Define the Lagrangian as: Note minus sign T has no x dependence
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But what does this tell us about mechanics? Define the Lagrangian as: Note minus sign 223 L “ T ´ U px, y, zq 1 1 2 1 2 T “ mv “ m9r “ mpx9 2 ` y9 2 ` z9 2 q 2 2 2 BL BU T has no “´ “ Fx U has no Bx Bx x dependence BL BT dependence “ “ mx9 “ px B x9 B x9 on xdot d BL d “ px “ p9x “ Fx Ñ dt B x9 dt BL d BL “ Newton Bx dt B x9 But what does this tell us about mechanics? BL d BL “ Bx dt B x9 BL d BL “ By dt B y9 BL d BL “ Bz dt B z9 224 From last slide. Repeat to get the other two equations. These are in form of EulerLagrange equations! Hamilton’s principle: S “ Action integral “ ª t2 t1 Ldt “ ª t2 t1 pT ´ U qdt is stationary 225 Keys to using Lagrange’s Equations BL d BL “ Bx dt B x9 BL d BL “ By dt B y9 BL d BL “ Bz dt B z9 Can use more generalized coordinates qi (generalized coordinates) can really be any coordinates in any frame, but remember to write down Lagrangian in an inertial frame BL d BL “ Bq1 dt B q91 BL d BL “ Bq2 dt B q92 BL d BL “ Bq3 dt B q93 Why is this? 226 Keys to using Lagrange’s Equations BL d BL “ Bq1 dt B q91 BL d BL “ Bq2 dt B q92 BL d BL “ Bq3 dt B q93 BL d BL “ Bp1 dt B p91 BL d BL “ Bp2 dt B p92 BL d BL “ Bp3 dt B p93 Additional generalized coordinates for every new particle. Can use, for example, center of mass position and relative position from CM Some examples (following Taylor) A particle moves in a conservative force field in two dimensions. What are the equations of motion, using Cartesian coordinates? 1 9 yq 9 “ T ´ U “ mpx9 2 ` y9 2 q ´ U px, yq L “ Lpx, y, x, 2 d BL BL “ Bx dt B x9 BL BU “´ “ Fx Bx Bx BL “ mx9 B x9 BL d BL “ Ñ Fx “ m: x Bx dt B x9 y similarly Fy “ m: 227 228 From last slide ith component of generalized force BL BU “´ “ Fx Bx Bx BL “ mx9 B x9 Generalize BL Bqi BL B q9i ith component of generalized momentum And Lagrange says the generalized BL d BL force = rate of change of generalized “ Bqi dt B q9i momentum 229 Another examples (following Taylor) A particle moves in a conservative force field in two dimensions. What are the equations of motion, using Polar coordinates? 1 9 9 q “ T ´ U “ mpr9 2 ` r2 9 2 q ´ U pr, q L “ Lpr, , r, 2 BL d BL “ Br dt B r9 BU d 2 9 mr ´ “ mr9 “ m: r Br dt d 2 9 mr ` Fr “ mr9 “ m: r dt r ´ r 9 2q Fr “ mp: Fr = mar 230 Continuing on BL d BL “ B dt B 9 BU d ´ “ pmr2 9 q B dt F “ rU F “ prU q BU 1 BU ˆ rU “ rΜ ` Br rB 1 BU prU q “ rB BU “ rprU q B I! “ L BU “ rprU q “ ´rF B d d rF “ I! “ L dt dt Torque “ dL “ dt A useful thing to have spotted BL d BL “ Bqi dt B q9i BL “0Ñ Bqi d BL BL “0Ñ conserved dt B q9i B q9i 231 232 Go ahead And work on problem 7.1 yourself, and then we’ll work on 7.2 and 7.17 together More examples from Taylor (chapter 7.2) x O π y=0 l m Two dimensional system, but really need only one coordinate to describe the system due to constraints of the rod (since rod length is constant) 233 234 More examples from Taylor (chapter 7.2) y=0 1 L “ T ´ U “ mpl2 9 2 q ´ mgy 2 O x 1 L “ T ´ U “ mpl2 9 2 q ´ mglp1 ´ cos q 2 BL d BL “ π l B dt B 9 d ´mgl sin “ pml2 9 q “ ml2 : m dt Torque on mass about O “ I↵ ml2 “ I :“↵ More examples from Taylor (chapter 7.3), just to see... O θ1 x 235 1 1 2 2 L “ T ´ U “ m1 px9 1 ` y9 1 q ` m2 px9 22 ` y9 22 q ´ m1 gy1 ´ m2 gy2 2 2 x1 “ l1 sin β1 , y1 “ l1 p1 ´ cos β1 q x2 “ x1 ` l2 sin β2 “ l1 sin β1 ` l2 sin β2 l1 y2 “ y1 ` l2 p1 ´ cos β2 q “ l1 p1 ´ cos β1 q ` l2 p1 ´ cos β2 q m1 y=0 l2 θ2 m2 x91 “ l1 cos β1 β91 y91 “ l1 sin β1 β91 x91 2 ` y91 2 “ l12 β912 x92 “ l1 cos β1 β91 ` l2 cos β2 β92 y92 “ l1 sin β1 β91 ` l2 sin β2 β92 x92 2 ` y92 2 “ l12 β912 ` l22 β922 ` 2l1 l2 β91 β92 pcos β1 cos β2 ` sin β1 sin β2 q x92 2 ` y92 2 “ l2 β92 ` l2 β92 ` 2l1 l2 β91 β92 cospβ1 ´ β2 q 1 1 2 2 More examples from Taylor (chapter 7.3) x O θ1 236 1 1 m1 px9 21 ` y9 12 q ` m2 px9 22 ` y9 22 q ´ m1 gy1 ´ m2 gy2 2 2 ´ ¯ 1 1 2 92 2 92 2 92 L “ m1 l1 β1 ` m2 l1 β1 ` l2 β2 ` 2l1 l2 β91 β92 cospβ1 ´ β2 q 2 2 ´m1 gl1 p1 ´ cos β1 q ´ m2 g pl1 p1 ´ cos β1 q ` l2 p1 ´ cos β2 qq L“T ´U “ l1 m1 y=0 l2 θ2 BL d BL “ Bβ1 dt B β91 ´m2 l1 l2 β91 β92 sinpβ1 ´ β2 q ´ pm1 ` m2 qgl1 sin β1 “ Δ± d ” m1 l12 β91 ` m2 l12 β91 ` m2 l1 l2 β92 cospβ1 ´ β2 q dt BL d BL “ Bβ2 dt B β92 m2 l1 l2 β91 β92 sinpβ1 ´ β2 q ´ m2 gl2 sin β2 “ ” Δ± d m2 m2 l22 β92 ` m2 l1 l2 β91 cospβ1 ´ β2 q dt More examples from Taylor (chapter 7.3) No full solutions, but we’ll get to small-angle solutions in Chapter 11. Imagine doing this with Newton, though? x O θ1 237 l1 m1 y=0 l2 θ2 BL d BL “ Bβ1 dt B β91 ´m2 l1 l2 β91 β92 sinpβ1 ´ β2 q ´ pm1 ` m2 qgl1 sin β1 “ Δ± d ” m1 l12 β91 ` m2 l12 β91 ` m2 l1 l2 β92 cospβ1 ´ β2 q dt BL d BL “ Bβ2 dt B β92 m2 l1 l2 β91 β92 sinpβ1 ´ β2 q ´ m2 gl2 sin β2 “ ” Δ± d m2 m2 l22 β92 ` m2 l1 l2 β91 cospβ1 ´ β2 q dt 238 Now for some terminology BL d BL “ Bqi dt B q9i Let’s say we chose q1 = θ1, q2 = θ2 r1 “ x1 xˆ1 ` y1 yˆ1 r2 “ x2 xˆ2 ` y2 yˆ2 with x1 “ l1 sin β1 y1 “ l1 cos β1 x2 “ l1 sin β1 ` l2 sin β2 y2 “ l1 cos β1 ` l2 cos β2 These relationships do not depend on time, so they are referred to as being natural 239 More terminology x O π l m y=0 y Two dimensional system, but really need only one coordinate to describe the system due to constraints of the rod. We call such systems constrained More terminology 240 This system, with a ball moving along this bead, is also constrained And yet a bit more terminology O x π1 l1 m1 y=0 l2 y 241 π2 System has two degrees of freedom, and is described by two generalized coordinates. If these two numbers are equal, the system is holonomic Why does this work for problems like this? FN Ball has (for example) gravity pulling it down, but also a constraint force/normal force that keeps it moving along the parabola. Why doesn’t this matter? 242 243 Lagrange’s Equations with constraints Constraining force Total force Other forces Fcstr Ftot “ Fcstr ` F F “ ´rU Other forces (ie gravity, etc) are conservative L and derivable from a potential energy “T ´U 244 Follow similar strategy as before Recall Hamilton’s principle: S “ Action integral “ ª t2 t1 Ldt “ ª t2 t1 pT ´ U qdt is stationary Particle passes through r1 and r2 at t1 and t2 Another (potentially wrong) path Rptq “ rptq ` βptq path difference Right path βpt1 q “ βpt2 q “ 0 245 Define action for right/wrong paths S0 “ ª t2 t1 S0 “ S“ ª t2 t1 S“ S“ S ´ S0 “ S “ ª t2 t1 ª t2 t1 t1 action for the correct path m 2 r9 ´ U pr, r9 , tqdt 2 9 tq ´ U pR, R, 9 tqdt T pR, R, ª t2 ª t2 T pr, r9 , tq ´ U pr, r9 , tqdt action for arbitrary path m 92 9 tqdt R ´ U pR, R, 2 m 9 tqdt 9 2 ´ U pR, R, p9r ` βq t1 2 ´ ¯ m 2 9 tq ´ U pr, r9 , tq dt β9 ` m9r ¨ β9 ´ U pR, R, 2 change in action between paths Follow similar strategy as before ª t2 ´ ¯ m 2 9 ´ U pr, r9 , tq , tqdt β9 ` m9r ¨ β9 ´ U pR, R S“ t1 2 Zero for small ε 9 tq ´ U pr, r9 , tq “ U pr ` β, r9 ` β, U pR, R, 9 tq ´ U pr, r9 , tq 9 tq ´ U pr, r9 , tq “ rU pr, r9 , tq ¨ β U pR, R, S“ ª t2 t1 Recall definition of gradient rm9r ¨ β9 ´ rU pr, r9 , tq ¨ βs dt 246 And once again integrate by parts and use boundary conditions S“ Integrate by parts: ª t2 t1 ªb a rm9r ¨ β9 ´ rU pr, r9 , tq ¨ βs dt udv “ ruvsba ´ ª vdu u “ m9r, dv “ βdt 9 ª t2 ª t2 t2 9 “ rm9r ¨ βst1 ´ pm9r ¨ βqdt pmβ ¨ :rqdt t1 t1 Zero because π(t1) = π(t2) = 0 S“ ª t2 t1 r´mβ ¨ :r ´ rU ¨ βs dt 247 Working out the math ª t2 S“ r´mβ ¨ :r ´ rU ¨ βs dt t1 S“´ ª t2 t1 β ¨ rm:r ` rU s dt m:r “ Ftot “ Fcstr ` F “ Fcstr ´ rU ª t2 S“´ β ¨ rFcstr ´ rU ` rU s dt t1 S“´ ª t2 t1 pβ ¨ Fcstr qdt So we are left thinking about the dot product between π and the constraining force 248 What are the potential variations in path? So we are left thinking about the dot product between π and the constraining force Fcstr π is a variation on the path that the particle could take. BUT it can only take a path along the surface allowed by the constraint 249 What does it mean to have a constraining force? 250 So we are left thinking about the dot product between π and the constraining force Fcstr The constraining force is a normal force, ie perpendicular to the volume/surface in which the particle can move. So ... β ¨ Fcstr “ 0 Ñ S “ 0 Putting it together The action integral is stationary at the right path, which means that Euler-Lagrange equations still apply Remember that this only works if the constraining force is a normal force to the surface of motion 251 More problems You should try and become as comfortable as possible working out problems in the Lagrangian formalism, so let’s solve examples 7.4-7.7 on the board together, and then problems 7.16,7.20,7.23,7.34 252 253 Homework Taylor 7.3, 7.8, 7.14, 7.18, 7.29
