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

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1 CHAPTER 7 VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University Plane Motion of Rigid Bodies: Energy and Momentum Methods 003 The McGraw-Hill Companies, Inc. All rights reserved.

2 Contents Introduction Principle of Work and Energy for a Rigid Body Work of Forces Acting on a Rigid Body Kinetic Energy of a Rigid Body in Plane Motion Systems of Rigid Bodies Conservation of Energy Power Sample Problem 7. Sample Problem 7. Sample Problem 7.3 Sample Problem 7.4 Sample Problem 7.5 Principle of Impulse and Momentum Systems of Rigid Bodies Conservation of Angular Momentum Sample Problem 7.6 Sample Problem 7.7 Sample Problem 7.8 Eccentric Impact Sample Problem 7.9 Sample Problem 7.0 Sample Problem The McGraw-Hill Companies, Inc. All rights reserved. 7 -

3 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 Principle of impulse and momentum is appropriate for problems involving velocities and time. t t L Fdt L H M dt H t t ( O ) O ( O ) Problems involving eccentric impact are solved by supplementing the principle of impulse and momentum with the application of the coefficient of restitution. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-3

4 Principle of Work and Energy for a Rigid Body Method of work and energy is well adapted to problems involving velocities and displacements. Main advantage is that the 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. Internal forces between particles A and B are equal and opposite. In general, small displacements of the particles A and B are not equal but the components of the displacements along AB are equal. Therefore, the net work of internal forces is zero. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-4

5 Work of Forces Acting on a Rigid Body Work of a force during a displacement of its point of application, A s U F dr ( F cosα ) ds A Consider the net work of two forces F and forming a couple of moment M during a displacement of their points of application. du F dr F dr F dr U F ds M dθ θ M dθ θ M s Fr dθ ( θ θ ) if M is constant. F 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-5

6 Work of Forces Acting on a Rigid Body Forces acting on rigid bodies which do no work: Forces applied to fixed points: - reactions at a frictionless pin when the supported body rotates about the pin. Forces acting in a direction perpendicular to the displacement of their point of application: - reaction at a frictionless surface to a body moving along the surface - weight of a body when its center of gravity moves horizontally Friction force at the point of contact of a body rolling without sliding on a fixed surface. du ( v ) 0 F ds F dt C c 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-6

7 Kinetic Energy of a Rigid Body in Plane Motion Consider a rigid body of mass m in plane motion. T mv mv mv ( r i m v Iω i i m i ) ω 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. Consider a rigid body rotating about a fixed axis through O. T ( rω ) mivi mi i ( ri mi ) I O ω ω 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-7

8 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. We may also apply the principle of work and energy to the entire system, T U T T,T arithmetic sum of the kinetic energies of all bodies forming the system U work of all forces acting on the various bodies, whether these forces are internal or external to the system as a whole. 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 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-8

9 Conservation of Energy mass m released with zero velocity determine ω at θ Expressing the work of conservative forces as a change in potential energy, the principle of work and energy becomes T V T T V Consider the slender rod of mass m. T, V 0 V 0 T T mv m V ml 0 ω 3 3g ω sinθ l ( ) ( lω ml ) ω ω Iω V Wl sinθ mgl sinθ mgl sinθ ml The McGraw-Hill Companies, Inc. All rights reserved. 7-9

10 Power Power rate at which work is done For a body acted upon by force F and moving with velocity, du Power F v dt For a rigid body rotating with an angular velocity and ω acted upon by a couple of moment M parallel to the axis of rotation, Power du dt M dθ dt Mω v 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-0

11 Sample Problem 7. SOLUTION: Consider the system of the flywheel and block. The work done by the internal forces exerted by the cable cancels. Note that the velocity of the block and the angular velocity of the drum and flywheel are related by v rω For the drum and flywheel, I 5 kg m. The bearing friction is equivalent to a couple of 80 N m. At the instant shown, the block is moving downward at.8 m/s. Determine the velocity of the block after it has moved. m downward. Apply the principle of work and kinetic energy to develop an expression for the final velocity. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7 -

12 Sample Problem 7. SOLUTION: Consider the system of the flywheel and block. The work done by the internal forces exerted by the cable cancels. Note that the velocity of the block and the angular velocity of the drum and flywheel are related by v.8m s v v rω ω 4.5rad s ω r 0.4m r Apply the principle of work and kinetic energy to develop an expression for the final velocity. v 0.4 T mv I ω (0 kg)(.8 m/s) 330J (5 kg m )(4.5rad s) T mv Iω (0) v v (5) v 003 The McGraw-Hill Companies, Inc. All rights reserved. 7 -

13 Sample Problem 7. T mv Iω 330J T 0 v mv Iω. 9 Note that the block displacement and pulley rotation are related by θ Then, U.m 0.4m r s 3.0 rad W ( s s ) M ( θ θ) ( 079 N)(. N) ( 80 N m)( 3.0 rad) J Principle of work and energy: T U 300 J J 0.9 v v T 3.7 m s v 3.7 m s 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-3

14 Sample Problem 7. m m A B 0kg 3kg k k A B 00mm 80mm 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. 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. 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. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-4

15 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. ω ω B A ( 600rpm)( π rad rev) ω B r r B A 60s min rad 5.rad s s I I A B m m A B k k A B ( 0kg)( 0.00m) 0.400kg m ( 3kg)( 0.080m) 0.09kg m T I A ω A I Bω B ( 0.400)( 5.) ( 0.09)( 6.8) 63.9J 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-5

16 Sample Problem 7. Apply the principle of work and energy. Calculate the number of revolutions required for the work. T U 0 θ B ( 6θ ) B T J 63.9J 7.3rad θ B 7.3 π 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. r 0.00 θ B A θb rad r 0.50 T A I A ω A ( 0.400)( 5.) 6.0J 4.35rev T U 0 M M A A r T ( 0.93rad) A F 6.0J.5 N m F N 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-6

17 Sample Problem 7.3 SOLUTION: A sphere, cylinder, and hoop, each having the same mass and radius, are released from rest on an incline. Determine the velocity of each body after it has rolled through a distance corresponding to a change of elevation h. The work done by the weight of the bodies is the same. From the principle of work and energy, it follows that each body will have the same kinetic energy after the change of elevation. Because each of the bodies has a different centroidal moment of inertia, the distribution of the total kinetic energy between the linear and rotational components will be different as well. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-7

18 003 The McGraw-Hill Companies, Inc. All rights reserved. Seventh 7-8 Sample Problem 7.3 SOLUTION: The work done by the weight of the bodies is the same. From the principle of work and energy, it follows that each body will have the same kinetic energy after the change of elevation. r v ω With v r I m r v I mv I mv T ω 0 mr I gh r I m Wh v v r I m Wh T U T

19 Sample Problem 7.3 Because each of the bodies has a different centroidal moment of inertia, the distribution of the total kinetic energy between the linear and rotational components will be different as well. gh v I mr Sphere Cylinder Hoop : : : I I I 5 mr mr mr v v v gh gh gh NOTE: For a frictionless block sliding through the same distance, ω 0, v gh The velocity of the body is independent of its mass and radius. The velocity of the body does depend on I k mr r 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-9

20 Sample Problem 7.4 A 5 kg slender rod pivots about the point O. The other end is pressed against a spring (k 34 kn/m) until the spring is compressed one inch 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. 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. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-0

21 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 I ml.8kg m ( 5 kg)(.5m) Evaluate the initial and final potential energy. V V V V 0 kx g e 0.5J 0.5J V V Wh 0 g 66.J e ( 34kN m)( 0.05m) ( 47 N)( 0.45m) Express the final kinetic energy in terms of the angular velocity of the rod. T ( ω ) mv Iω m r Iω (5) ( 0.45ω ) (.8) ω.9ω 003 The McGraw-Hill Companies, Inc. All rights reserved. 7 -

22 Sample Problem 7.4 From the principle of work and energy, T V 0 0.5J T V.9ω 66. ω 3.46 rad s Based on the free-body-diagram equation, solve for the reactions at the pivot. a a n t ( 0.45m)( 3.46rad s) rω 5.39m s rα O ( M O ) eff Iα F ( ) m( rα ) n 5.39m s a a t rα M 0 m( rα )r α 0 x F x F y F y ( ) eff eff R R x 0 y 47 N ma R y n 47 N 5 m s 66.5 N R x ( ) 5.39 m R 66.5 s 003 The McGraw-Hill Companies, Inc. All rights reserved. 7 -

23 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. Each of the two slender rods has a mass of 6 kg. The system is released from rest with β 60 o. Determine a) the angular velocity of rod AB when β 0 o, and b) the velocity of the point D at the same instant. 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. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-3

24 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. V Wy 38.6 J ( N)( 0.35 m) V Wy 5.0 J ( N)( 0.83m) W mg N ( )( ) 6 kg 9.8m s 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-4

25 Sample Problem 7.5 Express the final kinetic energy of the system in terms of the angular velocities of the rods. v AB ( m)ω Since v Bis perpendicular to AB and v is Dhorizontal, the instantaneous center of rotation for rod BD is C. BC 0.75 m CD ( 0.75 m) sin m and applying the law of cosines to CDE, EC 0.5 m Consider the velocity of point B vb ( AB) ω ( BC ) ω AB ω BD ω v BD ( 0.5 m)ω For the final kinetic energy, I AB T I ml BD ( 6 kg)( 0.75 m) 0.8 kg m mvab I ABω AB mvbd I BDωBD ( )( ) ( ) ( )( ) ω 0.8 ω 6 0.5ω ( 0.8).50ω ω 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-5

26 Sample Problem 7.5 Solve the energy equation for the angular velocity, then evaluate the velocity of the point D. T V T J.50ω V ω 3.90rad 5.0 J s ω AB 3.90 rad s v D ( CD) ( 0.53m)( 3.90rad s) ω.00m s v D.00 m s 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-6

27 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 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-7

28 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 m mv G i i i and a couple equal to the sum of their moments about the mass center, H r v m i i For the plane motion of a rigid slab or of a rigid body symmetrical with respect to the reference plane, H G Iω 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-8

29 Principle of Impulse and Momentum 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 expressed as a free-body-diagram equation, 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 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-9

30 Principle of Impulse and Momentum Noncentroidal rotation: - The angular momentum about O I O ω Iω ( mv ) r ( mrω ) Iω r ( I mr )ω - Equating the moments of the momenta and impulses about O, t I ω M dt IOω O t O 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-30

31 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. 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 do not generate nonzero net impulses. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-3

32 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 ) ( H 0 ) 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 0 ) ( H 0 ) Two additional 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. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-3

33 Sample Problem 7.6 SOLUTION: Considering each gear separately, apply the method of impulse and momentum. m m A B 0kg 3kg k k A B 00mm 80mm The system is at rest when a moment of M 6 N mis 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. Solve the angular momentum equations for the two gears simultaneously for the unknown time and tangential force. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-33

34 Sample Problem 7.6 SOLUTION: Considering each gear separately, apply the method of impulse and momentum. FtrA I A( ω A ) ( 0.50 m) ( kg m)( 5.rad s) 0 Ft moments about A: Ft 40. N s moments about B: 0 Mt FtrB I B ( ω B ) ( 6 N m) t Ft ( 0.00 m) ( 0.09 kg m )( 6.8 rad s) Solve the angular momentum equations for the two gears simultaneously for the unknown time and tangential force. t 0.87 s F 46. N 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-34

35 Sample Problem 7.7 SOLUTION: Apply principle of impulse and momentum to find variation of linear and angular velocities with time. Uniform sphere of mass m and radius r is projected along a rough horizontal surface with a linear velocity and no angular v velocity. The coefficient of kinetic friction is µ 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. 003 The McGraw-Hill Companies, Inc. All rights reserved. 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. 7-35

36 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 N W mg 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. x components: mv Ft mv mv k mgt µ mv v v µ gt moments about G: Ftr Iω ( ) ( µ k mg tr mr ) ω 5 ω k 5 µ k g r t v v µ gt k rω 5 µ g r k r t t 7 v µ k g 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-36

37 Sample Problem 7.7 Evaluate the linear and angular velocities at that instant. v v v g 7 µ k g µ k v 5 v 7 Sys Momenta Sys Ext Imp - Sys Momenta y components: x components: moments about G: N W v ω v 5 mg µ k µ k g r gt t 5 µ k g v r 7 µ k g ω ω 5 7 v r v v µ gt k rω 5 µ g r k r t t 7 v µ k g 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-37

38 Sample Problem 7.8 SOLUTION: Observing that none of the external forces produce a moment about the y axis, the angular momentum is conserved. Two solid spheres (radius 75 mm, W kg) are mounted on a spinning horizontal rod ( I R 0.3 kg m, ω 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. Equate the initial and final angular momenta. Solve for the final angular velocity. The energy lost due to the plastic impact is equal to the change in kinetic energy of the system. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-38

39 Sample Problem 7.8 Sys Momenta Sys Ext Imp - Sys Momenta [( m sr ω) r I Sω] I Rω [ ( msrω ) r I Sω ] I R ω ω ω ω 6 rad m m s s s r r I I S S I R I I R R 0.3 kg m 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 S 3 ( kg)( m).5 0 kg 5 ma m 5 m S r 3 3 ( )( 0.5) m ( )( 0.65) S r 0 ω.08 rad s 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-39

40 Sample Problem 7.8 The energy lost due to the plastic impact is equal to the change in kinetic energy of the system. m S r ω 6rad I R s 0.3kg m kg m ω.08rad s I S m S r kg m kg m ( ) ( ) m v I ω I ω m r I I T ω S S R S S R T T T T ( )( 6) ( )(.85) T 6.04J J T 4.8 J 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-40

41 Eccentric Impact Period of deformation Impulse Rdt ( u A ) n ( ub ) n Period of restitution Impulse Pdt Principle of impulse and momentum is supplemented by Rdt e coefficien t of restitutio n Pdt ( v B ) n ( v A ) n ( v A ) n ( vb ) n 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-4

42 Sample Problem 7.9 SOLUTION: A 0 g bullet is fired into the side of a 0 kg square panel which is initially at rest. 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. 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. The reaction at A is found from the horizontal and vertical momenta and impulses. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-4

43 Sample Problem 7.9 moments about A: B B ( 0.35 m) 0 m P v( 0.5 m) I P ω m v SOLUTION: v ( 0.5 m) ω I ( )( ) m P 6 mpb 0 kg 0.45 m 3375kg ( 0.0)( 450)( 0.35) ( 0 kg)( 0.5ω )( 0.5) ω 6 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. ω v 4.67 rad s ( 0.5) ω m s ω 4.67 rad 003 The McGraw-Hill Companies, Inc. All rights reserved. s 7-43

44 Sample Problem 7.9 The reactions at A are found from the horizontal and vertical momenta and impulses. ( 0.5) m s ω 4.67 rad s v ω x components: m B v B A t x ( 0.0)( 450) A ( ) ( 0)( 0.839) x m p v Ax 07 N y components: A x 07 0 A y t 0 A y 0 N 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-44

45 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. 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. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-45

46 Sample Problem 7.0 SOLUTION: Consider the sphere and rod as a single system. Apply the principle of impulse and momentum. moments about A: m s v s (. m) m v (. m) m v ( 0.6 m) Iω v R I rω s s ( 0.6m) ω R R ( 8kg)(.m) 0.96kg ml m 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. ( kg)( 5m s)(.m) ( kg) v (.m) ( 8kg)( 0.6m) ω ( 0.6m) s.4v s 3. 84ω ( 0.96kg m )ω 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-46

47 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. Moments about A:.4v s 3. 84ω Relative velocities: vb vs e( vb vs ) (. m) ω v 0.8( 5m s) s Solve the two relations simultaneously for the angular velocity of the rod and velocity of the sphere. Solving, ω 3.rad/s ω 3.rad/s v s 0.43 m s v s 0.43 m s 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-47

48 Sample Problem 7. SOLUTION: 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 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. 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. 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-48

49 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: ( )( ) ( )( ) m v a 0 mv a Iω ( ) ( )( ) ( )( ) ( m v a 0 m aω a ma ) ω 4 aω 3 v 6 v ω a I 6 ma 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-49

50 Sample Problem 7. 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 V T3 V3 T mv Iω h ( GB) sin( 45 5 ) ( ) a sin a m V Wh ( ) ( ) aω ma ω ma ω T 0 (solving for the minimum ω ) V 3 Wh 3 3 ω ma ω 3W ma Wh 0 Wh 3g ( h h ) ( 0.707a 0.6a) 0.85 g a 3 a 3 3 h a a 4 4 v aω a g a 3 3 v 0. 7 ga 003 The McGraw-Hill Companies, Inc. All rights reserved. 7-50