PHYS 321A Lecture Notes 27 University of Victoria Lecture 27: Generalized Coordinates and Lagrange's Equations of Motion Calculating T and V in terms of generalized coordinates. Example: Pendulum attached to a movable support 6 Cartesian Coordinates: (X; Y; Z) and (x; y; z). 4 Holonomic constrains: ( Z = 0 ; Y = 0 z = 0 ; [(x − X)2 + (y − Y )2] − r2 = 0 =) two generalized coordinates! Choose X and θ. (Note: we can change θ without changing X; therefore θ and X are independent). 1 1 T = Mx_ 2 + m(_x2 +y _2) 2 2 V = mgy 8 8 x = X + r sin θ x_ = X_ + rθ_ cos θ <> <> y = −r cos θ =) y_ = +rθ_ sin θ :>X = X :>X_ = X_ 1 1 h i T = mX_ 2 + m (X_ + rθ_ cos θ)2 + (rθ_ sin θ)2 2 2 1 1 h i = mX_ 2 + m X_ 2 + 2Xr_ θ_ cos θ + r2θ_2 cos2 θ + r2θ_2 sin2 θ 2 2 Page 1 PHYS 321A Lecture Notes 27 University of Victoria 8 h i >T = 1 (M + m)X_ 2 + 1 m r2θ_2 + 2rX_ θ_ cos θ <> 2 2 > :>V = −mgr cos θ Notice how 2rX_ θ_ cos θ corresponds to a \cross term." This plays an important role, and it is difficult to get it correctly without starting from Cartesian coordinates. In addition, note that V only depends on θ and is independent of X! This has important consequences as we will see later. In general, if there are N particles each with cartesian coordinates (xi; yi; zi); i = 1; :::; N and m holonomic constraints, we can reduce the system to n = 3N − m generalized coordinates (# of degrees of freedom). We specify these as ~q = (q1; q2; :::; qn) We have 8 Pn @xi _ x_ i = q_j =x _ i(~q; ~q) > j=1 @qj 8 > x = x (~q) > <> i i <> Pn @yi _ yi = yi(~q) =) y_i = j=1 @q q_j =y _i(~q; ~q) > > j :zi = zi(~q) > > > Pn @zi _ :z_i = q_j =z _i(~q; ~q) j=1 @qj Lagrange's Equations of Motion for a Conservative System Hamilton's principle: Z t2 Z t2 δJ = δ L(qi; q_i)dt = δLdt = 0 t1 t1 Z t2 X @L @L = (δq ) + (δq_ ) dt = 0 @q i @q_ i t1 i i i Page 2 PHYS 321A Lecture Notes 27 University of Victoria We integrate the second term by parts. Let u = @L and dv = d(@qi) : @q_i dt Z t2 t Z t2 d @L d(@qi) @L 2 d @L δ(_qi) = (δqi) =) dt = (δqi) − (δqi)dt t dt t1 @q_i dt @q_i 1 t1 dt @qi and since @L t2 (δqi) = 0 @q_i t1 we have that Z t2 X @L d @L δJ = dt − (δq ) = 0 @q dt @q_ i t1 i i i (δqi) is completely arbitrary; the only requirement is that the endpoints are fixed. Con- sequently, the only way δJ can vanish, given the infinite varieties of (δq)i is that each component variable, i.e,: @L d @L − = 0 i = 1; :::; n @qi dt @q_i (Conservative, subject only to holonomic constraints). n Lagrangian equations of motion for n degrees of freedom. Applications (i): Find suitable set of independent general coordinates qi (ii): Find Cartesian coordinates as one of general coordinates: xi(~q); yi(~q); zi(~q) _ (iii): Find T and V as a function of ~q, ~qi; Find L(~q; ~q) = T − V (iv): Use @L − d ( dL ) = 0 to find the equations of motion. @qi dt dq_i Page 3 PHYS 321A Lecture Notes 27 University of Victoria Example 1: Harmonic Oscillator 1 1 L(x; x_) = T − V = mx_ 2 − kx2 2 2 ) @L = −kx @L d dL d @x − = 0 =) −kx − (mx_ = 0) @L @q dt dq_ dt @x_ = mx_ i i =) −kx − mx¨ = 0 =) −kx = mx¨ Example 2: Particle in a Central Force Field x = r cos θ =) x_ =r _ cos θ − rθ_ sin θ y = r sin θ =) y_ =r _ sin θ + rθ_ cos θ z = 0 =) z_ = 0 Page 4 PHYS 321A Lecture Notes 27 University of Victoria 1 T = m(_x2 +y _2) 2 1 h i = r_2 cos2 θ − 2rr_θ_ sin θ cos θ + r2θ_2 sin2 θ +r _2 sin2 θ + 2rr_θ_ sin θ cos θ + r2θ_2 cos2 θ 2 1 = m[r2θ_2 +r _2] ; and V = V (r) 2 and hence we have 1 L = m[r2θ_2 +r _2] − V (r) 2 Two equations of motion: (i): For r (ii): For θ @L d @L @L d @L − = 0 − = 0 @r dt @r_ @θ dt @θ_ d 0 − (mr2θ_) = 0 @V d dt mθ_2r − − (mr_) = 0 @r dt d (mr2θ_) = 0 dt @V mr¨ = − + mrθ_2 @r Note: Conservation law becomes \auto- matic"! =) we get a \constant of motion" for each qi that does not appear explicitly in Note that −@V=@r is the central force and L. mrθ_2 is the centrifugal force. Page 5 PHYS 321A Lecture Notes 27 University of Victoria Example 3: Atwood's Machine Pulley has moment of inertia I; hence we have two Cartesian coordinates plus the constraint x1 + πa + x2 = l (l is the length of the cord). Use x = x1 as coordinate: 1 1 x_ 2 1 T = m x_ 2 + I + m x_ 2 2 1 2 a 2 2 Note that (x=a _ )2 = θ_2. We also have that V = − m1gx1 − m2gx2 = − m1gx − m2g(l − πa − x) V = − (m1 − m2)gx − m2g(l − πa) Hence 1 1 I L = (m + m )_x2 + x_ 2 + (m − m )gx + m g(l − πa) 2 1 2 2 a2 1 2 2 Now d @L @L I = =) m + m + x¨ = (m − m )g dt @x_ @x 1 2 a2 1 2 ! m1 − m2 =) x¨ = I g m1 + m2 + a2 We see here that a massive pulley reduces acceleration. Page 6 PHYS 321A Lecture Notes 27 University of Victoria Example 4: The Double Atwood Machine (neglect pulley radii) Let xi denote the distance to the mass mi from the upper pulley and let xp denote the distance to the pulley which moves (the lower pulley). Let l be the length of the upper rope and l0 be the length of the lower rope. Note that 8 x = x <> 1 0 x2 = (l − x) + x > 0 0 :x3 = (l − x) + (l − x ) We thus have the following constraints: x1 + xp = l 0 0 (x2 − xp) + (x3 − xp) = l =) x2 + x3 − 2(l − x1) = l Now 1 1 1 T = m x_ 2 + m (−x_ +x _ 0)2 + m (−x_ − x_ 0)2 2 1 2 2 2 3 0 0 0 V = −m1gx − m2g(l − x + x ) − m3g(l − x + l − x ) Page 7 PHYS 321A Lecture Notes 27 University of Victoria 1 1 1 =)L = T − V = m x_ 2 + m (_x2 − 2x _ 0x_ +x _ 02) + m (_x2 + 2x _x_ 0 +x _ 02) 2 1 2 2 2 3 0 +(m1 − m2 − m3)gx + (m2 − m3)gx + constant (i) d @L @L = dt @x_ @x 0 (m1 + m2 + m3)¨x + (m3 − m2)¨x = (m1 − m2 − m3)g (ii) d @L @L = dt @x_ 0 @x0 0 (m2 + m3)¨x + (m3 − m2)¨x = (m2 − m3)g We obtain the accelerations by solving this system of two equations and two unknowns (¨x andx ¨0) 2 (m3 − m2) + (m2 + m3)(m1 − m2 − m3) =) x¨ = −g 2 (m3 − m2) − (m2 + m3)(m1 + m2 + m3) In the homework, be wary of the \double-double" Atwood machine (m1 is replaced by a pulley which itself holds two masses). Page 8 PHYS 321A Lecture Notes 27 University of Victoria Example 5: Particle Sliding on a movable inclined plane Let V~ be the velocity of the inclined plane and ~v be the velocity of the block. Then V~ = ^ix_ 0 ~v = ^ix_ +e ^θx_ wheree ^θ = (cos θ; − sin θ). We thus have 1 1 1 1 T = MV 2 + mv2 = Mx_ 2 + m(^ix_ +e ^ x_ 0)2 2 2 2 2 θ 1 1 = Mx_ 2 + m (_x + cos θx_ 0)2 + (sin θx_ 0)2 2 2 1 1 = Mx_ 2 + m x_ 2 + 2 cos θx_x_ 0 + cos2 θx_ 02 + sin2 θx_ 02 2 2 1 1 = Mx_ 2 + m x_ 2 + 2 cos θx_x_ 0 +x _ 02 2 2 and V = mg[L sin θ − x0 sin θ] = (constant) − mg sin θx0 where L is the length of the plane. Hence 1 1 L = T − V = (M + m)_x2 + m cos θx_x_ 0 + mx_ 02 + mg sin θx0 2 2 Page 9 PHYS 321A Lecture Notes 27 University of Victoria Equations of Motion: (i) d @L @L = dt @x_ @x d =) ((M + m)_x + m cos θx_ 0) = 0 dt (since L is independent of x). Note that conservation of momentum comes out automatically. (ii) d @L @L = dt @x_ 0 @x0 d =) (m cos θx_ + mx_ 0) = mg sin θ dt We now have two equations and two unknowns: ( (M + m)¨x + m cos θx¨0 = 0 cos θx¨ +x ¨0 = g sin θ Solving forx ¨ andx ¨0 yields g sin θ g sin θ cos θ x¨0 = x¨ = − 1 − m cos2 θ m+M 2 m+M m − cos θ Page 10.