2. Physics 207: Lecture 16, Pg 1 Physics 207, Lecture 16, Oct. 29 Agenda: Chapter 13 l Center of Mass l Torque l Moment of Inertia l Rotational Energy l Rotational Momentum Assignment: l Wednesday is an exam review session, Exam will be held in rooms B102 & l Wednesday is an exam review session, Exam will be held in rooms B102 & B130 in Van Vleck at 7:15 PM l MP Homework 7, Ch. 11, 5 problems, NOTE: Due Wednesday at 4 PM NOTE: Due Wednesday at 4 PM l MP Homework 7A, Ch. 13, 5 problems, available soon

3. Physics 207: Lecture 16, Pg 2 Chap. 13: Rotational Dynamics l Up until now rotation has been only in terms of circular motion with a c = v 2 / R and | a T | = d| v | / dt l Rotation is common in the world around us. l Many ideas developed for translational motion are transferable.

4. Physics 207: Lecture 16, Pg 3 Conservation of angular momentum has consequences How does one describe rotation (magnitude and direction)?

5. Physics 207: Lecture 16, Pg 4 Rotational Dynamics: A child’s toy, a physics playground or a student’s nightmare l A merry-go-round is spinning and we run and jump on it. What does it do? l We are standing on the rim and our “friends” spin it faster. What happens to us? l We are standing on the rim a walk towards the center. Does anything change?

6. Physics 207: Lecture 16, Pg 5 Rotational Variables l Rotation about a fixed axis:  Consider a disk rotating about an axis through its center:] l How do we describe the motion: (Analogous to the linear case )  

7. Physics 207: Lecture 16, Pg 6 Rotational Variables... l Recall: At a point a distance R away from the axis of rotation, the tangential motion:  x =  R  v =  R  a =  R   R v =  R x 

8. Physics 207: Lecture 16, Pg 7 Summary (with comparison to 1-D kinematics) AngularLinear And for a point at a distance R from the rotation axis: x = R  v =  R  a =  R

9. Physics 207: Lecture 16, Pg 8 Lecture 15, Exercise 5 Rotational Definitions A. The wheel is spinning counter-clockwise and slowing down. B. The wheel is spinning counter-clockwise and speeding up. C. The wheel is spinning clockwise and slowing down. D. The wheel is spinning clockwise and speeding up A friend at a party (perhaps a little tipsy) sees a disk spinning and says “Ooh, look! There’s a wheel with a negative  and positive  !” l Which of the following is a true statement about the wheel? 

10. Physics 207: Lecture 16, Pg 9 Lecture 15, Exercise 5 Rotational Definitions A goofy friend sees a disk spinning and says “Ooh, look! There’s a wheel with a negative  and with antiparallel  and  !” l Which of the following is a true statement about the wheel? (A) (A) The wheel is spinning counter-clockwise and slowing down. (B) (B) The wheel is spinning counter-clockwise and speeding up. (C) (C) The wheel is spinning clockwise and slowing down. (D) The wheel is spinning clockwise and speeding up 

11. Physics 207: Lecture 16, Pg 10 Example: Wheel And Rope l A wheel with radius r = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4m/s 2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2  radians)a r

12. Physics 207: Lecture 16, Pg 11 Example: Wheel And Rope l A wheel with radius r = 0.4 m rotates freely about a fixed axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4 m/s 2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2  radians) Revolutions = R = (    and a =  r      t + ½  t    R = (    + ½  a/r  t    R = (0.5 x 10 x 100) / 6.28a r

13. Physics 207: Lecture 16, Pg 12 System of Particles (Distributed Mass): l Until now, we have considered the behavior of very simple systems (one or two masses). l But real objects have distributed mass ! l For example, consider a simple rotating disk and 2 equal mass m plugs at distances r and 2r. l Compare the velocities and kinetic energies at these two points. 1 2 

14. Physics 207: Lecture 16, Pg 13 System of Particles (Distributed Mass): l An extended solid object (like a disk) can be thought of as a collection of parts. l The motion of each little part depends on where it is in the object! l The rotation axis matters too! 1 K= ½ m v 2 = ½ m (  r) 2 2 K= ½ m (2v) 2 = ½ m (  2r) 2 

15. Physics 207: Lecture 16, Pg 14 System of Particles: Center of Mass l If an object is not held then it rotates about the center of mass. l Center of mass: Where the system is balanced !  Building a mobile is an exercise in finding centers of mass. m1m1 m2m2 + m1m1 m2m2 + mobile

16. Physics 207: Lecture 16, Pg 15 System of Particles: Center of Mass l How do we describe the “position” of a system made up of many parts ? Center of Mass l Define the Center of Mass (average position):  For a collection of N individual pointlike particles whose masses and positions we know: (In this case, N = 2) y x rr2rr2 rr1rr1 m1m1 m2m2 R R CM

17. Physics 207: Lecture 16, Pg 16 Sample calculation: l Consider the following mass distribution: (24,0) (0,0) (12,12) m 2m m R CM = (12,6) X CM = (m x 0 + 2m x 12 + m x 24 )/4m meters Y CM = (m x 0 + 2m x 12 + m x 0 )/4m meters X CM = 12 meters Y CM = 6 meters

18. Physics 207: Lecture 16, Pg 17 System of Particles: Center of Mass l For a continuous solid, convert sums to an integral. y x dm r where dm is an infinitesimal mass element.

19. Physics 207: Lecture 16, Pg 18 Rotational Dynamics: What makes it spin?  TOT = |r| |F Tang | ≡ |r| |F| sin  l Torque is the rotational equivalent of force Torque has units of kg m 2 /s 2 = (kg m/s 2 ) m = N m l A constant torque gives constant angular acceleration iff the mass distribution and the axis of rotation remain constant. F F Tangentiala r F F randial F  A force applied at a distance from the rotation axis

20. Physics 207: Lecture 16, Pg 19 Lecture 16, Exercise 1 Torque A. Case 1 B. Case 2 C. Same l In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same. Remember torque requires F, r and sin  or the tangential force component times perpendicular distance L L F F axis case 1case 2

21. Physics 207: Lecture 16, Pg 20 Lecture 16, Exercise 1 Torque l In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same. Remember torque requires F, r and sin  or the tangential force component times perpendicular distance (A) (A) case 1 (B) (B) case 2 (C) (C) same L L F F axis case 1case 2

22. Physics 207: Lecture 16, Pg 21 Rotational Dynamics: What makes it spin?  TOT = |r| |F Tang | ≡ |r| |F| sin  l Torque is the rotational equivalent of force Torque has units of kg m2/s2 = (kg m/s2) m = N m F F Tangentiala r F F randial F  A force applied at a distance from the rotation axis   TOT = r F Tang = r m a = r m r  = m r 2  For every little part of the wheel

23. Physics 207: Lecture 16, Pg 22  TOT = m r 2  and inertia  TOT = m r 2  and inertia l This is the rotational version of F TOT = ma Moment of inertia, m r 2, (here I is just a point on the wheel) is the rotational equivalent of mass. Moment of inertia, I ≡ m r 2, (here I is just a point on the wheel) is the rotational equivalent of mass. If I is big, more torque is required to achieve a given angular acceleration. F F Tangentiala r F F randial F  The further a mass is away from this axis the greater the inertia (resistance) to rotation

24. Physics 207: Lecture 16, Pg 23 Calculating Moment of Inertia where r is the distance from the mass to the axis of rotation. Example: Calculate the moment of inertia of four point masses (m) on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square: mm mm L

25. Physics 207: Lecture 16, Pg 24 Calculating Moment of Inertia... For a single object, I depends on the rotation axis! Example: I 1 = 4 m R 2 = 4 m (2 1/2 L / 2) 2 L I = 2mL 2 I 2 = mL 2 mm mm I 1 = 2mL 2

26. Physics 207: Lecture 16, Pg 25 Lecture 16, Exercise 2 Moment of Inertia A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is I a, I b, and I c respectively.  Which of the following is correct: ( A) ( A) I a > I b > I c (B) (B) I a > I c > I b (C) (C) I b > I a > I c a b c

27. Physics 207: Lecture 16, Pg 26 Lecture 16, Home Exercise Moment of Inertia I a = 2 m (2L) 2 I b = 3 m L 2 I c = m (2L) 2 l Which of the following is correct: ( A) ( A) I a > I b > I c (B) (B) I a > I c > I b (C) (C) I b > I a > I c a b c L L

28. Physics 207: Lecture 16, Pg 27 Calculating Moment of Inertia... l For a discrete collection of point masses we found: l For a continuous solid object we have to add up the mr 2 contribution for every infinitesimal mass element dm.  An integral is required to find I : r dm

29. Physics 207: Lecture 16, Pg 28 Moments of Inertia Solid disk or cylinder of mass M and radius R, about perpendicular axis through its center. I = ½ M R 2 Some examples of I for solid objects: R L r dr

30. Physics 207: Lecture 16, Pg 29 Moments of Inertia... Some examples of I for solid objects: Solid sphere of mass M and radius R, about an axis through its center. I = 2/5 M R2 R See Table 13.3, Moments of Inertia Thin spherical shell of mass M and radius R, about an axis through its center. Use the table… R

31. Physics 207: Lecture 16, Pg 30 Moments of Inertia Some examples of I for solid objects: Thin hoop (or cylinder) of mass M and radius R, about an axis through it center, perpendicular to the plane of the hoop is just MR 2 R Thin hoop of mass M and radius R, about an axis through a diameter. R

32. Physics 207: Lecture 16, Pg 31 Rotation & Kinetic Energy l Consider the simple rotating system shown below. (Assume the masses are attached to the rotation axis by massless rigid rods). l The kinetic energy of this system will be the sum of the kinetic energy of each piece: K = ½  m 1 v 1  + ½  m 2 v 2  + ½  m 3 v 3  + ½  m 4 v 4  rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3 

33. Physics 207: Lecture 16, Pg 32 Rotation & Kinetic Energy Notice that v 1 =  r 1, v 2 =  r 2, v 3 =  r 3, v 4 =  r 4 l So we can rewrite the summation: We recognize the quantity, moment of inertia or I, and write: rr1rr1 rr2rr2 rr3rr3 rr4rr4 m4m4 m1m1 m2m2 m3m3 

34. Physics 207: Lecture 16, Pg 33 Lecture 16, Exercise 2 Rotational Kinetic Energy A. ¼ B. ½ C. 1 D. 2 E. 4 l We have two balls of the same mass. Ball 1 is attached to a 0.1 m long rope. It spins around at 2 revolutions per second. Ball 2 is on a 0.2 m long rope. It spins around at 2 revolutions per second. l What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ? Ball 1 Ball 2

35. Physics 207: Lecture 16, Pg 34 Lecture 16, Exercise 2 Rotational Kinetic Energy K 2 /K 1 = ½ m  r 2 2 / ½ m  r 1 2 =  2 /  2 = 4 l What is the ratio of the kinetic energy of Ball 2 to that of Ball 1 ? (A) 1/4 (B) 1/2 (C) 1 (D) 2 (E) 4 Ball 1 Ball 2

36. Physics 207: Lecture 16, Pg 35 Rotation & Kinetic Energy... l The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle Rotating System v is “linear” velocity m is the mass.  is angular velocity I is the moment of inertia about the rotation axis.

37. Physics 207: Lecture 16, Pg 36 Moment of Inertia and Rotational Energy Notice that the moment of inertia I depends on the distribution of mass in the system.  The further the mass is from the rotation axis, the bigger the moment of inertia. l For a given object, the moment of inertia depends on where we choose the rotation axis (unlike the center of mass). In rotational dynamics, the moment of inertia I appears in the same way that mass m does in linear dynamics ! lSo where

38. Physics 207: Lecture 16, Pg 37 Work (in rotational motion) F Consider the work done by a force F acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d  :where dr =R d   dW = F Tangential dr  dW = (F Tangential R) d   dW =  d  and with a constant torque) We can integrate this to find: W =  (  f  i  l Analogue of W = F  r W will be negative if  and  have opposite sign ! axis of rotation R F dr =Rd  dd 

39. Physics 207: Lecture 16, Pg 38 Work & Kinetic Energy: Recall the Work Kinetic-Energy Theorem:  K = W NET l This is true in general, and hence applies to rotational motion as well as linear motion. l So for an object that rotates about a fixed axis:

40. Physics 207: Lecture 16, Pg 39 Lecture 16, Home exercise Work & Energy Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same linear distance. Disk 1 has a bigger radius, but both are identical material (i.e. their density  = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest.  Which disk has the biggest angular velocity after the pull? W =  F d = ½ I  2 ( A) ( A) Disk 1 (B) (B) Disk 2 (C) (C) Same FF 11 22 start finish d

41. Physics 207: Lecture 16, Pg 40 Lecture 16, Home exercise Work & Energy Strings are wrapped around the circumference of two solid disks and pulled with identical forces for the same linear distance. Disk 1 has a bigger radius, but both are identical material (i.e. their density  = M/V is the same). Both disks rotate freely around axes though their centers, and start at rest.  Which disk has the biggest angular velocity after the pull? W = F d = ½ I 1  1 2 = ½ I 2  2 2 and  1 = (I 2 / I 1 ) ½  2 and I 2 < I 1 ( A) ( A) Disk 1 (B) (B) Disk 2 (C) (C) Same FF 11 22 start finish d

42. Physics 207: Lecture 16, Pg 41 Example: Rotating Rod l A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. What is (A) its angular speed when it reaches the lowest point ? (B) its initial angular acceleration ? (C) initial linear acceleration of its free end ? L m

43. Physics 207: Lecture 16, Pg 42 Example: Rotating Rod l A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless hinge passing through one end as shown. The rod is released from rest in the horizontal position. What is (B) its initial angular acceleration ? 1. For forces you need to locate the Center of Mass CM is at L/2 ( halfway ) and put in the Force on a FBD 2. The hinge changes everything! L m mg  F = 0 occurs only at the hinge but  z = I  z = r F sin 90° at the center of mass and (I CM + m(L/2) 2 )  z = (L/2) mg and solve for  z

44. Physics 207: Lecture 16, Pg 43 Example: Rotating Rod l A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless hinge passing through one end as shown. The rod is released from rest in the horizontal position. What is (C) initial linear acceleration of its free end ? 1. For forces you need to locate the Center of Mass CM is at L/2 ( halfway ) and put in the Force on a FBD 2. The hinge changes everything! L m mg a =  L

45. Physics 207: Lecture 16, Pg 44 Example: Rotating Rod l A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless hinge passing through one end as shown. The rod is released from rest in the horizontal position. What is (A) its angular speed when it reaches the lowest point ? 1. For forces you need to locate the Center of Mass CM is at L/2 ( halfway ) and use the Work-Energy Theorem 2. The hinge changes everything! m mg W = mgh = ½ I  2 W = mgL/2 = ½ (I CM + m (L/2) 2 )  2 and solve for  mg L/2 L

46. Physics 207: Lecture 16, Pg 45 Connection with CM motion l If an object of mass M is moving linearly at velocity V CM without rotating then its kinetic energy is l What if the object is both moving linearly and rotating? If an object of moment of inertia I CM is rotating in place about its center of mass at angular velocity  then its kinetic energy is

47. Physics 207: Lecture 16, Pg 46 Connection with CM motion... l So for a solid object which rotates about its center of mass and whose CM is moving :  V CM

48. Physics 207: Lecture 16, Pg 47 Rolling Motion l Now consider a cylinder rolling at a constant speed. V CM CM The cylinder is rotating about CM and its CM is moving at constant speed (V CM ). Thus its total kinetic energy is given by :

49. Physics 207: Lecture 16, Pg 48 Lecture 16, Example: Lecture 16, Example: The YoYo l A solid uniform disk yoyo of radium R and mass M starts from rest, unrolls, and falls a distance h. (1) What is the angular acceleration? (2) What will be the linear velocity of the center of mass after it falls h meters? (3) What is the tension on the cord ?  M h T

50. Physics 207: Lecture 16, Pg 49 Lecture 16, Example: Lecture 16, Example: The YoYo l A solid uniform disk yoyo of radium R and mass M starts from rest, unrolls, and falls a distance h. l Conceptual Exercise: Which of the following pictures correctly represents the yoyo after it falls a height h? (A)(B)(C)  M h T h  M T  M h T

51. Physics 207: Lecture 16, Pg 50 Lecture 16, Example: Lecture 16, Example: The YoYo l A solid uniform disk yoyo of radium R and mass M starts from rest, unrolls, and falls a distance h. l Conceptual Exercise: Which of the following pictures correctly represents the yoyo after it falls a height h? (A) (B) No F x, no a x (C)  M h T h  M T  M h T Mg

52. Physics 207: Lecture 16, Pg 51 Lecture 16, Example: Lecture 16, Example: The YoYo l A solid uniform disk yoyo of radium R and mass M starts from rest, unrolls, and falls a distance h. (1) What is the angular acceleration? (2) What will be the linear velocity of the center of mass after it falls h meters? (3) What is the tension on the cord ?  M h T Choose a point and calculate the torque   = I  z = Mg R + T0 ( ½ MR 2 + MR 2 )  z = Mg R  z = Mg /(3/2 MR) = 2 g / (3R) X

53. Physics 207: Lecture 16, Pg 52 Lecture 16, Example: Lecture 16, Example: The YoYo l A solid uniform disk yoyo of radium R and mass M starts from rest, unrolls, and falls a distance h. (1) What is the angular acceleration? (2) What will be the linear velocity of the center of mass after it falls h meters? (3) What is the tension on the cord ?  M h T  z = Mg /(3/2 MR) = 2 g / (3R)  and a CM =  z R = 2g/3 (down)  Can use kinetics or work energy  Mgh = ½ mv 2 + ½ I CM  2  Mgh = ½  Mv 2 + ½ ( ½ MR 2 ) (v/R) 2  gh = ¾ v 2  v = 2 (gh/3) ½ X

54. Physics 207: Lecture 16, Pg 53 Lecture 16, Example: Lecture 16, Example: The YoYo l A solid uniform disk yoyo of radium R and mass M starts from rest, unrolls, and falls a distance h. (1) What is the angular acceleration? (2) What will be the linear velocity of the center of mass after it falls h meters? (3) What is the tension on the cord ?  M h T  a CM =  z R = -2g/3  Ma CM =- 2Mg/3 = T – Mg  T = Mg/3  or from torques  I  z’ = TR = ½ MR 2 (2g/3R)  T = Mg/3 X

55. Physics 207: Lecture 16, Pg 54 Rolling Motion l Again consider a cylinder rolling at a constant speed. V CM CM 2V CM

56. Physics 207: Lecture 16, Pg 55 Example : Rolling Motion l A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? M  h M v ? Ball has radius R M M M M M

57. Physics 207: Lecture 16, Pg 56 Example : Rolling Motion l A cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? l Use Work-Energy theorem M  h M v ? Ball has radius R M M M M M Mgh = ½ Mv 2 + ½ I CM  2 Mgh = ½ Mv 2 + ½ (½ M R 2 )(v/R) 2 = ¾ Mv 2 v = 2(gh/3) ½

58. Physics 207: Lecture 16, Pg 57 Rolling Motion l Now consider a cylinder rolling at a constant speed. V CM CM The cylinder is rotating about CM and its CM is moving at constant speed (V CM ). Thus its total kinetic energy is given by :

59. Physics 207: Lecture 16, Pg 58 Motion Motion l Again consider a cylinder rolling at a constant speed. V CM CM 2V CM CM V CM Sliding only CM Rotation only V Tang =  R Both with |V Tang | = |V CM |

60. Physics 207: Lecture 16, Pg 59 Angular Momentum: l We have shown that for a system of particles, momentum is conserved if l What is the rotational equivalent of this? angular momentum is conserved if

61. Physics 207: Lecture 16, Pg 60 Example: Two Disks A disk of mass M and radius R rotates around the z axis with angular velocity  0. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity  F. 00 z FF z

62. Physics 207: Lecture 16, Pg 61 Example: Two Disks A disk of mass M and radius R rotates around the z axis with initial angular velocity  0. A second identical disk, at rest, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity  F. 00 z FF z No External Torque so L z is constant L i = L f  I  i i = I  f  ½ mR 2  0 = ½ 2mR 2  f

63. Physics 207: Lecture 16, Pg 62 Lecture 16, Oct. 29 Assignment: l Wednesday is an exam review session, Exam will be held in rooms B102 & l Wednesday is an exam review session, Exam will be held in rooms B102 & B130 in Van Vleck at 7:15 PM l MP Homework 7, Ch. 11, 5 problems, NOTE: Due Wednesday at 4 PM NOTE: Due Wednesday at 4 PM l MP Homework 7A, Ch. 13, 5 problems, available soon

64. Physics 207: Lecture 16, Pg 63 Angular Momentum: Definitions & Derivations l We have shown that for a system of particles Momentum is conserved if l What is the rotational equivalent of this? F  The rotational analog of force F is torque  p l Define the rotational analog of momentum p to be angular momentum, L or p = mv

65. Physics 207: Lecture 16, Pg 64 Recall from Chapter 9: Linear Momentum l Newton’s 2 nd Law: Fa F = ma l Units of linear momentum are kg m/s. pv p ≡ mv pv (p is a vector since v is a vector) So p x = mv x etc. l Definition: p l Definition: For a single particle, the momentum p is defined as:

66. Physics 207: Lecture 16, Pg 65 Linear Momentum and Angular Momentum l Newton’s 2 nd Law: Fa F = ma l Units of angular momentum are kg m 2 /s. l So from:

67. Physics 207: Lecture 16, Pg 66 Putting it all together l l In the absence of external torques Total angular momentum is conserved

68. Physics 207: Lecture 16, Pg 67 Angular momentum of a rigid body about a fixed axis: l Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momentum of each particle: i rr1rr1 rr3rr3 rr2rr2 m2m2 m3m3 j m1m1  vv2vv2 vv1vv1 vv3vv3 L We see that L is in the z direction. Using v i =  r i, we get (since r i, v i, are perpendicular)

69. Physics 207: Lecture 16, Pg 68 Example: Two Disks A disk of mass M and radius R rotates around the z axis with angular velocity  0. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity  F. 00 z FF z

70. Physics 207: Lecture 16, Pg 69 Example: Two Disks A disk of mass M and radius R rotates around the z axis with initial angular velocity  0. A second identical disk, at rest, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity  F. 00 z FF z No External Torque so L z is constant L i = L f  I  i i = I  f  ½ mR 2  0 = ½ 2mR 2  f

71. Physics 207: Lecture 16, Pg 70 Demonstration: Conservation of Angular Momentum l Figure Skating : AA z BB z Arm I A I B  A  B L A = L B No External Torque so L z is constant even if internal work done.

72. Physics 207: Lecture 16, Pg 71 Demonstration: Conservation of Angular Momentum l Figure Skating : AA z BB z Arm I A < I B  A >  B ½ I A  A 2 > ½ I B  B 2 (work needs to be done) I A  A = L A = L B = I B  B No External Torque so L z is constant even if internal work done.

73. Physics 207: Lecture 16, Pg 72 Angular Momentum Conservation l A freely moving particle has a well defined angular momentum about any given axis. l If no torques are acting on the particle, its angular momentum remains constant (i.e., will be conserved). L l In the example below, the direction of L is along the z axis, and its magnitude is given by L Z = pd = mvd. y x v d m

74. Physics 207: Lecture 16, Pg 73 Example: Bullet hitting stick l A uniform stick of mass M and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v 1, and the final speed is v 2.  What is the angular speed  F of the stick after the collision? (Ignore gravity) v1v1 v2v2 M FF beforeafter m D D/4

75. Physics 207: Lecture 16, Pg 74 Example: Bullet hitting stick What is the angular speed  F of the stick after the collision? (Ignore gravity). l Process: (1) Define system (2) Identify Conditions (1) System: bullet and stick (No Ext. torque, L is constant) (2) Momentum is conserved (I stick = I = MD 2 /12 ) L init = L bullet + L stick = m v 1 D/4 + 0 = L final = m v 2 D/4 + I  f v1v1 v2v2 M FF beforeafter m D D/4

76. Physics 207: Lecture 16, Pg 75 Example: Throwing ball from stool A student sits on a stool, initially at rest, but which is free to rotate. The moment of inertia of the student plus the stool is I. They throw a heavy ball of mass M with speed v such that its velocity vector moves a distance d from the axis of rotation.  What is the angular speed  F of the student-stool system after they throw the ball ? Top view: before after d v M I I FF

77. Physics 207: Lecture 16, Pg 76 Example: Throwing ball from stool What is the angular speed  F of the student-stool system after they throw the ball ? l Process: (1) Define system (2) Identify Conditions (1) System: student, stool and ball (No Ext. torque, L is constant) (2) Momentum is conserved L init = 0 = L final = m v d + I  f Top view: before after d v M I I FF

78. Physics 207: Lecture 16, Pg 77 An example: Neutron Star rotation period of pulsar is 1.187911164 s Neutron star with a mass of 1.5 solar masses has a diameter of ~11 km. Our sun rotates about once every 37 days  f /  i = I i / I f = r i 2 / r f 2 = (7x10 5 km) 2 /(11 km) 2 = 4 x 10 9 gives millisecond periods!

79. Physics 207: Lecture 16, Pg 78 Angular Momentum as a Fundamental Quantity l The concept of angular momentum is also valid on a submicroscopic scale l Angular momentum has been used in the development of modern theories of atomic, molecular and nuclear physics l In these systems, the angular momentum has been found to be a fundamental quantity  Fundamental here means that it is an intrinsic property of these objects

80. Physics 207: Lecture 16, Pg 79 Fundamental Angular Momentum l Angular momentum has discrete values l These discrete values are multiples of a fundamental unit of angular momentum l The fundamental unit of angular momentum is h-bar  Where h is called Planck’s constant

81. Physics 207: Lecture 16, Pg 80 Intrinsic Angular Momentum photon

82. Physics 207: Lecture 16, Pg 81 Angular Momentum of a Molecule l Consider the molecule as a rigid rotor, with the two atoms separated by a fixed distance l The rotation occurs about the center of mass in the plane of the page with a speed of

83. Physics 207: Lecture 16, Pg 82 Angular Momentum of a Molecule (It heats the water in a microwave over) E = h 2 /(8  2 I) [ J (J+1) ]J = 0, 1, 2, ….

84. Physics 207: Lecture 16, Pg 83 Center of Mass Example: Astronauts & Rope l Two astronauts are initially at rest in outer space and 20 meters apart. The one on the right has 1.5 times the mass of the other (as shown). The 1.5 m astronaut wants to get back to the ship but his jet pack is broken. There happens to be a rope connected between the two. The heavier astronaut starts pulling in the rope. (1) Does he/she get back to the ship ? (2) Does he/she meet the other astronaut ? M = 1.5m m

85. Physics 207: Lecture 16, Pg 84 Example: Astronauts & Rope (1) There is no external force so if the larger astronaut pulls on the rope he will create an impulse that accelerates him/her to the left and the small astronaut to the right. The larger one’s velocity will be less than the smaller one’s so he/she doesn’t let go of the rope they will either collide (elastically or inelastically) and thus never make it. M = 1.5m m

86. Physics 207: Lecture 16, Pg 85 Example: Astronauts & Rope (2) However if the larger astronaut lets go of the rope he will get to the ship. (Too bad for the smaller astronaut!) In all cases the center of mass will remain fixed because they are initially at rest and there is no external force. To find the position where they meet all we need do is find the Center of Mass M = 1.5m m