Lecture 27: Generalized Coordinates and Lagrange s Equations of Motion

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1 Lecture 27: Generalize Coorinates an Lagrange s Equations of Motion Calculating T an V in terms of generalize coorinates. Example: Penulum attache to a movable support 6 Cartesian Coorinates: (X, Y, Z) an (x, y, z). 4 Holonomic constrains: { Z 0 ; Y 0 z 0 ; [(x X) 2 + (y Y ) 2 ] r 2 0 two generalize coorinates! Choose X an θ. (Note: we can change θ without changing X; therefore θ an X are inepenent). T 1 2 Mẋ m(ẋ2 + ẏ 2 ) V mgy x X + r sin θ y r cos θ X X ẋ Ẋ + r θ cos θ ẏ +r θ sin θ Ẋ Ẋ T 1 2 mẋ m [ (Ẋ + r θ cos θ) 2 + (r θ sin θ) 2 ] 1 2 mẋ2 + 1 [Ẋ2 2 m + 2Ẋr θ ] cos θ + r 2 θ2 cos 2 θ + r 2 θ2 sin 2 θ Page 1

2 T 1(M + 2 m)ẋ2 + [r 1m 2 2 θ2 + 2rẊ θ ] cos θ V mgr cos θ Notice how 2rẊ θ cos θ correspons to a cross term. This plays an important role, an it is ifficult to get it correctly without starting from Cartesian coorinates. In aition, note that V only epens on θ an is inepenent of X! This has important consequences as we will see later. In general, if there are N particles each with cartesian coorinates (x i, y i, z i ), i 1,..., N an m holonomic constraints, we can reuce the system to n 3N m generalize coorinates (# of egrees of freeom). We specify these as q (q 1, q 2,..., q n ) We have x i x i ( q) y i y i ( q) z i z i ( q) ẋ i n x i j1 q j q j ẋ i ( q, q) ẏ i n y i j1 q j q j ẏ i ( q, q) ż i n z i j1 q j q j ż i ( q, q) Lagrange s Equations of Motion for a Conservative System Hamilton s principle: t2 δj δ L(q i, q i ) t 1 t2 t 1 i t2 t 1 δl 0 [ (δq i ) + ] (δ q i ) 0 q i q i Page 2

3 We integrate the secon term by parts. Let u q i an v ( q i) : δ( q i ) (δq i) t2 t 1 ( q i ) q i (δq i ) q i t t2 2 t 1 t 1 ( q i ) (δq i ) an since we have that δj t2 t 1 i (δq i ) t 2 0 q i t 1 [ ( )] (δq i ) 0 q i q i (δq i ) is completely arbitrary; the only requirement is that the enpoints are fixe. Consequently, the only way δj can vanish, given the infinite varieties of (δq) i is that each component variable, i.e,: 0 i 1,..., n q i q i (Conservative, subject only to holonomic constraints). n Lagrangian equations of motion for n egrees of freeom. Applications (i): Fin suitable set of inepenent general coorinates q i (ii): Fin Cartesian coorinates as one of general coorinates: x i ( q), y i ( q), z i ( q) (iii): Fin T an V as a function of q, q i ; Fin L( q, q) T V (iv): Use q i ( L q i ) 0 to fin the equations of motion. Page 3

4 Example 1: Harmonic Oscillator L(x, ẋ) T V 1 2 mẋ2 1 2 kx2 kx } x mẋ ẋ q i ( ) L q i 0 kx (mẋ 0) kx mẍ 0 kx mẍ Example 2: Particle in a Central Force Fiel x r cos θ ẋ ṙ cos θ r θ sin θ y r sin θ ẏ ṙ sin θ + r θ cos θ z 0 ż 0 Page 4

5 T 1 2 m(ẋ2 + ẏ 2 ) 1 2 [ ṙ 2 cos 2 θ 2rṙ θ sin θ cos θ + r 2 θ2 sin 2 θ + ṙ 2 sin 2 θ + 2rṙ θ ] sin θ cos θ + r 2 θ2 cos 2 θ 1 2 m[r2 θ2 + ṙ 2 ], an V V (r) an hence we have Two equations of motion: L 1 2 m[r2 θ2 + ṙ 2 ] V (r) (i): For r (ii): For θ r 0 ṙ θ θ 0 m θ 2 r V r (mṙ) 0 0 (mr2 θ) 0 m r V r + mr θ 2 (mr2 θ) 0 Note that V/ r is the central force an mr θ 2 is the centrifugal force. Note: Conservation law becomes automatic! we get a constant of motion for each q i that oes not appear explicitly in L. Page 5

6 Example 3: Atwoo s Machine Pulley has moment of inertia I; hence we have two Cartesian coorinates plus the constraint x 1 + πa + x 2 l (l is the length of the cor). Use x x 1 as coorinate: T 1 2 m 1ẋ I (ẋ a ) m 2ẋ 2 Note that (ẋ/a) 2 θ 2. We also have that V m 1 gx 1 m 2 gx 2 m 1 gx m 2 g(l πa x) V (m 1 m 2 )gx m 2 g(l πa) Hence L 1 2 (m 1 + m 2 )ẋ ( ) I ẋ 2 + (m a 2 1 m 2 )gx + m 2 g(l πa) Now ẋ x (m 1 + m 2 + Ia 2 ) ẍ (m 1 m 2 )g ( ) m 1 m 2 ẍ g m 1 + m 2 + I a 2 We see here that a massive pulley reuces acceleration. Page 6

7 Example 4: The Double Atwoo Machine (neglect pulley raii) Let x i enote the istance to the mass m i from the upper pulley an let x p enote the istance to the pulley which moves (the lower pulley). Let l be the length of the upper rope an l be the length of the lower rope. Note that x 1 x x 2 (l x) + x x 3 (l x) + (l x ) We thus have the following constraints: x 1 + x p l (x 2 x p ) + (x 3 x p ) l x 2 + x 3 2(l x 1 ) l Now T 1 2 m 1ẋ m 2( ẋ + ẋ ) m 3( ẋ ẋ ) 2 V m 1 gx m 2 g(l x + x ) m 3 g(l x + l x ) Page 7

8 L T V 1 2 m 1ẋ m 2(ẋ 2 2ẋ ẋ + ẋ 2 ) m 3(ẋ 2 + 2ẋẋ + ẋ 2 ) +(m 1 m 2 m 3 )gx + (m 2 m 3 )gx + constant (i) ẋ x (m 1 + m 2 + m 3 )ẍ + (m 3 m 2 )ẍ (m 1 m 2 m 3 )g (ii) ( ) ( ) ẋ x (m 2 + m 3 )ẍ + (m 3 m 2 )ẍ (m 2 m 3 )g We obtain the accelerations by solving this system of two equations an two unknowns (ẍ an ẍ ) ẍ g [ ] (m3 m 2 ) 2 + (m 2 + m 3 )(m 1 m 2 m 3 ) (m 3 m 2 ) 2 (m 2 + m 3 )(m 1 + m 2 + m 3 ) In the homework, be wary of the ouble-ouble Atwoo machine (m 1 is replace by a pulley which itself hols two masses). Page 8

9 Example 5: Particle Sliing on a movable incline plane Let V be the velocity of the incline plane an v be the velocity of the block. Then where ê θ (cos θ, sin θ). We thus have V îẋ v îẋ + ê θ ẋ T 1 2 MV mv2 1 2 Mẋ m(îẋ + ê θ ẋ ) Mẋ m [ (ẋ + cos θẋ ) 2 + (sin θẋ ) 2] 1 2 Mẋ m [ ẋ cos θẋẋ + cos 2 θẋ 2 + sin 2 θẋ 2] an 1 2 Mẋ m [ ẋ cos θẋẋ + ẋ 2] V mg[l sin θ x sin θ] (constant) mg sin θx where L is the length of the plane. Hence L T V 1 2 (M + m)ẋ2 + m cos θẋẋ mẋ 2 + mg sin θx Page 9

10 Equations of Motion: (i) ẋ x ((M + m)ẋ + m cos θẋ ) 0 (since L is inepenent of x). Note that conservation of momentum comes out automatically. (ii) ( ) ( ) ẋ x (m cos θẋ + mẋ ) mg sin θ We now have two equations an two unknowns: { (M + m)ẍ + m cos θẍ 0 cos θẍ + ẍ g sin θ Solving for ẍ an ẍ yiels ẍ g sin θ 1 m m+m cos2 θ ẍ g sin θ cos θ cos2 θ m+m m Page 10