Friction in Lagrange’s Formulation

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 ∑ m ( &&x δ x + &&y δ y i =1 i i i i i p ( + && ziδ zi ) = ∑ Fxi δ xi + Fyi δ yi + Fzi δ zi i =1 )  ∂x  • 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: N  ∂xi ∂yi ∂zi && && && + + m x y z ∑  i i i ∂qr ∂qr i =1  ∂qr p p   ∂xi ∂yi ∂zi  + Fyi + Fzi δ qr = ∑  Fxi δ qr ∂ ∂ ∂ q q q i =1   r r r  • Applying lots of calculus on LHS and noting independence of the δ qi , for each DOF we get a Lagrange equation: d  ∂T  dt  ∂q&r p  ∂T  ∂xi ∂yi ∂zi  − = F + F + F ∑  xi   yi zi ∂qr ∂qr   ∂qr i =1  ∂qr • Further, we “moved” the conservative forces (those derivable from a potential function to the LHS: d  ∂L  dt  ∂q&r p  ∂L  ∂xi ∂yi ∂zi  − = F + F + F ∑  xi   yi zi ∂qr ∂qr   ∂qr i =1  ∂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 ∂y ∂z  Qqr = ∑  Fxi i + Fyi i + Fzi i  ∂qr ∂qr ∂qr  i =1  p • Recall that the RHS was derived from the virtual work: Qqr = δW 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. • Constraint equations: y = z = 0. • Generalized coordinate: q g k q(t) c 1 • Kinetic Energy: T = mq& 2 2 m θ µ 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 = mq& , & ∂q d  ∂L  = mq&&,   & dt  ∂q  ∂L = − kq + mg sin θ ∂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. o Since body in motion at the time of the virtual displacement, use the d’Alembert principle and include the inertia forces as well as the real external forces mq&& Fs mg Fd N Ff 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: Qqr = δW = ( −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:  ∂xi ∂yi ∂zi  Qqr = ∑  Fxi + Fyi + Fzi  ∂ q ∂ q ∂ q i =1  r r r  p o o For example, the gravity force: Fyi = − mg , yi = − q sin θ , ∂yi = − 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  dt  ∂q&r  ∂L N − = Q  qr ∂ q r  with L=T-V and Q N qr = Fx ∂x ∂y ∂z + 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 n ∂F = ∑ 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  dt  ∂q&r  ∂L ∂F + =0 −  ∂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