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Tenth Edition CHAPTER 17 VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Ferdinand P. Beer E. Russell Johnston, Jr. Phillip J. Cornwell Lecture Notes: Brian P. Self California State Polytechnic University Plane MotionofofRigid Rigid Plane Motion Bodies:Energy Bodies: Energy and Momentum Methods © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Introduction To predict the launch from a catapult, you must apply the principle of work-energy. To determine the forces acting on the stopper pin when the catapult reaches its final position, angular impulse momentum equations are used. 17 - 2 Introduction • Method of work and energy and the method of impulse and momentum will be used to analyze the plane motion of rigid bodies and systems of rigid bodies. • Principle of work and energy is well suited to the solution of problems involving displacements and velocities. T1 U12 T2 • Principle of impulse and momentum is appropriate for problems involving velocities and time. t2 t2 H O 1 M O dt H O 2 L1 Fdt L2 t1 t1 • Problems involving eccentric impact are solved by supplementing the principle of impulse and momentum with the application of the coefficient of restitution. 17 - 3 Introduction Approaches to Rigid Body Kinetics Problems Forces and Accelerations Velocities and Displacements Velocities and Time Newton’s Second Law (last chapter) Work-Energy ImpulseMomentum å F = ma åM = H t2 G G 2-4 G T1 U12 T2 mv1 F dt mv2 t1 t2 I G1 M G dt I G2 t1 Principle of Work and Energy for a Rigid Body •Work and kinetic energy are scalar quantities. • Assume that the rigid body is made of a large number of particles. T1 U12 T2 T1 , T2 initial and final total kinetic energy of particles forming body U12 total work of internal and external forces acting on particles of body. • Internal forces between particles A and B are equal and opposite. • Therefore, the net work of internal forces is zero. 17 - 5 Work of Forces Acting on a Rigid Body • Work of a force during a displacement of its point of application, A2 s 2 U12 F dr F cos ds A1 s1 • Consider the net work of two forces F and F forming a couple of moment M during a displacement of their points of application. dU F dr1 F dr1 F dr2 F ds2 Fr d M d 2 U12 M d 1 M 2 1 if M is constant. 17 - 6 Kinetic Energy of a Rigid Body in Plane Motion • Consider a rigid body of mass m in plane motion consisting of individual particles i. The kinetic energy of the body can then be expressed as: T 12 mv 2 12 Δmi vi2 12 mv 2 12 ri2 Δmi 2 12 mv 2 12 I 2 • Kinetic energy of a rigid body can be separated into: - the kinetic energy associated with the motion of the mass center G and - the kinetic energy associated with the rotation of the body about G. T 12 mv 2 17 - 7 1 2 I 2 Translation + Rotation Kinetic Energy of a Rigid Body in Plane Motion • Consider a rigid body rotating about a fixed axis through O. T 1 2 Δm v 2 i i 1 2 2 2 1 Δ m r r Δ m i i i 2 i 2 12 IO 2 • This is equivalent to using: T 12 mv 2 1 2 I 2 • Remember to only use T 12 I O 2 when O is a fixed axis of rotation 17 - 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, T1 U12 T2 T1 ,T2 = arithmetic sum of the kinetic energies of all bodies forming the system U12 = work of all forces acting on the various bodies, whether these forces are internal or external to the system as a whole. T T 17 - 9 W = 120 g Systems of Rigid Bodies • For problems involving pin connected members, blocks and pulleys connected by inextensible cords, and meshed gears, - internal forces occur in pairs of equal and opposite forces - points of application of each pair move through equal distances - net work of the internal forces is zero - work on the system reduces to the work of the external forces 17 - 10 Conservation of Energy • Expressing the work of conservative forces as a change in potential energy, the principle of work and energy becomes T1 V1 T2 V2 • Consider the slender rod of mass m. T1 0, V1 0 T2 12 mv22 12 I 22 12 m 12 l 2 2 1 ml 1 ml 12 12 2 2 3 2 2 V2 12 Wl sin 12 mgl sin T1 V1 T2 V2 • mass m • released with zero velocity • determine at 17 - 11 1 ml 2 2 1 0 mgl sin 2 3 2 3g sin l Power • Power = rate at which work is done • For a body acted upon by force F and moving with velocity v, dU Power F v dt • For a rigid body rotating with an angular velocity and acted upon by a couple of moment M parallel to the axis of rotation, dU M d Power M dt dt 17 - 12 Sample Problem 17.1 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 r For the drum and flywheel, I = 16kg ×m2 . The bearing friction is equivalent to a couple of 90 N ×m. At the instant shown, the block is moving downward at 2 m/s. Determine the velocity of the block after it has moved 1.25 m downward. 17 - 13 • Apply the principle of work and kinetic energy to develop an expression for the final velocity. Sample Problem 17.1 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 v1 2m s v2 v = = 5rad s = 2 ωw 2 = r 0.4m r 0.4m • Apply the principle of work and kinetic energy to develop an expression for the final velocity. v = rrω w w1 = ω T1 = 12 mv12 + 12 I w12 1 1 2 2 120kg 2m s + 16kg ×m 2 (5rad s ) ( )( ) 2 2 = 440J = ( ) T2 = 12 mv22 + 12 I w 22 2 æ ö v 1 1 = (120)v22 + (16) ç 2 ÷ = 110v22 2 2 è 0.4 ø 17 - 14 Sample Problem 17.1 T1 = 12 mv12 + 12 I w12 = 440J T2 = 12 mv22 + 12 I w22 = 110v22 • Note that the block displacement and pulley rotation are related by qT22 = s2 1.25m = = 3.125rad r 0.4m Then, U1®1 - 22 = W (s2 - s1 ) - M (q2 - q1 ) ( ) = (120kg ) 9.81m/s 2 (1.2m) - (90 N ×m )(3.125rad ) = 1190J • Principle of work and energy: T1 + U1®1 -22 = T2 1.25 m 1.25 m 440J + 1190J = 110 v22 v2 = 3.85m s 17 - 15 v2 = 3.85m s Sample Problem 17.2 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 mA 10 kg k A 200 mm required for the work of the applied mB 3 kg k B 80 mm moment to equal the final kinetic energy of the system. The system is at rest when a moment • Apply the principle of work and energy to of M 6 N m is applied to gear B. a system consisting of gear A. With the Neglecting friction, a) determine the final kinetic energy and number of number of revolutions of gear B revolutions known, calculate the moment before its angular velocity reaches 600 and tangential force required for the rpm, and b) tangential force exerted indicated work. by gear B on gear A. 17 - 16 Sample Problem 17.2 SOLUTION: • Consider a system consisting of the two gears. Noting that the gear rotational speeds are related, evaluate the final kinetic energy of the system. B 600 rpm2 rad rev 62.8 rad s 60 s min r 0.100 A B B 62.8 25.1rad s rA 0.250 I A m Ak A2 10kg 0.200m 2 0.400 kg m 2 I B mB k B2 3kg 0.080m 2 0.0192 kg m 2 T2 12 I A A2 12 I B B2 12 0.400 25.1 2 12 0.0192 62.82 163.9 J 17 - 17 Sample Problem 17.2 • Apply the principle of work and energy. Calculate the number of revolutions required for the work. T1 U12 T2 0 6 B J 163.9J B 27.32 rad B 27.32 4.35 rev 2 • 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 0.100 A B B 27.32 10.93 rad rA 0.250 T2 12 I A A2 12 0.40025.1 2 126.0 J T1 U12 T2 0 M A 10.93 rad 126.0J M A rA F 11.52 N m 17 - 18 F 11.52 46.2 N 0.250 Sample Problem 17.3 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. 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. 17 - 19 • 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. Sample Problem 17.3 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. v With r v T2 12 mv 12 I 12 mv 12 I r I 12 m 2 v 2 r 2 2 2 T1 U1 2 T2 I 0 Wh 12 m 2 v 2 r 2Wh 2 gh v2 m I r 2 1 I mr 2 17 - 20 2 Sample Problem 17.3 • 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. 2 gh v2 1 I mr 2 I 52 mr 2 v 0.845 2 gh Cylinder : I 12 mr 2 v 0.816 2 gh Sphere : Hoop : I mr 2 v 0.707 2 gh NOTE: • For a frictionless block sliding through the same distance, 0, v 2 gh 17 - 21 • The velocity of the body is independent of its mass and radius. • The velocity of the body does depend on k2 I mr 2 r2 Sample Problem 17.4 SOLUTION: • The weight and spring forces are conservative. The principle of work and energy can be expressed as T1 V1 T2 V2 A 15-kg slender rod pivots about the point O. The other end is pressed against a spring (k = 300 kN/m) until the spring is compressed 40 mm and the rod is in a horizontal position. • 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. 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. 17 - 22 Sample Problem 17.4 SOLUTION: • The weight and spring forces are conservative. The principle of work and energy can be expressed as T1 V1 T2 V2 • Evaluate the initial and final potential energy. 2 V1 = Vg + Ve = 0 + 12 kx12 = 12 (300,000 N m)(0.04m) = 240J V2 = Vg + Ve = Wh + 0 = (147.15 N )(0.75m ) I = 121 ml 2 1 (15kg )(2.5m )2 12 = 7.81kg ×m 2 = = 110.4J • Express the final kinetic energy in terms of the angular velocity of the rod. 2 T2 = 12 mv22 + 12 I w22 = 12 m (rw2 ) + 12 I w22 = 17 - 23 1 (15 )(0.75w2 )2 + 12 (7.81)w22 = 8.12w22 2 Sample Problem 17.4 From the principle of work and energy, T1 + V1 = T2 + V2 0 + 240 J = 8.12w22 + 110.4 J w2 = 3.995rad s • Based on the free-body-diagram equation, solve for the reactions at the pivot. 2 2 an = r w22 = (0.75m )(3.995rad s ) = 11.97 m s 2 an = 11.97 m s at = ra M O M O eff Fx Fx eff Fy Fy eff at = ra 0 I mr r 0 Rx mr Rx 0 Ry - 147.15 N = - man ( = - (15kg ) 11.97 m s 2 Ry = - 32.4 N R = 32.4 N 17 - 24 ) Sample Problem 17.5 SOLUTION: • Consider a system consisting of the two rods. With the conservative weight force, T1 V1 T2 V2 • Evaluate the initial and final potential energy. • Express the final kinetic energy of the Each of the two slender rods has a system in terms of the angular velocities of mass of 6 kg. The system is released the rods. from rest with b = 60o. • Solve the energy equation for the angular Determine a) the angular velocity of velocity, then evaluate the velocity of the o rod AB when b = 20 , and b) the point D. velocity of the point D at the same instant. 17 - 25 Sample Problem 17.5 SOLUTION: • Consider a system consisting of the two rods. With the conservative weight force, T1 V1 T2 V2 • Evaluate the initial and final potential energy. V1 2Wy1 258.86 N 0.325 m 38.26 J V2 2Wy2 258.86 N 0.1283 m 15.10 J W mg 6 kg 9.81m s 2 58.86 N 17 - 26 Sample Problem 17.5 • Express the final kinetic energy of the system in terms of the angular velocities of the rods. vAB 0.375m Since vB is perpendicular to AB and vDis horizontal, the instantaneous center of rotation for rod BD is C. CD 20.75 msin 20 0.513 m BC 0.75 m and applying the law of cosines to CDE, EC = 0.522 m Consider the velocity of point B vB AB BC AB BD vBD 0.522 m For the final kinetic energy, 1 ml 2 1 6 kg 0.75 m 2 0.281kg m 2 I AB I BD 12 12 1 mv 2 1 I 2 1 mv 2 1 I 2 T2 12 AB 2 AB AB 12 BD 2 BD BD 1 6 0.375 2 1 0.281 2 1 6 0.522 2 1 0.281 2 12 2 12 2 17 - 27 1.520 2 Sample Problem 17.5 • Solve the energy equation for the angular velocity, then evaluate the velocity of the point D. T1 V1 T2 V2 0 38.26 J 1.520 2 15.10 J 3.90 rad s AB 3.90 rad s vD CD 0.513 m 3.90 rad s 2.00 m s vD 2.00 m s 17 - 28 Exercise 1 Exercise 2 THE END