16.
61 Aerospace Dynamics Spring 2003
Lecture #10
Friction in Lagrange’s Formulation
Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 1
16.61 Aerospace Dynamics Spring 2003
Generalized Forces Revisited
• Derived Lagrange’s Equation from D’Alembert’s equation:
p p
∑ m ( &&x δ x + &&y δ y
i i i i i (
ziδ zi ) = ∑ Fxi δ xi + Fyi δ yi + Fzi δ zi
+ && )
i =1 i =1
∂x N
• Define virtual displacements δ xi = ∑ i δ q j
j =1 ∂q j
• Substitute in and noting the independence of the δ q j , for each
DOF we get one Lagrange equation:
p
∂xi ∂yi ∂zi p
∂xi ∂yi ∂zi
∑ m x
&&
i + y
&&i + z
&&i δ qr = ∑ Fxi + Fyi + Fzi δ qr
i =1 ∂qr ∂qr ∂qr i =1 ∂qr ∂qr ∂qr
• Applying lots of calculus on LHS and noting independence of
the δ qi , for each DOF we get a Lagrange equation:
d ∂T ∂T p
∂xi ∂yi ∂zi
dt ∂q&r
− = ∑ xiF + Fyi + Fzi
∂qr i =1 ∂qr ∂qr ∂qr
• Further, we “moved” the conservative forces (those derivable
from a potential function to the LHS:
d ∂L ∂L p
∂xi ∂yi ∂zi
dt ∂q&r
− = ∑ xiF + Fyi + Fzi
∂qr i =1 ∂qr ∂qr ∂qr
Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 2
16.61 Aerospace Dynamics Spring 2003
• Define Generalized Force:
∂x
p
∂y ∂z
Qqr = ∑ Fxi i + Fyi i + Fzi i
i =1 ∂qr ∂qr ∂qr
• Recall that the RHS was derived from the virtual work:
δW
Qqr =
δ qr
• Note, we can also find the effect of conservative forces using
virtual work techniques as well.
Example
• Mass suspended from linear spring and velocity proportional
damper slides on a plane with friction.
• Find the equation of motion of the mass.
• DOF = 3 – 2 = 1. g
k q(t)
• Constraint equations: y = z = 0.
• Generalized coordinate: q c m
µ
1 θ
• Kinetic Energy: T = mq& 2
2
1
• Potential Energy: V = kq 2 − mgq sin θ
2
Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 3
16.61 Aerospace Dynamics Spring 2003
1 1
• Lagrangian: L = T − V = mq& 2 − kq 2 + mgq sin θ
2 2
• Derivatives:
∂L d ∂L ∂L
= mq& , = mq&&, = − kq + mg sin θ
∂q
& dt ∂q
& ∂q
• Lagrange’s Equation:
d ∂L ∂L
− = mq&& + kq − mg sin θ = Qqr
dt ∂q ∂q
&
• To handle friction force in the generalized force term, need to
know the normal force Æ Lagrange approach does not
indicate the value of this force.
o Look at the free body diagram. Fs mg
mq&&
o Since body in motion at the time
of the virtual displacement, use Fd
the d’Alembert principle and Ff
include the inertia forces as well N
as the real external forces
o Sum forces perpendicular to the motion: N = mg cosθ
Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 4
16.61 Aerospace Dynamics Spring 2003
• Recall δ W = F ⋅ δ s . Two nonconservative components, look
at each component in turn:
o Damper: δ W = −cq&δ q
o Friction Force:
δ W = − sgn(q ) µ N δ q
= − sgn(q ) µ mg cosθδ q
• Total Virtual Work:
δ W = ( −cq& − sgn(q) µ mg cosθ ) δ q
• The generalized force is thus:
δW
Qqr = = ( −cq& − sgn(q ) µ mg cosθ )
δ qr
• And the EOM is:
mq&& + kq − mg sin θ = −cq& − sgn( q ) µ mg cosθ
⇒ mq&& + cq& + kq = mg ( sin θ − sgn(q ) µ cos θ )
Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 5
16.61 Aerospace Dynamics Spring 2003
• Note: Could have found the generalized forces using the
coordinate system mapping:
p
∂xi ∂yi ∂zi
Qqr = ∑ Fxi + Fyi + Fzi
i =1 ∂qr ∂qr ∂qr
o
o For example, the gravity force:
∂yi
Fyi = − mg , yi = − q sin θ , = − sin θ
∂q
⇒ Qqr = mg sin θ
Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 6
16.61 Aerospace Dynamics Spring 2003
Rayleigh's Dissipation Function
• For systems with conservative and non-conservative forces,
we developed the general form of Lagrange's equation
d ∂L ∂L
− = Q N
qr
dt ∂q&r ∂qr
with L=T-V and
∂x ∂y ∂z
Q N qr = Fx + Fy + Fz
∂qr ∂qr ∂qr
• For non-conservative forces that are a function of q& , there is
an alternative approach. Consider generalized forces
n
Q N i = − ∑ cij (q, t )q& j
j =1
where the cij are the damping coefficients, which are dissipative
in nature Î result in a loss of energy
• Now define the Rayleigh dissipation function
1 n n
F = ∑∑ cij q&i q& j
2 i =1 j =1
Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 7
16.61 Aerospace Dynamics Spring 2003
• Then we can show that
∂F n
= ∑ cqr j q& j = −Q N qr
∂q&r j =1
• So that we can rewrite Lagrange's equations in the slightly
cleaner form
d ∂L ∂L ∂F
− + =0
dt ∂q&r ∂qr ∂q&r
• In the example of the block moving on the wedge,
1 2
F = cq&
2
d ∂L ∂L ∂F
− + = mq&& + kq − mg sin θ + cq& = Qq′r
dt ∂q& ∂q ∂q&
where Qq′r now only accounts for the friction force.
Massachusetts Institute of Technology © How, Deyst 2003 (Based on Notes by Blair 2002) 8
