Chapter 13. Kinetics of Particles: Energy and Momentum Methods Introduction Work of a Force Kinetic Energy of a Particle. Principle of Work & Energy

Transcription

1 Chapter 3. Kinetics of Particles: Energy and Momentum Methods Introduction Work of a Force Kinetic Energy of a Particle. Principle of Work & Energy pplications of the Principle of Work & Energy Power and Efficiency Potential Energy Conservative Forces Conservation of Energy Motion Under a Conservative Central Force Principle of Impulse and Momentum Impulsive Motion Impact Direct Central Impact Oblique Central Impact Problems Involving Energy and Momentum

2 Energy and Momentum Methods The potential energy of the roller coaster car is converted into kinetic energy as it descends the track. Impact tests are often analyzed by using momentum methods

3 Introduction Previously, problems dealing with the motion of particles were solved through the fundamental equation of motion, Σ F ma. The current chapter introduces two additional methods of analysis. Method of work and energy: directly relates force, mass, velocity and displacement. Method of impulse and momentum: directly relates force, mass, velocity, and time.

4 pproaches to Kinetics Problems Forces and ccelerations Velocities and Displacements Velocities and Time Newton s Second Law Work-Energy Impulse-Momentum F ma G T + U T t mv + F dt mv t

5 3. Work of a Force Differential vector is the particle displacement. Work of the force is Work is a scalar quantity, i.e., it has magnitude and sign but not direction. du Dimensions of work are F dr F ds cos α dx + dy length force. ( joule) ( N)( m) ft lb.356 J J F F x y + F z dz Units are Fig.3.

6 Work of a Force Work of a force during a finite displacement, U F dr s ( F cosα ) s ( Fxdx + Fydy + Fzdz) ds s s F t ds Fig.3. Work is represented by the area under the curve of F t plotted against s. F t is the force in the direction of the displacement ds

7 What is the work of a constant force in rectilinear motion? a) b) c) d) U F x ( α ) U F cos x ( α ) U F sin x U 0 answer b)

8 Work of the force of gravity, U du F W dy x y y W dx + F W dy y dy + F dz ( y y ) W y z Work of the weight is equal to product of weight W and vertical displacement y. In the figure above, when is the work done by the weight positive? a) Moving from y to y b) Moving from y to y c) Never answer b)

9 Magnitude of the force exerted by a spring is proportional to deflection, F k kx spring constant ( N/m or lb/in. ) Work of the force exerted by spring, du U F dx kx dx x x kx dx kx kx

10 Work of the force exerted by spring is positive 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 negative of area under curve of F plotted against x, U ( F + F ) x Work of a Force s the block moves from 0 to, is the work positive or negative?

11 Positive? Negative? s the block moves from to o, is the work positive or negative? Positive? Negative? answer ; negative, positive

12 Work of a gravitational force (assume particle M occupies fixed position O while particle m follows path shown), U du Mm Fdr G dr r r r Mm G dr G r Mm r G Mm r

13 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 Positive or Negative work? nswer ; No, Yes, Positive

14 Work of a Force Forces which do not do work (ds 0 or cosα 0): Reaction at frictionless pin supporting rotating body, Reaction at frictionless surface when body in contact moves along surface, Reaction at a roller moving along its track, and Weight of a body when its center of gravity moves horizontally.

15 3.B Principle of Work & Energy The work of the force Integrating from to, s v Ft ds m v dv mv s v Units of work and kinetic energy are the same: m m T mv kg kg m N m s F Consider a particle of mass m acted upon by force dv Ft mat m dt dv ds dv m mv ds dt ds F ds mv dv t U T T T mv mv kinetic energy is equal to the change in kinetic energy of the particle. s J

16 3..C pplications of the Principle of Work and Energy The bob is released from rest at position. Determine the velocity of the pendulum bob at using work & kinetic energy. Force P acts normal to path and does no work. T + U T 0 + Wl v W g v 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.

17 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. s the bob passes through, F n P W m a W g P W n v l W + g gl l 3W If you designed the rope to hold twice the weight of the bob, what would happen?

18 3..D Power and Efficiency Power du F dr dt dt F v rate at which work is done. Dimensions of power are work/time or force*velocity. Units for power are J m ft lb W (watt) N or hp 550 s s s η efficiency output work input work power output power input 746 W

19 Sample Problem 3. n automobile of mass 000 kg is driven down a 5 o incline at a speed of 7 km/h when the brakes are applied causing a constant total breaking force of 5000 N. Determine the distance traveled by the automobile as it comes to a stop. STRTEGY: Evaluate the change in kinetic energy. Determine the distance required for the work to equal the kinetic energy change.

20 MODELING and NLYSIS: Evaluate the change in kinetic energy. v 0 T 0 Determine the distance required for the work to equal the kinetic energy change. x 48.3 m

21 REFLECT and THINK Solving this problem using Newton s second law would require determining the car s deceleration from the free-body diagram and then integrating this to use the given velocity information. Using the principle of work and energy allows you to avoid that calculation.

22 Sample Problem 3. Two blocks are joined by an inextensible cable as shown. If the system is released from rest, determine the velocity of block after it has moved m. ssume that the coefficient of friction between block and the plane is m k 0.5 and that the pulley is weightless and frictionless. STRTEGY: pply the principle of work and energy separately to blocks and B. When the two relations are combined, the work of the cable forces cancel. Solve for the velocity.

23 MODELING and NLYSIS pply the principle of work and energy separately to blocks and B. W F T W T B + U 0 + F F ( )( ) 00 kg 9.8m s µ N C + U 0 F F k C T µ W : k ( m) F ( m) 96 N ( ) N ( m) ( 490 N)( m) ( 00kg) v ( )( ) 300 kg 9.8m s c c T : ( m) + W ( m) B m v 490 N ( m) + ( 940 N)( m) ( 300kg) v 940 N m B v

24 REFLECT and THINK: When the two relations are combined, the work of the cable forces cancel. Solve for the velocity. This problem can also be solved by applying the principle of work and energy to the combined system of blocks. When using the principle of work and energy, it usually saves time to choose your system to be everything that moves. F ( m) ( 490 N)( m) ( 00 kg) v C F ( m) ( 940 N)( m) ( 300kg) v c + ( 940 N)( m) ( 490 N)( m) ( 00kg + 300kg) 4900 J ( 500kg) v v v 4.43 m s

25 Sample Problem 3.3 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 velocity of.5 m/s in the position shown and the maximum deflection of the spring is 40 mm. Determine (a) the coefficient of kinetic friction between the package and surface and (b) the velocity of the package as it passes again through the position shown. STRTEGY: pply the principle of work and energy between the initial position and the point at which the spring is fully compressed and the velocity 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 velocity at the final position.

26 ( U ) P min f T µ W x k MODELING and NLYSIS: pply principle of work and energy between initial position and the point at which spring is fully compressed. ( kg)(.5m s) 87.5J 0 mv 60 T µ 60kg 9.8m s m 377J kx0 ( 0kN m)( 0.0 m) 400 N k x + x 0kN m 0.60 m 300 N k ( )( )( ) ( ) k Pmax ( 0 ) ( )( ) ( U ) e ( Pmin + Pmax ) x ( 400 N N)( m).0j µ ( ) ( ) ( ) U U U 377 J µ + k J T + U T : f ( ) 87.5 J- 377 J µ J 0 k e µ k 0.0

27 *pply the principle of work and energy for the rebound of the package. T 0 T mv 60kg v U µ 3 ( ) 3 3 ( U ) + ( U ) ( 377J) 3 3 f 3 e J k + J T + U T 3 3 : ( ) J 60 kg v 3 v 3.03m s REFLECT and THINK: You needed to break this problem into two segments. From the first segment you were able to determine the coefficient of friction. Then you could use the principle of work and energy to determine the velocity of the package at any other location. Note that the system does not lose any energy due to the spring; it returns all of its energy back to the package. You would need to design something that could absorb the kinetic energy of the package in order to bring it to rest.

28 Sample Problem 3.6 curvature at point kg car starts from rest at point and moves without friction down the track shown. Determine: a) the force exerted by the track on the car at point, and b) the minimum safe value of the radius of STRTEGY: pply principle of work and energy to determine velocity at point. pply Newton s second law to find normal force by the track at point. pply principle of work and energy to determine velocity at point 3. pply Newton s second law to find minimum radius of curvature at point 3 such that a positive normal force is exerted by the track.

29 MODELING and NLYSIS: a. pply principle of work and energy to determine velocity at point. + F ma n n : pply Newton s second law to find normal force by the track at point. N kn

30 b. pply principle of work and energy to determine velocity at point 3. pply Newton s second law to find minimum radius of curvature at point 3 such that a positive normal force is exerted + F n ma n : by the track.

31 REFLECT and THINK This is an example where you need both Newton s second law and the principle of work and energy. Work energy is used to determine the speed of the car, and Newton s second law is used to determine the normal force. normal force of 5W is equivalent to a fighter pilot pulling 5g s and should only be

32 experienced for a very short time. For safety, you would also want to make sure your radius of curvature was quite a bit larger than 5 m. Sample Problem 3.7 The dumbwaiter D and its load have a combined mass of 300 kg, while the counterweight C has a mass of 400 kg. Determine the power delivered by the electric motor M when the dumbwaiter (a) is moving up at a constant speed of.5 m/s and (b) has an instantaneous velocity of.5 m/s and an acceleration of m/s, both directed upwards.

33 STRTEGY: Force exerted by the motor cable has same direction as the dumbwaiter velocity. Power delivered by motor is equal to Fv D, v D.5 m/s. In the first case, bodies are in uniform motion. Determine force exerted by motor cable from conditions for static equilibrium. In the second case, both bodies are accelerating. pply Newton s second law to each body to determine the required motor cable force. MODELING and NLYSIS: In the first case, bodies are in uniform motion. Determine force exerted by motor cable from conditions for static equilibrium. Free-body C: + F y 0 : T 400g 0 T 00g 96 N Free-body D: + F y 0 : F T 300 g 0 F 300 g T 300 g 00 g 00 g 98 N

34 Power Fv D 45 W ( 98 N)(.5m s) Power 450 W In the second case, both bodies are accelerating. pply Newton s second law to each body to determine the required motor cable force. Free-body C: + F y mc ac : 400g T T 86 N

35 Free-body D: F y mdad : + F T 300g 300 F F 38 N ( )( ) Power Fv D 38 N.5m s 345 W Power 3450 W

36 REFLECT and THINK s you might expect, the motor needs to deliver more power to produce accelerated motion than to produce motion at constant velocity.

37 3. Conservation of Energy The potential energy stored at the top of the ball s path is transferred to kinetic energy as the ball meets the ground. Why is the ball s height reducing?

38 3. 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 YES NO Friction YES NO Normal force YES NO Spring force YES NO YES NO NO YES

39 Work of the force of gravity, U W y W y Work is independent of path followed; depends only on the initial and final values of Wy. V g Wy force of gravity. U potential energy of the body with respect to ( V g ) ( V ) g W Choice of datum from which the elevation y is measured is arbitrary. Units of work and potential energy are the same: V g Wy N m J

40 Previous expression for potential energy of a body with respect to gravity is only valid when the weight of the body can be assumed constant. For a space vehicle, the variation of the force of gravity with distance from the center of the earth should be considered. Work of a gravitational force, U GMm GMm r r Potential energy V g when the variation in the force of gravity can not be neglected, WR V g GMm r r

41 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 e kx U ( Ve ) ( Ve )

42 Note that the preceding expression for V e is valid only if the deflection of the spring is measured from its undeformed position. 3.B Conservative 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 conservative forces. For any conservative force applied on a closed path,

43 F dr 0 Elementary work corresponding to displacement between two neighboring points, du V ( x, y, z) V ( x + dx, y + dy, z + dz) dv ( x, y, z) V V V Fxdx + Fydy + Fzdz dx + dy + dz x y z V V V F + + gradv x y z

44 3.C The Principle of Conservation of Energy Work of a conservative force, T T 0 + V V W W W T mv ( g ) W V 0 g V V Concept of work and energy, Follows that T + V T When a particle moves under the action of conservative forces, the total mechanical energy is constant. Friction forces are not conservative. Total T + V W U U T T E T + V + V constant mechanical energy of a system involving friction decreases. Mechanical energy is dissipated by friction into thermal energy. Total energy is constant.

45 3.D Motion Under a Conservative Central Force When a particle moves under a conservative central force, both the principle of conservation of angular momentum r 0 mv 0 sinφ 0 rmvsinφ and the principle of conservation of energy mv 0 T 0 + V GMm r 0 0 T + V mv GMm r Given r, the equations may be solved for v.

46 lso, using Eqns (5),(6) t minimum and maximum r, φ 90 o. Given the launch conditions, the equations may be solved for r min, r max, v min, and v max. Sample Problem 3.8 Sample Problem 3.0 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 travel around the loop and remain in contact with the loop at all times.

47 STRTEGY: 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, solve for the minimum velocity at D. pply the principle of conservation of energy between points and D. Solve for the spring deflection required to produce the required velocity and kinetic energy at D. MODELING and NLYSIS: Setting the force exerted by the loop to zero, solve for the minimum velocity at D. + F n ma n : D W ma mg mv r v n D rg 0.6 m 9.8 m s m s

48 pply the principle of conservation 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 9.8 m s. m.943 N m T e g 0.5 kg m s mvd N m ( )( ) T + V T + V REFLECT and THINK x x 0.07 m common misconception in problems like this is assuming that the speed of the particle is zero at the top of the loop, rather than that the normal force is equal to or greater than zero. If the pellet had a speed of zero at the top, it would clearly fall straight down, which is impossible.

49 Sample Problem 3. satellite is launched in a direction parallel to the surface of the earth with a velocity 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 STRTEGY: For motion under a conservative central force, the principles of conservation of energy and conservation of angular momentum may be applied simultaneously. 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.

50 MODELING and NLYSIS: pply the principles of conservation of energy and conservation of angular momentum to the points of minimum and maximum altitude to determine the maximum altitude. Conservation of energy: T V T Conservation of angular momentum: r r 0 0mv0 r mv v v0 r Combining, + V + mv 0 GMm r r 0 GM r0 r v r r0 r r r0 r v km km 6870 km km h m s ( )( ) 0 GM v 0 mv 6 3 GM gr 9.8m s m m s 6 r m km GMm r

51 pply the principles to the orbit insertion point and the point of minimum altitude to determine maximum allowable orbit insertion angle error. T 0 Conservation of energy: + V 0 T + V Conservation of angular momentum: mv max Combining and solving for sin 0 j, sinφ ±.5 allowable error ±. 5 0 GMm r 0mv0 sinφ 0 rminmvmax vmax v0 sin r ϕ 0 0 mv r φ0 r 0 min GMm r min REFLECT and THINK: Space probes and other long-distance vehicles are designed with small rockets to allow for midcourse corrections. Satellites launched from the Space Station usually do not need this kind of fine-tuning.

52 Impulsive Motion The thrust of a rocket acts over a specific time period to give the rocket linear momentum.

53 The impulse applied to the railcar by the wall brings its momentum to zero. Crash tests are often performed to help improve safety in different vehicles. 3.3 Principle of Impulse and Momentum From Newton s second law, linear momentum ( mv ) F d dt ( mv) (3.7)

54 Dimensions of the impulse of a force are force*time. Units for the impulse of a force are t t mv Fdt d mv t t Fdt Imp ( ) Fdt mv mv N s impulse of the force F + Imp mv...(3.8) ( ) kg m s s kg m s The final momentum of the particle can be obtained by adding vectorially its initial momentum and the impulse of the force during the time interval. Fig.3.7 Fig.3.8

55 3.3B Impulsive Motion Force acting on a particle during a very short time interval that is large enough to cause a significant change in momentum is called an impulsive force. When impulsive forces act on a particle, mv + F t m v (3.35) When a baseball is struck by a bat, contact occurs over a short time interval but force is large enough to change sense of ball motion. Nonimpulsive forces are forces for which F t neglected an example of this is the weight of the baseball. is small and therefore, may be

56 Sample Problem 3.3 n automobile weighing 800 kg is driven 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. STRTEGY: pply the principle of impulse and momentum. The impulse is equal to the product of the constant forces and the time interval.

57 MODELING and NLYSIS: pply the principle of impulse and momentum. mv Imp mv + Taking components parallel to the incline, REFLECT and THINK t 9.6s You could use Newton s second law to solve this problem. First, you would determine the car s deceleration, separate variables, and then integrate a dv/dt to relate the velocity, deceleration, and time. You could not use conservation of energy to solve this problem, because this principle does not involve time.

58 Sample Problem g baseball is pitched with a velocity of 4 m/s. fter the ball is hit by the bat, it has a velocity of 36 m/s in the direction shown. If the bat and ball are in contact for 0.05 s, determine the average impulsive force exerted on the ball during the impact. STRTEGY: pply the principle of impulse and momentum in terms of horizontal and vertical component equations.

59 MODELING and NLYSIS: pply the principle of impulse and momentum in terms of horizontal and vertical component equations. mv Imp m + v x component equation: y component equation: F 43 N i + 85.N j, F 45 N ( ) ( )

60

61 Sample Problem kg package drops from a chute into a 4 kg cart with a velocity of 3 m/s. Knowing that the cart is initially at rest and can roll freely, determine (a) the final velocity 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. STRTEGY: pply the principle of impulse and momentum to the package-cart system to determine the final velocity. pply the same principle to the package alone to determine the impulse exerted on it from the change in its momentum.

62 MODELING and NLYSIS pply the principle of impulse and momentum to the package-cart system to determine the final velocity. y x mv+ Imp m + m v ( ) p p c x components: m p v cos ( m + m ) ( 0 kg)( 3 m/s) cos30 ( 0 kg + 5 kg) v p c v v 0.74 m/s

63 pply the same principle to the package alone to determine the impulse exerted on it from the change in its momentum. y mv p + Imp mv p x x components: m p v cos30 + F x t m ( 0 kg)( 3 m/s) cos30 + F t ( 0 kg) v p x v F x t 8.56 N s y components: m p v sin30 + F t 0 ( 0 kg)( 3 m/s) sin30 + F t 0 y y F y t 5N s Imp F t ( 8.56 N s) i + ( 5 N s) j F t 3.9 N s

64 To determine the fraction of energy lost, T p ( )( ) mv 0 kg 3m s 45 J ( p c) ( )( ) T m + m v 0 kg + 5 kg 0.74 m s 9.63 J T T T 45 J 9.63 J J REFLECT and THINK: Except in the purely theoretical case of a perfectly elastic collision, mechanical energy is never conserved in a collision between two objects, even though linear momentum may be conserved. Note that, in this problem, momentum was conserved in the x direction but was not conserved in the y direction because of the vertical impulse on the wheels of the cart. Whenever you deal with an impact, you need to use impulse-

65 momentum methods. 3.4 Impact 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

66 Impact: Collision between two bodies which occurs during a small time interval and during which the bodies exert large forces on each other. Line of Impact: Common normal to the surfaces in contact during impact. Central Impact: Impact for which the mass centers of the two bodies lie on the line of impact; otherwise, Direct Central Impact it is an eccentric impact. Direct Impact: Impact for which the velocities of the two bodies are directed along the line of impact. Oblique Impact: Impact for which one or both of the bodies move along a line other than the line of impact.

67 3.4 Direct Central Impact Bodies moving in the same straight line, v > v B. Upon impact the bodies undergo a period of deformation, at the end of which, they are in contact and moving at a common velocity. period of restitution follows during which the bodies either regain their original shape or remain permanently deformed. Wish to determine the final velocities of the two bodies. The total momentum of the two body system is preserved, m v + m v m v + m v B B B B B B second relation between the final velocities is required. e coefficient of Rdt Pdt 0 e u v v u restitution

68 Period of deformation: Period of restitution: similar analysis of particle B yields Combining the relations leads to the desired second relation between the final velocities. Perfectly plastic impact, e 0: Perfectly elastic impact, e : Total energy and total momentum conserved. u m Pdt v m v m Rdt u m B B v u u v e ( ) B B v v e v v v v v B ( )v m m v m v m B B B + + B B v v v v

69 3.4B Oblique Central Impact Final velocities are unknown in magnitude and direction. Four equations are required. No tangential impulse component; tangential component of momentum for each particle is conserved. ( v ) t ( v ) t ( vb ) t ( v B ) t Normal component of total momentum of the two particles is conserved. ( v ) + mb ( vb ) m( v ) + mb ( vb ) n m n Normal components of relative velocities before and after impact are related by the coefficient of restitution. ( v ) ( v ) e ( v ) ( v ) B n n n n [ ] n B n

70 Block constrained to move along horizontal surface. Impulses from internal forces F ext F and F along the n axis and from external force exerted by horizontal surface and directed along the vertical to the surface. Final velocity of ball unknown in direction and magnitude and unknown final block velocity magnitude. Three equations required.

71 Tangential momentum of ball is conserved. ( v B ) ( v B ) t t Total horizontal momentum of block and ball is conserved. ( v ) + mb ( vb ) m( v ) + mb ( vb ) x m Normal component of relative velocities of block and ball are related by coefficient of restitution. ( v ) ( v ) e ( v ) ( v ) B n n x [ ] n B n Note: Validity of last expression does not follow from previous relation for the coefficient of restitution. similar but separate derivation is required.

72 3.4C Problems Involving Multiple Principles 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.

73 Sample Problem 3.9 ball is thrown against a frictionless, vertical wall. Immediately before the ball strikes the wall, its velocity has a magnitude v and forms angle of 30 o with the horizontal. Knowing that e 0.90, determine the magnitude and direction of the velocity of the ball as it rebounds from the wall. STRTEGY: Resolve ball velocity into components normal and tangential to wall. Impulse exerted by the wall is normal to the wall. Component of ball momentum tangential to wall is conserved. ssume that the wall has infinite mass so that wall velocity before and after impact is zero. pply coefficient of restitution relation to find change in normal relative velocity between wall and ball, i.e., the normal ball velocity.

74 MODELING and NLYSIS: Resolve ball velocity into components parallel and perpendicular to wall. vn vcos v vt vsin v Component of ball momentum tangential to wall is conserved. vt vt v pply coefficient of restitution relation with zero wall velocity. 0 v e v 0 v n n 0.9 ( n ) ( 0.866v) 0.779v v 0.779vλ n vλ v 0.96v tan t

75 REFLECT and THINK: Tests similar to this are done to make sure that sporting equipment such as tennis balls, golf balls, and basketballs are consistent and fall within certain specifications. Testing modern golf balls and clubs shows that the coefficient of restitution actually decreases with increasing club speed (from about 0.84 at a speed of 45 kmph to about 0.80 at club speeds of 0 kmph).

76 Sample Problem 3.0 The magnitude and direction of the velocities of two identical frictionless balls before they strike each other are as shown. ssuming e 0.9, determine the magnitude and direction of the velocity of each ball after the impact. STRTEGY: Resolve the ball velocities into components normal and tangential to the contact plane. Tangential component of momentum for each ball is conserved. Total normal component of the momentum of the two ball system is conserved. The normal relative velocities of the balls are related by the coefficient of restitution. Solve the last two equations simultaneously for the normal velocities of the balls after the impact.

77 MODELING and NLYSIS: Resolve the ball velocities into components normal and tangential to the contact plane. v vcos m s v sin m s n v t v v cos60 6 m s v v sin m s B n B B t B Tangential component of momentum for each ball is conserved. Total normal component of the momentum of the two ball system is conserved.

78 The normal relative velocities of the balls are related by the coefficient of restitution. Solve the last two equations simultaneously for the normal velocities of the balls after the impact.

79 REFLECT and THINK: Rather than choosing your system to be both balls, you could have applied impulsemomentum along the line of impact for each ball individually. This would have resulted in two equations and one additional unknown, FΔt. To determine the impulsive force F, you would need to be given the time for the impact, Δt.

80 Sample Problem 3. 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 velocity v 0 before striking ball B. ssuming perfectly elastic impact (e ) and no friction, determine the velocity of each ball immediately after impact. STRTEGY: Determine orientation of impact line of action. The momentum component of ball tangential to the contact plane is conserved. The total horizontal momentum of the two ball system is conserved. The relative velocities along the line of action before and after the impact are related by the coefficient of restitution. Solve the last two expressions for the velocity of ball along the line of action and the velocity of ball B which is horizontal.

81 r sinθ 0.5 r θ 30 MODELING and NLYSIS: Determine orientation of impact line of action. The momentum component of ball tangential to the contact plane is conserved. mv mv 0 + F t mv ( v ) 0.5v0 ( ) sin m v t t The total horizontal (x component) momentum of the two ball system is conserved. mv + T t mv + mvb 0 m( v ) t cos30 m( v ) n sin 30 mv 0 ( 0.5v0 ) cos30 ( v ) n sin 30 v B 0.5( v ) + v B 0.433v0 n B

82 The relative velocities along the line of action before and after the impact are related by the coefficient of restitution. v v e v v Solve the last two expressions for the velocity of ball along the line of action and the velocity of ball B which is horizontal. 0.50v0 v B v v 0.5v 0.50v ( v ) 0 n v v 0 0.7v 0.693v [ ] ( B ) n ( ) n ( ) n ( B ) v B sin30 ( v ) n v0 cos30 0.5v ( v ) 0.866v0 α B λ t 0 0 B 0 n β tan λ n n 0

83 REFLECT and THINK Since e, the impact between and B is perfectly elastic. Therefore, rather than using the coefficient of restitution, you could have used the conservation of energy as your final equation.

84 Sample Problem kg block is dropped from a height of m onto 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. STRTEGY: pply the principle of conservation of energy to determine the velocity of the block at the instant of impact. Since the impact is perfectly plastic, the block and pan move together at the same velocity after impact. Determine that velocity from the requirement that the total momentum of the block and pan is conserved. pply the principle of conservation of energy to determine the maximum deflection of the spring.

85 MODELING and NLYSIS: pply principle of conservation of energy to determine velocity of the block at instant of impact. Determine velocity after impact from requirement that total momentum of the block and pan is conserved. ( ) ( ) ( ) ( )( ) ( ) s 4.70m v v v m m v m v m B B B ( )( )( ) ( ) ( )( ) ( )( ) ( ) s 6.6m J J v v V T V T V v v m T y W V T

86 pply the principle of conservation of energy to determine the maximum deflection of the spring. deflection due to pan weight: W ( 0)( 9.8) 3 x3 B m k T V T V V V ( m + m ) v ( )( 4.7) g g + V + V e kx e 3 T + V T + V B ( 0 0 )( ) ( W + W )( h) B 3 ( x ) 4 x3 + ( 0 0 ) x ( x ) ( 0 0 ) x4 + kx 4 44 J 0.4 J Initial ( x ) ( ) x 0.30 m spring x h x x h 0.5 m m m

87 REFLECT and THINK: The spring constant for this scale is pretty large, but the block is fairly massive and is dropped from a height of m. From this perspective, the deflection seems reasonable. We included the spring in the system so we could treat it as an energy term rather than finding the work of the spring force.

88 Sample Problem 3.3 -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 k800 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 α 40 o, find (a) the speed of B (b) the tension in the rope. STRTEGY: This is a multiple step problem. Formulate your overall approach. Use work-energy to find the velocity of the block just before impact Use conservation of momentum to determine the speed of ball B after the impact Use work energy to find the velocity at a Use Newton s nd Law to find tension in the rope

89 MODELING and NLYSIS: Given: m -kg m B -kg, k 800 N/m, m 0., e 0.8 Find (a) v B (b) T rope Use work-energy to find the velocity of the block just before impact Determine the friction force acting on the block Sum forces in the y-direction N m gcosθ 0 F y 0: f Solve for N N m g F cosθ ()(9.8) cos N µ N (0.)(8.4368) k N

90 Set your datum, use work-energy to determine v at impact. T + ( V ) + ( V ) + U T + ( V ) + ( V )...() e g e g Determine values for each term. T 0, ( V) e kx (800)(0.) 4.00 J ( V ) m gh m g( x+ d)sin θ ()(9.8)(.6)sin ?J g U F ( x+ d) (3.6874)(.6) ?J f T mv ()( v ).000? v V 0 Substitute into the Work-Energy equation and solve for v T V U T V v : x d Datum m /s v v.97 m/s

91 Use conservation of momentum to determine the speed of ball B after the impact Draw the impulse diagram pply conservation of momentum in the x direction mv cosθ + 0 mv cosθ + mv B B ()(.977) cos 0 v cos 0 + (.00) v Use the relative velocity/coefficient of restitution equation ( v ) ( v ) e[( v ) ( v ) ] B n n B n n v cos θ v ev [ 0] B v cos 0 v (0.8)(.977) B (3) Solve () and (3) simultaneously B () v.038 m/s v m/s B

92 Use work energy to find the velocity at a Set datum, use Work-Energy to determine v B at a 40 o T + ( V ) + ( V ) + U T + ( V ) + ( V ) e g e g Determine values for each term. T mb( vb ) V 0 T mbv V mbgh mbgl( cos α) Datum Substitute into the Work-Energy equation and solve for v T + V T + V: mb( vb ) + 0 mv B + mg B ( cos α) v ( v ) gl( cos α) B (3.6356) ()(9.8)( cos 40 ) m /s v.94?m/s

93 Use Newton s nd Law to find tension in the rope Draw your free-body and kinetic diagrams e n e t Sum forces in the normal direction Fn mb an : T mg B cosα ma B n Determine normal acceleration ρ.00 m a n v m/s ρ.00 Substitute and solve T m ( a + gcos α) B n

94 T (.0)( cos 40 ) T 6.4 Summary pproaches to Kinetics Problems Forces and ccelerations -> Newton s Second Law F ma G Velocities and Displacements -> Work-Energy Velocities and Time -> Impulse-Momentum T + U T t mv + F dt mv t