Physics 339 Gibbs-Appell November 2017

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1 Physics 339 Gibbs-Appell November 2017 In1879JosiahWillardGibbspublished 1 analternativeformulationfornewtonianmechanics. Just as Lagrange s formulation produces equations identical to those from F = ma but deals more easily with constraints, so Gibbs formulation produces the same equations as Newton s but deals more easily with non-holonomic constraints. Gibb s formulation looks quite simple. Form what looks like total kinetic energy of all N particles but use accelerations not velocities: G = N i=1 1 2 m ia 2 i (1) Express the result in terms of generalize coordinate accelerations ( q r for r {1...k}). Find the work done for displacements in those generalized coordinates: dw = (The Q r are called generalized forces.) Then k Q r dq r (2) r=1 G q r = Q r (3) Note that contributions to G that do not depend on the q r will play no role in the equations of motion and will be dropped often without comment. Examples Example 1: A particle responding to a potential V(x,y,z): G = 1 2 m(ẍ2 +ÿ 2 + z 2 ) (4) Q = V (5) ma = V (6) Example 2a: A particle responding to a central potential V(r) (polar coordinates): G = 1 2 m(ẍ2 +ÿ 2 ) = 1 ( ) 2 m ( r r θ 2 ) 2 +(r θ+2ṙ θ) 2 (7) = 1 ) ( r 2 m 2 2r r θ+r 2 θ2 +4rṙ θ θ (8) m(r 2 θ d ( +2rṙ θ) = 0 = mr 2 θ ) = dl (9) ) dt dt m ( r r θ 2 = r V (10) m r = r V + L2 mr 3 (11) 1 Gibbs, JW (1879). On the Fundamental Formulae of Dynamics.. American Journal of Mathematics. 2: At the time Gibb s work was mostly ignored; Paul Émile Appell independently rediscovered this formulation in 1900.

2 Example 2b: Instead of using the holonomic coordinate θ we can use the non-holonomic coordinate dq = x dy y dx. (dq is twice the area swept by r as the particle goes from (x,y) to (x+dx,y+dy) and is therefore closely related to Kepler s second law and angular momentum note: m q = L, but clearly q depends on history not current configuration.) { G = 1 2 m(ẍ2 +ÿ 2 ) = 1 ) } 2 ( r 2 m q2 + q2 (12) = 1 ( ) 2 m r 2 2 r q2 + q2 (13) r 3 r 2 m q = 0 i.e., q=constant=l/m (14) r2 ) m ( r q2 = r V (15) r 3 Note that while the Lagrangian m r = r V + L2 mr 3 (16) is KE PE, it produces incorrect equations of motion: because q is non-holonomic. L = 1 ( ) 2 m ṙ 2 + q2 V(r) (17) r 2 r 3 m q r 2 = constant (18) m r = r V m q2 r 3 (19) Example 3: Pseudo-forces on a rotating plane. Let r = (x,y) be the coordinates in a plane that is rotating at Ω = Ωẑ relative to the inertial frame. v = ṙ+ω r (20) a = r+ω ṙ+ω (ṙ+ω r) (21) = r+2ω ṙ Ω 2 r (22) G = 1 2 m( r 2 +4 r Ω ṙ 2Ω 2 r r ) (23) m ( r+2ω ṙ Ω 2 r ) = F (24) which displays the Coriolis and centrifugal pseudo-forces. m r = F 2mΩ ṙ+mω 2 r (25) r 2 The Rolling Penny We have from the Appendix (particularly Example 2): G rot = 1 2 ( ) I1 ( ω ω 2)+I 2 3 ω 3 2 +(I1 Ω 3 I 3 ω 3 )( ω 2 ω 1 ω 1 ω 2 ) (26)

3 3 θ z 2 3 ψ θ z φ 3 R 2 2h C 1 C C r P P F P P Figure 1: Coordinate frame for rolling disk. Axis 3 is in the direction of the disk s axle; 1 is always parallel to the plane and in the trailing direction if ψ > 0; 2 points away from the contact point P. We must add to this G CM. Following rolling.pdf Eq. (7) and using r P = Re 2 we have: A CM = ω r P Ω ω r P (27) = R{ ω e 2 +Ω ω e 2 } (28) = R( ω 3 +Ω 2 ω 1, Ω 1 ω 1 Ω 3 ω 3, ω 1 +Ω 2 ω 3 ) (29) So: G CM = 1 2 MR2( ( ω 3 ω 2 ω 1 ) 2 +( ω 1 +ω 2 ω 3 ) 2) (30) The resulting equations of motion (with V = MgRsinθ) MR 2 ( ω 1 +ω 2 ω 3 )+I 1 ω 1 (I 1 Ω 3 I 3 ω 3 )ω 2 = MgRcosθ (31) I 1 ω 2 +(I 1 Ω 3 I 3 ω 3 )ω 1 = 0 (32) MR 2 ( ω 3 ω 2 ω 1 )+I 3 ω 3 = 0 (33) Making the usual rescaling: I I/MR 2, g g/r: (I 1 +1) ω 1 I 1 Ω 3 ω 2 +(I 3 +1)ω 2 ω 3 = gcosθ (34) I 1 ω 2 +(I 1 Ω 3 I 3 ω 3 )ω 1 = 0 (35) (I 3 +1) ω 3 ω 2 ω 1 = 0 (36) (I 1 +1) ω 1 I 1 ω 2 (I 3 +1) ω 3 = I 1 Ω 3 ω 2 (I 3 +1)ω 2 ω 3 gcosθ (I 3 ω 3 I 1 Ω 3 )ω 1 ω 2 ω 1 (37)

4 The Rolling Ring Following Fig. 1 RHS, r P = R(e 2 +he 3 ) (note that h has already been scaled: h h/r), we have: So: A CM = ω r P Ω ω r P (38) A CM /R = ω (e 2 +he 3 )+Ω ω (e 2 +he 3 ) (39) ω 3 +h ω 2 +ω 1 ω 2 +hω 1 Ω 3 = h ω 1 ω1 2 ω 3 Ω 3 +hω 2 Ω 3 (40) ω 1 +ω 2 ω 3 h(ω1 2 +ω2 2 ) G CM = 1 2 MR2( ω 2 1(1+h 2 )+h 2 ω ω ω 1 (ω 3 hω 2 )(ω 2 +hω 3 )+2 ω 2 ω 1 h(ω 2 +hω 3 ) 2 ω 3 ( ω 2 h+ω 1 ω 2 +hω 1 Ω 3 ) ) (41) The resulting LHS equations of motion MR 2 ( ω 1 (1+h 2 )+(hω 3 ω 2 )(ω2+hω 3 ))+I 1 ω 1 (I 1 Ω 3 I 3 ω 3 )ω 2 MR 2 (h( ω 3 + ω 2 h+ω 1 (ω 2 +hω 3 )))+I 1 ω 2 +(I 1 Ω 3 I 3 ω 3 )ω 1 MR 2 ( ω 3 ω 2 h ω 1 (ω 2 +hω 3 ))+I 3 ω 3 (42) With V = MgR(sinθ+hcosθ) the RHS is quite simple: MgR(cosθ hsinθ) 0 0 (43) Scaling the variables as usual and following the division of LHS/RHS as in rolling.pdf Eq. (50): (I 1 +1+h 2 ) ω 1 (I 1 +1+h 2 ) θ (I 1 +h 2 ) ω 2 h ω 3 = (I 1 +h 2 ) ( φsinθ + φcosθ θ) h( ψ + φcosθ φsinθ θ) = (I 3 +1) ω 3 h ω 2 (I 3 +1) ( ψ + φcosθ φsinθ θ) h( φsinθ + φcosθ θ) ((I 1 +h 2 )Ω 3 (I 3 +1)ω 3 +hω 2 )ω 2 hω 3 Ω 3 g(cosθ hsinθ) (I 3 ω 3 (I 1 +h 2 )Ω 3 hω 2 )ω 1 ω 1 (ω 2 +hω 3 ) (44) Hurricane Balls Following Fig. 2, r P = a(sinθ e 2 +(1+cosθ)e 3 ) A CM = ω r P ω ṙ P Ω ω r P (45) A CM /a = ω (sinθ e 2 +(1+cosθ)e 3 )+ω (cosθ e 2 sinθ e 3 )ω 1 + (46) = Ω ω (sinθ e 2 +(1+cosθ)e 3 ) (47) ω 2 +ω 1 Ω 3 +( ω 2 +ω 1 ( ω 3 +Ω 3 ))cosθ ω 3 sinθ ω 1 +ω 2 Ω 3 +( ω 1 +ω 2 Ω 3 )cosθ ω 3 Ω 3 sinθ (48) ω1 2 ω2 2 ω2 2 cosθ+( ω 1 +ω 2 ω 3 )sinθ

5 ψ φ 3 θ C Figure 2: Coordinate system for Hurricane Balls r P 2 P a mg N F α Figure 3: Coordinate system for rolling on a tilted plane y z So: G CM = ma 2 (2 ω 2 1 (1+cosθ)+ ω2 2 (1+cosθ)2 + ω 2 3 sin2 θ+ 2 ω 1 ( ω 2 (Ω 3 (1+cosθ) 2 ω 3 sin 2 θ) sinθ(ω1 2 +(ω2 2 ω 3Ω 3 )(1+cosθ)))+ 2 ω 2 (1+cosθ)(ω 1 Ω 3 (1+cosθ) ω 1 ω 3 cosθ ω 3 sinθ)+ ) 2 ω 3 (ω 1 sinθ(ω 3 cosθ Ω 3 (1+cosθ))) (49) The resulting LHS equations of motion (scaling the variables as usual: I I/2ma 2 and letting c = cosθ,s = sinθ): 2 ω 1 (1+c) s(ω1 2 +(ω2 2 ω 3Ω 3 )(1+c)) ω 2 (Ω 3 (1+c) 2 ω 3 s 2 )+I 1 ω 1 (I 1 Ω 3 I 3 ω 3 )ω 2 ω 2 (1+c) 2 +(1+c)(ω 1 Ω 3 (1+c) ω 1 ω 3 c ω 3 s)+i 1 ω 2 +(I 1 Ω 3 I 3 ω 3 )ω 1 ω 2 (1+c)s+ω 1 (ω 3 c Ω 3 (1+c))s+ ω 3 s 2 +I 3 ω 3 (50) With V = 2mga(1+cosθ), scaling by 2ma 2 and g g/a the RHS is quite simple: gsinθ 0 0 (51) Following the division of LHS/RHS as in rolling.pdf Eq. (80 81), the RHS becomes: s(g +ω1 2 +(ω2 2 ω 3Ω 3 )(1+c))+ω 2 (Ω 3 (I 1 +(1+c) 2 ) ω 3 (I 3 +s 2 )) ω 1 (1+c)(Ω 3 (1+c) ω 3 c) ω 1 (I 1 Ω 3 I 3 ω 3 ) (52) ω 1 (ω 3 c Ω 3 (1+c))s and the LHS: ω 1 (I 1 +2(1+c)) ω 2 (I 1 +(1+c) 2 ) ω 3 (1+c)s ω 3 (s 2 +I 3 ) ω 2 (1+c)s (53)

6 Appendix: Finding G rot for a Rigid Body We seek a formula for G rot calculated in a frame where the 123 axes are aligned with the principal axes. So ( I r r r) dm = 0 I 2 0 (54) 0 0 I 3 hence, for example, xy dm = 0 and (y 2 +z 2 ) dm = I 1. This principal axes frame is rotating at Ω relative to the inertial frame, the rigid body has angular velocity ω, and we let r, u, a denote respectively the location, velocity, and acceleration of a piece of the rigid body relative to the CM. u = ω r = ṙ+ω r hence: ṙ = (ω Ω) r (55) a = ω ṙ+ ω r+ω ω r (56) = ω (ω Ω) r+ ω r+ω ω r (57) = ω ω r+ ω r+ω ω r ω Ω r (58) = (ω r) ω ω 2 r+ ω r+(ω r) ω (ω r) Ω (59) = (ω r) ω ω 2 r+ ω r r Ω ω (60) = (ω r) ω ω 2 r r ( ω +Ω ω) (61) (ω r) ω ω 2 r r φ (62) Two important points: (A) only φ contains acceleration ω, so in calculating G rot we can drop terms that do not contain φ and (B) when dotted with itself the term r φ is nicely connected with I: (r φ) (r φ) = φ (r φ r) = φ (r 2 1 r r) φ (63) So (r φ) 2 dm = φ I φ (64) The non-zero cross term in a a that includes φ: 2(ω r) ω (r φ) looks to be a mess, but notethat the rcomponent inω (r φ) must match the rcomponent inω rasnon-matching terms will vanish when integrated as, e.g., xy dm = 0. Dropping terms that will vanish on integration yields: 2(ω r) ω 1 ω 2 ω 3 x 1 x 2 x 3 φ 1 φ 2 φ 3 = 2(ω 2 2ω 1 φ 3 x2 +ω 3ω 2 φ 1 x 2 3 +ω 1ω 3 φ 2 x 2 1 ) +2(ω 3 ω 1 φ 2 x 2 3 +ω 1ω 2 φ 3 x 2 1 +ω 2ω 3 φ 1 x 2 2 ) (65) = 2ω 2 ω 1 φ 3 (x 2 2 x 2 1) 2ω 3 ω 2 φ 1 (x 2 3 x 2 2) 2ω 1 ω 3 φ 2 (x 2 1 x 2 3) (66) 2ω 2 ω 1 ω 3 (I 1 I 2 ) 2ω 3 ω 2 ω 1 (I 2 I 3 ) 2ω 1 ω 3 ω 2 (I 3 I 1 ) (67) So G rot = 1 2 φ I φ+ω 2 ω 1 ω 3 (I 2 I 1 )+ω 3 ω 2 ω 1 (I 3 I 2 )+ω 1 ω 3 ω 2 (I 1 I 3 ) (68)

7 Example 1: In the general case I 1 I 2 I 3 we must evaluate in the body fixed frame so Ω = ω and φ = ω, so G rot = 1 2 ( I1 ω 1 2 +I 2 ω 2 2 +I ) 3 ω ω 2 ω 1 ω 3 (I 2 I 1 )+ω 3 ω 2 ω 1 (I 3 I 2 )+ω 1 ω 3 ω 2 (I 1 I 3 ) (69) The equations of motion are Euler s equations: where Γ is the torque in the body-fixed frame. I 1 ω 1 +ω 3 ω 2 (I 3 I 2 ) = Γ 1 (70) I 2 ω 2 +ω 1 ω 3 (I 1 I 3 ) = Γ 2 (71) I 3 ω 3 +ω 2 ω 1 (I 2 I 1 ) = Γ 3 (72) Example 2: Consider a top (I 1 = I 2 I 3 ), where the calculational frame has the top spinning along the 3-axis (inclined at θ from z axis) with the 1-axis horizontal. We then have: ) Ω = ( θ, φsinθ, φcosθ (73) ω = ( θ, φsin(θ), φcos(θ)+ ψ) (74) (cf. rolling.pdf Eqs. (13-16)) Note: ω = Ω+(0, 0, ω 3 Ω 3 ) So φ = ω +Ω ω = ω +Ω (0, 0, ω 3 Ω 3 ) = ω +(ω 2, ω 1, 0){ω 3 Ω 3 } (75) G rot = 1 2 = 1 2 = 1 2 ( I1 ( ω ω2 2 )+I ) 3 ω 3 2 +I1 ω 1 ω 2 {ω 3 Ω 3 } I 1 ω 2 ω 1 {ω 3 Ω 3 }+ ω 3 ω 2 ω 1 (I 3 I 1 )+ω 1 ω 3 ω 2 (I 1 I 3 ) (76) ( I1 ( ω ω2 2 )+I ) 3 ω 3 2 +I1 Ω 3 ( ω 2 ω 1 ω 1 ω 2 )+I 3 ω 3 (ω 2 ω 1 ω 1 ω 2 ) (77) ( ) I1 ( ω ω 2)+I 2 3 ω 3 2 +(I1 Ω 3 I 3 ω 3 )( ω 2 ω 1 ω 1 ω 2 ) (78) The resulting equations of motion: I 1 ω 1 (I 1 Ω 3 I 3 ω 3 )ω 2 = Mglsinθ (79) I 1 ω 2 +(I 1 Ω 3 I 3 ω 3 )ω 1 = 0 (80) I 3 ω 3 = 0 (81) Fromthe last equation we conclude ω 3 =constant, which relates to p ψ =constant in the usual treatment. Substituting our Euler angle expression for ω into the second equation: I 1 ( φsinθ +2 φcosθ θ) I 3 ω 3 θ = 0 (82) d { } I 1 φsin 2 θ +I 3 ω 3 cosθ = 0 (83) dt I 1 φsin 2 θ+i 3 ω 3 cosθ p φ = constant (84)

8 where in the second line we used sinθ as an integrating factor. If the first equation is multiplied by ω 1 and the second by ω 2 and the two are added: I 1 (ω 1 ω 1 +ω 2 ω 2 ) = Mglsinθ θ (85) { } d 1 dt 2 I 1 (ω1 2 +ω2)+mglcosθ 2 = 0 (86) 1 2 I 1 (ω1 2 +ω2 2 )+Mglcosθ E = constant (87) which is conservation of energy (aside from the conserved energy associated with ω 3 ). The above are the usual starting point to describe gyroscopic motion. Example 3: In the case of a sphere I 1 = I 2 = I 3 I; any coordinate system will have principal axes aligned with coordinate axes so use the initial inertial frame (Ω = 0), then: G rot = 1 2 I( ω ω2 2 + ) ω2 3 (88) Consider the rolling without slipping motion of a sphere on a plane tilted at an angle α with they coordinatepointinguphill, z perpendiculartotheplane(seefig.3). SoV = Mgsinα y. We have: G = 1 2 M ( ẍ 2 +ÿ 2) I ( ω ω2 2 + ) ω2 3 (89) The rolling without slipping condition gives: The resulting equations of motion: v CM = Rẑ ω (90) a CM = Rẑ ω (91) ẍ CM = R ω 2 (92) ÿ CM = R ω 1 (93) We can compare this to the Newtonian solution: G = 1 2 (M +I/R2 ) ( ẍ 2 +ÿ 2) I ω2 3 (94) (M +I/R 2 ) ẍ = 0 (95) (M +I/R 2 ) ÿ = Mgsinα (96) I ω 3 = 0 (97) Ma CM = Mgsinα ŷ+f (98) a CM = Rẑ ω (99) I ω = Rẑ F (100) = Rẑ (Ma CM +MRgsinα ŷ) (101) = MR 2 (ẑ ẑ ω)+mrgsinα ˆx (102) = MR 2 ( ω 3 ẑ ω)+mrgsinα ˆx (103) (I +MR 2 ) ω 1 = MgRsinα (104) (I +MR 2 ) ω 2 = 0 (105) I ω 3 = 0 (106)