DYNAMICS VECTOR MECHANICS FOR ENGINEERS: Plane Motion of Rigid Bodies: Energy and Momentum Methods. Tenth Edition CHAPTER

Transcription

1 Tenth E CHAPTER 7 VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr. Phillip J. Cornwell Lecture Notes: Brian P. Self California State Polytechnic University Plane Motion of Rigid Bodies: Energy and Momentum Methods 03 The McGraw-Hill Companies, Inc. All rights reserved.

2 Introduction Method of work and energy and the method of impulse and momentum will be used to analyze the plane motion of rigid bodies and systems of rigid bodies. Principle of work and energy is well suited to the solution of problems involving displacements and velocities. T U T 03 The McGraw-Hill Companies, Inc. All rights reserved. 7 -

3 Introduction Principle of impulse and momentum is appropriate for problems involving velocities and time. L t t Fdt L 03 The McGraw-Hill Companies, Inc. All rights reserved. t H O M Odt H O Problems involving eccentric impact are solved by supplementing the principle of impulse and momentum with the application of the coefficient of restitution. t 7-3

4 Introduction Approaches to Rigid Body Kinetics Problems Forces and Accelerations Velocities and Displacements Velocities and Time Newton s Second Law (last chapter) F M G ma G H G Work-Energy U T 03 The McGraw-Hill Companies, Inc. All rights reserved. T Impulse- Momentum t mv F dt mv t t I M dt I t G G G - 4

5 Kinetic Energy of Rigid Bodies Here, we express kinetic energy in terms of velocity of C.M., Inertia and angular velocity. 03 The McGraw-Hill Companies, Inc. All rights reserved. - 5

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10 Principle of Work and Energy for a Rigid Body Work and kinetic energy are scalar quantities. Assume that the rigid body is made of a large number of particles. T U T, T U T initial and final total kinetic energy of particles forming body total work of internal and external forces acting on particles of body. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-0

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14 The answer is zero as f and r are in the same direction and so the cross product is zero. 03 The McGraw-Hill Companies, Inc. All rights reserved. 4

15 So in a rigid body work is only due to external forces. Internal forces do not contribute to work. 03 The McGraw-Hill Companies, Inc. All rights reserved. 5

16 How we use these two equations is shown in next slide. 03 The McGraw-Hill Companies, Inc. All rights reserved. 6

17 03 The McGraw-Hill Companies, Inc. All rights reserved. Soln.

18 03 The McGraw-Hill Companies, Inc. All rights reserved. f=4.n

19 Principle of Work and Energy for a Rigid Body Assume that the rigid body is made of a large number of particles. T U T, T U T initial and final total kinetic energy of particles forming body total work of external forces acting on particles of body. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-9

20 Work of Forces Acting on a Rigid Body Work of a force during a displacement of its point of application, U A A F dr s F cos s ds 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-0

21 Work of Forces Acting on a Rigid Body Consider the net work of two forces F and F forming a couple of moment during a displacement of their points of application. du F dr F ds M d M F dr Fr d F dr U M d M 03 The McGraw-Hill Companies, Inc. All rights reserved. if M is constant. 7 -

22 Kinetic Energy of a Rigid Body in Plane Motion Consider a rigid body of mass m in plane motion consisting of individual particles i. The kinetic energy of the body can then be expressed as: T mv Δm v i i mv r Δm i i mv I 03 The McGraw-Hill Companies, Inc. All rights reserved. 7 -

23 Kinetic Energy of a Rigid Body in Plane Motion Kinetic energy of a rigid body can be separated into: - the kinetic energy associated with the motion of the mass center G and - the kinetic energy associated with the rotation of the body about G. T mv I Translation + Rotation 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-3

24 Kinetic Energy of a Rigid Body in Plane Motion Consider a rigid body rotating about a fixed axis through O. T Δm v Δm r r Δm i i i i i i I O This is equivalent to using: T mv I Remember to only use T O I when O is a fixed axis of rotation 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-4

25 Systems of Rigid Bodies For problems involving systems consisting of several rigid bodies, the principle of work and energy can be applied to each body. T 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-5

26 Systems of Rigid Bodies We may also apply the principle of work and energy to the entire system, T,T = arithmetic sum of the kinetic energies of all bodies forming the T U T U system = work of all forces acting on the various bodies, whether these forces are internal or external to the system as a whole. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-6

27 Work of Forces Acting on a Rigid Body Do the pin forces at point A do work? YES NO Does the force P do work? YES NO 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-7

28 Work of Forces Acting on a Rigid Body Does the normal force N do work on the disk? YES NO Does the weight W do work? YES NO If the disk rolls without slip, does the friction force F do work? YES NO du F ds C F v dt 0 03 The McGraw-Hill Companies, Inc. All rights reserved. c - 8

29 Systems of Rigid Bodies For problems involving pin connected members, blocks and pulleys connected by inextensible cords, and meshed gears, - internal forces occur in pairs of equal and opposite forces - points of application of each pair move through equal distances - net work of the internal forces is zero - work on the system reduces to the work of the external forces 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-9

30 Conservation of Energy Expressing the work of conservative forces as a change in potential energy, the principle of work and energy becomes T V T V Consider the slender rod of mass m. mass m released with zero velocity determine at 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-30

31 Conservation of Energy Consider the slender rod of mass m. T mv m T 0, V 0 l ml I ml 3 V Wl sin mgl sin 03 The McGraw-Hill Companies, Inc. All rights reserved. T V T V ml 0 3 3g sin l mgl sin 7-3

32 Power Power = rate at which work is done For a body acted upon by force velocity, du v Power dt F and moving with F v For a rigid body rotating with an angular velocity and acted upon by a couple of moment M parallel to the axis of rotation, Power du dt M d dt M 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-3

33 Sample Problem 7.4 A 5-kg slender rod pivots about the point O. The other end is pressed against a spring (k = 300 kn/m) until the spring is compressed 40 mm and the rod is in a horizontal position. If the rod is released from this position, determine its angular velocity and the reaction at the pivot as the rod passes through a vertical position. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-33

34 Sample Problem 7.4 SOLUTION: The weight and spring forces are conservative. The principle of work and energy can be expressed as T V T V Evaluate the initial and final potential energy. Express the final kinetic energy in terms of the final angular velocity of the rod. Based on the free-body-diagram equation, solve for the reactions at the pivot. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-34

35 Sample Problem 7.4 SOLUTION: The weight and spring forces are conservative. The principle of work and energy can be expressed as T V T V 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-35

36 Sample Problem 7.4 Evaluate the initial and final potential energy. = g + e = + = ( )( ) V V V 0 kx 300,000 N m 0.04m = 40J ( )( ) V = Vg + Ve = Wh + 0 = 47.5 N 0.75m = 0.4J 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-36

37 Sample Problem 7.4 Express the final kinetic energy in terms of the angular velocity of the rod. ( ) = + w = w + w T mv I m r I I = ml = 5kg.5m ( )( ) = w + ( 7.8 ) w = 8. w ( )( ) = 7.8kg m 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-37

38 Sample Problem 7.4 From the principle of work and energy, T + V = T + V J = 8.w + 0.4J w = 3.995rad s 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-38

39 03 The McGraw-Hill Companies, Inc. All rights reserved. enth Sample Problem 7.4 Based on the free-bodydiagram equation, solve for the reactions at the pivot. a a n t ( )( ) = rw = 0.75m 3.995rad s =.97 m s = ra a a n t =.97m s = ra

40 Sample Problem 7.4 M O M O eff 0 I m r r 0 F x F x eff F y F y eff R x m r 0 R x N = - =- 03 The McGraw-Hill Companies, Inc. All rights reserved. R y R y =- ma n ( )( 5kg.97 m s ) 3.4 N R = 3.4 N 7-40

41 One can use the rate form of energy equation to calculate. 03 The McGraw-Hill Companies, Inc. All rights reserved. - 4

42 Sample Problem 7.5 Each of the two slender rods has a mass of 6 kg. The system is released from rest with b = 60 o. Determine a) the angular velocity of rod AB when b = 0 o, and b) the velocity of the point D at the same instant. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-4

43 Sample Problem 7.5 SOLUTION: Consider a system consisting of the two rods. With the conservative weight force, T V T V Evaluate the initial and final potential energy. Express the final kinetic energy of the system in terms of the angular velocities of the rods. Solve the energy equation for the angular velocity, then evaluate the velocity of the point D. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-43

44 Sample Problem 7.5 SOLUTION: Consider a system consisting of the two rods. With the conservative weight force, T V T V 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-44

45 Sample Problem 7.5 Evaluate the initial and final potential energy. V Wy 38.6J 58.86N 0.35m V Wy 5.0J 58.86N 0.83m W mg 58.86N 6kg 9.8m s 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-45

46 Sample Problem 7.5 Express the final kinetic energy of the system in terms of the angular velocities of the rods. vab 0.375m 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-46

47 Sample Problem 7.5 Since is perpendicular to AB and is horizontal, the instantaneous center of rotation for rod BD is C. BC 0.75m v B v D We can guess this because velocity of B is perpendicular to AB and also BC. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-47

48 Sample Problem 7.5 CD 0.75msin m and applying the law of cosines to CDE, EC = 0.5 m 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-48

49 Sample Problem 7.5 Consider the velocity of point B v B 03 The McGraw-Hill Companies, Inc. All rights reserved. AB BC AB v BD 0.5m BD 7-49

50 Sample Problem 7.5 For the final kinetic energy, I AB I BD 6kg0.75m 0.8kg ml m T mv AB AB 03 The McGraw-Hill Companies, Inc. All rights reserved. I AB mv BD I BD BD 7-50

51 Sample Problem 7.5 Solve the energy equation for the angular velocity, then evaluate the velocity of the point D. 0 T V 38.6J T V J 3.90rad s AB 3.90rad s 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-5

52 Sample Problem 7.5 v D CD 0.53m.00m s 3.90rad s v D.00m s 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-5

53 Sample Problem 7. m m A B 0kg 3kg k k A B 00mm 80mm 03 The McGraw-Hill Companies, Inc. All rights reserved. The system is at rest when a moment of M 6 N mis applied to gear B. Neglecting friction, a) determine the number of revolutions of gear B before its angular velocity reaches 600 rpm, and b) tangential force exerted by gear B on gear A. 7-53

54 Sample Problem 7. SOLUTION: Consider a system consisting of the two gears. Noting that the gear rotational speeds are related, evaluate the final kinetic energy of the system. Apply the principle of work and energy. Calculate the number of revolutions required for the work of the applied moment to equal the final kinetic energy of the system. Apply the principle of work and energy to a system consisting of gear A. With the final kinetic energy and number of revolutions known, calculate the moment and tangential force required for the indicated work. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-54

55 Sample Problem 7. SOLUTION: Consider a system consisting of the two gears. Noting that the gear rotational speeds are related, evaluate the final kinetic energy of the system. B A 600rpm rad rev 60s min The McGraw-Hill Companies, Inc. All rights reserved. B r r B A 6.8rad 5.rad s s 7-55

56 Sample Problem 7. I I A B m m A B k k A B 0kg 3kg 0.00m 0.080m 0.400kg 0.09kg m m T A J I A I B 5. B The McGraw-Hill Companies, Inc. All rights reserved. 7-56

57 Sample Problem 7. Apply the principle of work and energy. Calculate the number of revolutions required for the work. T 0 B U 6 B J T 7.3rad 63.9J B 03 The McGraw-Hill Companies, Inc. All rights reserved rev 7-57

58 Sample Problem 7. Apply the principle of work and energy to a system consisting of gear A. Calculate the moment and tangential force required for the indicated work. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-58

59 Sample Problem 7. A B r r B A rad 0.50 T J I A A T U T Transfer F to the centre of the gear. You get a moment and a force. The force does no work and work of moment is calculated as M 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-59

60 Sample Problem 7. T 0 M U A M A r A 0.93rad F T 6.0J.5N m F N 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-60

61 Team Problem Solving A slender 4-kg rod can rotate in a vertical plane about a pivot at B. A spring of constant k = 400 N/m and of unstretched length 50 mm is attached to the rod as shown. Knowing that the rod is released from rest in the position shown, determine its angular velocity after it has rotated through 90 o. 03 The McGraw-Hill Companies, Inc. All rights reserved. - 6

62 Team Problem Solving SOLUTION: Because the problem deals with positions and velocities, you should apply the principle of work energy. Draw out the system at position and position and define your datum Use the work-energy equation to determine the angular velocity at position 03 The McGraw-Hill Companies, Inc. All rights reserved. - 6

63 Draw your diagrams, set your datum and apply the work energy equation T V U T V Are any of the T V U T V terms zero? 03 The McGraw-Hill Companies, Inc. All rights reserved. - 63

64 Team Problem Solving Determine the spring energy at position x Unstretched Length CD (50 mm ) mm 0. m V e kx (400 N/m)(0. m) 9.68 J Determine the potential energy due to gravity at position Vg Wh mgh (4 kg)(9.8 m/s )( 0. m) J 03 The McGraw-Hill Companies, Inc. All rights reserved. - 64

65 Team Problem Solving Determine the spring energy at position x 30 mm 50 mm 80 mm 0.08 m V (400 N/m)(0.08 m).8 J e kx Determine the potential energy due to gravity at position Vg 0 03 The McGraw-Hill Companies, Inc. All rights reserved. - 65

66 Team Problem Solving Determine an expression for T T mv I Can you relate v and? v r (0.8 m) 03 The McGraw-Hill Companies, Inc. All rights reserved. - 66

67 Team Problem Solving Find I and substitute in to T (4 kg)(0.6 m) 0. kg m I ml T mv I (4 kg)(0.8 ) (0.) 0.48 Substitute into T + V = T + V J The McGraw-Hill Companies, Inc. All rights reserved rad/s - 67

68 Concept Question For the previous problem, how would you determine the reaction forces at B when the bar is horizontal? a) Apply linear-momentum to solve for B x Dt and B y Dt b) Use work-energy to determine the work done by the moment at C c) Use sum of forces and sum of moments equations when the bar is horizontal 03 The McGraw-Hill Companies, Inc. All rights reserved. - 68

69 Each of the two wheels in the mechanism of Fig. has mass m and axial radius of gyration k 0. Each link OB has mass m and is modelled as a thin rod of length L. The collar of mass m 3 slides on the fixed vertical shaft with a constant friction force F. The spring has stiffness k and is contacted by the collars when the links reach the horizontal position. The system is released from rest at position Ѳ = Ѳ. Assume that the friction is sufficient to prevent the wheels from slipping. 03 The McGraw-Hill Companies, Inc. All rights reserved. - 69

70 Find (a) the acceleration of the collar at the instant of release, (b) the velocity and acceleration of the collar when Ѳ = Ѳ >0, (c) the velocity and acceleration of the collar when Ѳ =0 and (d) The maximum compression of the spring. (e) Work out parts a to d if the rollers are smooth. 03 The McGraw-Hill Companies, Inc. All rights reserved. - 70

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79 Angular Impulse Momentum When two rigid bodies collide, we typically use principles of angular impulse momentum. We often also use linear impulse momentum (like we did for particles). 03 The McGraw-Hill Companies, Inc. All rights reserved. - 79

80 Introduction Approaches to Rigid Body Kinetics Problems Forces and Accelerations Velocities and Displacements Velocities and Time Newton s Second Law (last chapter) F M ma G G H G Work-Energy T U T Impulse- Momentum t mv F dt mv t t I M dt I t G G G 03 The McGraw-Hill Companies, Inc. All rights reserved. - 80

81 Principle of Impulse and Momentum Method of impulse and momentum: - well suited to the solution of problems involving time and velocity - the only practicable method for problems involving impulsive motion and impact. Sys Momenta + Sys Ext Imp - = Sys Momenta 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-8

82 Principle of Impulse and Momentum The momenta of the particles of a system may be reduced to a vector attached to the mass center equal to their sum, L v Δm i i mv and a couple equal to the sum of their moments about the mass center, H G r i v i Δm i 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-8

83 Principle of Impulse and Momentum For the plane motion of a rigid slab or of a rigid body symmetrical with respect to the reference plane, H G I 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-83

84 Principle of Impulse and Momentum For plane motion problems, draw out an impulse-momentum diagram, (similar to a free-body diagram) 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-84

85 Principle of Impulse and Momentum This leads to three equations of motion: - summing and equating momenta and impulses in the x and y directions - summing and equating the moments of the momenta and impulses with respect to any given point (often choose G) 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-85

86 Impulse Momentum Diagrams A sphere S hits a stationary bar AB and sticks to it. Draw the impulse-momentum diagram for the ball and bar separately; time is immediately before the impact and time is immediately after the impact. 03 The McGraw-Hill Companies, Inc. All rights reserved. - 86

87 Principle of Impulse and Momentum Fixed axis rotation: - The angular momentum about O I O I mv r I I mr mr r 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-87

88 Principle of Impulse and Momentum I - Equating the moments of the momenta and impulses about O, O t M dt IO The pin forces at point O now contribute no moment to the equation t O 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-88

89 Systems of Rigid Bodies Motion of several rigid bodies can be analyzed by applying the principle of impulse and momentum to each body separately. For problems involving no more than three unknowns, it may be convenient to apply the principle of impulse and momentum to the system as a whole. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-89

90 Systems of Rigid Bodies For each moving part of the system, the diagrams of momenta should include a momentum vector and/or a momentum couple. Internal forces occur in equal and opposite pairs of vectors and generate impulses that cancel out. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-90

91 Practice From the previous problem, notice that the impulse acting on the sphere is equal and opposite to the impulse acting on the bar. We can take advantage of this by drawing the impulse-momentum diagram of the entire system, as shown on the next slide. F imp Dt F imp Dt 03 The McGraw-Hill Companies, Inc. All rights reserved. - 9

92 Impulse Momentum Diagrams Momentum of the ball before impact Impulse on ball Momentum of the ball after impact F imp Dt 03 The McGraw-Hill Companies, Inc. All rights reserved. - 9

93 Impulse Momentum Diagrams Momentum of the bar before impact Impulse on bar Momentum of the bar after impact F imp Dt 03 The McGraw-Hill Companies, Inc. All rights reserved. - 93

94 Practice Diagram for combined system 03 The McGraw-Hill Companies, Inc. All rights reserved. - 94

95 Conservation of Angular Momentum The moments acting through the skater s center of gravity are negligible, so his angular momentum remains constant. He can adjust his spin rate by changing his moment of inertia. t I M dt I G G G t I G 03 The McGraw-Hill Companies, Inc. All rights reserved. G I - 95

96 Conservation of Angular Momentum When no external force acts on a rigid body or a system of rigid bodies, the system of momenta at t is equipollent to the system at t. The total linear momentum and angular momentum about any point are conserved, L L H 0 H0 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-96

97 Conservation of Angular Momentum When the sum of the angular impulses pass through O, the linear momentum may not be conserved, yet the angular momentum about O is conserved, H H 0 0 Two adal equations may be written by summing x and y components of momenta and may be used to determine two unknown linear impulses, such as the impulses of the reaction components at a fixed point. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-97

98 Concept Question For the problem we looked at previously, is the angular momentum about G conserved? YES NO For the problem we looked at previously, is the angular momentum about point A conserved? For the problem we looked at previously, is the linear momentum of the system conserved? 03 The McGraw-Hill Companies, Inc. All rights reserved. YES YES NO NO - 98

99 Sample Problem 7.6 m m A B 0kg 3kg k k A B 00mm 80mm 03 The McGraw-Hill Companies, Inc. All rights reserved. The system is at rest when a moment of M 6 N m is applied to gear B. Neglecting friction, a) determine the time required for gear B to reach an angular velocity of 600 rpm, and b) the tangential force exerted by gear B on gear A. 7-99

100 Sample Problem 7.6 SOLUTION: Considering each gear separately, apply the method of impulse and momentum. Solve the angular momentum equations for the two gears simultaneously for the unknown time and tangential force. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-00

101 Sample Problem 7.6 SOLUTION: Considering each gear separately, apply the method of impulse and momentum. moments about A: FtrA I A A 0.50m 0.400kg m5.rad s 0 Ft Ft 40. N s 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-0

102 Sample Problem Mt moments about B: Ftr B I 6 N mt Ft0.00m 0.09kg m 6.8rad s B B 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-0

103 Sample Problem 7.6 Solve the angular momentum equations for the two gears simultaneously for the unknown time and tangential force. t 0.87s F 46. N 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-03

104 Sample Problem 7.7 Uniform sphere of mass m and radius r is projected along a rough horizontal surface with a linear velocity v and no angular velocity. The coefficient of kinetic friction is k. Determine a) the time t at which the sphere will start rolling without sliding and b) the linear and angular velocities of the sphere at time t. 03 The McGraw-Hill Companies, Inc. All rights reserved. SOLUTION: Apply principle of impulse and momentum to find variation of linear and angular velocities with time. Relate the linear and angular velocities when the sphere stops sliding by noting that the velocity of the point of contact is zero at that instant. Substitute for the linear and angular velocities and solve for the time at which sliding stops. Evaluate the linear and angular velocities at that instant. 7-04

105 Sample Problem 7.7 SOLUTION: Apply principle of impulse and momentum to find variation of linear and angular velocities with time. Sys Momenta + Sys Ext Imp - = Sys Momenta y components: Nt Wt 0 x components: mv Ft mv mv mgt mv k moments about G: Ftr I kmg tr mr 5 N W v gt 03 The McGraw-Hill Companies, Inc. All rights reserved. v mg k 5 k g r t Relate linear and angular velocities when sphere stops sliding by noting that velocity of point of contact is zero at that instant. Substitute for the linear and angular velocities and solve for the time at which sliding stops. v v gt k r 5 g r k t r t 7 v g k 7-05

106 Sample Problem 7.7 Evaluate the linear and angular velocities at that instant. v v v g 7 k g k Sys Momenta + Sys Ext Imp - = Sys Momenta y components: x components: moments about G: N W v v mg gt k 5 k g r t 5 k g v r 7 k g 5 v v 7 5 v 7 r v v gt k r 5 g r k t r t 7 v g k 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-06

107 Sample Problem 7.8 Two solid spheres (radius = 00 mm, m = kg) are mounted on a spinning horizontal rod (of rod I R and pivot = 0.4 kg m, = 6 rad/sec) as shown. The balls are held together by a string which is suddenly cut. Determine a) angular velocity of the rod after the balls have moved to A and B, and b) the energy lost due to the plastic impact of the spheres and stops. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-07

108 Sample Problem 7.8 SOLUTION: Observing that none of the external forces produce a moment about the y axis, the angular momentum is conserved. Equate the initial and final angular momenta. Solve for the final angular velocity. I R The energy lost due to the plastic impact is equal to the change in kinetic energy of the system. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-08

109 Sample Problem 7.8 SOLUTION: Observing that none of the external forces produce a moment about the y axis, the angular momentum is conserved. Equate the initial and final angular momenta. Solve for the final angular velocity. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-09

110 Sample Problem 7.8 Sys Momenta + Sys Ext Imp - =SysMomenta 03 The McGraw-Hill Companies, Inc. All rights reserved. m r r I I s S R m r r I I s S R m m s s r r I I S S I I R R I R 6rad s 0.5 lb ft s 7-0

111 Sample Problem 7.8 IS ma kg 0.m 0.004kg m 5 5 ( )( ) = = = mr S mr S kg 0.m 0.0kg m kg 0.6m 0.36 kg m w =.8rad s 03 The McGraw-Hill Companies, Inc. All rights reserved. 7 -

112 Sample Problem 7.8 The energy lost due to the plastic impact is equal to the change in kinetic energy of the system. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7 -

113 Sample Problem 7.8 T m v I I S S R m r I I S S R T T ( )( ) = = J ( )( ) 03 The McGraw-Hill Companies, Inc. All rights reserved. =.8.8 =.93 J Energy Lost = T - T = T lost = 4.77 J 7-3

114 Team Problem Solving An 8-kg wooden panel P is suspended from a pin support at A and is initially at rest. A - kg metal sphere S is released from rest at B and falls into a hemispherical cup C attached to the panel at the same level as the mass center G. Assuming that the impact is perfectly plastic, determine the angular velocity of the panel immediately after the impact. 03 The McGraw-Hill Companies, Inc. All rights reserved. - 4

115 Team Problem Solving SOLUTION: Consider the sphere and panel as a single system. Apply the principle of impulse and momentum. The moments about A of the momenta and impulses provide a relation between the angular velocity of the panel and velocity of the sphere. Use the principle of work-energy to determine the angle through which the panel swings. 03 The McGraw-Hill Companies, Inc. All rights reserved. - 5

116 Team Problem Solving Draw the impulse momentum diagram 03 The McGraw-Hill Companies, Inc. All rights reserved. - 6

117 Team Problem Solving Apply the angular impulse momentum equation about point A Given: m S = kg, m P = 8 kg, h S = 0.50 m, e= 0. Find: Angle through which the panel and sphere swing after the impact m ( v ) (0. m) 0 m ( v ) ( AC ) I m v (0.5 m) S C S C P H A of sphere before impact H A of sphere after impact 03 The McGraw-Hill Companies, Inc. All rights reserved. H A of panel after impact - 7

118 Team Problem Solving m S ( v ) (0. m) 0 C m ( v ) ( AC) I m v (0.5 m) S C P Determine velocity of sphere at impact (v S ) You can apply work-energy or kinematics ( v ) gy S (9.8 m/s )(0.5 m) 3.3 m/s 03 The McGraw-Hill Companies, Inc. All rights reserved. - 8

119 Team Problem Solving Determine velocity of sphere after impact in terms of ( v ) S AC AC (0.) (0.5) m ( v ) S (perpendicular to AC.) 03 The McGraw-Hill Companies, Inc. All rights reserved. - 9

120 Team Problem Solving m S ( v ) (0. m) 0 C m ( v ) ( AC) I m v (0.5 m) S C P Determine mass moment of inertia for panel I m P (0.5 m) (8)(0.5) kg m The McGraw-Hill Companies, Inc. All rights reserved. - 0

121 Team Problem Solving Substitute into H equation and solve for m ( v ) (0. m) 0 m ( v ) ( AC ) I m v (0.5 m) S C S C P ( kg)(3.3 m/s)(0. m) ( kg)(0.306 )(0.306 m) (8 kg)(0.5 m).584 ( ).066 rad/s 03 The McGraw-Hill Companies, Inc. All rights reserved. -

122 Concept Question For the previous problem, what would you do if you wanted to determine how high up the panel swung after the impact? a) Apply linear-momentum to solve for mv G b) Use work-energy and set T final equal to zero c) Use sum of forces and sum of moments equations 03 The McGraw-Hill Companies, Inc. All rights reserved. -

123 Concept Question For the previous problem, what if the ball was dropped closer to point A (e.g., at x= 00 mm instead of 00 mm)? a) The angular velocity after impact would be bigger b) The angular velocity after impact would be smaller c) The angular velocity after impact would be the same d) Not enough information to tell 03 The McGraw-Hill Companies, Inc. All rights reserved. - 3

124 Eccentric Impact Period of deformation Impulse u A u n B n Rdt Period of restitution Impulse Pdt 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-4

125 Eccentric Impact Principle of impulse and momentum is supplemented by e coefficient of vb n va n va n vb n restitution 03 The McGraw-Hill Companies, Inc. All rights reserved. Rdt Pdt These velocities are for the points of impact 7-5

126 Sample Problem 7.0 A -kg sphere with an initial velocity of 5 m/s strikes the lower end of an 8-kg rod AB. The rod is hinged at A and initially at rest. The coefficient of restitution between the rod and sphere is 0.8. Determine the angular velocity of the rod and the velocity of the sphere immediately after impact. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-6

127 Sample Problem 7.0 SOLUTION: Consider the sphere and rod as a single system. Apply the principle of impulse and momentum. The moments about A of the momenta and impulses provide a relation between the final angular velocity of the rod and velocity of the sphere. The definition of the coefficient of restitution provides a second relationship between the final angular velocity of the rod and velocity of the sphere. Solve the two relations simultaneously for the angular velocity of the rod and velocity of the sphere. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-7

128 Sample Problem 7.0 SOLUTION: Consider the sphere and rod as a single system. Apply the principle of impulse and momentum. The moments about A of the momenta and impulses provide a relation between the final angular velocity of the rod and velocity of the rod. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-8

129 Sample Problem 7.0 moments about A: m s v s.m m v.m m v 0.6m I s s R R v R I r 0.6m ml 8kg.m 0.96kg m 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-9

130 Sample Problem 7.0 kg5m s.m kgvs.m 8kg0.6m 0.6m 0.96kg m.4vs The McGraw-Hill Companies, Inc. All rights reserved. 7-30

131 Sample Problem 7.0 The definition of the coefficient of restitution provides a second relationship between the final angular velocity of the rod and velocity of the sphere. Solve the two relations simultaneously for the angular velocity of the rod and velocity of the sphere. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-3

132 Sample Problem 7.0 Moments about A: Relative velocities:.4vs v B v s.m v s e v 0.8 B v s 5m s Solving, 3.rad/s v s 0.43m s v s 03 The McGraw-Hill Companies, Inc. All rights reserved. 3.rad/s 0.43m s 7-3

133 Concept Questions The cars collide, hitting at point P as shown. Which of the following can you use to help analyze the collision? a) The linear momentum of car A is conserved. b) The linear momentum of the combined two cars is conserved c) The total kinetic energy before the impact equals the total kinetic energy after the impact d) The angular momentum about the CG of car B is conserved P B A P 03 The McGraw-Hill Companies, Inc. All rights reserved. - 33

134 Sample Problem 7.9 SOLUTION: Consider a system consisting of the bullet and panel. Apply the principle of impulse and momentum. The final angular velocity is found from the moments of the momenta and impulses about A. A 5 g bullet is fired into the side of a 0-kg square panel which is initially at rest. The reaction at A is found from the horizontal and vertical momenta and impulses. Determine a) the angular velocity of the panel immediately after the bullet becomes embedded and b) the impulsive reaction at A, assuming that the bullet becomes embedded in s. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-34

135 Sample Problem 7.9 SOLUTION: Consider a system consisting of the bullet and panel. Apply the principle of impulse and momentum. moments about A: ( ) ( ) m v 0.4m + 0 = m v 0.5m + I w B B P P The final angular velocity is found from the moments of the momenta and impulses about A. v = ( 0.5m) w I m b ( )( ) ( )( )( ) ( )( w )( ) P = 6 P = 0 kg 0.5 m = 0.47kg m = w w v = 4.3rad s ( ) = 0.5 w =.08m s w = 4.3rad s 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-35

136 Sample Problem 7.9 The reactions at A are found from the horizontal and vertical momenta and impulses. w ( ) = 4.3rad s v = 0.5 w =.08 m s x components: m v + A D t = m v B B x p ( 0.5)( 450) + A ( ) = ( 0)(.08) A x =- 750N y components: x A x = 0 A y Dt 0 A y 0 750N 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-36

137 Sample Problem 7. SOLUTION: Apply the principle of impulse and momentum to relate the velocity of the package on conveyor belt A before the impact at B to the angular velocity about B after impact. A square package of mass m moves down conveyor belt A with constant velocity. At the end of the conveyor, the corner of the package strikes a rigid support at B. The impact is perfectly plastic. Derive an expression for the minimum velocity of conveyor belt A for which the package will rotate about B and reach conveyor belt C. Apply the principle of conservation of energy to determine the minimum initial angular velocity such that the mass center of the package will reach a position directly above B. Relate the required angular velocity to the velocity of conveyor belt A. 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-37

138 Sample Problem 7. SOLUTION: Apply the principle of impulse and momentum to relate the velocity of the package on conveyor belt A before the impact at B to angular velocity about B after impact. Moments about B: mv a 0 mv a I v a I m mv a 0 m a a ma 4 a 3 v 6 a 6 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-38

139 Sample Problem 7. h GBsin45 5 a sin a Apply the principle of conservation of energy to determine the minimum initial angular velocity such that the mass center of the package will reach a position directly above B. T 3 V T 3 V3 ma T mv m V Wh V3 Wh 3 a ma ma I T 3 0 (solving for the minimum ) 3W ma Wh 0 Wh 3g 6 h h 0.707a 0.6a 0.85g a 3 a h a a v a a 0. 85g a v 0. 7 ga 03 The McGraw-Hill Companies, Inc. All rights reserved. 7-39

140 Team Problem Solving SOLUTION: Consider the projectile and bar as a single system. Apply the principle of impulse and momentum. The moments about C of the momenta and impulses provide a relation between the final angular velocity of the rod and velocity of the projectile. Use the principle of work-energy to determine the angle through which the bar swings. A projectile of mass 40 g is fired with a horizontal velocity of 50 m/s into the lower end of a slender 7.5-kg bar of length L= 0.75 m. Knowing that h = 0.3 m and that the bar is initially at rest, determine the angular velocity of the bar when it reaches the horizontal position. 03 The McGraw-Hill Companies, Inc. All rights reserved. - 40

141 Team Problem Solving Draw the impulse momentum diagram Apply the angular impulse momentum equation about point C m v ( L h) m v ( L h) I B C Given: m o = 0.04 kg, v o = 50 m/s m AB = 7.5 kg L= 0.75 m h = 0.3 m Find: AB when = 90 o Or you could use the relationship: L m0v0 ( L h) m0vb ( L h) mv 0 h I 03 The McGraw-Hill Companies, Inc. All rights reserved. - 4

142 Team Problem Solving m v ( L h) m v ( L h) I B C () Relate v B and (after the impact) v ( L h) B Substitute into equation () and solve for m0v0 ( L h) m0 ( L h) IC Find I C L = 0.75 m m = 7.5 kg h = 0.3 m m0v0( L h) 0.75 m0 ( L h) I I (7.5)(0.75) C C = ml + m - h = + - è ø è ø Substitute and solve I C æl ö æ ö = kg m m0v0( L h) (0.04)(50)( ) m ( L h) I (0.04)( ) C rad/s 03 The McGraw-Hill Companies, Inc. All rights reserved. - 4

143 Team Problem Solving Draw position and, set your datum (datum set at ) and apply the conservation of energy equation T V T V Find T T IC ( )(6.789 ) T J Find V V m ABgy AB mogyo L V (7.5)(9.8) h (0.04)(9.8)( L h) 0.75 (7.5)(9.8) 0.3 (0.04)(9.8)( ) 5.345J Solve for 3 T I T V C rad/s ( ) The McGraw-Hill Companies, Inc. All rights reserved. B DATUM - 43

144 Concept Question For the previous problem, how would you determine the reaction forces at C when the bar is horizontal? a) Apply linear-momentum to solve for C x Dt and C y Dt b) Use work-energy to determine the work done by the moment at C c) Use sum of forces and sum of moments equations when the bar is horizontal 03 The McGraw-Hill Companies, Inc. All rights reserved. - 44

145 Concept Question For the previous problem, what would happen if the coefficient of restitution between the projectile and bar was.0 instead of zero? a) The angular velocity after impact would be bigger b) The angular velocity after impact would be smaller c) The angular velocity after impact would be the same d) Not enough information to tell 03 The McGraw-Hill Companies, Inc. All rights reserved. - 45