Lecture 8: Rolling Constraints II

Transcription

1 Lecture 8: Rolling Constraints II Generaliza3ons and review Nonhorizontal surfaces Coin rolling on a slope Small sphere rolling on a larger sphere s surface Hoop rolling inside a hoop 1

2 What can we say in general? vector from the iner<al origin to the center of mass ( ( ) 2 + I ZZ ( ψ + φ cosθ ) 2 ) L = 1 2 I θ cosψ + φ sinθ sinψ XX ( ) 2 + I YY θ sinψ + φ sinθ cosψ ( z 2 ) mg R m x 2 + y 2 + ω = ( θ cosφ + ψ sinθ sinφ)i + ( θ sinφ ψ sinθ cosφ)j + ( φ + ψ cosθ)k v = ω r vector from the contact point to the center of mass 2

3 We can restrict our aaen3on to axisymmetric wheels and we can choose K to be parallel to the axle without loss of generality ( ( ) 2 + B( θ sinψ + φ sinθ cosψ ) 2 + C( ψ + φ cosθ ) 2 ) L = 1 2 A θ cosψ + φ sinθ sinψ ( z 2 ) mg R m x 2 + y 2 + ( ( ) + C( ψ + φ cosθ ) 2 ) m ( x 2 + y 2 + z 2 ) mg R L = 1 2 A θ 2 + φ 2 sin 2 θ mgz 3

4 If we don t put in any simple holonomic constraint (which we ohen can do) x y z q = φ θ ψ 4

5 We know v and ω in terms of q any difficulty will arise from r v = ω r r = aj 2 Actually, it s something of a ques3on as to where the difficul3es will arise in general This will depend on the surface flat, horizontal surface we ve been doing this flat surface we can do this today general surface: z = f(x, y) this can be done for a rolling sphere 5

6 d dt d dt d dt d dt L j + mg x i x = λ j C 1 L j y i + mg y = λ j C 2 L j + mg z i z = λ j C 3 d dt L j φ i = λ j C 4 L θ L θ = λ C j j 5 d dt L j ψ = λ j C 6 6

7 We have the usual Euler Lagrange equa3ons d dt L q i L = λ q i j C i j and we can write out the six equa3ons 7

8 The key to the problem lies in the constraint matrix The analysis is preay simple for flat surfaces, whether horizontal or 3lted Let s play with the 3lted surface Choose a Cartesian iner3al system such that i and j lie in the 3lted plane and choose i to to be horizontal, so that j points down hill (This is a rota3on of the usual system about i) 8

9 y = cosα y + sinαz' z = cosα z sinα y 9

10 so the poten3al energy in the primed coordinates is ( ) V = mg cosα z sinα y the kine3c energy is unchanged We can go forward from here exactly as before everything is the same except for gravity 10

11 This has the same body system as before but the angle θ can vary (it s equal to 0.65π here) r remains equal to aj 2 but we need the whole ω 11

12 ?? Let s look at a rolling coin on a 3lted surface in Mathema3ca 12

13 Curved surfaces Spherical surface: spherical ball on a sphere Two d surface: wheel inside a wheel General surface 13

14 Spherical ball on a sphere R a holonomic constraint x 2 + y 2 + z 2 = ( R + a) 2 14

15 The Lagrangian simplifies because of the spherical symmetry ( ) m ( x 2 + y 2 + z 2 ) mgz L = 1 2 A θ 2 + φ 2 + ψ φ ψ cosθ We have a constraint, which we can parameterize x 2 + y 2 + z 2 = ( R + a) 2 x = ( R + a)sinξ cosχ, y = ( R + a)sinξ sinχ, z = ( R + a)cosξ 15

16 which transforms the Lagrangian ( ) L = 1 2 A θ 2 + φ 2 + ψ φ ψ cosθ ( ) m R + a ( ξ 2 + χ 2 ( 1 cos( 2ξ) )) m( R + a)gcos ξ ( ) φ θ We can now assign generalized coordinates q = ψ ξ χ 16

17 We have rolling constraints ω is unchanged, and r is as shown on the figure and we recalculate v ω = ( θ cosφ + ψ sinθ sinφ)i + ( θ sinφ ψ sinθ cosφ)j + ( φ + ψ cosθ)k ( ) v = a + R asinξ cosχ r C = asinξ sinχ acosξ ξ cosξ cosχ χ sinξ sinχ ξ cosξ sinχ + χ cosξ sinχ ξ sinξ 17

18 The rolling constraint appears to have three components but the normal component has already been sa3sfied v ω r = 0 The normal is parallel to r, so I need two tangen3al vectors ( k r) v ω r ( k k r ) v ω r ( ) = 0 ( ) ( ) = 0 18

19 We have the usual Euler Lagrange equa3ons d dt L q i L = λ q i j C i j and we can write out the five equa3ons 19

20 d dt d dt d dt L j = λ j C 1 φ L θ L θ = λ C j j 2 d dt L ξ L ξ = λ C j j 4 d dt L j ψ = λ j C 3 L j = λ j C 5 χ 20

21 The constraint matrix is a 2 sin 2 χ 1 2 a2 sin2χ cos( φ ξ) a 2 sinχ sinθ cos( χ θ)sin( φ ξ) 0 R + a a2 sin2χ sin( φ ξ) ( )sinχ 1 2 a2 sin2χ sinθ cos φ ξ ( ) 1 2 a R + a ( )sin2χ 0 The last two Euler Lagrange equa3ons are suitable for elimina3ng the Lagrange mul3pliers 21

22 AHer some algebra λ 2 = m R + a a ( ) λ 1 = m R + a asinχ 2cosχ χ ξ + sinχ χ ξ 2 m R + a acosχ sinχ χ + mg acosχ I have three remaining Euler Lagrange equa3ons and two constraint equa3ons that I need to differen3ate to give me five equa3ons for the generalized coordinates We need to go to Mathema3ca to see how this goes. QUESTIONS FIRST?? 22

23 Wheel within a wheel Treat them both as hoops radii r 1 > r 2 y 2 = ( r 1 r 2 )sinχ z 2 = r 1 ( r 1 r 2 )cosχ χ 23

24 We have holonomic constraints Put us in two dimensions x 1 = 0 = x 2 φ 1 = π 2 = φ 2 θ 1 = π 2 = θ 2 realize that z 1 = r 1 24

25 We have an interes3ng connec3vity constraint define the posi3on of the small wheel in terms of the angle χ y 2 = y 1 + ( r 1 r 2 )sinχ, z 2 = r 1 + ( r 1 r 2 )cosχ Pueng all this in gives us a Lagrangian L = 1 ( 2 m + m 1 2) y m r 1 12 ψ m r 2 22 ψ m r 2( r 1 2) 2 χ m r 2( r 1 2) 2 cosχ + gm 2 ( r 1 r 2 )cosχ gr 1 ( m 1 + m 2 ) χ y 1 25

26 y 1 ψ We define a vector of generalized coordinates q = 1 ψ 2 χ 26

27 There will be two nonholonomic constraints y 1 = r 1 ψ 1, ( r 1 r 2 ) χ = r 2 ψ 2 The corresponding constraint matrix is C = 1 r r 2 r 1 r 2 27

28 The second and third Euler Lagrange equa3ons are fairly simple so I will use those to find the two Lagrange mul3pliers λ 1 = m 1 r 1 ψ 1, λ 2 = m 2 r 2 ψ 2 To solve the problem we use the first and fourth Euler Lagrange equa3ons and the differen3ated constraints The solu3on is numerical and we need to go to Mathema3ca to look at it. QUESTIONS FIRST?? 28

29 That s All Folks 29