#### Transcript dyn-part3 - An

CHAPTER 18 PLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY (Sections 18.1-18.4) Objectives: a) Define the various ways a force and couple do work. b) Apply the principle of work and energy to a rigid body. APPLICATIONS The work of the torque developed by the driving gears on the two motors on the mixer is transformed into the rotational kinetic energy of the mixing drum. The work done by the compactor's engine is transformed into the translational kinetic energy of the frame and the translational and rotational kinetic energy of its roller and wheels KINETIC ENERGY The kinetic energy of a rigid body in general plane motion is given by T = 1/2 m (vG)2 + 1/2 IG ω2 Several simplifications can occur. 1. Pure Translation: the rotational kinetic energy is zero (ω = 0): T = 0.5 m (vG)2 KINETIC ENERGY (continued) 2. Rotation: When a rigid body is rotating about a fixed axis passing through point O, the body has both translational and rotational kinetic energy: T = 0.5m(vG)2 + 0.5IGw2 Since vG = rGw, T = 0.5(IG + m(rG)2)w2 = 0.5IOw2 If the rotation occurs about the mass center (pure rotation), G, then vG=0 T = 0.5 IG w2 WORK OF A FORCE The work done by a force is UF = F•dr = s (F cos θ) ds Constant force : UFc = (Fc cos θ)s Work of a weight: Uw = -WΔy. Work of a spring force: Us = -0.5k[(s2)2 – (s1)2] FORCES THAT DO NO WORK 1. Reactions at fixed supports because the displacement at their point of application is zero. 2. Normal and frictional forces acting on bodies as they roll without slipping over a rough surface since there is no instantaneous displacement of the point in contact with ground. 3. Internal forces because they always act in equal and opposite pairs. THE WORK OF A COUPLE A body subjected to a couple produces work only when it undergoes rotation. If the body rotates through an angular displacement dq, the work of the couple q2 moment, M, is UM = M dq constant M: UM = M (q2 – q1) q1 PRINCIPLE OF WORK AND ENERGY The principle states: SU1-2 = T12 i.e. the work done by all external forces and couple moments equals the change in body’s kinetic energy (translational and rotational). This equation is a scalar equation. It can be applied to a system of rigid bodies by summing contributions from all bodies. EXAMPLE Given:The disk weighs 40 N, has a radius of gyration (kG) 0.6 m. A 15 N.m moment is applied. The spring constant is 10N/m. Find: Angular velocity of the disk when point G moves 0.5 m. The disk starts from rest and rolls without slipping. The spring is initially unstretched. EXAMPLE (solution) Free body diagram of the disk: Only the spring force and couple moment M do work. Spring will stretch twice the amount of displacement of G, or 1 m. Why? EXAMPLE (continued) Work: U1-2 = -0.5k[(s2)2 – (s1)2] + M(q2 – q1) U1-2 = -0.5(10)(12 – 0) + 15(0.5/0.8) = 4.4N.m Kinematic relation: vG = r w = 0.8w Kinetic energy: T1 = 0 T2 = 0.5m (vG)2 + 0.5 IG w2 T2 = 0.5(40/9.8)(0.8w)2 + 0.5(40/9.8)(0.6)2w2 T2 = 2.0 w2 Work and energy: U1-2 = T12 4.4 = 2.0 w2 w = 1.5 rad/s GROUP PROBLEM SOLVING Given: A sphere weighing 10N rolls along a semicircular hoop. Its w equals 0 when q = 0. Find: The angular velocity of the sphere when q = 45° if the sphere rolls without slipping. GROUP PROBLEM SOLVING (continued) Solution: Draw a FBD and calculate the vertical distance the mass center moves. Now calculate the _______________ : GROUP PROBLEM SOLVING (continued) A kinematic equation for finding the velocity of the mass center is needed. It is _________ energy: Now apply the principle of work and energy equation: w = 13.1 rad/s PLANAR KINETICS OF A RIGID BODY: CONSERVATION OF ENERGY (Section 18.5) Objectives: a) Determine the potential energy of conservative forces. b) Apply the principle of conservation of energy. APPLICATIONS Torsional spring at the top of the door winds up as the door is lowered. When the door is raised, the spring potential energy is transferred into gravitational potential energy of the door’s weight, thus making it easy to open. CONSERVATION OF ENERGY The conservation of energy theorem is a “simpler” energy method for solving problems. Once again, the problem parameter of distance is a key indicator of when conservation of energy is a good method for solving the problem. If it is appropriate, conservation of energy is easier to use than the principle of work and energy. CONSERVATIVE FORCES A force F is conservative if the work done by the force is independent of the path. Typical conservative forces encountered in dynamics are gravitational forces (i.e., weight) and elastic forces (i.e., springs). What is a common force that is not conservative? CONSERVATION OF ENERGY When a rigid body is acted upon by a system of conservative forces, the work done by these forces is conserved. Thus, the sum of kinetic energy and potential energy remains constant: T1 + V1 = T2 + V2 = Constant i.e. as a rigid body moves from one position to another when acted upon by only conservative forces, kinetic energy is converted to potential energy and vice versa. GRAVITATIONAL POTENTIAL ENERGY The gravitational potential energy is a function of the height of the body’s center above or below a datum. The gravitational potential energy is found by: Vg = W yG Gravitational potential energy is +ve when yG is +ve, since the weight has the ability to do +ve work when the body is moved back to the datum. ELASTIC POTENTIAL ENERGY Spring forces are also conservative forces. The potential energy of a spring force (F = ks) is found by the equation Ve = ½ ks2 Notice that the elastic potential energy is always +ve EXAMPLE 1 Given:The rod AB has a mass of 10 kg. Piston B is attached to a spring of constant k = 800 N/m. The spring is un-stretched when q = 0°. Neglect the mass of the pistons. Find: The angular velocity of rod AB at q = 0° if the rod is released from rest when q = 30°. EXAMPLE 1 (solution) Initial Position Final Position Potential Energy: put datum in line with rod when q=0°. Gravitational and elastic energy=0 at 2. => V2 = 0 Gravitational energy at 1: - (10)( 9.81) ½ (0.4 sin 30) Elastic energy at 1: ½ (800) (0.4 sin 30)2 So V1 = - 9.81J + 16.0 J = 6.19 J EXAMPLE 1 (continued) Initial Position Final Position Kinetic Energy: The rod released from rest (vG1=0, w1=0). Thus, T1 = 0. At 2, the angular velocity is w2 and the velocity is vG2 . EXAMPLE 1 (continued) Thus, T2 = ½ (10)(vG2)2 + ½ (1/12)(10)(0.42)(w2)2 At 2, vA=0. Hence, vG2 = r w = 0.2 w2 . Then, T2 = 0.2 w22 + 0.067 w22 = 0.267 w22 conservation of energy: T1 + V1 = T2 + V2 0 + 6.19 = 0.267w22 + 0 => w2 = 4.82 rad/s EXAMPLE 2 Given: The weight of the disk is 30 N and its kG equals 0.6 m. The spring has a stiffness of 2 N/m and an unstretched length of 1m. Find: The angular velocity at the instant G moves 3 m to the left. The disk is released from rest in the position shown and rolls without slipping. EXAMPLE 2 (solution) Potential Energy: No changes in the gravitational potential energy The elastic potential energy at 1: V1 = 0.5 k (s1)2 where s1 = 4 m. Thus, V1 = ½ 2 (4)2 = 16 N.m. The elastic potential energy at 2 is V2 = ½ 2 (3)2 = 9 N.m. EXAMPLE 2 (continued) Kinetic Energy: Released from rest: vG1=w1=0, T1=0 At 2, w2 and vG2: T2 = ½ m (vG2)2 + ½ IG (w2) 2 = ½ (30/9.8) (vG2)2 + ½ (30/9.8) 0.62 (w2) 2 Disk is rolling without slipping: vG2 = (0.75 w2) T2 = ½(30/9.8)(0.75 w2)2 + ½(30/9.8) 0.62 (w2)2 = 1.41 (w2)2 EXAMPLE 2 (continued) conservation of energy: T1 + V1 = T2 + V2 0 +16.0 = 1.41 w22 + 9 Solving , w2 = 2.23 rad/s GROUP PROBLEM SOLVING Given:A 50 N bar is rotating downward at 2 rad/s. The spring has an unstretched length of 2m and a spring constant of 12 N/m Find: The angle (measured down from the horizontal) to which the bar rotates before it stops its initial downward movement. GROUP PROBLEM SOLVING (solution) Potential Energy: Put the datum in line with the rod when q = 0. Gravitational potential energy at 2: Elastic potential energy at 2: So, V2 = GROUP PROBLEM SOLVING (continued) Kinetic Energy: At 1 (when q = 0): At 2: GROUP PROBLEM SOLVING (continued) conservation of energy Thus, q = 49.9 deg. CHAPTER 19 PLANAR KINETICS: IMPULSE AND MOMENTUM (Sections 19.1-19.2) Today’s Objectives: a) Develop formulations for the linear and angular momentum of a body. b) Apply the principle of linear and angular impulse and momentum. APPLICATIONS As the pendulum swings downward, its angular momentum and linear momentum both increase. By calculating its momenta in the vertical position, we can calculate the impulse the pendulum exerts when it hits the test specimen. The space shuttle has several engines that exert thrust on the shuttle when they are fired. By firing different engines, the pilot can control the motion and direction of the shuttle. LINEAR AND ANGULAR MOMENTUM The linear momentum of a rigid body is: L = m vG The angular momentum of a rigid body is: HG = IG w LINEAR AND ANGULAR MOMENTUM (continued) Translation. When a rigid body undergoes rectilinear or curvilinear translation, its angular momentum is zero because ω = 0. Therefore: L = m vG HG = 0 LINEAR AND ANGULAR MOMENTUM (continued) Rotation about a fixed axis. The body’s linear momentum and angular momentum are: L = m vG HG = IGw HO = ( rG x mvG) + IGw = IO w LINEAR AND ANGULAR MOMENTUM (continued) General plane motion. The linear and angular momentum computed about G are required. L = m vG HG = IGω The angular momentum about point A is HA = IGω + (d)mvG PRINCIPLE OF IMPULSE AND MOMENTUM The principle is developed by combining the equation of motion with kinematics. The resulting equations allow a direct solution to problems involving force, velocity, and time. Linear impulse-linear momentum equation: t2 L1 + t F dt = L2 t2 or (mvG)1 + 1 t F dt = (mvG)2 1 Angular impulse-angular momentum equation: t2 (HG)1 + t MG dt = (HG)2 1 t2 or IGw1 + MG dt = IGw2 t1 PRINCIPLE OF IMPULSE AND MOMENTUM (continued) The previous relations can be represented graphically by drawing the impulse-momentum diagram. + = EXAMPLE Given: A disk weighing 50 N has a rope wrapped around it. The rope is pulled with a force P equaling 2 N. Find: The angular velocity of the disk after 4 seconds if it starts from rest and rolls without slipping. EXAMPLE (solution) y Impulse-momentum diagram: W t x Pt G (m vG)1 + IGw1 Kinematics: (vG)2 = r w2 = r G A Nt Ft (m vG)2 IG w2 t2 Impulse & Momentum: (HA)1 + MA dt = (HA)2 t1 (P t) 2 r = (mvG)2 r + IGw2 = m r2 w2 + 0.5 m r2 w2 = 1.5 m r2 w2 w2 = 4 Pt 4(2)(4) = = 3.5 rad/s 3mr 3(50/9.81)(0.6) GROUP PROBLEM SOLVING Given: A gear set with: WA = 15 N WB = 10 N kA = 0.5 m kB = 0.35 m M = 2(1 – e-0.5t) N.m Find: The angular velocity of gear A after 5 seconds if the gears start turning from rest. Plan: Time is a parameter, thus GROUP PROBLEM SOLVING (continued) Solution: Impulse-momentum diagrams: Gear A: y x Gear B: GROUP PROBLEM SOLVING (continued) Kinematics: Angular impulse & momentum relation for gear A about point A yields: For gear B: GROUP PROBLEM SOLVING (continued) and wA = 14.4 rad/s CONSERVATION OF MOMENTUM (Section 19.3) Objectives: a) Understand the conditions for conservation of linear and angular momentum b) Use the condition of conservation of linear/ angular momentum APPLICATIONS A skater spends a lot of time either spinning on the ice or rotating through the air. To spin fast, or for a long time, the skater must develop a large amount of angular momentum. If the skater’s angular momentum is constant, can the skater vary his rotational speed? How? The skater spins faster when the arms are drawn in and slower when the arms are extended. Why? CONSERVATION OF LINEAR MOMENTUM Recall that the linear impulse and momentum relationship is t2 L1 + 0 t F dt = L2 1 t2 or (m vG)1 + t 0 F dt = (m vG)2 1 If the sum of all the linear impulses acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, the linear momentum for a rigid body (or system) is constant, or conserved. So L1 = L2. This equation is referred to as the conservation of linear momentum. The conservation of linear momentum equation can be used if the linear impulses are small or non-impulsive. CONSERVATION OF ANGULAR MOMENTUM The angular impulse-angular momentum relationship is: t2 (HG)1 + 0 t MG dt = (HG)2 1 t2 or IGw1 + 0 t MG dt = IGw2 1 Similarly, if the sum of all the angular impulses due to external forces acting on the rigid body (or a system of rigid bodies) is zero, all the impulse terms are zero. Thus, angular momentum is conserved . The resulting equation is referred to as the conservation of angular momentum or (HG)1 = (HG)2 . If the initial condition of the rigid body (or system) is known, conservation of momentum is often used to determine the final linear or angular velocity of a body just after an event occurs. EXAMPLE Given: A 10 kg wheel (IG = 0.156 kg·m2) rolls without slipping and does not bounce at A. Find: The minimum velocity, vG, of the wheel to just roll over the obstruction at A. Note: Since no slipping or bouncing occurs, the wheel pivots about point A. The force at A is much greater than the weight, and since the time of impact is very short, the weight can be considered non-impulsive. The reaction force at A, since we don’t know either its direction or magnitude, can be eliminated by applying the conservation of angular momentum equation about A. EXAMPLE (solution) y Impulse-momentum diagram: + x = Conservation of angular momentum: (HA)1 = (HA)2 r ' (mvG)1 + IG w1 = r (mvG)2 + IG w2 (0.2 - 0.03)(10)(vG)1 + 0.156 w1 = 0.2(10)(vG)2 + 0.156 w2 Kinematics: no slip, w = vG / r = 5 vG .Substituting and solving: (vG)2 = 0.892 (vG)1 EXAMPLE (continued) To complete the solution, conservation of energy can be used. Since it cannot be used for the impact (why?), it is applied just after the impact. In order to roll over the bump, the wheel must go to position 3 from 2. When (vG)2 is a minimum, (vG)3 is zero. Why? T2 + V2 = T3 + V3 {½(10)(vG)22 + ½(0.156)w22 } + 0 = 0 + 98.1 (0.03) Substituting w2 = 5 (vG)2 and (vG)2 = 0.892 (vG)1 and solving yields: (vG)1 = 0.729 m/s, (vG)2 = 0.65 m/s GROUP PROBLEM SOLVING Given: A slender rod (Wr = 5 N) has a wood block (Ww = 10 N) attached. A bullet (Wb = 0.2 N) is fired into the center of the block at 1000 m/s. Assume the pendulum is initially at rest and the bullet embeds itself into the block. Find: The angular velocity of the pendulum just after impact. GROUP PROBLEM SOLVING (continued) Solution: First draw a FBD. To use conservation of angular momentum, IA = ? IA = 7.2kg.m2 GROUP PROBLEM SOLVING (continued) Apply the conservation of angular momentum equation: Solving yields: w2 = 7.0 rad/s CHAPTER 22 VIBRATIONS (section 22.1) Objectives: Discuss Undamped single degree of freedom (SDOF) vibration of a rigid body Vibration definition Vibration is the periodic motion of a body or system of connected bodies displaced from a position of equilibrium Types of vibrations Free vibrations: motion is maintained by conservative forces (like gravitational or elastic restoring forces). Forced vibrations: motion is caused by external periodic or intermittent forces. Undamped free response for SDOF systems If you displace mass m a distance u from equilibrium position and release it, vibration occurs Undamped free response for SDOF systems (cont): Equilibrium eq: A linear, constant coefficients, homogeneous, second order ordinary differential equation u Ku = m 2 t 2u m 2 Ku = 0 t 2 u m 2 ku = 0 t 2u A and B are found from 2 wn u = 0 2 initial conditions t u = A cos wnt sin wnt 2 Undamped free response for SDOF systems: ωn =natural circular frequency of vibration Tn =natural period of vibration k wn = m Tn = 2 wn fn =1/Tn= natural cyclic frequency of vibration in hertz (cycles per second) Solving initial conditions u = A cos wnt sin wnt u (t = 0) = u0 = A . . u (t = 0) = u 0 = Bwn . u = u0 cos wnt u0 wn sin wnt Example 1 Example 1 (continued) Example 1 (continued) By analogy By analogy Example 2 Example 2 (continued) From equilibrium Example 2 (continued) Thus: Group Solving problem Draw FBD Write equation of motion Use analogy Energy method Example 1 Energy methods: Example 2 Undamped forced vibrations Free Body Diagram Equilibrium Equation u F (t ) ku = m 2 t 2 u m 2 ku = F (t ) t 2 Solution for constant loading u m ku = P 2 t 2 u 2 wn u = P / m 2 t u = A cos wn t sin wn t P / k 2 Solving initial conditions u = A cos . wt sin wt P / k t = 0, u0 = 0 B = 0 t = 0, u0 = 0 A = P / k P u= (1 cos wn t ) K umax = 2 P / K = 2u static Sine curve .. m u ku = F sin wot F 1 u = C cos wnt D sin wnt sin wot k 1 ( wo ) 2 wn DMF = 1 1 ( wo wn ) 2 Viscous Damped free vibration : k F = kx cx = mx m c x Define: viscous damping factor (no unit) ( zeta) c 2mw n c : viscous damping constant (in N s/m) mx cx kx = 0 (dashpot ) x=0 x 2wn x wn2 x = 0 Use a trial solution x = A et : 2 2w n wn2 = 0 Obtain two possible ’s : 1 = w n ( e λ1t and e λ2t 2 1) and 2 = wn ( 2 1) are two solutions. General solution is a linear combination x = A1e1t A2e2t 90 Three cases: x >1, overdamped Critical = 1 damped t <1 underdamping Case 1 : > 1 (overdamping) : 1 and 2 < 0 x = A1e 1 t A2e 2 t x decays to zero without oscillation 91 Case 2: = 1 (critical damping) : 1 = 2 = -wn A1 exp(-wnt) is a solution A2 t exp(-wnt) is also a solution (prove !) The general solution is x = (A1 + A2 t )exp(-wnt) x approaches zero quickly without oscillation. 92 Case 3: < 1 (underdamping) : x = ( A1e i 1 2 wn t w n t ) e Define damped natural frequency w d = w n 1 2 x = ( A1e iwd t A2 e i 1 2 wn t A2e iwd t )e w n t [ x is real, so that A1 and A2 are complex conjucate. i i C C Let A1 = e and A2 = e ] 2 2 w nt x = (Ce ) sin(wd t ) x Period d = 2/wd 93 x1 t1 d x2 t2 t The ratio of the displaceme nts of two successive cycles x = Ce w = ew x Ce w 1 2 n t1 n d n ( t1 d ) Definelogarithmic decrement n x1 / x2 2 2 = w n d = w n = wn 1 2 1 2 It is because = c/(2mwn), measurement of determines the viscous damping constant c. 94 B. Viscous Damped Forced vibration The equation of motion : F = kx cx F sin wt = mx o mx cx kx = F sin wt o F x 2w x w x = sin wt m o 2 n n F = F sin w t O c (dashpot ) x x=0 95 m k (spring) Let x = C sin( wt ), then x = Cw cos(wt ) and x = Cw sin( wt ) p 2 p p F x 2w x w x = sin wt m n C= o 2 n Fo / k 1 2ww/ w/ w 2w / w 2 2 2 n tan = 1/ 2 n n 1 w / wn 2 Defineamplit uderatioor magnification fact or 1/ 2 C 2 2 2 M= = 1 / 1 w / wn 2w / wn Fo / k 96 2 w 2 d w 2 2 Maximum M occurs at: {[ 1( ) ] ( ) }= 0 dw wn wn The resonance frequency is w res = 12 2 =0 M = 0.1 = 0.2 =1 1 1 97 w / wn wn wwn tan = (wwn ) 2 =0 = 0.2 =1 /2 0 1 (resonance) w / wn Consider the following regions: (1) w is small, tan > 0, 0+, xp in phase with the driving force (2) w is large, tan < 0, 0-, = , xp lags the driving force by 90o (3) w wn-, tan +, /2(-) w wn+, tan -, /2(+) 98 If the driving force is not applied to the mass, but is applied to the base of the system: cx kx = m( x x ) c B mx cx kx = mx x=0 B m x x 2wn x w x = bw sin wt 2 n 2 k If bw2 is replaced by Fo/m: Fo x 2w n x w x = sin wt m 2 n x = b sin wt B This can be used as a device to detect earthquake. 99 Example m =45 kg, k = 35 kN/m, c = 1250 N.s/m, p = 4000 sin (30 t) Pa, A= 50 x 10-3 m2. Determine : (a) steady-state displacement (b) max. force transmitted to the base. A p m k c wn = k / m = 35103 / 45 = 27.9 rad / s = c /(2mwn ) = 1250/(2 45 27.9) = 0.498 (underdam pe d) 100 The amplitude of the steady-state vibration is: X = = Fo / k 1 w / wn 2 2 2w / w n 4000 50 10 = tan 1 30 / 27.9 1 2 2 3 2 1/ 2 / 35 10 3 2 0.498 30 / 2.79 2 1/ 2 = 0.00528 m 1 2 0.498 30 / 27.9 2 = tan 2 = 1.716 rad 1 30 / 27.9 1 w / w n 2w / w n xP = X cos(wt ) = 0.00528cos(30t 1.716) m # 101 The force transmitted to the base is : F = kx cx = kX sin (wt ) cwX cos (wt ) tr p p For max Ftr : dF = 0 wt = tan cw d wt k 1 tr wt = tan 1250 3035 10 = 46.9 35 10 3 1 o 3 F tr max = 35 10 0.00528 cos 46.9 1250 30 0.00528 sin 46.9 = 271 N 3 o # 102 o Newton Laws, Energy Theorems and Momentum Principles Integration Principles Group Working Problems Problem 1: Chapters 17, 18 and 19 Given: The 30-kg disk is pin connected at its center. If it starts from rest, Find: the number of revolutions it must make and the time needed to attain an angular velocity of 20rad/s? Problem 1: Newton Laws Find angular acceleration: (11.7rad/s2) Find angle (17.1 radians) Find number of revolutions: (2.73) Find time (1.71 sec) Problem 1: Energy Theorems Find angle (17.1 radians) Find number of revolutions: (2.73) Find angular acceleration: (11.7rad/s2) Find time (1.71 sec) Problem 1: Momentum Principle Find time (1.71 sec) Find angular acceleration: (11.7rad/s2) Find angle (17.1 radians) Find number of revolutions: (2.73) Problem 2: Chapter 22 in view of Chapters 17, 18 and 19 Given: a 10-kg block suspended from a cord wrapped around a 5kg disk. Find: natural period of vibration Problem 2: Newton Laws Find differential equation Find period (1.57s) Problem 2: Energy Theorems: use conservation of energy Write conservation of energy equation Differentiate equation Find period (1.57s) Problem 2: Momentum Principle: use M=d(HG)/dt Write M=d(HG)/dt for a displacement θ from equilibrium position Differentiate equation Find period (1.57s)