DYNAMICS. Kinetics of Particles: Energy and Momentum Methods VECTOR MECHANICS FOR ENGINEERS: Tenth Edition CHAPTER

Transcription

1 Tenth E CHPTER 3 VECTOR MECHNICS FOR ENGINEERS: DYNMICS Ferdinand P. Beer E. Russell Johnston, Jr. Phillip J. Cornwell Lecture Notes: Brian P. Self California Polytechnic State Uniersity Kinetics of Particles: Energy and Momentum Methods 03 The McGraw-Hill Companies, Inc. ll rights resered.

2 Contents Introduction Work of a Force Principle of Work & Energy pplications of the Principle of Work & Energy Power and Efficiency Sample Problem 3. Sample Problem 3. Sample Problem 3.3 Sample Problem 3.4 Sample Problem 3.5 Potential Energy Conseratie Forces Conseration of Energy Motion Under a Conseratie Central Force 03 The McGraw-Hill Companies, Inc. ll rights resered. Sample Problem 3.6 Sample Problem 3.7 Sample Problem 3.9 Principle of Impulse and Momentum Impulsie Motion Sample Problem 3.0 Sample Problem 3. Sample Problem 3. Impact Direct Central Impact Oblique Central Impact Problems Inoling Energy and Momentum Sample Problem 3.4 Sample Problem 3.5 Sample Problems 3.6 Sample Problem

3 Energy and Momentum Methods The pogo stick allows the boy to change between kinetic energy, potential energy from graity, and potential energy in the spring. ccidents are often analyzed by using momentum methods. 03 The McGraw-Hill Companies, Inc. ll rights resered. - 3

4 Introduction Preiously, problems dealing with the motion of particles were soled through the fundamental equation of motion, F ma. The current chapter introduces two adal methods of analysis. Method of work and energy: directly relates force, mass, elocity and displacement. Method of impulse and momentum: directly relates force, mass, elocity, and time. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-4

5 Introduction pproaches to Kinetics Problems Forces and ccelerations Velocities and Displacements Velocities and Time Newton s Second Law (last chapter) Work-Energy Impulse- Momentum F ma G T U T t m F dt m t 03 The McGraw-Hill Companies, Inc. ll rights resered. - 5

6 Work of a Force Differential ector dr is the particle displacement. Work of the force is du F F ds cos F x dr dx F y dy F z dz Work is a scalar quantity, i.e., it has magnitude and sign but not direction. Dimensions of work are J joule N m length ft force. Units are lb.356 J 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-6

7 Work of a Force Work of a force during a finite displacement, U s s F F cos F x dr dx F ds y dy s s F F t z ds dz Work is represented by the area under the cure of F t plotted against s. F t is the force in the direction of the displacement ds 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-7

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9 Work of a Force What is the work of a constant force in rectilinear motion? a) b) c) d) U F x U F cos x U F sin x U 0 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-9

10 Work of a Force Work of the force of graity, U du F x W dy y y W dx W dy y F y dy y F z W dz y Work of the weight is equal to product of weight W and ertical displacement y. In the figure aboe, when is the work done by the weight positie? a) Moing from y to y b) Moing from y to y c) Neer 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-0

11 Work of a Force Magnitude of the force exerted by a spring is proportional to deflection, F kx k spring constant N/m or lb/in. Work of the force exerted by spring, du U F dx x x kx dx kx dx Work of the force exerted by spring is positie when x < x, i.e., when the spring is returning to its undeformed position. Work of the force exerted by the spring is equal to negatie of area under cure of F plotted against x, U F F kx x kx 03 The McGraw-Hill Companies, Inc. ll rights resered. 3 -

12 Work of a Force s the block moes from 0 to, is the work positie or negatie? Positie Negatie s the block moes from to o, is the work positie or negatie? Positie Negatie Displacement is in the opposite direction of the force Displacement is in the same direction of the force 03 The McGraw-Hill Companies, Inc. ll rights resered. 3 -

13 Work of a Force Work of a graitational force (assume particle M occupies fixed position O while particle m follows path shown), Mm du Fdr G dr r U r r Mm G dr r G Mm r G Mm r 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-3

14 Does the normal force do work as the block slides from B to? YES NO Does the weight do work as the block slides from B to? YES NO Positie or Negatie work? 03 The McGraw-Hill Companies, Inc. ll rights resered. - 4

15 Work of a Force Forces which do not do work (ds = 0 or cos : Reaction at frictionless pin supporting rotating body, Reaction at frictionless surface when body in contact moes along surface, Reaction at a roller moing along its track, and Weight of a body when its center of graity moes horizontally. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-5

16 Particle Kinetic Energy: Principle of Work & Energy ds 03 The McGraw-Hill Companies, Inc. ll rights resered. Consider a particle of mass m acted upon by force F F t F t ma t m d ds m ds dt m d d dt d m ds Integrating from to, s s F ds t m d m m U T T T m kinetic energy The work of the force F is equal to the change in kinetic energy of the particle. Units of work and kinetic energy are the same: m m T m kg kg m N m J s s 3-6

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18 Th The work-kinetic energy is a scalar equation. lthough in Cartesian coordinates we find change in kinetic energy in x direction is equal to work done in that direction. This is not true for cylindrical coordinates. 03 The McGraw-Hill Companies, Inc. ll rights resered. - 8

19 pplications of the Principle of Work and Energy The bob is released from rest at position. Determine the elocity of the pendulum bob at using work & kinetic energy. Force P work. T acts normal to path and does no U 0 Wl T W g gl Velocity is found without determining expression for acceleration and integrating. ll quantities are scalars and can be added directly. Forces which do no work are eliminated from the problem. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-9

20 pplications of the Principle of Work and Energy Principle of work and energy cannot be applied to directly determine the acceleration of the pendulum bob. Calculating the tension in the cord requires supplementing the method of work and energy with an application of Newton s second law. gl s the bob passes through, P F n W P m a W g W n l W g gl l 3W If you designed the rope to hold twice the weight of the bob, what would happen? 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-0

21 Power and Efficiency Power rate at which work is done. du F dr dt dt F Dimensions of power are work/time or force*elocity. Units for power are W (watt) J s N m s or hp ft 550 lb s 746 W efficiency output work input work power output power input 03 The McGraw-Hill Companies, Inc. ll rights resered. 3 -

22 Sample Problem 3.3 SOLUTION: spring is used to stop a 60 kg package which is sliding on a horizontal surface. The spring has a constant k = 0 kn/m and is held by cables so that it is initially compressed 0 mm. The package has a elocity of.5 m/s in the position shown and the maximum deflection of the spring is 40 mm. pply the principle of work and energy between the initial position and the point at which the spring is fully compressed and the elocity is zero. The only unknown in the relation is the friction coefficient. pply the principle of work and energy for the rebound of the package. The only unknown in the relation is the elocity at the final position. Determine (a) the coefficient of kinetic friction between the package and surface and (b) the elocity of the package as it passes again through the position shown. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3 -

23 Sample Problem 3.3 SOLUTION: pply principle of work and energy between initial position and the point at which spring is fully compressed. T m T 60kg.5m s 87.5J 0 U f k W x k 60kg 9.8m s m 377J k P P min max U kx k e 0 x 0 0kN x P min m 400 N 0kN P 0.0 m max m x 300 N 400 N 0.60 m m 300 N.0J U U U 377J f e k J T U 87.5J - T : 377 J k J 0 k The McGraw-Hill Companies, Inc. ll rights resered. 3-3

24 Sample Problem 3.3 pply the principle of work and energy for the rebound of the package. T T 3 m 3 60kg 0 3 U U 36.5J U 377J 3 3 f 3 e k J T 0 U J T 3 : 60kg m s 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-4

25 Group Problem Soling Packages are thrown down an incline at with a elocity of m/s. The packages slide along the surface BC to a coneyor belt which moes with a elocity of m/s. Knowing that k= 0.5 between the packages and the surface BC, determine the distance d if the packages are to arrie at C with a elocity of m/s. 03 The McGraw-Hill Companies, Inc. ll rights resered. SOLUTION: The problem deals with a change in position and different elocities, so use work-energy. Draw FBD of the box to help us determine the forces that do work. Determine the work done between points and C as a function of d. Find the kinetic energy at points and C. Use the work-energy relationship and sole for d. - 5

26 Group Problem Soling SOLUTION: Gien: = m/s, C = m/s, k= 0.5 Find: distance d Will use: T U B UB C TC Draw the FBD of the block at points and C Determine work done B N B mg cos30 F N 0.5 mg cos 30 B k B U mg d sin 30 F d B B mg d(sin 30 cos 30 ) Determine work done B C k U N mg x F BC BC k mg mg x BC B C k BC 7m 03 The McGraw-Hill Companies, Inc. ll rights resered. - 6

27 Group Problem Soling Determine kinetic energy at and at C T m and m/s T m and C C C m/s Substitute alues into T U B UB C TC m mg d mg x m (sin30 k cos30 ) k BC 0 Diide by m and sole for d d C k BC / g x /g (sin 30 cos30 ) k () /()(9.8) (0.5)(7) () /()(9.8) sin cos30 d 6.7 m 03 The McGraw-Hill Companies, Inc. ll rights resered. - 7

28 k= 0.5 If you wanted to bring the package to a complete stop at the bottom of the ramp, would it work to place a spring as shown? No, because the potential energy of the spring would turn into kinetic energy and push the block back up the ramp Would the package eer come to a stop? Yes, eentually enough energy would be dissipated through the friction between the package and ramp. 03 The McGraw-Hill Companies, Inc. ll rights resered. - 8

29 The potential energy stored at the top of the roller coaster is transferred to kinetic energy as the cars descend. The elastic potential energy stored in the trampoline is transferred to kinetic energy and graitational potential energy as the girl flies upwards. 03 The McGraw-Hill Companies, Inc. ll rights resered. - 9

30 Potential Energy If the work of a force only depends on differences in position, we can express this work as potential energy. Can the work done by the following forces be expressed as potential energy? Weight Friction Normal force Spring force Yes Yes Yes Yes No No No No 03 The McGraw-Hill Companies, Inc. ll rights resered. - 30

31 Potential Energy Work of the force of graity W, U W y W y Work is independent of path followed; depends only on the initial and final alues of Wy. V g Wy U potential energy of the body with respect to force of graity. V g V g Choice of datum from which the eleation y is measured is arbitrary. Units of work and potential energy are the same: Wy N m J V g 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-3

32 Potential Energy Preious expression for potential energy of a body with respect to graity is only alid when the weight of the body can be assumed constant. For a space ehicle, the ariation of the force of graity with distance from the center of the earth should be considered. Work of a graitational force, U GMm r GMm r Potential energy V g when the ariation in the force of graity can not be neglected, GMm r WR r V g 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-3

33 Potential Energy Work of the force exerted by a spring depends only on the initial and final deflections of the spring, U kx kx The potential energy of the body with respect to the elastic force, V U e kx V e V e Note that the preceding expression for V e is alid only if the deflection of the spring is measured from its undeformed position. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-33

34 Conseratie Forces Concept of potential energy can be applied if the work of the force is independent of the path followed by its point of application. U V x, y, z V x, y z, Such forces are described as conseratie forces. For any conseratie force applied on a closed path, F dr 0 Elementary work corresponding to displacement between two neighboring points, du F x F dx V x, y, z dv F y V x x, y, z dy F V y z V dz x V z dx, y V x dx dy, z grad V V y dz dy V z dz 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-34

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40 Conseration of Energy T T 0 V V W W W T m g W V 0 g T V W Work of a conseratie force, U V V Concept of work and energy, U T T Follows that T E V T T V V constant When a particle moes under the action of conseratie forces, the total mechanical energy is constant. Friction forces are not conseratie. Total mechanical energy of a system inoling friction decreases. Mechanical energy is dissipated by friction into thermal energy. Total energy is constant. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-40

41 Motion Under a Conseratie Central Force When a particle moes under a conseratie central force, both the principle of conseration of angular momentum r 0 m 0 sin 0 rmsin and the principle of conseration of energy m 0 T 0 GMm r may be applied. V 0 0 T V m GMm r Gien r, the equations may be soled for and t minimum and maximum r, 90 o. Gien the launch cons, the equations may be soled for r min, r max, min, and max. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-4

42 Sample Problem 3.7 SOLUTION: Since the pellet must remain in contact with the loop, the force exerted on the pellet must be greater than or equal to zero. Setting the force exerted by the loop to zero, sole for the minimum elocity at D. The 50 g pellet is pushed against the spring and released from rest at. Neglecting friction, determine the smallest deflection of the spring for which the pellet will trael around the loop and remain in contact with the loop at all times. pply the principle of conseration of energy between points and D. Sole for the spring deflection required to produce the required elocity and kinetic energy at D. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-4

43 Sample Problem 3.7 SOLUTION: 03 The McGraw-Hill Companies, Inc. ll rights resered. Setting the force exerted by the loop to zero, sole for the minimum elocity at D. F n ma n : n D rg 0.5 m 9.8m s 4.905m s D W ma mg m r pply the principle of conseration of energy between points and D. e g N/m 300 V V V kx x x T 0 V V V 0 Wy 0.5 kg 9.8m/s ( m).45j T e m g T V T V 0.5 kg 4.905m s 0.63J D 0 300x.45J 0.63J x 0.0 m 0/mm 3-43

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47 Sample Problem 3.9 SOLUTION: For motion under a conseratie central force, the principles of conseration of energy and conseration of angular momentum may be applied simultaneously. satellite is launched in a direction parallel to the surface of the earth with a elocity of km/h from an altitude of 500 km. Determine (a) the maximum altitude reached by the satellite, and (b) the maximum allowable error in the direction of launching if the satellite is to come no closer than 00 km to the surface of the earth pply the principles to the points of minimum and maximum altitude to determine the maximum altitude. pply the principles to the orbit insertion point and the point of minimum altitude to determine maximum allowable orbit insertion angle error. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-47

48 Sample Problem 3.9 pply the principles of conseration of energy and conseration of angular momentum to the points of minimum and maximum altitude to determine the maximum altitude. Conseration of energy: T V T V m Conseration of angular momentum: r r 0 0m0 r m 0 r Combining, 0 GMm r r0 GM r0 r 0 0 r r0 r r r0 r 0 0 GM 6370 km km gr 500km h 9.8m 0.5 s km m s 0 6 m m GM GMm r 0 m 3 s r m km 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-48

49 Sample Problem 3.9 pply the principles to the orbit insertion point and the point of minimum altitude to determine maximum allowable orbit insertion angle error. Conseration of energy: T 0 V 0 T V m 0 GMm r 0 m max GMm r min Conseration of angular momentum: r 0m0 sin 0 rmin mmax max 0 sin 0 r r 0 min Combining and soling for sin, sin allowableerror.5 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-49

50 Group Problem Soling SOLUTION: This is two part problem you will need to find the elocity of the car using work-energy, and then use Newton s second law to find the normal force. section of track for a roller coaster consists of two circular arcs B and CD joined by a straight portion BC. The radius of CD is 80 m. The car and its occupants, of total mass 300 kg, reach Point with practically no elocity and then drop freely along the track. Determine the normal force exerted by the track on the car at point D. Neglect air resistance and rolling resistance. 03 The McGraw-Hill Companies, Inc. ll rights resered. Draw a diagram with the car at points and D, and define your datum. Use conseration of energy to sole for D Draw FBD and KD of the car at point D, and determine the normal force using Newton s second law. - 50

51 Group Problem Soling SOLUTION: Gien: = 0 m/s, r CD = 800 m, W=300 kg Find: N D Datum Define your datum, sketch the situation at points of interest Use conseration of energy to find D T V T D V D Find T Find V Find T D V mgy 0 T 0 Find V y 0 0 D D VD (300 kg)(9.8m/s )(30 m 0 m) =4750 J T m (300) 50 D D D D Sole for D D D D m/s 03 The McGraw-Hill Companies, Inc. ll rights resered. - 5

52 Group Problem Soling Draw FBD and KD at point D e n e t W ma n ma t Use Newton s second law in the normal direction D F n N mg m N m g D ma 03 The McGraw-Hill Companies, Inc. ll rights resered. n D R D R N D N D N D 660 N - 5

53 Group Problem Soling What happens to the normal force at D if. we include friction? a) N D gets larger b) N D gets smaller c) N D stays the same we moe point higher? a) N D gets larger b) N D gets smaller c) N D stays the same 03 The McGraw-Hill Companies, Inc. ll rights resered. the radius is smaller? a) N D gets larger b) N D gets smaller c) N D stays the same - 53

54 Impulsie Motion The thrust of a rocket acts oer a specific time period to gie the rocket linear momentum. The impulse applied to the railcar by the wall brings its momentum to zero. Crash tests are often performed to help improe safety in different ehicles. 03 The McGraw-Hill Companies, Inc. ll rights resered. - 54

55 Principle of Impulse and Momentum Dimensions of the impulse of a force are force*time. Units for the impulse of a force are N s kg m s s kg m s From Newton s second law, d F m m linear momentum dt Fdt d m t t t t Fdt Fdt m m Imp Imp m m impulse of the force F The final momentum of the particle can be obtained by adding ectorially its initial momentum and the impulse of the force during the time interal. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-55

56 Impulsie Motion Force acting on a particle during a ery short time interal that is large enough to cause a significant change in momentum is called an impulsie force. When impulsie forces act on a particle, m F t m When a baseball is struck by a bat, contact occurs oer a short time interal but force is large enough to change sense of ball motion. Nonimpulsie forces are forces for which F t is small and therefore, may be neglected an example of this is the weight of the baseball. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-56

57 Sample Problem 3.0 SOLUTION: pply the principle of impulse and momentum. The impulse is equal to the product of the constant forces and the time interal. n automobile of mass 800 kg is drien down a 5 o incline at a speed of 00 km/h when the brakes are applied, causing a constant total braking force of 7000 N. Determine the time required for the automobile to come to a stop. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-57

58 Sample Problem 3.0 SOLUTION: pply the principle of impulse and momentum. m Imp m Taking components parallel to the incline, m mgsin5 t Ft 0 km 000 m h 00 km/h m/s h km 3600 s 800 kg 778 m/s 800 kg 9.8 m/s (sin 5 ) t (7000 N) t 0 t 9.6s 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-58

59 Sample Problem 3. SOLUTION: pply the principle of impulse and momentum in terms of horizontal and ertical component equations. 0-g baseball is pitched with a elocity of 4 m/s. fter the ball is hit by the bat, it has a elocity of 36 m/s in the direction shown. If the bat and ball are in contact for 0.05 s, determine the aerage impulsie force exerted on the ball during the impact. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-59

60 Sample Problem 3. F SOLUTION: pply the principle of impulse and momentum in terms of horizontal and ertical component equations. m Imp m x component equation: m Fx t m cos40 (0. kg)(4 m/s) F (0.05 s) (0. kg)(36 m/s) cos 40 x 4.6 N y component equation: x y x 0 F t m sin 40 F F y y y 0.05 (0. kg)(36 m/s) sm40 85.N F (43 N) i (85. N) j, F 45N 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-60

61 Sample Problem 3. SOLUTION: 0 kg package drops from a chute into a 4 kg cart with a elocity of 3 m/s. Knowing that the cart is initially at rest and can roll freely, determine (a) the final elocity of the cart, (b) the impulse exerted by the cart on the package, and (c) the fraction of the initial energy lost in the impact. pply the principle of impulse and momentum to the package-cart system to determine the final elocity. pply the same principle to the package alone to determine the impulse exerted on it from the change in its momentum. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-6

62 Sample Problem 3. SOLUTION: pply the principle of impulse and momentum to the package-cart system to determine the final elocity. y x m p Imp mp mc x components: m p cos30 0 kg 3 m/s 0 cos30 m p m c 0 kg 5 kg 0.74 m/s 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-6

63 Sample Problem 3. pply the same principle to the package alone to determine the impulse exerted on it from the change in its momentum. y x m p Imp mp x components: m p cos30 0 kg 3 m/s F x t cos30 m p F x t 0 kg F x t 8.56 N s y components: m p sin30 F y t 0 0 kg 3 m/s sin30 Fy t 0 F y t 5N s Imp F t 8.56 N s i 5 N s j F t 3.9 N s The t is of order 0-4 or 0-5 s, F is much larger than weight, i.e. weight is neglected. Impulse due to weight is 00*0-4 Ns 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-63

64 Sample Problem 3. To determine the fraction of energy lost, T m p T m m p c 0 kg 3m s 45 J 0 kg 5 kg 0.74 m s 9.63 J T T T 45 J 9.63 J 45 J The McGraw-Hill Companies, Inc. ll rights resered. 3-64

65 SOLUTION: Draw impulse and momentum diagrams of the jumper. pply the principle of impulse and momentum to the jumper to determine the force exerted on the foot. The jumper approaches the takeoff line from the left with a horizontal elocity of 0 m/s, remains in contact with the ground for 0.8 s, and takes off at a 50 o angle with a elocity of m/s. Determine the aerage impulsie force exerted by the ground on his foot. Gie your answer in terms of the weight W of the athlete. Here, dt is quite large and weight should not be neglected 03 The McGraw-Hill Companies, Inc. ll rights resered. - 65

66 Group Problem Soling Gien: = 0 m/s, = m/s at 50º, Δt= 0.8 s Find: F ag in terms of W Draw impulse and momentum diagrams of the jumper m m + = 50º y W t Fag t x Use the impulse momentum equation in y to find F ag m ( P W) t m t 0.8 s 03 The McGraw-Hill Companies, Inc. ll rights resered. - 66

67 Group Problem Soling m m + = 50º W t Fag t y x m ( F W) t m t 0.8 s ag Use the impulse momentum equation in x and y to find F ag W W (0) ( Fag x )(0.8) ()(cos 50 ) g g F ag x 0 ()(cos 50 ) (9.8)(0.8) W W 0 ( Fag y W)(0.8) ()(sin 50 ) g ()(sin 50 ) Fag y W W (9.8)(0.8) Fag.95 W i 6.W j F ag-x is positie, which means we guessed correctly (acts to the left) 03 The McGraw-Hill Companies, Inc. ll rights resered. - 67

68 Group Problem Soling Car and B crash into one another. Looking only at the impact, which of the following statements are true? The total mechanical energy is the same before and after the impact True False If car weighs twice as much as car B, the force exerts on car B is bigger than the force B exerts on car. True False The total linear momentum is the same immediately before and after the impact True False 03 The McGraw-Hill Companies, Inc. ll rights resered. - 68

69 The coefficient of restitution is used to characterize the bounciness of different sports equipment. The U.S. Golf ssociation limits the COR of golf balls at 0.83 Ciil engineers use the coefficient of restitution to model rocks falling from hillsides 03 The McGraw-Hill Companies, Inc. ll rights resered. - 69

70 Impact Impact: Collision between two bodies which occurs during a small time interal and during which the bodies exert large forces on each other. Line of Impact: Common normal to the surfaces in contact during impact. Direct Central Impact Central Impact: Impact for which the mass centers of the two bodies lie on the line of impact; otherwise, it is an eccentric impact.. Direct Impact: Impact for which the elocities of the two bodies are directed along the line of impact. Oblique Central Impact Oblique Impact: Impact for which one or both of the bodies moe along a line other than the line of impact. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-70

71 Direct Central Impact Bodies moing in the same straight line, > B. Upon impact the bodies undergo a period of deformation, at the end of which, they are in contact and moing at a common elocity. period of restitution follows during which the bodies either regain their original shape or remain permanently deformed. Wish to determine the final elocities of the two bodies. The total momentum of the two body system is presered, m m B B m B B m second relation between the final elocities is required. B B 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-7

72 Direct Central Impact Period of deformation: m Pdt m u Period of restitution: m u Rdt m e coefficient of restitution Rdt u Pdt u 0 e 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-7

73 Direct Central Impact similar analysis of particle B yields e u B u B Combining the relations leads to the desired second relation between the final elocities. B e B Perfectly plastic impact, e = 0: B m m B B m m B Perfectly elastic impact, e = : Total energy and total momentum consered. B B 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-73

74 03 The McGraw-Hill Companies, Inc. ll rights resered. enth Oblique Central Impact 3-74 Final elocities are unknown in magnitude and direction. Four equations are required. t B t B t t n B B n n B B n m m m m n B n n n B e

75 In deriing aboe equations the balls nothing was said as to how they were supported. They could be supported in z- direction (lying on a table) but impact was in x-y plane, so no external forces due to reactions were there in x-y plane. In the next problem, reaction forces are in the plane of impact. 03 The McGraw-Hill Companies, Inc. ll rights resered. - 75

76 Oblique Central Impact Block constrained to moe along horizontal surface. Impulses from internal forces F and F along the n axis and from external force F ext exerted by horizontal surface and directed along the ertical to the surface. Final elocity of ball unknown in direction and magnitude and unknown final block elocity magnitude. Three equations required. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-76

77 Oblique Central Impact Tangential momentum of ball is consered. B t B t Total horizontal momentum of block and ball is consered. m m B B x m m B B x Normal component of relatie elocities of block and ball are related by coefficient of restitution. B n n e n B n Note: Validity of last expression does not follow from preious relation for the coefficient of restitution. similar but separate deriation is required. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-77

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82 Problems Inoling Energy and Momentum Three methods for the analysis of kinetics problems: - Direct application of Newton s second law - Method of work and energy - Method of impulse and momentum Select the method best suited for the problem or part of a problem under consideration. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-8

83 Sample Problem 3.6 Ball B is hanging from an inextensible cord. n identical ball is released from rest when it is just touching the cord and acquires a elocity 0 before striking ball B. ssuming perfectly elastic impact (e = ) and no friction, determine the elocity of each ball immediately after impact. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-83

84 Sample Problem 3.6 SOLUTION: Determine orientation of impact line of action. The momentum component of ball tangential to the contact plane is consered. The total horizontal momentum of the two ball system is consered. The relatie elocities along the line of action before and after the impact are related by the coefficient of restitution. Sole the last two expressions for the elocity of ball along the line of action and the elocity of ball B which is horizontal. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-84

85 Sample Problem 3.6 sin 30 r r 0.5 SOLUTION: Determine orientation of impact line of action. The momentum component of ball tangential to the contact plane is consered. m m 0 sin30 t F t m 0 m t 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-85

86 Sample Problem 3.6 The total horizontal (x component) momentum of the two ball system is consered. m m 0.5 T t n 0 t cos30 cos30 B m m m B 0 n n sin30 sin30 B m B 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-86

87 Sample Problem 3.6 The relatie elocities along the line of action before and after the impact are related by the coefficient of restitution. B B 0.5 n sin30 B n n e n 0 n cos30 0 B n 0 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-87

88 Sample Problem 3.6 Sole the last two expressions for the elocity of ball along the line of action and the elocity of ball B which is horizontal. n B t tan n B The McGraw-Hill Companies, Inc. ll rights resered. 3-88

89 Sample Problem kg block is dropped from a height of m onto the the 0 kg pan of a spring scale. ssuming the impact to be perfectly plastic, determine the maximum deflection of the pan. The constant of the spring is k = 0 kn/m. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-89

90 Sample Problem 3.7 SOLUTION: pply the principle of conseration of energy to determine the elocity of the block at the instant of impact. Since the impact is perfectly plastic, the block and pan moe together at the same elocity after impact. Determine that elocity from the requirement that the total momentum of the block and pan is consered. pply the principle of conseration of energy to determine the maximum deflection of the spring. 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-90

91 Sample Problem 3.7 SOLUTION: pply principle of conseration of energy to determine elocity of the block at instant of impact. T T T 0 0 V m 588 J T V V 30 W y V 588 J 0 6.6m s 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-9

92 Sample Problem 3.7 Determine elocity after impact from requirement that total momentum of the block and pan is consered. m m The McGraw-Hill Companies, Inc. ll rights resered. B B m m B m s 3-9

93 Sample Problem 3.7 x 3 Initial spring deflection due to pan weight: W k B m pply the principle of conseration of energy to determine the maximum deflection of the spring. T m m 3 B 3 V V V T J 0 kx J 0 g e The McGraw-Hill Companies, Inc. ll rights resered. 3-93

94 Sample Problem 3.7 V V V W W h kx 4 g e B 4 39 x x 0 0 x x x h x 4 x m m h 0.5 m 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-94

95 Sample Problem 3.7 T 44 x 3 4 V 3 T m V x x 4 03 The McGraw-Hill Companies, Inc. ll rights resered. 3-95

96 Group Problem Soling 03 The McGraw-Hill Companies, Inc. ll rights resered. - 96

97 Group Problem Soling -kg block is pushed up against a spring compressing it a distance x= 0. m. The block is then released from rest and slides down the 0º incline until it strikes a -kg sphere B, which is suspended from a m inextensible rope. The spring constant k=800 N/m, the coefficient of friction between and the ground is 0., the distance slides from the unstretched length of the spring d=.5 m, and the coefficient of restitution between and B is 0.8. When =40 o, find (a) the speed of B (b) the tension in the rope. 03 The McGraw-Hill Companies, Inc. ll rights resered. - 97

98 SOLUTION: This is a multiple step problem. Formulate your oerall approach. Use work-energy to find the elocity of the block just before impact 03 The McGraw-Hill Companies, Inc. ll rights resered. - 98

99 Use conseration of momentum to determine the speed of ball B after the impact Use work energy to find the elocity at Use Newton s nd Law to find tension in the rope 03 The McGraw-Hill Companies, Inc. ll rights resered. - 99

100 Gien: m = -kg m B = -kg, k= 800 N/m, =0., e= 0.8 Find (a) B (b) T rope 03 The McGraw-Hill Companies, Inc. ll rights resered. - 00

101 Use work-energy to find the elocity of the block just before impact Determine the friction force acting on the block 03 The McGraw-Hill Companies, Inc. ll rights resered. - 0

102 Sum forces in the y-direction F y N m g cos 0 N mg cos 0: Sole for N ()(9.8) cos N N (0.)(8.4368) N 03 The McGraw-Hill Companies, Inc. ll rights resered. - 0 F f k

103 Group Problem Soling Set your datum, use work-energy to determine at impact. T ( V ) ( V ) U T ( V ) ( V ) e g e g Determine alues for each term. T 0, ( V ) e k x (800)(0.) 4.00 J ( V ) m gh m g( x d)sin ()(9.8)(.6)sin J g U Ff ( x d) (3.6874)(.6) J x d Datum ()( ) T m V 03 The McGraw-Hill Companies, Inc. ll rights resered. - 03

104 Group Problem Soling Determine alues for each term. T 0, ( V ) e k x (800)(0.) 4.00 J ( V ) m gh m g( x d)sin ()(9.8)(.6)sin J g U Ff ( x d) (3.6874)(.6) J T m ()( ).000 V 0 03 The McGraw-Hill Companies, Inc. ll rights resered. - 04

105 Use conseration of momentum to determine the speed of ball B after the impact Note that the ball is Draw the constrained to moe only impulse horizontally immediately after the impact. diagram 03 The McGraw-Hill Companies, Inc. ll rights resered. - 05

106 Group Problem Soling pply conseration of momentum in the x direction Note that the ball is constrained to moe only horizontally immediately after the impact. m cos 0 m cos m B B ()(.977)cos0 cos0 (.00) () B 03 The McGraw-Hill Companies, Inc. ll rights resered. - 06

107 Group Problem Soling Use the relatie elocity/coefficient of restitution equation Note that the ball is constrained to moe only horizontally immediately after the impact. ( ) ( ) e[( ) ( ) ] B n n B n n B cos e[ 0] B cos 0 (0.8)(.977) () Sole () and () simultaneously.038 m/s m/s 03 The McGraw-Hill Companies, Inc. ll rights resered. B - 07

108 Group Problem Soling Use work energy to find the elocity at Set datum, use Work-Energy to determine B at = 40 o T ( V ) ( V ) U T ( V ) ( V ) e g e g Datum Determine alues for each term. T mb( B) V 0 T mb V mbgh mbgl( cos ) 03 The McGraw-Hill Companies, Inc. ll rights resered. - 08

109 Group Problem Soling Substitute into the Work- Energy equation and sole for Datum T V T V: mb ( B ) 0 mb mbg( cos ) B ( ) gl( cos ) (3.6356) ()(9.8)( cos 40 ) m /s.94 m/s 03 The McGraw-Hill Companies, Inc. ll rights resered. - 09

110 Group Problem Soling Use Newton s nd Law to find tension in the rope Draw your freebody and kinetic diagrams e n e t Sum forces in the normal direction Fn mb an : 03 The McGraw-Hill Companies, Inc. ll rights resered. T m g cos m a B B n T m ( a g cos ) B n - 0

111 Group Problem Soling Determine normal acceleration a n.00 m m/s Substitute and sole T (.0)( cos40 ) T m g cos m a B B n T m ( a g cos ) B T 6.4 N n 03 The McGraw-Hill Companies, Inc. ll rights resered. -

112 Concept Question Compare the following statement to the problem you just soled. If the coefficient of restitution is smaller than the 0.8 in the problem, the tension T will be Smaller Bigger 03 The McGraw-Hill Companies, Inc. ll rights resered. -

113 Concept Question If the rope length is smaller than the m in the problem, the tension T will be Smaller Bigger 03 The McGraw-Hill Companies, Inc. ll rights resered. - 3

114 Concept Question If the coefficient of friction is smaller than 0. gien in the problem, the tension T will be Smaller Bigger 03 The McGraw-Hill Companies, Inc. ll rights resered. - 4

115 Concept Question If the mass of is smaller than the kg gien in the problem, the tension T will be Smaller Bigger 03 The McGraw-Hill Companies, Inc. ll rights resered. - 5

116 Summary pproaches to Kinetics Problems Forces and ccelerations Velocities and Displacements Velocities and Time Newton s Second Law (last chapter) T U T Work- Energy Impulse- Momentum t m F dt m t F ma G 03 The McGraw-Hill Companies, Inc. ll rights resered. - 6

117 Summary pproaches to Kinetics Problems Forces and ccelerations Velocities and Displacements Velocities and Time Newton s Second Law (last chapter) Work-Energy Impulse- Momentum F ma G T U T t m F dt m t 03 The McGraw-Hill Companies, Inc. ll rights resered. - 7

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