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MEAM 535
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Lagrange’s Equations of Motion with Constraint Forces
Kane’s equations do not incorporate constraint forces
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Review: Linear Algebra
A is m × n of rank r The row space, Col (AT), dimension = r Col (AT) is spanned by:
The null space, N (A), dimension = n – r
http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/detail/lecture14.htm
€
r1 =
a11
a12
a1n
, r2 =
a21
a22
a2n
, , rm =
am1
am2
amn
.
r1 r2
rm
€
x ∈ N A( )
N (A)
Col (AT) orthogonal
Rn
Ax=0
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Review: Linear Algebra A is m × n of rank r
The column space, Col (A), dimension = r Col (A) is spanned by columns
The left null space, N (AT), dimension = m – r
yTA =0
c1 c2
cn
€
y ∈ N AT( )
€
c1 =
a11
a21
am1
, c2 =
a12
a22
am2
, , cn =
a1n
a2n
amn
.
N (AT)
Col(A) orthogonal
dim Col(A) = r dim Col(AT) = r dim N(A) = n – r dim N(AT) = m – r
Rm
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Application to Velocity Constraints
C is m × n of rank m The row space, Col (CT), dimension = m Col (CT) is spanned by:
The null space, N (C), dimension = n – m
€
r1 =
c11
c12
c1n
, r2 =
c21
c22
c2n
, , rm =
cm1
cm2
cmn
.
r1 r2
rm
€
˙ q ∈ N C( )
N (C)
Col (CT) orthogonal
Rn
€
Cm×n ˙ q n×1 = 0
N (C) set of admissible velocities (that don’t violate constraints)
Physical Interpretation of N(C)
Col (CT) is the set of constraint forces orthogonal to admissible velocities!
= Col (Γ)
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Example 2: Rolling Disk (Simplified)
(x, y) φ
θ radius R
C
τd
τs
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Example: Rolling Disk Simplified
(x, y) φ
θ radius R
C
τd
τs
C Γ 2×4 4×2 €
˙ q n×1 = Γ n× p up×1
€
Cm×n Γ n× p = 0
Two equations of motion
Nonholonomic constraints provide two additional equations
€
Cm×n ˙ q n×1 = 0
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Example 2: Rolling Disk (Simplified)
(x, y) φ
θ radius R
C
€
′ Q * =
−m˙ q 1−m˙ q 2−Ia˙ q 3−It˙ q 4
, ′ Q =
00τ d
τ s
τd
τs
PT Γ
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Application to Generalized Forces
Γ is n× p of rank p The column space, Col (Γ), dimension = p The left null space, N (ΓT), dimension = n – p = m
PTΓ =0
c1 c2
cp
€
P ∈ N ΓT( )
N (ΓT)
Col(Γ) orthogonal
Rn
€
c1 =
Γ11
Γ21
Γn1
, c2 =
Γ12
Γ22
Γn2
, , c p =
Γ1p
Γ2p
Γnp
.
Physical Interpretation of N(ΓT) set of admissible velocities (that don’t violate constraints)
= Col (Γ)
set of constraint forces
= N(ΓΤ) = Col (CT)
= N(C)
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Lagrange’s Equations with Multipliers
Pk
p Equations of Motion
m Constraints
m Columns of CT span the null space of ΓT
There exist a vector of m constants (multipliers) λ, such that
P lies in the null space of ΓT
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Example: Rolling Disk Simplified
(x, y) φ
θ radius R
C
τd
τs
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p+m+m equations in p+m+m unknowns
p+m equations in p+m unknowns
Rolling Disk Simplified: Comparison
(x, y) φ
θ radius R
C
τd
τs
C Γ m×n n×p
p eq
uatio
ns
n eq
uatio
ns
m e
quat
ions
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Multipliers are Constraint Forces!
m Columns of CT span the null space of ΓT
1. CTλ are generalized forces (associated with the derivatives of generalized coordinates)
2. The m constants (multipliers) are coefficients for vectors that are orthogonal to the allowable directions of motion
(x, y) φ
θ radius R
C
τd
τs
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Constraint Forces for Holonomic Systems
What if you choose more generalized coordinates than necessary? n is the number of generalized coordinates (more than necessary) p is the number of degrees of freedom (i.e., only p gen. coords. necessary) n > p
Notice the parallel with nonholomic systems! n speeds, but only p independent speeds
€
′ Q k + ′ Q k*( )Wkj
k=1
n
∑ = 0
p Equations of Motion
m Constraints €
˙ q n×1
= W[ ]n× pup×1+ X
n×1
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Example: Particle in circular hoop
θ R x
y
where
Pr =d
dt
!!L!r
"! !L
!r= mr !mr"2 + mg sin "
P! =d
dt
!!L!"
"! !L
!"= mr2" + mgr cos ".
Because of the hoop, there are constraints:
C(q)q = 0 (1)
or
[1 0]
#
$r
"
%
& = 0,
so that
CT (q) =
#
$1
0
%
& (2)
The equations of motion are given by:
P = CT #,
or
Pr = 1.#
P! = 0.#,
where # is the Lagrange multiplier.
From (1), r = r = 0. substituting into the equations of motion we get:
!mr"2 + mg sin " = # (3)
mr2" + mgr cos " = 0. (4)
From (3), it is clear that # is the outward pointing normal force acting on the particle.
The Kane Lagrange equations of motion are obtained by recognizing that
q = !u =
#
$0
1
%
& " (5)
and writing
PT ! =0 ,
or
mr2" + mgr cos " = 0 (6)
2
LHS of Lagrange’s equations of motion for unconstrained problem
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Example Generalized speed:
u=dθ/dt Velocities
Generalized Active Forces -Fa1
τa3
Generalized Active Inertial Forces -m AaP = -m x2dot a1
B
P
x
Find the constraint forces at the pin joint Q
Q
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Example (continued)
θ d e
φ
p=1
n=3
€
˙ q n×1
= W[ ]n× pup×1+ X
n×1
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Constraint Forces for Holonomic Systems
p Equations of Motion
m Constraints €
˙ q n×1
= W[ ]n× pup×1+ X
n×1
There exist a vector of m constants (multipliers) λ, such that
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Example: Normal Force at P Generalized speed:
u=dθ/dt Velocities
Generalized Active Forces -Fa1
τa3
Generalized Active Inertial Forces
B
P
x
What if we relax the constraint that keeps the piston moving horizontally?
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Example (continued)
n=1
p=1, n=2
€
˙ q n×1
= W[ ]n× pup×1+ X
n×1
θ y φ r l
C
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Example: Normal Force at P Generalized speeds
Partial Velocities
Generalized Active Forces -Fa1 τa3
Generalized Active Inertial Forces
B
P
θ y φ r l
x