A load of balls in Newton s cradle

Size: px

Start display at page:

Download "A load of balls in Newton s cradle"

Adelia York
5 years ago
Views:

1 A load of balls in Newton s cradle or Fragmentation of a line of balls by an impact John Hinch DAMTP, Cambridge March 26, 2010

2 t = 0 V=1 others at rest...

3 t = 0 V=1 others at rest... Contacts: linear springs in compression, zero force in tension No non-dimensional groups ẍ n = (x n+1 x n ) (x n x n 1 )

4 t = 0 V=1 others at rest... Contacts: linear springs in compression, zero force in tension No non-dimensional groups ẍ n = (x n+1 x n ) (x n x n 1 ) Granular media?

5 Position of particles 10 n + x n t Few forwards Most backwards

6 Position of particles 10 n + x n t Few forwards Most backwards If in contact: ẍ n = x n+1 2x n + x n 1 Continuum approximation: x tt = x nn, with wave speed 1.

7 Position of particles 10 n + x n t Few forwards Most backwards If in contact: ẍ n = x n+1 2x n + x n 1 Continuum approximation: x tt = x nn, with wave speed 1. Non-continuum effect: ẋ n (t = ) < 0

8 Questions How many fly off at the far end? At what velocities?

9 Questions How many fly off at the far end? At what velocities? How many rebound? At what velocities?

10 Questions How many fly off at the far end? At what velocities? How many rebound? At what velocities? Simple mechanics Simple questions Answers more complicated

11 Rebound velocities Chain of 25 particle, 20 rebound 0.1 ẋ n ( ) n n 3/4?

12 Rebound velocities Chain of 25 particle, 20 rebound 0.1 ẋ n ( ) n n 3/4? If so Rebound energy n 3/2 is finite Rebound momentum n 3/4 is infinite

13 Rebound velocities Chain of 25 particle, 20 rebound 0.1 ẋ n ( ) n n 3/4? If so Rebound energy n 3/2 is finite Rebound momentum n 3/4 is infinite But why n 3/4? Not diffusion.

14 Key: propagation of an impulse wave 0.5 t = 13.5, 24.5, 35.2, ẋ n n

15 Key: propagation of an impulse wave 0.5 t = 13.5, 24.5, 35.2, ẋ n n Peak velocity decreases slowly. Width of pulse increases slowly. How? How?

16 Spreading wave which conserves energy Slowly varying amplitude and wavelength of propagation wave of constant form f (.) ( ) n t x n = a(t) f λ(t)

17 Spreading wave which conserves energy Slowly varying amplitude and wavelength of propagation wave of constant form f (.) ( ) n t x n = a(t) f λ(t) Then ẋ n a λ f Kinetic energy (potential energy similar) ẋ 2 n a2 λ 2 λ

18 Spreading wave which conserves energy Slowly varying amplitude and wavelength of propagation wave of constant form f (.) ( ) n t x n = a(t) f λ(t) Then ẋ n a λ f Kinetic energy (potential energy similar) ẋ 2 n a2 λ 2 λ Conservation gives λ a 2, and so ẋ n a 1

19 Spreading wave which conserves energy Slowly varying amplitude and wavelength of propagation wave of constant form f (.) ( ) n t x n = a(t) f λ(t) Then ẋ n a λ f Kinetic energy (potential energy similar) ẋ 2 n a2 λ 2 λ Conservation gives λ a 2, and so ẋ n a 1 Replot for different t, ẋ n a against (n n 1 )/a 2, where n 1 is last in contact, and a = x n1.

20 Key: propagation of an impulse wave 0.5 t = 13.5, 24.5, 35.2, ẋ n n Replot for different t, ẋ n a against (n n 1 )/a 2,

21 Self similar impulse wave 0.7 t = 13.5, , 35.2, ẋ n a (n n )/a2 Here a = x n1 at last in contact.

22 Self similar impulse wave 0.7 t = 13.5, , 35.2, ẋ n a (n n )/a2 Here a = x n1 at last in contact. But how to predict a(t)?

23 Scaling for spreading wave : λ(t)? If touching ẍ n = x n+1 2x n + x n 1 Second approximation for continuum limit with numerical diffusion. x tt = x nn x nnnn

24 Scaling for spreading wave : λ(t)? If touching ẍ n = x n+1 2x n + x n 1 Second approximation for continuum limit with numerical diffusion. x tt = x nn x nnnn Transform to moving coordinate N = n t x tt + 2x Nt = 1 12 x NNNN

25 Scaling for spreading wave : λ(t)? If touching ẍ n = x n+1 2x n + x n 1 Second approximation for continuum limit with numerical diffusion. x tt = x nn x nnnn Transform to moving coordinate N = n t x tt + 2x Nt = 1 12 x NNNN To balance last two terms, use similarity variable N/t 1/3. Hence λ t 1/3

26 Results for spreading wave wavelength λ t 1/3 displacements x n = a t 1/6 velocities ẋ n = a/λ t 1/6

27 Results for spreading wave Forward moving momentum wavelength λ t 1/3 displacements x n = a t 1/6 velocities ẋ n = a/λ t 1/6 P = ẋ n (a/λ)λ t 1/6

28 Results for spreading wave Forward moving momentum wavelength λ t 1/3 displacements x n = a t 1/6 velocities ẋ n = a/λ t 1/6 P = ẋ n (a/λ)λ t 1/6 Rate of ejecting momentum backwards in rocket effect ẋ n ( ) t = Ṗ t 5/6

29 Results for spreading wave Forward moving momentum wavelength λ t 1/3 displacements x n = a t 1/6 velocities ẋ n = a/λ t 1/6 P = ẋ n (a/λ)λ t 1/6 Rate of ejecting momentum backwards in rocket effect ẋ n ( ) t = Ṗ t 5/6 t time between particles rebounding = 1 (wave speed = 1) Hence rebound velocities ẋ n ( ) t 5/6 = n 5/6

30 Rebound velocities ẋ n /n 3/ ẋ n /n 5/ /n ẋ n ( ) = 0.158n 5/6

31 Similarity solution Try x(n, t) = t 1/6 f ( ) ξ = n t t 1/3 in x tt = x nn x nnnn

32 Similarity solution Try x(n, t) = t 1/6 f ( ) ξ = n t t 1/3 in x tt = x nn x nnnn So f = 8ξf 4f

33 Similarity solution Try x(n, t) = t 1/6 f ( ) ξ = n t t 1/3 in x tt = x nn x nnnn So f = 8ξf 4f ẋ n a (n n1 1 4 )/a2

34 Similarity solution Try x(n, t) = t 1/6 f ( ) ξ = n t t 1/3 in x tt = x nn x nnnn So f = 8ξf 4f ẋ n a (n n1 1 4 )/a2 Solution f = ξ Ai 2 ( 2 1/3 y) dy

35 The 1 4 shift Near to back of the wave ξ = ξ 0 f f (ξ 0 ) ( (ξ ξ 0) ) Correction for ejected velocities at t 5/6 x n (t) t 1/6 f (ξ) + t 1/2 β(ξ ξ 0 ) Ball n detaches at t n where ξ = ξ 0 + δ if x n (t n ) = x n+1 (t n ), i.e. t 5/6 [ f (ξ 0 ) ( 2δ 2 + 2δ 3) 2 ] β = 0 and detaches with known velocity 1 6 f (ξ 0)t 5/6 = ẋ n = t 5/6 [ f (ξ 0 )2δ 2 β ] Hence δ = 1 4.

36 Finite chain of N: number fly off and their velocities When wave reaches end at t = N. width of wave 1.5N 1/3 and velocities 1.4N 1/6 and less.

37 Finite chain of N: number fly off and their velocities When wave reaches end at t = N. width of wave 1.5N 1/3 and velocities 1.4N 1/6 and less. N = 100, 200 +, 400, V n N 1/ (N n)n 1/3

38 Questions: Answers How many fly off at the far end?

39 Questions: Answers How many fly off at the far end? 1.5N 1/3

40 Questions: Answers How many fly off at the far end? 1.5N 1/3 At what velocities? V N = 1.4N 1/6

41 Questions: Answers How many fly off at the far end? 1.5N 1/3 At what velocities? V N = 1.4N 1/6 How many rebound? Most At what velocities? V n = 0.16n 5/6

42 Questions: Answers How many fly off at the far end? 1.5N 1/3 At what velocities? V N = 1.4N 1/6 How many rebound? Most At what velocities? V n = 0.16n 5/6 Simple mechanics Simple questions Answers more complicated

43 Questions: Answers How many fly off at the far end? 1.5N 1/3 At what velocities? V N = 1.4N 1/6 How many rebound? Most At what velocities? V n = 0.16n 5/6 Simple mechanics Simple questions Answers more complicated For linear force law. Next nonlinear Hertz contacts

44 Hertz contacts R a δ Radius R Contact radius a Overlap δ

45 Hertz contacts R a δ Radius R Contact radius a Overlap δ Geometry: a = 2Rδ

46 Hertz contacts R a δ Radius R Contact radius a Overlap δ Geometry: a = 2Rδ Strain = δ a

47 Hertz contacts R a δ Radius R Contact radius a Overlap δ Geometry: a = 2Rδ Strain = δ a Stress = E δ a, elastic modulus E

48 Hertz contacts R a δ Radius R Contact radius a Overlap δ Geometry: a = 2Rδ Strain = δ a Stress = E δ a, elastic modulus E Force = πa 2 E δ a = 2E 3(1 ν 2 ) R1/2 δ 3/2

49 Impulse wave propagating down a Hertzian chain x n n No spreading: nonlinearity balances diffusion

50 Impulse wave propagating down a Hertzian chain x n n No spreading: nonlinearity balances diffusion Two fly off end: V N = 0.984, V N 1 = 0.149, (V N 2 = 0.004, V N 3 = 10 5, V n 4 = )

51 Rebound velocities for Hertzian chain x n ( ) e n x n (t = ) = 0.084e 0.55n

52 Nesterenko s (1984) solitary wave (long wave approx) ẍ = D [ (Dx) 3/2] where Du = u n+ 1 u 2 n 1 2 u n u 24 n 3

53 Nesterenko s (1984) solitary wave (long wave approx) ẍ = D [ (Dx) 3/2] where Du = u n+ 1 u 2 n 1 2 Travelling wave solution x n (t) = f (ξ = n V t) ( ) V 2 f = f 3/ f 3/ f 1/2 f u n u 24 n 3

54 Nesterenko s (1984) solitary wave (long wave approx) ẍ = D [ (Dx) 3/2] where Du = u n+ 1 u 2 n 1 2 Travelling wave solution x n (t) = f (ξ = n V t) with solution ( ) V 2 f = f 3/ f 3/ f 1/2 f V 2 = 4 5 A1/2, f = A sin ξ 0 ξ π u n u 24 n 3

55 Nesterenko s (1984) solitary wave (long wave approx) ẍ = D [ (Dx) 3/2] where Du = u n+ 1 u 2 n 1 2 Travelling wave solution x n (t) = f (ξ = n V t) with solution ( ) V 2 f = f 3/ f 3/ f 1/2 f V 2 = 4 5 A1/2, f = A sin ξ 0 ξ π u n u 24 n 3 Set A = for energy = Predict V = Max ẋ n = [x n ] = Numerical V = Max ẋ n = [x n ] = 1.354

56 Nesterenko s (1984) solitary wave (long wave approx) ẍ = D [ (Dx) 3/2] where Du = u n+ 1 u 2 n 1 2 Travelling wave solution x n (t) = f (ξ = n V t) with solution ( ) V 2 f = f 3/ f 3/ f 1/2 f V 2 = 4 5 A1/2, f = A sin ξ 0 ξ π u n u 24 n 3 Set A = for energy = Predict V = Max ẋ n = [x n ] = Numerical V = Max ẋ n = [x n ] = Wave not so long with just 4 balls

57 Finished?

58 Impact by two Two solitary waves x n n

59 Impact by two Two solitary waves x n n Balls fly of end: from first wave at 1.234, 0.186, from second wave at 0.647, 0.090

60 Impact by three Three solitary waves x n From first wave at 1.336, 0.203, second at 0.951, 0.141, third at 0.469, n

61 Impact by K 2K fly off at far end 1.6 K = 3, 6 +, 9, ẋ n ( ) (N n )/K Open question

62 Impact by K 1 2K rebound at velocities > K = 10, 20 +, 30, 40, 60, 80 1 ẋ n ( ) n/k Open question

63 Newton s cradle Final velocities of position n after impact by K Extra forward n K = 1 K = 2 K = 3 K =

64 Granular media

65 Granular media Barchan dunes Aeolian sand transport

66 Splash experiment in Rennes Oger & Valance Apparatus

67 Splash experiment in Rennes Oger & Valance Apparatus Coefficient of restitution

68 Splash experiment in Rennes Splash of particles Oger & Valance

69 Splash experiment in Rennes Splash of particles Oger & Valance V = 2 gd, all V i, θ i

70 Splash experiment in Rennes Splash of particles Oger & Valance V = 2 gd, all V i, θ i # ejected V i

71 Spin in billiards Before V = 1 0

72 Spin in billiards Before After V =

73 Spin in billiards Before After V = But spinning when rolling

74 Spin in billiards Before V = 1 0 ω = 1 0

75 Spin in billiards Before After if µ > 0.2 V = 1 0 ω =

76 Spin in billiards Before After if µ > 0.2 Later, rolling again V = 1 0 ω =

77 Spin in billiards Before After if µ > 0.2 Later, rolling again V = 1 0 ω = Energy 1 2 mv 2 = (86%) (35%)

78 Spin in billiards Before After if µ > 0.2 Later, rolling again V = 1 0 ω = Energy 1 2 mv 2 = (86%) (35%) Energy lost through spinning rather than restitution

79 Newton s cradle Final velocities of position n after impact by K Extra forward n K = 1 K = 2 K = 3 K =

-: MECHANICS:- 2. Motion in one dimension & 3. Newton s laws of motion

-: MECHANICS:- 2. Motion in one dimension & 3. Newton s laws of motion -: MECHANICS:- Chapters: 1.Units and Dimensions, 2. Motion in one dimension & 3. Newton s laws of motion 01. The quantities L/R and RC (where L, C and R stand for inductance, capacitance and resistance