Lecture 7 Rolling Constraints

Transcription

1 Lecture 7 Rolling Constraints The most common, and most important nonholonomic constraints They cannot be wri5en in terms of the variables alone you must include some deriva9ves The resul9ng differen9al is not integrable 1

2 The major assump9on/simplifica9on is point contact This is true of a rigid hoop on a rigid surface or a rigid sphere on a rigid surface I will also generally assume that the surface is horizontal so that gravity is normal to the surface 2

3 rigid hoop on a rigid surface 3

4 rigid sphere on a rigid surface 4

5 It s not a bad approxima9on for a well inflated bicycle or motorcycle 9re on hard pavement It s not so good for an automobile 9re, which has a pre5y big patch on the ground Cylindrical wheels, like those on grocery carts and children s toys must slip over parts of their surfaces when turning We aren t going to care about any of this We re going to suppose that all the wheels we consider have point contact 5

6 Two fundamental ideas: The speed of a point rota9ng about a fixed point is v = ω r The contact point between a wheel and the ground is not moving 6

7 For a sphere of radius a r = ak ω y v = aω k = a ω x 0 This is iner9al, so we have ω y v = a ω x 0 θ sinφ ψ cosφ sinθ = a θ cosφ ψ sinφ sinθ 0 7

8 We can write the constraints as x = a( θ sinφ ψ cosφ sinθ), y = a θ cosφ + ψ sinφ sinθ ( ) Of course, we also have the simple constraint for this problem: z = a. 8

9 An erect wheel has the same r If we choose to let the axle be the K body coordinate then θ = ±π/2, fixed, and the constraint becomes ω y v = a ω x 0 θ sinφ ψ cosφ sinθ = a θ cosφ ψ sinφ sinθ 0 ψ cosφ v = a ψ sinφ 0 x = aψ cosφ, y = aψ sinφ The general wheel is fancier; I ll get to that shortly BUT FIRST 9

10 How do we impose these? The classical method is through Lagrange mul9pliers, and that s what we ll do today (There s an alternate method that we will see later in the semester) To get at this we need to go back to first principles and it helps to note that we can write the constraints in matrix form 10

11 x = a( θ sinφ + ψ cosφ sinθ), y = a θ cosφ + ψ sinφ sinθ x y z q = φ θ ψ ( ) = 0 = q 1 a q 5 sinq 4 + q 6 cosq 4 sinq 5 ( ) ( ) q 2 + a q 5 cosq 4 + q 6 sinq 4 sinq 5 11

12 ( ) = 0 = q 1 a q 5 sinq 4 + q 6 cosq 4 sinq 5 ( ) q 2 + a q 5 cosq 4 + q 6 sinq 4 sinq asinq 4 acosq 4 sinq acosq 4 asinq 4 sinq 5 q 1 q 2 q 3 q 4 q 5 q 6 = 0 C q = 0 C ji q j 12

13 We can view the 9me deriva9ve of q as a surrogate for δq C j i δq j = 0 Now let s go back to lecture two and the ac9on integral it is sta9onary if di( η) dη = 0 = t L * q d L * 2 k dt q k qk η dt = t 1 t 2 t 1 L * q k d dt L * q k δq k dt Consider the final version 13

14 t L * q d L * 2 k dt q k δq k dt = 0 t 1 and let s add mul9ples of the constraints to this They are zero, so there s no problem t L * q d L * 2 k dt q k δq k dt = 0 t 1 mul9ply each row by its own constant λ i C j i δq j = 0 change the dummy index j to k λ i C k i δq k = 0 14

15 t L * q d L * i 2 k dt q k + λ i C k δq k dt = 0 t 1 drop the asterisk and incorporate external forces directly the constrained Euler Lagrange equa9ons d dt L q k L q = λ C i k i k + Q k I have added as many variables as there are nonholonomic constraints I need as many extra equa9ons, which are the constraints themselves 15

16 d dt L q k L q = λ C i k i k + Q k We have a physical interpreta9on of the Lagrange mul9pliers They are the forces (and torques) of constraint what the world does to make the system conform How do we actually apply this? The explana9on is on pp of chapter three, perhaps a bit telegraphically I d like to go through the erect coin, example 3.5, pp

17 First let s see how we can set up a wheel I like the axle to be the K axis; let s see how we can erect the wheel Star9ng with all Euler angles equal to zero 17

18 Pick a direc9on by adjus9ng φ 18

19 erect the wheel with θ = π/2 (the text example uses + π/2) 19

20 r = aj 2, ω = ψ K ω has a k component, but it s perpendicular to r cosφ v = ω r = ψ K aj 2 = aψ I 2 = aψ sinφ 0 x aψ cosφ = 0 = y aψ sinφ Variables are x, y, φ and ψ z and θ are fixed 20

21 The Lagrangian is simple L = 1 2 m x 2 + ( y 2 ) A φ C ψ 2 I ve lek out the gravity term because the center of mass cannot move ver9cally The constraint matrix is C = acosφ asinφ 21

22 The product { } acosφ λc = λ 1 λ asinφ = λ 1 λ 2 0 a λ 1 cosφ + λ 2 sinφ { ( )} The Lagrange equa9ons are then m x = λ 1, m y = λ 2, φ = 0, ψ = a ( C λ 1cosφ + λ 2 sinφ) and the constraints are x a ψ cosφ = 0 = y a ψ sinφ 22

23 So I have six equa9ons in six unknowns: the variables and the mul9pliers Solve the first two dynamical equa9ons for the mul9pliers λ 1 = m x, λ 2 = m y The remaining dynamical equa9ons become φ = 0, ψ = am C ( x cosφ + y sinφ ) and I can use the constraints to eliminate the accelera9on terms 23

24 x aψ cosφ = 0 = y aψ sinφ x = aψ cosφ aψ φ sinφ, y = aψ sinφ + aψ φ cosφ subs9tute all this back in ψ = am C (( ψ cosφ ψ φ sinφ )cosφ + ( ψ sinφ + ψ φ cosφ )sinφ) = am ψ C ψ = 0 so, we have an exact solu9on for the two angles φ = φ 0 + φ 0 t, ψ =ψ 0 + ψ 0 t 24

25 and we can subs9tute this into the constraints and integrate to get the posi9ons x = aψ 0 cos( φ 0 + φ 0 t), y = aψ 0 sin( φ 0 + φ 0 t) ψ x = a 0 sin φ 0 + ψ ( φ 0 t), y = a 0 cos φ φ φ 0 t φ 0 ( ) The coin rolls around in a circle of constant radius, which can be infinite, in which case x = a ψ 0 t sinφ 0, y = a ψ 0 t cosφ 0 25

26 Suppose we consider a general rolling coin, one not held ver9cal The book doesn t deal with this un9l chapter five, but we can certainly set it up here. It s a hard problem and does not have a closed form solu9on 26

27 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 27

28 cosθ sinφ cosφθ + sinφ sinθψ J 2 = cosθ cosφ, ω = sinφ θ cosφ sinθψ sinθ φ + cosθψ and, aker some simplifica9on x v = y z cosφ cosθ φ sinφ sinθ θ + cosφψ = ω aj 2 = a sinφ cosθ φ + cosφ sinθ θ + sinφψ cosθ θ 28

29 ( ) = 0 ( φ + cosφ sinθ θ + sinφ ψ ) = 0 x a cosφ cosθ φ sinφ sinθ θ + cosφ ψ y a sinφ cosθ z + acosθ θ = 0 The last one is actually integrable, but I won t do that We have a constraint matrix that appears to disagree with the one on p. 15 of chapter five. The difference stems from the assump9on in the book that θ 0 whereas in the picture here, θ 0. Choose the one that fits your idea of the ini9al condi9ons. 29

30 I will use the book s picture, and I will go to Mathema9ca to analyze this system 30