PHY1004W 2014 Modern Mechanics Part 4

Transcription

1 PHY1004W 014 Modern Mechanics Part 4 Prof Andy Buffler Room 503 RW James andy.buffler@uct.ac.za These slides have benefited from significant guidance from the notes of Roger Fearick (UCT Physics) and the resources provided by the textbook authors. These slides are available on

2 M&I Chapter 9 Multiparticle Systems

3 M&I 3E 9.1 The motion of the centre of mass Two basic principles for forces acting on particles: Momentum principle: dp dt sys F net... leads to: conservation of momentum. Energy principle: Esystem Wsurr other energy transfers... leads to: conservation of energy. We want to extend these to multiparticle systems. 3

4 Centre of mass We define the centre of mass for a system of particles as: r CM m r M i i total total mass = mi x CM mx M i i total my ; i i ycm ; M total z CM mz M i i total then r x ˆi y ˆj z kˆ CM CM CM CM For an extended body ( a continuous mass distribution) : r CM 1 M total rdm 1 ˆ 1 ˆ 1 xdm i ydm j zdm kˆ M M M total total total 4

5 Velocity of the centre of mass The importance of centre of mass lies in the fact that the motion of the centre of mass for a system of particles (or extended body) can often be described simply since it is related to net force on the system. Consider n particles of total mass M which remains constant. Then: MtotalrCM miri M total dr CM dt m i dri dt p M v mv sys total CM i i Velocity of centre of mass Velocity of i th particle of mass m 5

6 6

7 Application: Pull on two hockey pucks One puck is pulled by a string attached to its centre. The other puck is pulled by a string wrapped around its edge, which unrolls as the puck is pulled. See twopucks.py 7

8 M&I 3E 9. Separation of multiparticle system energy For a complex system Ktot Ktrans Krot Kvib K M v 1 trans total CM The total energy of a translating, rotating, vibrating oxygen molecule can be written as E K K K k s mc 1 tot trans rot vib s K relative (to centre of mass) p M sys total 8

9 See krel.py Which object has the greater total momentum (magnitude)? (1) Top object (blue) () Bottom object (red) (3) Their total momentum is the same 9

10 See krel.py Which object has the greater total kinetic energy? (1) Top object (blue) () Bottom object (red) (3) Their total kinetic energy is the same 10

11 See krel.py Which object has the greater translational kinetic energy (K TRANS )? (1) Top object (blue) () Bottom object (red) (3) Their total translational kinetic energy is the same 11

12 See RotateVibrateTranslate.py CASE = 0: left ball fixed -- think about vcm & vrel CASE = 1: v_cm=0, vibration, zero external force CASE = : initially in motion, rotation + translation, zero external force CASE = 3: initially in motion, rot + vib + trans, zero external force CASE = 4: weaker spring, 0 initial stretch, const force on b CASE = 5: initially slightly stretched, at angle, const force CASE = 6: vibration perp to translation; zero net force 1

13 Gravitational energy of a multiparticle system How does the centre of mass influence energy? e.g. the change in potential energy near the surface of the Earth is... U gm y gm y gm y g g( m y m y m y...) gmy CM near the Earth s surface (the centre of mass is sometimes called the centre of gravity in this context). 13

14 M&I 3E 9.3 Rotational kinetic energy Consider a rigid disk rotating at constant period T around a fixed axis of rotation. Then the angular speed T (radians per second) The translational speed of an atom on the disk can be expressed as ri vi ri ri T T 14

15 Now consider four point masses held rigid as shown, rotating at constant... 1 Krot m r m r m r m r m r m r m r m r I where I m r m r m r m r is the moment of inertia In general for a collection of i point particles I mir i i 15

16 Moment of inertia of a thin rod Slice up the rod of length L and total mass M into N segments. L Since x N M M x then M M N L x L Then x M I M x n M xn xnx L L Total moment of inertia M I I x x N N n n1 L n1 Let x 0 then L M 1 I x dx... ML 1 L L 16

17 x Moment of inertia of a ring (bicycle wheel) Slice up the ring of radius R and total mass M into N segments. Since x R N and x R then Then M M M M N R x I M R M R Let 0 then MR MR I d MR 0 17

18 Moment of inertia of a thin disk R Slice up the disk of radius R and total mass M and area A into N concentric rings, each of thickness r. r r Then M M M A r r A R Then M I M r r r r R Let r 0 then R 4 3 M R R 4 0 M 1 I r dr MR R 18

19 So we can see that the moment of inertia I is a measure of the rotational inertia of a body, and plays the same role for rotational motion that the mass does for translational motion. We can see that I depends both on the mass of the body and how that mass is distributed. Moments of inertia can be calculated for any shape of body for rotation about any axis from the formula I r dm 19

20 Some moments of inertia for rigid bodies axis of rotation a thin rod of length L a thin rod of length L a thin hoop of radius R a solid cylinder of radius R I a solid sphere of radius R 1 1 ML I 1 3 a hollow sphere of radius R ML I a thin plate MR I 1 MR a hollow cylinder of inner radius R 1 b a I 5 MR I 3 MR I M a b 1 1 ( ) I M R R 1 ( 1 ) 0

21 Demonstration A solid cylinder and a hollow cylinder are raced down an incline. If the outer radii are the same and the masses are the same, then which reaches the bottom first, and why? 1

22 Rotational Kinetic Energy K m v m v m v rot CM m1r 1 mr m3r mr K ( ) rot 1 I For a rigid body rotating about fixed axis A body that rotates while its CM undergoes translational motion will have K Total : K K K Mv I 1 1 Total trans rot CM CM

23 y h 0 v = 0 v Consider a solid sphere (mass m and radius r 0 ) rolling without slipping down an incline of height h. 1 1 Total energy at height y is K trans + K rot + U g mv I mgy Total energy at the top (v = 0, = 0) = m g h Total energy at the bottom mv I 1 1 I for a solid sphere for axis of rotation through centre = and since v vtangential r0 Conservation of energy: giving... v gh r0 mgh mv ( mr ) v mr If the ball slipped (no friction): v gh v gh roll 7 3

24 Rigid rotation about a point which is not the centre of mass I CM A object with known is connected to a low mass rod and rotates about an axle. K Mv = M r Mr trans CM CM CM K K K Mr I 1 1 Total trans rot CM CM 1 = MrCM ICM I Mr I parallel axis theorem CM CM 4

25 Example A thin rod of mass 140 g and 60 cm long rotates at an angular speed of 5 radians per second about an axle that is 0 cm from one end. K K K Mr I 1 Total trans rot CM CM M r 1 1 CM 1 L rcm 0.1 m Get KTotal 1.75 J 5

26 A pair of upright metre sticks, with lower ends against a wall, are allowed to fall to the floor. One is bare, and the other has a heavy weight attached to its upper end. The stick to hit the floor first is the... (A) bare stick (B) weighted stick (C) both the same 6

27 Roll a pair of identical cans of carbonated cooldrink down an incline. You won t be surprised to find they roll at the same rate. Now shake one of them so bubbles form inside, then repeat the experiment. Now... (A) the shaken can wins the race (B) the shaken can loses the race (C) both cans still roll together. 7

28 Two balls of mass 0.7 kg are connected by a low mass rigid rod of length 0.4 m. The object rotates around a pivot at its center, with angular speed 13 radians/s. What is the rotational kinetic energy of this object? (1) 484 J () 4.73 J (3).37 J (4) J (5) 0 J 8

29 A diatomic molecule such as molecular nitrogen (N ) consists of two atoms each of mass M, whose nuclei are a distance d apart. What is the moment of inertia of the molecule about its center of mass? (1) Md d () Md (3) (4) (5) 1 Md 1 Md 4 4Md 9

30 M&I 3E 9.4 The point particle system A point particle system has the same mass as a real multiparticle system, but all its mass is concentrated into a point particle located at the centre of mass of the real system. For a point particle system: dp dt sys f F net K F d r trans net CM i 30

31 The point particle system : jumping up f 1 0 trans CM net CM N F r i K Mv d F mg h 31

32 M&I Chapter 10 Collisions

33 M&I 3E 10.1 Internal interactions in collisions A collision is when two bodies interact over a short time interval. The forces that the bodies exert on each other are usually so strong during the collision that all forces acting on a body may be ignored. During a collision between two bodies (1 and ), the contact force exerted by one body on the other jumps from zero to a very large value and then abruptly drops to zero again. t i t f t The time interval t t t is usually very small. f i F 1 Note that F F 1 1 for the collision 33

34 Whenever an interaction occurs in a system, forces occur in equal and opposite pairs. Which of the following do not always occur in equal and opposite pairs? (A) Impulses (B) Accelerations (C) Momentum changes 34

35 A ball bounces off a wall. mass 0.1 kg v i = 6 m/s v f = 6 m/s What is the change in p x of the ball? (1) 0 kg m/s () 1. kg m/s (3) +1. kg m/s (4) +0.6 kg m/s (5) 0.6 kg m/s 35

36 A ball bounces off a wall. mass 0.1 kg v i = 6 m/s v f = 6 m/s What is the change in p x of the Earth? (1) 0 kg m/s () 1. kg m/s (3) +1. kg m/s (4) +0.6 kg m/s (5) 0.6 kg m/s 36

37 M&I 3E 10. Elastic and inelastic collisions Elastic collision: no change in the internal energy of the interacting objects... i.e. no thermal energy rise, no lasting deformations, no new vibrations, etc. E int 0 therefore K f Ki inelastic collision: Eint 0 and K f Ki maximally inelastic collision: Maximum energy dissipation and the objects stick together In all cases above, momentum still conserved: p 0 sys since F net 0 37

38 M&I 3E 10.3 A head-on collision of equal masses in 1D Two extreme cases: elastic collision: p p K xf 1xi K f 1i Maximally inelastic collision: p p p xf 1xf 1xi K K K 1 1 f f 1i 38

39 An elastic collision of unequal masses in 1D p p p 1i f 1 f K K K 1i 1 f f 1f f p1 i p p m m m solve to find... p m m p 1 1f 1i m1 m p m p 1 f 1i m1 m 39

40 M&I 3E 10.4 A head-on collision of unequal masses p p p p 1i i 1 f f... find... p p f 1i and v m v 1 f 1i m 40

41 M&I 3E 10.5 Frame of reference 41

42 See: Reference_frames.py 4

43 M&I 3E 10.6 Scattering: collisions in D and 3D p p p p p cos p cos 0 p sin p sin 3 4 K K K p1 p3 p4 m m m

44 p p p hence Elastic scattering: identical particles, one at rest p p p p p p p p p p p cos A divide by m... p1 p3 p4 p3 p4 cos A m m m m p3p4cos A K1 K3 K4 m cos A 0 or A 90 44

45 See alpha_on_alpha.py alpha_on_electron.py alpha_on_gold.py 45

46 Demonstration: Newton s cradle Five spheres of equal mass hang at the end of strings. One ball is pulled back and released to strike the other stationary balls and one ball flies off on the other side. Kinetic energy is conserved so the collision is elastic. But why don t two balls fly out with half the speed? 46

47 If two balls fly off with half the speed of the incoming ball, then that would conserve momentum, since mv mv mv 1 1 But it wouldn t conserve kinetic energy. The incoming ball has kinetic energy 1 mv and two outgoing balls with half the speed have kinetic energy: m( v) m( v) mv mv

48 Two lead bricks moving in the +x and x directions, each with kinetic energy K, smash into each other and come to a stop. What happened to the energy? (1) The kinetic energy of the system remained constant. () The kinetic energy changed into thermal energy. (3) The total energy of the system decreased by an amount K. (4) Since the blocks were moving in opposite directions, the initial kinetic energy of the system was zero, so there was no change in energy. 48

49 A squishy clay ball collides in midair with a baseball, and sticks to the baseball. The stuck-together objects keep moving. Initial kinetic energies: K i = K 1clay +K 1baseball Final kinetic energy of stuck clay+ball: K f = K (clay+ball) Which must be true for this collision? (1) K f = K i () K f > K i (3) K f < K i 49

50 Which of the following is a property of all elastic collisions? (1) The colliding objects interact through springs. () The kinetic energy of one of the objects doesn t change. (3) The total kinetic energy is constant at all times - before, during, and after the collision. (4) The total kinetic energy after the collision is equal to the total kinetic energy before the collision. (5) The elastic spring energy after the collision is greater than the elastic spring energy before the collision. 50

51 Which of the following is true for both elastic and inelastic collisions? (1) The internal energy of the system after the collision is different from what it was before the collision. () The total momentum of the system doesn t change. (3) The total kinetic energy of the system doesn t change. 51

52 A ping-pong ball bounces elastically off a bowling ball which is initially at rest After the collision the ping-pong ball s kinetic energy is K p. What is the kinetic energy of the bowling ball? (1) K p () K p (3) much greater than K p (4) negligibly small (nearly zero) 5

53 A ball of mass m 1 hits a stationary target of mass m head-on. The total initial and final kinetic energies are the same. Which of the following statements is false? (1) If m 1 << m, the momentum of the ball hardly changes. () If m 1 < m, the ball bounces straight back. (3) If m 1 < m, the ball bounces straight back with less kinetic energy than it had originally. (4) If m 1 >> m, the ball keeps going without change of direction. 53

54 A bullet of mass m traveling horizontally at a very high speed v embeds itself in a block of mass M that is sitting at rest on a nearly frictionless surface Which must be true for the system of bullet + block in this collision? (1) K f = K i () K f > K i (3) K f < K i 54

55 Demonstration: The ballistic pendulum 1. A bullet is fired into a pendulum, which is initially at rest.. The bullet lodges in the pendulum, which moves to the right. 3. The bullet and pendulum swing to a height h. u b m u m u ( m m ) v b b p p b p bp m u 0 ( m m ) v b b b p bp v bp 1 ( ) ( ) b p bp b p m m v m m gh h 55

56 M&I 3E 10.7 Discovering the nucleus inside atoms JJ Thomson (Nobel Prize in Physics, 1906) Plum pudding model. Poor agreement with experiment. 56

57 The Rutherford model Ernest Rutherford (Nobel Prize in Chemistry, 1908) 57

58 Scattering of a 4 He nucleus off (a) Thomson s atom Simulation of Rutherford scattering off a gold nucleus (b) Rutherford s atom See Rutherford.py 58

59 Rutherford: It was the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15 inch shell at a piece of tissue paper and it came back and hit you. Rutherford model (1911): The atom has a small hard central core (nucleus) where all the positive charge is concentrated. The negative charge inhabitants the nearly empty space around the nucleus. Thus most of the alpha particles migrate through the gold foil with some or no (Coulomb) interaction, but some will experience a head-on collision with a nucleus and return in a backwards direction. 59

60 But there were still unanswered questions why does the nucleus (all positive charge) not fly apart due to Coulomb repulsion? why do the negative charges not radiate energy, spiral inwards and collapse into the nucleus due to Coulomb attraction? the model did also not explain existing experimental observations. 60

61 M&I 3E 10.9 Relativistic momentum and energy (collisions in a particle accelerator) p p p p p cos p cos 0 p sin p sin E m c E E p 1c m1c mc p3c m3c p4c m4c 61

62 M&I Chapter 11 Angular Momentum

63 M&I 3E 11.1 Translational angular momentum Consider the momentum of the Earth at four positions as it moves around the Sun. p The magnitude of the translational angular momentum is defined as L, r p sin trans Sun 63

64 Translational angular momentum: direction L trans Sun, r p sin What about the direction of L trans?, Sun The direction of Lˆ trans depends on the direction of the rotation and is perpendicular to the plane defined by r and p. 64

65 The vector cross product A A ˆi A ˆj A kˆ x y z B B ˆi B ˆj B kˆ x y z AB ( A B A B ) ˆi + ( A B A B ) ˆj + ( A B A B ) kˆ y z z y z x x z x y y x easy to remember: ˆi ˆj kˆ A A A x y z B B B x y z always 65

66 AB ( A B A B ) ˆi + ( A B A B ) ˆj + ( A B A B ) kˆ y z z y z x x z x y y x In polar form in D: AB ABsin where is the angle between the tails of A and B. AB B A AA 0 ˆi ˆi ˆj ˆj kˆ kˆ 0 ˆi ˆj kˆ ˆj kˆ ˆi kˆ ˆi ˆj Use right hand rule 66

67 What is the direction of < 0, 0, 3> < 0, 4, 0>? (1) +x () x (3) +y (4) y (5) +z (6) z (7) zero magnitude 67

68 What is the direction of < 0, 4, 0> < 0, 0, 3>? (1) +x () x (3) +y (4) y (5) +z (6) z (7) zero magnitude 68

69 What is the direction of < 0, 0, 6> < 0, 0, -3>? (1) +x () x (3) +y (4) y (5) +z (6) z (7) zero magnitude 69

70 Translational angular momentum of an object relative to location A L r p trans, A A Ltrans, A ra p sin rpsin A psin is the component of perpendicular to. p r See important worked example in M&I Calculate L trans for a particle moving relative to locations A, B and C. 70

71 A ball falls straight down in the xy plane. Its momentum is shown by the red arrow. What is the direction of the ball s angular momentum about location A? A (1) +x () x (3) +y (4) y (5) +z (6) z (7) zero magnitude 71

72 A planet orbits a star, in a circular orbit in the xy plane. Its momentum is shown by the red arrow. What is the direction of the angular momentum of the planet? 7

73 A comet orbits the Sun in the xz plane. Its momentum is shown by the red arrow. What is the direction of the comet s angular momentum about the Sun? (1) +x () x (3) +y (4) y (5) +z (6) z (7) zero magnitude 73

74 M&I 3E 11. Rotational angular momentum Consider a bicycle wheel spinning about its centre of mass with angular speed (in radians per second). Nearly all mass M is in rim. Divide rim into 0 pieces each of mass M/0. LCM R M 0 v sin 90 R M 0 R where v R LCM 0 M 0 R MR I For whole wheel where I is the moment of inertia of the wheel (rim). 74

75 L CM I Rotational angular momentum L rot or Iω where ω is the angular velocity vector. T and direction. v r r T In general v ωr 75

76 Rotational angular momentum: general case Consider a collection of 4 masses rotating with the same about a common COM. L rot r 1m1v r mv r mv r mv m r m r m r m r I L rot Iω Also K rot I L 1 1 I I I rot 76

77 M&I 3E 11.3 Translational plus rotational angular momentum For a system which is both translating (relative to A) and rotating about a centre of mass L L L A trans, A rot rcm, A P tot r 1, CM p1 r, CM p r3, CM p3 where Ptot p1 p p3 L trans, A L rot :translational (or orbital ) angular momentum :rotational (or spin ) angular momentum 77

78 Translational plus rotational angular momentum: examples 78

79 M&I 3E 11.4 The angular momentum principle d dt L r p trans, A d Ltrans, A ra p dt dra p ra dt dra Now = A A = 0 dt p v m v A dp dt dl dt dp dt A r A r A F τ A where the torque τ A raf 79

80 Torque : tau τ r F Also called the moment of the force F about the turning point (axis of rotation). e.g. a metre stick is free to rotate about a fixed axis at one end as shown. F r 60º τ r F axis of rotation τ rf sin is the angle between the tails of the r and F r 10º F (1)(4)sin10 = 3.46 N m τ 3.46 N m into the page r 80

81 Examples For each situation below, determine the resultant torque acting on the axis of rotation O. Use a coordinate system with the ˆk - axis out of the page. (a) (b) 37º 10º 3 m 100 N 40º 100 N 160 N 60º 3 m 5 m 140 N 10 N 81

82 A yo-yo is in the xy plane. You pull up on the string with a force of magnitude 0.6 N. What is the direction of the torque you exert on the yo-yo? (1) +x () x (3) +y (4) y (5) +z (6) z (7) zero magnitude 8

83 The angular momentum principle dl dt A τ A finite time form: L τ, A net A t Principle of conservation of angular momentum L + L 0 A, system A, surroundings Angular momentum update formula for a closed system L τ t A net, A L L τ A, f A, i + t net, A 83

84 M&I 3E 11.5 Angular momentum in multiparticle systems L τ t tot, A net, A 84

85 Child runs and jumps on playground merry-go-round. For the system of the child + disk (excluding the axle and the Earth), which statement is true from just before to just after impact? 85

86 What is the initial angular momentum of the child + disk about the axle? (1) < 0, 0, 0 > () < 0, Rmv, 0 > (3) < 0, Rmv, 0 > (4) < 0, 0, Rmv > (5) < 0, 0, Rmv > 86

87 The disk has moment of inertia I, and after the collision it is rotating with angular speed ω. The rotational angular momentum of the disk alone (not counting the child) is (1) < 0, 0, 0 > () < 0, Iω, 0 > (3) < 0, Iω, 0 > (4) < 0, 0, Iω > (5) < 0, 0, Iω > 87

88 After the collision, what is the speed (in m/s) of the child? (1) ω R () ω (3) ω R (4) ω / R (5) ω R 88

89 After the collision, what is the translational angular momentum of the child about the axle? (1) < 0, 0, 0 > () < 0, Rmω, 0 > (3) < 0, Rmω, 0 > (4) < 0, Rm(ωR), 0 > (5) < 0, Rm(ωR), 0 > 89

90 Example A playground ride consists of a uniform-density disk of mass 300 kg and radius m mounted on a low friction axle. Starting from a distance of 5 m from the edge of the disk, a child of mass 40 kg runs at 3 m s -1 on a line tangential to the disk and jumps onto the outer edge of the disk. If the disk was initially at rest, how fast does it rotate just after the collision? L L τ A, f A, i + t net, A τ net, A 0-1 LAi, rpsin = Rmv (40)(3) 40 kg m s L I mr MR L 40 kg m s 1-1 A, f A, i radians s error in textbook 90

91 M&I 3E 11.6 Three fundamental principles of mechanics Momentum Angular momentum Energy dp dt dl dt F A net τnet, A E W Q If there are external forces, then momentum changes. Location of object does not matter. If there are external torques, angular momentum changes. Location of object relative to point A is important. If there are energy inputs, then energy changes. Location of object does not matter. 91

92 M&I 3E 11.7 Systems with zero torques L A 0 L i L I I i i f f f 9

93 Example A man stands on a rotating stool which is free to rotate without friction. He holds 3 kg in each hand at a distance of 1 m from the centre of this body. Say that he is rotating at an angular speed of 10 radians per second. 3 kg 3 kg If the man brings the masses straight towards his chest until they are a distance of 0. m from the centre of his body, what will now be his angular speed? 93

94 M&I 3E 11.8 Systems with nonzero torques Example 1 A beam is balanced on a fulcrum (triangle) with three masses hung from the positions shown. What is the mass of m? Assume that the beam is massless. 1 m m 1 m m 3 1 kg? 3 kg 94

95 The broom balances at its centre of mass. If you cut the broom into two parts through the centre of mass and then weigh each part on a sale, which part will weigh more? (A) the longer piece (B) the shorter piece (C) both the same 95

96 Example 45º 90º 15 kg A beam of mass 5 kg and length 10 m is held at an angle of 45º by a cable as shown. If a 15 kg mass hangs from the end of the beam, determine the force of the wall on the beam and the tension in the cable. dp dt dl dt A 0 F 0 and net 0 τ 0 net, A 96

97 Example 3 A 10 kg block is attached to a light rope which is wound around a 0 kg cylindrical pulley of radius r =.5 m. How long will it take for the speed of the 10 kg block to increase from zero to 4.0 m s -1 after the system is released? Use I cylinder = 6 kg m. 10 kg r =.5 m 0 kg 97

98 M&I 3E Angular momentum quantization The Bohr Model Niels Bohr (Noble Prize in Physics 19) proposed a revolutionary model which broke from classical theory. For his model of the hydrogen atom, Bohr postulated that: the electron follows circular orbits around the positive nucleus. (Centripetal force = Coulomb attraction) e r mv r the electron moves in certain allowed orbits without radiating energy. The electron is said to be in a stationary state. 98

99 the electron jumps (undergoes a transition) between stationary states by absorbing or emitting a quantum, hf, of energy. Absorption: hf = E f E i Emission: hf = E i E f E f E i E i E f 99

100 Allowed radii of electron orbits Bohr: angular momentum is quantized : L rp N C N = 1,, 3, and Then combining with e r h mv r gives rn N em For N = 1, r = m 100

101 For the hydrogen atom, Bohr was able to propose a formula for the allowed energy levels for the electron: E 13.6 ev ; 1,,3,... N N N... which allowed the calculation of the frequencies of the photon emitted when an electron undergoes a transition from an outer orbit to an inner one... which in turn explained the experimental observations perfectly. Bohr s model of the hydrogen atom was a great triumph. He extended his model to other (ionized) single-electron atoms, e.g. He,Li,Be However... the Bohr model was not satisfactory for other multielectron atoms and did not explain the quantum postulates. 101

102 10