z
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Spring 2003
• Called D’Alembert’s equation:
m ( ""
xδ x + ""
yδ y + ""
zδ z ) = Fxδ x + Fyδ y + Fzδ z
• 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
x = f1 ( q1 , q2 ) ,
y = f 2 ( q1 , q2 ) , z = f 3 ( q1 , q2 )
• Define virtual displacements in terms of generalized
coordinates:
∂x
∂x
δx=
δ q1 +
δ q2
∂q1
∂q2
n
 ∂x j
δx j = ∑ 
i =1  ∂qi

δqi

!
δy=
∂y
∂y
δ q1 +
δ q2
∂q1
∂q2
δz =
∂z
∂z
δ q1 +
δ q2
∂q1
∂q1
• Note: these virtual displacements conform to the constraints,
because the mapping of the generalized coordinates conforms
to the surface of the sphere.
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16.61 Aerospace Dynamics
Spring 2003
• Substitute virtual displacements into D’Alembert’s equation
δx=
∂x
∂x
δ q1 +
δ q2
∂q1
∂q2
δy=
∂y
∂y
δ q1 +
δ q2
∂q1
∂q2
δz =
∂z
∂z
δ q1 +
δ q2
∂q1
∂q1
m ( ""
xδ x + ""
yδ y + ""
zδ z ) = Fxδ x + Fyδ y + Fzδ z
 ∂x
 ∂x
∂y
∂z 
∂y
∂z 
""
""
""
δ
m  ""
x
y
z
q
m
x
y
z
+ ""
+ ""
+
+
+
 1

 δ q2
∂q1
∂q1 
∂q2
∂q2 
 ∂q1
 ∂q2
 ∂x
 ∂x
∂y
∂z 
∂y
∂z 
δ
q
F
F
F
=  Fx
+ Fy
+ Fz
+
+
+
 1  x
 δ q2
y
z
q
q
q
q
q
q
∂
∂
∂
∂
∂
∂
1
1
1 
2
2
2 


• Facts:
o Virtual displacements δ q1 and δ q2 conform to constraints
o Virtual work δ W is work that conforms to constraints
o δ q1 and δ q2 are independent and can be independently
moved without violating constraints
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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 δ q1 and δ q2 .
 ∂x
∂y
∂z   ∂x
∂y
∂z 
+ ""
+ ""
=
+
+
m  ""
x
y
z
F
F
F
  x

y
z
∂
∂
∂
∂
∂
∂
q
q
q
q
q
q


1
1
1 
1
1
1 
 ∂x
∂y
∂z   ∂x
∂y
∂z 
+ ""
+ ""
=
+
+
m  ""
x
y
z
F
F
F
  x

y
z
∂
∂
∂
∂
∂
∂
q
q
q
q
q
q


2
2
2 
2
2
2 
5. Finally, eliminate acceleration terms
• Consider the total derivative
d  ∂x 
∂x
d  ∂x 
"
""
"
x
=
x
+
x




dt  ∂q1 
dt  ∂q1 
∂q1
• Rearrange
""
x
∂x d  ∂x 
d  ∂x 
"
=  x"
−
x



∂q1 dt  ∂q1 
dt  ∂q1 
Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002)
(1)
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16.61 Aerospace Dynamics
Spring 2003
• Recall
x = f1 ( q1 , q2 ) ∴ x" =
d
 f1 ( q1 , q2 ) 
dt 
• Perform the derivative (chain rule):
x" =
∂x
∂x
q"1 +
q"2
∂q1
∂q2
(2)
• Partial derivative of (2) with respect to q"1 gives
∂x"
∂x
=
∂q"1 ∂q1
(3)
∂x
= g1 ( q1 , q2 ) is a ftn of both q1 and q2
∂q1
∂x
the time derivative of
gives (chain rule again)
∂q1
• Since x = f1 ( q1 , q2 ) ,
d  ∂x  ∂  ∂x 
∂  ∂x 
"
q
=
+



 1

 q"2
dt  ∂q1  ∂q1  ∂q1 
∂q2  ∂q1 
(4)
• Partial derivative of x" (2) with respect to q1 gives
∂x"
∂  ∂x 
∂  ∂x 
"
q
=
+

 1

 q"2
∂q1 ∂q1  ∂q1 
∂q1  ∂q2 
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(5)
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16.61 Aerospace Dynamics
Spring 2003
• Note RHS of 4 and 5 are the same, thus
!
∂x"
d  ∂x 
= 

∂q1 dt  ∂q1 
(6)
• Now, insert (3) and (6) into (1):
""
x
d  ∂x 
∂x d  ∂x 
"
x
=  x"
−



∂q1 dt  ∂q1 
dt  ∂q1 
∂x
∂x"
=
∂q1 ∂q"1
d  ∂x  ∂x"

=
dt  ∂q1  ∂q1
(1)
(3 and 6)
• Results in:
""
x
∂x d  ∂x" 
∂x"
"
x
=  x"
−

∂q1 dt  ∂q"1 
∂q1
(7)
• Note that
 x" 2 
∂ 
2
"
∂x
=  
x"
∂q"1
∂q"1
 x" 2 
∂ 
2
"
∂x
and x"
=  
∂q1
∂q1
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16.61 Aerospace Dynamics
Spring 2003
• Finally
  x" 2  
 x" 2 
 ∂   ∂ 
2
∂x d   2  
""
x
=
−  
∂q1 dt  ∂q"1 
∂q1




(8)
• The above process is identical for y and z.
• Recall our virtual work equation for q1 :
 ∂x
∂y
∂z   ∂x
∂y
∂z 
m  ""
x
+ ""
y
+ ""
z
=
F
+
F
+
F
  x

y
z
∂
q
∂
q
∂
q
∂
q
∂
q
∂
q
1
1
1 
1
1
1 


• Insert equations (8) for x, y and z and collect terms to
eliminate acceleration terms. (Reason 3 for Lagrange, pg 24)
d  ∂  x" 2 + y" 2 + z" 2   ∂  x" 2 + y" 2 + z" 2 
m
m

−




2
2
dt  ∂q"1 
  ∂q1 

 ∂x
∂y
∂z 
=  Fx
+ Fy
+ Fz

∂
∂
∂
q
q
q

1
1
1 
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16.61 Aerospace Dynamics
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• Observe that:
1 2 1
T = mv = m ( x" 2 + y" 2 + z" 2 )
2
2
which is the kinetic energy of the particle
• Finally:
 ∂x
d ∂  ∂
∂y
∂z 
T =  Fx
+ Fy
+ Fz
 " T −

∂
∂
∂
dt  ∂q1  ∂q1
q
q
q
1
1
1 

• Similarly:
 ∂x
∂y
∂z 
d ∂  ∂
+ Fy
+ Fz
T =  Fx
 " T −

dt  ∂q2  ∂q2
∂
q
∂
q
∂
q

2
2
2 
• The general form of Lagrange’s equation is thus:
d ∂  ∂
T = Qqr
 " T −
dt  ∂qr  ∂qr

∂x
∂y
∂z 
Qqr =  Fx
+ Fy
+ Fz

∂
q
∂
q
∂
q
r
r
r 

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• 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 Qq only contains external forces
r
• 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 Qqr
a. If forces are known in inertial coordinates, transform
them to generalized coordinates
b. Apply generalized force equation for each force

∂x
∂y
∂z 
Qqr =  Fx
+ Fy
+ Fz

∂
q
∂
q
∂
q
r
r
r 

5. Substitute into Lagrange’s equation
6. Solve analytically or numerically
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Example: Projectile Problem:
z
y
x
1, 2, 3. DOF = 3, no constraints
4.
5.
T=
1 2 1
mv = m ( x" 2 + y" 2 + z" 2 )
2
2
∂T
∂T
∂T
= mx" ,
= my" ,
= mz"
∂x"
∂y"
∂z"
d  ∂T 
d  ∂T 
d  ∂T 
""
""
mx
,
my
,
=
=





 = mz""
dt  ∂x" 
dt  ∂y" 
dt  ∂z" 
∂T ∂T ∂T
=
=
=0
∂x ∂y ∂z
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6. Generalized forces:

∂x
∂y
∂z 
Qqr =  Fx
+ Fy
+ Fz

q
q
q
∂
∂
∂
r
r
r


F = − mg zˆ
Qqx = Qq y = 0,
Qqz = Fz
∂z
= −mg
∂z
7. EOMs: mx"" = 0, = my"" = 0, mz"" = −mg
8. Solve differential equations.
• Comments:
o Method is overkill for this problem
o Inspection shows agreement with Newton
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Example: Mass moving along a frictionless track.
Z
ρ
θ
z
Y
φ
X
• Track geometry defined such that:
ρ = az , and φ = −bz
DOF = 3 – 2 =1
• Constraint equations: ρ = az , and φ = −bz
• Generalized coordinate: z
• Find T =
1 2
mv , what is v?
2
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• Define rotating coordinate frame such that mass remains in
xˆ − zˆ plane.
ẑ
Z
ρ
ρ
z
x̂
z
ŷ
Y
x̂
φ
X
r = ρx# +zz# = azx# + zz#
and
ω = φ" z# = −bzz#
v 2 = r" • r"
" # + zz
"# + ( −bzz# ) × azx# + zz#
r" = azx
!
" # − abzzy
" # + zz
"#
= azx
" )2 + z" 2
= (az")2 + (abzz
= (1 + a 2 + a 2 b 2 z 2 )z" 2
∂T
m
2
2 2 2 2
= m(1 + a 2 + a 2b2 z 2 ) z"
T = (1 + a + a b z )z" and
∂z"
2
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16.61 Aerospace Dynamics
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d  ∂T 
2
2 2 2
z + 2m a 2b2 z z" 2
 "  = m(1 + a + a b z )""
dt  ∂z 
(
)
∂T
= m(a 2b2 z ) z" 2
∂z
• External force is gravity

∂x
∂y
∂z 
Qqr =  Fx
+ Fy
+ Fz

∂
q
∂
q
∂
q
r
r
r 

F = −mg zˆ
Qqx = Qq y = 0,
Qqz = Fz
∂z
= −mg
∂z
• Equation of Motion:
2
2 2 2
2 2
2
""
"
a
+
a
b
z
+
1
z
+
a
b
z
z
= −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)
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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:
Fx1 = mx""1 , Fy1 = my""1 , Fz1 = mz""1
$
$
$
Fx p = mx""p , Fy p = my""p , Fz p = mz""p
• 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.
p
∑ m ( ""x δ x + ""y δ y
i =1
i
i
i
i
p
(
+ ""
ziδ zi ) = ∑ Fxi δ xi + Fyi δ yi + Fzi δ zi
i =1
)
• Recall that this is D’Alembert’s equation
• Assume the system has n DOF, n ≤ 3 p
• Select generalized coordinates, qi that enforce the constraints:
xi = f i ( q1 , q2 , …, qn , t )
yi = gi ( q1 , q2 , …, qn , t )
zi = hi ( q1 , q2 , …, qn , t )
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:
δ xi =
∂xi
∂x
∂x
δ q1 + i δ q2 + … + i δ qn
∂q1
∂q2
∂qn
δ yi =
∂yi
∂y
∂y
δ q1 + i δ q2 + … + i δ qn
∂q1
∂q2
∂qn
∂zi
∂z
∂z
δ q1 + i δ q2 + … + i δ qn
∂q1
∂q2
∂qn
• Substitute the relations into D’Alembert’s equation
δ zi =
 ∂xi
∂y
∂z
m
xi
yi i + ""
zi i
+ ""
∑
 ""
∂qr
∂qr
i =1
 ∂qr
p

δ qr


∂x
∂y
∂z 
= ∑  Fxi i + Fyi i + Fzi i  δ qr
∂qr
∂qr
∂qr 
i =1 
• 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.
p
• Using the calculus relations (chain rule), one can show that
d ∂  ∂
T = Qqr
 " T −
dt  ∂qr  ∂qr
• 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
A P
(
V = − ∫ ∑ Fxi dxi + Fyi dyi + Fzi dzi
B i =1
)
• 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:
∂V
Fxi = −
∂xi
∂V
Fyi = −
∂yi
∂V
Fzi = −
∂zi
(C1)
• Observe that:
∂Fx3
∂  ∂V 
∂ 2V
=
−
=−
∂y4 ∂y4  ∂x3 
∂x3∂y4
∂Fy4
∂  ∂V 
∂ 2V
=
−
=−
∂x3 ∂x3  ∂y4 
∂x3∂y4
• Thus, in general
∂Fxi
∂yr
=
∂Fyr
∂xi
(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)
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16.61 Aerospace Dynamics
Spring 2003
• Recall expression for generalized forces:

∂xi
∂yi
∂zi 
Qqr = ∑  Fxi
+ Fyi
+ Fzi

∂
∂
∂
q
q
q
i =1 
r
r
r 
p
o Separate forces into conservative and non-conservative
 ∂V ∂xi ∂V ∂yi ∂V ∂zi 
N
Qqr = −∑ 
Q
+
+
+

qr
∂yi ∂qr ∂yi ∂qr 
i =1  ∂xi ∂qr
∂V
=−
+ Q N qr
∂qr
p
• Lagrange’s Equation:
d ∂  ∂
T = Qqr
 " T −
dt  ∂qr  ∂qr
• Substitute in generalized force:
d ∂  ∂
∂V
N
T
−
T
=
−
+
Q


qr
dt  ∂q"r  ∂qr
∂qr
⇒
d ∂  ∂
N
T
T
V
Q
−
−
=
(
)


qr
dt  ∂q"r  ∂qr
• Since conservative forces are not functions of
∂
velocities:
V =0
∂q"r
Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002)
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16.61 Aerospace Dynamics
Spring 2003
• Thus, can define the Lagrangian L = T − V to obtain the
final form of Lagrange’s equation:
d  ∂L

dt  ∂q"r
 ∂L
= Fqr
−
∂
q
r

Example: Planar pendulum with an inline spring.
y
k
r
θ
m
x
•
•
•
•
DOF = 3 – 1 = 2
Constraint equation: z = 0
Generalized coordinates: r, θ
Coordinate mapping: x = r cosθ ,
y = r sin θ
Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002)
32
16.61 Aerospace Dynamics
Spring 2003
• Kinetic energy
1 2 1
mv = m ( x" 2 + y" 2 )
2
2
• Derivatives of coordinates:
T=
x" = r" cosθ − rθ" sin θ , y" = r" sin θ + rθ" cosθ
• Substitute into kinetic energy equation
T=
• Potential energy
V=
• Lagrangian
1
m r" 2 + r 2θ" 2
2
(
)
1
2
k ( r − ro ) − mgr cosθ
2
1
1
2
m r" 2 + r 2θ" 2 − k ( r − ro ) + mgr cosθ
2
2
• Derivatives of L (note need to do this for each GC)
L = T −V =
∂L
= mr",
∂r"
∂L
= mr 2θ",
∂θ"
(
d  ∂L 
 "  = mr"",
dt  ∂r 
)
∂L
= mrθ" 2 − k ( r − ro ) + mg cosθ
∂r
d  ∂L 
2 ""
θ + 2mrr"θ",
=
mr
 "
dt  ∂θ 
∂L
= −mgr sin θ
∂θ
• Substitute into Lagrange’s Equation:
mr"" − mrθ" 2 + k ( r − ro ) = mg cosθ
mr 2θ"" + 2mrr"θ" − mgr sin θ = 0
Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002)
33