Chapter 12: Rotation of Rigid Bodies. Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics

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1 Chapter 12: Rotation of Rigid Bodies Center of Mass Moment of Inertia Torque Angular Momentum Rolling Statics

2 Translational vs Rotational 2 / / 1/ 2 m x v dx dt a dv dt F ma p mv KE mv Work Fd P Fv 2 / / 1/ 2 I d dt d dt I L I KE I Work P 2 c t s r v r a r a r Fr L pr Connection

3 More

4 Center of Mass The geometric center or average location of the mass.

5 Rotational & Translational Motion Objects rotate about their Center of Mass. The Center of Mass Translates as if it were a point particle. v d r CM dt CM

6 Center of Mass The Center of Mass Translates as if it were a point particle and, if no external forces act on the system, momentum is then conserved. This means: EVEN if the bat EXPLODED into a thousand pieces, all the pieces would move so that the momentum of the CM is conserved that is, the CM continues in the parabolic trajectory!!!! THIS IS VERY VERY IMPORTANT!

7 System of Particles Center of Mass A projectile is fired into the air and suddenly explodes With no explosion, the projectile would follow the dotted line After the explosion, the center of mass of the fragments still follows the dotted line, the same parabolic path the projectile would have followed with no explosion! If no external forces act on the system, then the velocity of the CM doesn t change!!

8 Center of Mass: Stability If the Center of Mass is above the base of support the object will be stable. If not, it topples over.

9 Balance and Stability This dancer balances en pointe by having her center of mass directly over her toes, her base of support. Slide 12-88

10 Center of Mass The geometric center or average location of the mass. Extended Body: System of Particles: r CM 1 M r dm x CM i i mx M i

11 CM Lecture Problem 1. Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. A) What is the CM of the system? B) What is the torque acting on the system? C) What is the rotational inertia of the system about the end? D) If this body is released from rest in a horizontal position, what is the angular acceleration at the release? E) What is the angular speed of the meter stick as it swings through its lowest position?

12 Prelab A meter stick has a mass of 75.0 grams and has two masses attached to it 50.0 grams at the 20.0cm mark and grams at the 75.0 cm mark. (a) Find the center of mass of the system - that is, at what mark on the meter stick should the fulcrum be placed so that the system balances? (b) A fulcrum is then placed at the CM. Sketch and label the system. (c) Show that the net torque about the cm is zero. (d) Calculate the rotational inertia of the system about the CM axis. Box answers, make it neat.

13 Center of Mass The geometric center or average location of the mass. Extended Body: r CM 1 M r dm

14 Center of Mass of a Solid Object Divide a solid object into many small cells of mass m. As m 0 and is replaced by dm, the sums become Before these can be integrated: dm must be replaced by expressions using dx and dy. Integration limits must be established. Slide 12-41

15 Example 12.2 The Center of Mass of a Rod Slide 12-42

16 Example 12.2 The Center of Mass of a Rod Slide 12-43

17 Example Extended Body: 1 rcm rdm M You must generate an expression for the density and the mass differential, dm, from geometry and by analyzing a strip of the sign. We assume the sign has uniform density. If M is the total mass then the total volume and density is given by: x CM V 1 M abt, 2 1 abt 2 M 2My dm dv ( ytdx) ytdx dx 1 abt ab 2 1 M x dm Where a, b and t are the width, height and thickness of the sign, respectively. Then the mass element for the strip shown is: a 3 a 1 2My 2 b 2 x 2 x dx x ( x) dx a M ab ab a a

18 Newton s 2 nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration. I Acceleration thing Force thing Inertia thing The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.

19 Torque: Causes Rotations Fr sin Fd lever arm: d rsin The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force The horizontal component of F (F cos ) has no tendency to produce a rotation

20 Torque: Causes Rotations Fr sin Fd The direction convention is: Counterclockwise rotations are positive. Clockwise rotations are negative.

21 Newton s 1 st Law for Rotation If the sum of the torques is zero, the system is in rotational equilibrium. boy 500N 1.5m 750Nm 0 girl 250N 3m 750Nm

22 Torque Is there a difference in torque? (Ignore the mass of the rope) NO! In either case, the lever arm is the same! What is it? 3m

23 CM Lecture Problem 1. Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. A) What is the CM of the system? B) What is the torque acting on the system? C) What is the rotational inertia of the system about the end? D) If this body is released from rest in a horizontal position, what is the angular acceleration at the release? E) What is the angular speed of the meter stick as it swings through its lowest position?

24 Newton s 2 nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration. I Acceleration thing Force thing Inertia thing The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.

25 Torque is a Vector! = r x F The direction is given by the right hand rule where the fingers extend along r and fold into F. The Thumb gives the direction of.

26 The Vector Product The magnitude of C is AB sin and is equal to the area of the parallelogram formed by A and B The direction of C is perpendicular to the plane formed by A and B The best way to determine this direction is to use the right-hand rule A B ˆ i ˆ j k ˆ A B A B A B A B A B A B y z z y x z z x x y y x

27 COMPARE! rf Fr sin Fd CROSS PRODUCT F and d must be mutually perpendicular! L r p L mvr sin CROSS PRODUCT L and p must be mutually perpendicular! DOT PRODUCT W F d Fd cos F and d must be mutually PARALLEL!

28 A B ˆ i ˆ j k ˆ A B A B A B A B A B A B y z z y x z z x x y y x Two vectors lying in the xy plane are given by the equations A = 5i + 2j and B = 2i 3j. The value of AxB is a. 19k b. 11k c. 19k d. 11k e. 10i j

29 CM Lecture Problem 1. Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. A) What is the CM of the system? B) What is the torque acting on the system? C) What is the rotational inertia of the system about the end? D) If this body is released from rest in a horizontal position, what is the angular acceleration at the release? E) What is the angular speed of the meter stick as it swings through its lowest position?

30 Net external torques

31 Find the Net Torque Fr sin Fd Fd F d ( 20 N)(.5 m) (35 N)(1.10 msin60) OR ( 20 N)(.5 m) (35N cos30)(1.10 m) F and d must be mutually perpendicular! 23.3 Nm CCW

32 Newton s 2 nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration. I Acceleration thing Force thing Inertia thing The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.

33 NEW: Rotational Inertia The resistance of an object to rotate. The further away the mass is from the axis of rotation, the greater the rotational inertia.

34 Rotational Inertia The resistance of an object to rotate. Extended Body: System of Particles: I 2 r dm I 2 i i m r

35 Moments of Inertia of Various Rigid Objects

36 Rotational Inertia Depends on the axis.

37 Calculate Rotational Inertia For a system of point particles: 2 I m r I i i i Where r is the distance to the axis of rotation. SI units are kg. m 2

38 CM Lecture Problem 1. Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. A) What is the CM of the system? B) What is the torque acting on the system? C) What is the rotational inertia of the system about the end? D) If this body is released from rest in a horizontal position, what is the angular acceleration at the release? E) What is the angular speed of the meter stick as it swings through its lowest position?

39 Moment of Inertia of a Rigid Object: The Rod I 2 r dm

40 Calculate: Moment of Inertia of a Rigid Object 2 I r dm R dm 2 The trick is to write dm in terms of the density: dm dv Divide the cylinder into concentric shells with radius r, thickness dr and length L: R I r dm r 2 Lr dr LR 2 0 But M / 2 R L dv 2 rl dr I z 1 2 MR 2

41 Newton s 2 nd Law for Rotation The net external torques acting on an object around an axis is equal to the rotational inertia times the angular acceleration. I Force thing Inertia thing Acceleration thing The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.

42 Tangential and Angular a t dv dt Acceleration d dt d ( r ) dt d r dt at r

43 A 50 N m torque acts on a wheel of moment of inertia 150 kg m 2. If the wheel starts from rest, how long will it take the wheel to make a quarter turn (90 degrees)? 1 Use : 0t t 2 I t t t 2 / I 2 I t 2 / 2rad 50 N m /150kg m 2 3.1s

44 a)? b)? m R 24.3kg.314m F r F r 1 2 ( 90N 125 N).314m 11Nm = I 1 mr 2 11Nm = 1/2 24.3kg(.314m) 2 2 = 9.2 rad / s 2

45 CM Lecture Problem 1. Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. A) What is the CM of the system? B) What is the torque acting on the system? C) What is the rotational inertia of the system about the end? D) If this body is released from rest in a horizontal position, what is the angular acceleration at the release? E) What is the angular speed of the meter stick as it swings through its lowest position?

46 Superposition of Inertia The Parallel Axis Th m Superposition: Inertia ADD I I i The theorem states I = I CM + MD 2 I is about any axis parallel to the axis through the center of mass of the object I CM is about the axis through the center of mass D is the distance from the center of mass axis to the arbitrary axis

47 Moment of Inertia for a Rod Rotating The moment of inertia of the rod about its center is The position of the CM is D=½ L Therefore, Around One End CM ICM I I MD ML 2 2 L I ML M ML It is easier to rotate a rod about its center than about an end. 2

48 Rotational Inertia: Parallel Axis Theorem A uniform rod (mass = 2.0 kg, length = 0.60 m) is free to rotate about a frictionless pivot at one end. The rod is released from rest in the horizontal position. What is the magnitude of the angular acceleration of the rod at the instant it is 60 below the horizontal? a. 15 rad/s2 b. 12 rad/s2 c. 18 rad/s2 d. 29 rad/s2 e. 23 rad/s2

49 Rotational Energy A rotating object has kinetic energy because all particles in the object are in motion. The kinetic energy due to rotation is called rotational kinetic energy. Adding up the individual kinetic energies, and using v i = r i : 2013 Pearson Education, Inc. Slide 12-47

50 Rotational Energy Define the object s moment of inertia: Then the rotational kinetic energy is simply The units of moment of inertia are kg m 2. Moment of inertia depends on the axis of rotation. Mass farther from the rotation axis contributes more to the moment of inertia than mass nearer the axis. This is not a new form of energy, merely the familiar kinetic energy of motion written in a new way Pearson Education, Inc. Slide 12-48

51 CM Lecture Problem 1. Identical particles are placed at the 50-cm and 80-cm marks on a meter stick of negligible mass. This rigid body is then mounted so as to rotate freely about a pivot at the 0-cm mark on the meter stick. A) What is the CM of the system? B) What is the torque acting on the system? C) What is the rotational inertia of the system about the end? D) If this body is released from rest in a horizontal position, what is the angular acceleration at the release? E) What is the angular speed of the meter stick as it swings through its lowest position?

52 Rotational Inertia Which reaches the bottom first? (Same mass and radius)

53 Why Solid Cylinder? PROVE IT! ICM mr 2 ICM 1 2 mr 2

54 Rolling The red curve shows the path moved by a point on the rim of the object. This path is called a cycloid The green line shows the path of the center of mass of the object which moves in linear motion.

55 Rolling without Slipping All points on a rolling object move in a direction perpendicular to an axis through the instantaneous point of contact P. In other words, all points rotate about P. The center of mass of the object moves with a velocity v CM, and the point P moves with a velocity 2v CM. (Why 2?) Fig , p.317

56 For pure rolling motion, (no slipping) as the cylinder rotates through an angle, its center moves a linear distance s = R with speed v CM. At any instant, the point on the rim located at point P is at rest relative to the surface since no slipping occurs.

57 The motion of a rolling object can be modeled as a combination of pure translation and pure rotation. Translation: CM moves with v CM. Rotation: All points rotate about P with angular speed. Fig , p.318

58 Rolling Without Slipping Friction If an object rolls without slipping then the frictional force that causes the rotation is a static force. If no slipping occurs, the point of contact is momentarily at rest and thus friction is static and does no work on the object. No energy is dissipated and Mechanical Energy is Conserved. If the object slips, then friction is not static, does work on the object and dissipates energy.

59 Rolling Without Slipping K I P Kinetic Energy Use Parallel-axis Th : I I MR m 2 P CM 2 2 K 1 ( I ) CM MR K I mv CM CM The total kinetic energy of a rolling object is the sum of the rotational kinetic energy about the center of mass and the translational kinetic energy of the center of mass.

60 Rolling Without Slipping Conservation of Energy vcm 10 7 gh E i E f Mgh MvCM ICM vcm MvCM ICM ( ) 2 2 R v ( )( CM Mv MR ) R MvCM MvCM Mv CM 2 CM

61 Rolling Without Slipping Conservation of Energy Compare to Slipping Only: What about little ball big ball? vcm 10 7 gh E i Mgh vcm E 1 2 f Mv 2 2gh Some of the gravitational potential energy goes into rolling the sphere so the translational velocity of the cm is less when rolling!

62 Rolling Without Slipping Conservation of Energy Compare to Slipping Only: Rolling down: vcm 10 7 gh E i Mgh vcm E 1 2 f Mv 2 2gh What about the acceleration down the incline? Does change while in flight?

63 Rolling Down the Incline

64 Rotational Inertia Which reaches the bottom first? (Same mass and radius)

65 Why Solid Cylinder? PROVE IT! ICM mr 2 ICM 1 2 mr 2

66 Total Energy E KE KE PE trans rot 1 1 mgh mv I v mv I ( ) 2 2 r 2 2 v 2mgh 0 m I / r 2

67 General equation for the total final velocity at the end of the ramp: v 2mgh m I / r 2 v gh 0 v 4 3 gh 0 Solid Disk has the greatest velocity at the bottom of the ramp! Note: the velocity is independent of the radius!

68 Total Kinetic Energy A 1.0-kg wheel in the form of a solid disk rolls without slipping along a horizontal surface with a speed of 6.0 m/s. What is the total kinetic energy of the wheel? Ktotal Ktrans Krot mv I v ( )( ) 2 mv mr r mv mv (1 )(6 / ) 2 4 mv 4 kg m s 27J What I to use?

69 Rolling Ball Problem A tennis ball starts from rest and rolls without slipping down the hill. Treat the ball as a solid sphere. Find the range x. R v t v cos t x mgh mvcm ICM ( 2 v 10 )( ) 2 mv Mr v gh r v sin 2 y v 0 v sin t gt t yt a yt 2 2 g v R vxt v cos t 2 2sincos (10 / 7 gh)2sin cos g g 10h sin 2 R =.25m 7g Does change while in flight?

70 Angular Momentum L I

71 If no NET external Torques act on a system then Angular Momentum is Conserved. Linitial I Lfinal I

72 Angular Momentum L I

73 Angular Momentum L I

74 A Skater spins with an angular speed of 2 rev/s. If she brings her arms in and decreases her rotational inertia by a factor of 5, what is her new angular speed in rev/s? L I I L f f f f 0 I I 0 2 rev / s I f I /5 10 rev / s L I L I f f f

75 Angular Momentum for a Point Particle Single mass, a distance r from the axis of rotation. L I r v mr 2 v mvr r L mvr

76 6 rp 8.37x10 m 6 r 25.1x10 m A vp 8450 m / s v A? What is the velocity of the satellite at apogee?

77 6 rp 8.37x10 m 6 r 25.1x10 m A vp 8450 m / s v A? Is the angular momentum of the system CONSERVED? L A L P

78 6 rp 8.37x10 m 6 r 25.1x10 m A vp 8450 m / s v A? mr v mr v A A P P v A r P v 2820m P / s r A

79 Angular Momentum is a Vector Angular Speed is a Vector When a rigid object rotates about an axis, the angular momentum L is in the same direction as the angular velocity, according to the expression L = I, both directions given by the RIGHT HAND RULE.

80 Angular Momentum is a Vector L r p L mvr sin Only the perpendicular component of p contributes to L. The angular momentum L of a particle of mass m and linear momentum p located at the vector position r is a vector given by L = r p. The value of L depends on the origin about which it is measured and is a vector perpendicular to both r and p.

81 Torque Changes Angular Momentum ext L t S ext = dl/dt Looks like SF ext = dp/dt S and L must be measured about the same origin This is valid for any origin fixed in an inertial frame

82 A particle whose mass is 2 kg moves in the xy plane with a constant speed of 3 m/s along the direction r = i + j. What is its angular momentum (in kg m 2 /s) relative to the origin? a. 0 k b. 6k c. 6k d. 6 k e. 6 k Only the perpendicular component of p contributes to L! L mvr sin p = 0

83 Torque Changes Angular Momentum ext L t S ext = dl/dt S and L must be measured about the same origin This is valid for any origin fixed in an inertial frame = r x F = dl/dt The torque is perpendicular to both the applied force and the lever arm, but parallel to the angular speed, angular acceleration and angular momentum.

84 COMPARE! rf Fr sin Fd CROSS PRODUCT F and d must be mutually perpendicular! L r p L mvr sin CROSS PRODUCT L and p must be mutually perpendicular! DOT PRODUCT W F d Fd cos F and d must be mutually PARALLEL!

85 Colliding Probems L r p L mvr sin

86 A particle of mass m = 0.10 kg and speed v0 = 5.0 m/s collides and sticks to the end of a uniform solid cylinder of mass M = 1.0 kg and radius R = 20 cm. If the cylinder is initially at rest and is pivoted about a frictionless axle through its center, what is the final angular velocity (in rad/s) of the system after the collision? a. 8.1 b. 2.0 c. 6.1 d. 4.2 e. 10 L r p L mvr sin

87 System of Particles Center of Mass v CM dr d( m r / M ) m v dt dt M i i i i CM i i CM i M v m v p p i i i i tot v CM 1 M i m i v i v CM 1 M p tot p tot Mv CM Note: If the velocity of the CM is zero, the total momentum is zero! p p Mv tot,final tot,initial CM

88 Chapter 12: Statics Newton s 1 st Law: Conditions for Equilibrium If the sum of the net external torques is zero, the system is in rotational equilibrium. 0 If the sum of the net external forces is zero, the system is in translational equilibrium. SF = 0

89 The Ladder Problem The ladder is 8m long and weighs 355 N. The weight of the firefighter is 875N and he stands 2.30m from the center of mass of the ladder. If the wall is frictionless, find the minimum coefficient of friction of the floor so that the ladder doesn t slip.

90 F 0 0

91 What Keeps a Spinning Gyroscope from Falling Down?

92 Precession Keeps it from Falling! The Couple are equal and opposite forces so SF = 0. Since the normal force is through the axis of rotation P (the support) it produces no torque. Only the weight of the CM produces a torque with lever arm r that changes the angular momentum, L. P L t r L is in the direction of! This causes the Precession!

93 Spinning Wheel What happens if you rotate the wheel while sitting in a spin stool? The stool will spin in the direction of the torque. Conservation of Angular Momentum!

94 The Gyroscopic Effect Once you spin a gyroscope, its axle wants to keep pointing in the same direction. If you mount the gyroscope in a set of gimbals so that it can continue pointing in the same direction, it will. This is the basis of the gyro-compass which is used in navigation systems.

95 What is the best gyroscope on Earth?

96 Earth s Precession The Earth s precession is due to the gravitational tidal forces of the Moon and Sun applying torque as they attempt to pull the Equatorial Bulge into the plane of Earth s orbit.

97 The Earth s Precession causes the position of the North Pole to change over a period of 26,000 years.

98 TOTAL Angular Momentum Orbital and Spin Angular Momentum The TOTAL angular momentum includes both the angular momentum due to revolutions (orbits) and rotations (spin). It is the TOTAL angular momentum that is conserved.

99 Earth- Moon System: Total Angular Momentum is Conserved! Earth Rotation Slowing due to friction of ocean on bottom.0023 s per century: 900 Million yrs ago, Earth day was 18 hrs! Decrease of Earth s spin angular momentum, increases the orbital angular momentum of the Moon by increasing the distance, r, in order to keep L conserved! Earth is slowing down and Moon is moving further away!

100 Bouncing laser beams off the Moon demonstrates that it slowly moving away from the Earth ~.25 cm/month

101 Angular Momentum as a Fundamental Quantity The concept of angular momentum is also valid on a submicroscopic scale Angular momentum has been used in the development of modern theories of atomic, molecular and nuclear physics 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 It is a part of their nature

102 Orbital Angular Momentum & its Z-component are Quantized Angular momentum magnitude is quantized: h L l( l 1) ( l 0,1,2,... n 1) 2 Angular momentum direction is quantized: h L m ( m l,.. 1,0,1,2,... l) Z l 2 l Note: for n = 1, l = 0. This means that the ground state angular momentum in hydrogen is zero, not h/2, as Bohr assumed. What does it mean for L = 0 in this model?? Standing Wave

103 Intrinsic Spin is Quantized Fine Line Splitting: B field due to intrinsic spin interacts with B field due to orbital motion and produces additional energy states. Spin magnitude is quantized: h 1 S s( s 1) ( s ) 2 2 Spin direction is quantized: h 1 1 S m ( m, ) Z s 2 s 2 2

104 Cosmic Rotations: Conservation of Angular Momentum

105 Does the Universe have NET Angular Momentum? According the General Relativity, The Universe is ISOTROPIC it looks the same in every direction. Net Angular Momentum would define a direction.

106 Isotropic Universe

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