Anotações Cinemática Plana de Corpos Rígidos-2023 1

1 Parte 5 - Cinemática Plana de Corpos Rígidos Conceito: Geometria apenas, sem considerar a causa do movimento. Corpo Rígido: é um sistema de partículas para o qual as distâncias entre elas permanecem inalteradas. Tipos de Movimento Plano: Translação, rotação em torno de um eixo fixo e movimento plano geral. Vide Figura 5/1 abaixo. 2 5.1 Rotação em Torno de Um Eixo Fixo e Método do Movimento Absoluto 3 Lista de Exercícios 11: Problema resolvido 5/2, Problema Resolvido 5/3, Problema Resolvido 5/26, Problema 5/17, Problema 5/28, Problema Resolvido 5/4 e Problema Resolvido 5/57. SAMPLE PROBLEM 5/5 The load L is being hoisted by the pulley and cable arrangement shown. Each cable is wrapped securely around its respective pulley so it does not slip. The two pulleys to which L is attached are fastened together to form a single rigid body. Calculate the velocity and acceleration of the load L and the corresponding angular velocity ω and angular acceleration α of the double pulley under the following conditions: Case (a) Pulley 1: ω1 = 0; α1 = 0 (pulley at rest) Pulley 2: ω2 = 2 rad/sec, α2 = 3 rad/sec2 Case (b) Pulley 1: ω1 = 1 rad/sec, α1 = -4 rad/sec2 Pulley 2: ω2 = 2 rad/sec, α2 = -2 rad/sec2 Solution. The tangential displacement, velocity, and acceleration of a point on the rim of pulley 1 or 2 equal the corresponding vertical motions of point A or B since the cables are assumed to be inextensible. Case (a). With A momentarily at rest, line AB rotates to A'B' through the angle dө during time dt. From the diagram we see that the displacements and their time derivatives give dAB = AB dө νB = ABω a(b)t = ABα dA = 0 νA = 0 aA = 0 With vD = rAω1 = 4(2) = 8 in/sec and ω = rAω1 = 4(3) = – 12 in/sec2, we have for the angular motion of the double pulley ν = vB / AB = 8 / 12 = 2/3 rad/sec (CCW) Ans. a = (a(b)t - a(a)t)/AB = -12/12 = -1 rad/sec2 (CW) Ans. The corresponding motion of O and the load L is νO = vO = 4(2/3) = 83 in/sec αO = aO = 4(–1) = –4 in/sec2 Case (b). With point C, and hence point A, in motion, line AB moves to A'B' during time dt... Helpful Hint Recognize that the inner pulley is a wheel rolling along the fixed line of the left-hand cable. Thus, the expressions of Sample Problem 5/4 hold... 8 Faça um programa computacional para expressar graficamente 𝜔𝐴𝐵 e 𝛼𝐴𝐵 em função da posição angular da manivela 0 ≤ θ ≤ 3600. 9 5.2 Método do Movimento Relativo para Eixos Transladados. Análise de Velocidades Conceito: Movimento Plano Geral = Translação Pura + Rotação em Torno de Um Eixo Fixo (Figura 5/5 (a)) XY: sistema inercial A Figura 5/5 (a) mostra que, tomando dois pontos A e B de um corpo rígido, pode se escrever a seguinte equação, ∆𝐫A = ∆𝐫B + ∆𝐫A B ∆𝐫A: deslocamento absoluto do ponto A, ∆𝐫B: deslocamento absoluto do ponto B e ∆𝐫A B: deslocamento relativo do ponto A com relação ao ponto B. Dividindo a equação acima por ∆t e fazendo o limite quando ∆t tende a zero, resulta, lim ∆𝑡→0 ∆𝐫𝐴 ∆𝑡 = lim ∆𝑡→0 ∆𝐫𝐵 ∆𝑡 + lim ∆𝑡→0 ∆𝐫𝐴 𝐵 ∆𝑡 Ou, 𝐯A = 𝐯B + 𝐯A B 10 𝐯A: velocidade absoluta do ponto A, 𝐯B: velocidade absoluta do ponto B e 𝐯A B: velocidade relativa do ponto A com relação ao ponto B. Conceito: O movimento relativo do ponto A com relação ao ponto B é a rotação do ponto A com relação a um eixo fixo em B. Vide Figura 5/5 (b) Portanto, 𝐯A B = 𝛚 × 𝐫 𝛚: rotação absoluta do corpo rígido que contém os dois pontos A e B e 𝐫: posição relativa do ponto A com relação ao ponto B. Portanto, a equação de velocidades fica como, 𝐯A = 𝐯B + 𝐯A B = 𝐯B + 𝛚 × 𝐫 Vide Figura 5/6 11 Lista de Exercícios 12: Problema resolvido 5/7, Problema Resolvido 5/8, Problema Resolvido 5/9, Problema 5/60, Problema 5/64, Problema 5/88 e Problema 5/89. SAMPLE PROBLEM 5/8 Crank CB oscillates about O through a limited arc, causing crank OA to oscillate about O. When the linkage passes the position shown with CB horizontal and OA vertical, the angular velocity of CB is 2 rad/s counterclockwise. For this instant, determine the angular velocities of OA and AB. Solution I (Vector). The relative-velocity equation νA = νB + νAB is rewritten as ωOA x rA = ωCA x rC + ωAB x rAB where ωOA = ωOA k ωCA = 2k rad/s ωAB = ωABk rA = 100j mm rC = 75i mm rAB = 175i + 50j mm Substitution gives ωOA x 100j = 2k x (−75i) + ωABk x (−175i + 50j) −100ωOA i = −150j − 175ωAB j = 50ωAB i Matching coefficients of the respective i - and j -terms gives −100ωOA + 50ωAB = 0 25(6 + 7aAB) = 0 The solutions of which are ωAB = −6/7 rad/s and ωOA = −3/7 rad/s Ans. Solution II (Scalar-Geometric). Solution by the scalar geometry of the vector triangle is particularly simple here since νA and νB are at right angles for this special position of the linkages. First, we compute νA which is νA = rω νB = 0.075(2) = 0.150 m/s and represent it in its correct direction as shown. The vector νAB must be perpendicular to AB, and the angle θ between νAB and νA is also the angle made by AB with the horizontal direction. This angle is given by tan θ = 100−50/250−75 = ?/? The horizontal vector νA completes the triangle for which we have νAB = νA(cos θ) = 0.150(cos θ) νA = νB tan θ = 0.150(2/7) = 0.307 m/s The angular velocities become [ω = ν/r] ωAB = νAB/AB = 0.150/cos θ cos θ/0.250 = 0.075 ωOA = νA/OA = 0.30/1 = 3/7 rad/s CW Ans. SAMPLE PROBLEM 5/9 The common configuration of a reciprocating engine is that of the slider-crank mechanism shown. If the crank OB has a clockwise rotational speed of 1500 rev/min, determine for the position where Ө = 60° the velocity of the piston A, the velocity of point G on the connecting rod, and the angular velocity of the connecting rod. Solution. The velocity of the crank pin B as a point on AB is easily found, so that B will be used as the reference point for determining the velocity of A. The relative-velocity equation may now be written νA = νG + νAB The crank-pin velocity is νB = ωr νOB = 5, (1500/12)(2π) / 60 = 65.4 ft/sec and is normal to OB. The direction of νA is, of course, along the horizontal cylinder axes. The direction of VGM must be perpendicular to the line AB as explained in the present article and as indicated in the lower diagram, where the reference point B is shown as fixed. We obtain question by computing angle β from the law of sines, which gives 5/sin 18.02° = 14/sin 60° = sin–1 0.309 = 18.02° We now complete the sketch of the velocity triangle, where the angle between νAB and νA is g' = 180°−2 = 7° and the third angle is 180°−30°−72°=78.0°. Vectors νA and νAB are thus represented with their proper sense such that the head-to-tail sum of νA and νGM equals νB. The angular velocity of AB is counterclockwise, as revealed by the sense of νAB, and is ωAB = (νAB/AB = 34.4/14)=29.5 rad/sec Ans. We now determine the velocity of G by writing νG = νA + νGM where νGM = VBMωAB = GB/AB = EA/n =1.4 (34.4) = 9.63 ft/sec νG = 6.1 ft/sec Ans. 5/64 The circular disk of radius 0.2 m is released very near the horizontal surface with a velocity of its center v_0 = 0.7 m/s to the right and a clockwise angular velocity ω = 2 rad/s. Determine the velocities of points A and P of the disk. Describe the motion upon contact with the ground. Problem 5/64 5/74 For an interval of its motion the piston rod of the hydraulic cylinder has a velocity v_A = 4 ft/sec as shown. At a certain instant θ = β = 60°. For this instant determine the angular velocity ω_BC of link BC. 16 5.3 Método do Movimento Relativo para Eixos Transladados. Análise de Aceleração. Derivando a equação de velocidade 𝐯A = 𝐯B + 𝐯A B = 𝐯B + 𝛚 × 𝐫 com relação ao tempo dá, a_A = a_B + (a_A/B)_n + (a_A/B)_t (a_A/B)_n = ω × (ω × r) (a_A/B)_t = α × r 5/147 The four-bar linkage of Prob. 5/88 is repeated here. If the angular velocity and angular acceleration of drive link OA are 10 rad/s and 5 rad/s², respectively, both counterclockwise, determine the angular accelerations of bars AB and BC for the instant represented. Problem 5/147 5/153 The elements of a power hacksaw are shown in the figure. The saw blade is mounted in a frame which slides along the horizontal guide. If the motor turns the flywheel at a constant counterclockwise speed of 60 rev/min, determine the acceleration of the blade for the position where θ = 90°, and find the corresponding angular acceleration of the link AB. Problem 5/153 5/155 An oil pumping rig is shown in the figure. The flexible pump rod D is fastened to the sector at E and is always vertical as it enters the fitting below D. The link AB causes the beam BCE to oscillate as the weighted crank OA revolves. If OA has a constant clockwise speed of 1 rev every 3 s, determine the acceleration of the pump rod D when the beam and the crank OA are both in the horizontal position shown. 22 5.4 Método do Movimento Relativo para Eixos Girantes. Análises de Velocidade e Aceleração. Desenvolvimento das equações no quadro. Figure 5/10 Figure 5/13 SAMPLE PROBLEM 5/16 At the instant represented, the disk with the radial slot is rotating about O with a counterclockwise angular velocity of 4 rad/sec which is decreasing at the rate of 10 rad/sec². The motion of slider A is separately controlled, and at this instant, r = 6 in., ṙ = 15 in/sec, and ṙ = 81 in/sec². Determine the absolute velocity and acceleration of A for this position. Solution. We have motion relative to a rotating path, so that a rotating coordinate system with origin at O is indicated. We attach xy axes to the disk and use the unit vectors i and j. Velocity. With the origin at O, the term v₀ of Eq. 5/12 disappears and we have vₐ = ω × r + vₛₑₗ The angular velocity as a vector is ω = 4k rad/sec, where k is the unit vector normal to the x-y plane in the +z-direction. Our relative-velocity equation becomes vₐ = 4k × 6i + 5i = 24j + 5i in/sec Ans. in the direction indicated and has the magnitude vₐ = √(24)² + (5)² = 24.5 in/sec Ans. Acceleration. Equation 5/14 written for zero acceleration of the origin of the rotating coordinate system is ₐₐ = α × r + ω × (ω × r) + 2ω × vₛₑₗ + ₐₛₑₗ The terms become α × r = – 10k × 6i = – 60j in/sec² ω × (ω × r) = 4k × (4k × 6i) = 4k × 24j = – 96i in/sec² 2ω × vₛₑₗ = 2(4k) × 5i = 40j in/sec² ₐₛₑₗ = 81i in/sec² The total acceleration is, therefore, ₐₐ = (81 – 96)i + (40 – 60)j = – 15i – 20j in/sec² Ans. in the direction indicated and has the magnitude ₐₐ = √(15)² + (20)² = 25 in/sec² Ans. Vector notation is certainly not essential to the solution of this problem. The student should be able to work out the steps with scalar notation just as easily. The correct direction of the Coriolis-acceleration term can always be found by the direction in which the head of the vₛₑₗ vector would move if rotated about its tail in the sense of ω as shown. Helpful Hints 1 The equation is the same as vₐ = vₚ + ω × rₚ/ₐ where P is a point attached to the disk coincident with A at this instant. 2 Note that the x-y-z axes chosen constitute a right-handed system. 3 Be sure to recognize that ω × (ω × r) and ω × r represent the normal and tangential components of acceleration of a point P on the disk coincident with A. This description becomes that of Eq. 5/14. SAMPLE PROBLEM 5/18 For the conditions of Sample Problem 5/17, determine the angular acceleration of AC and the acceleration of A relative to the rotating slot in arm OD. Solution. We attach the rotating coordinate system xyz to arm OD and use Eq. 5/14. With the origin at the fixed point O, the term ar becomes zero so that BarA= ar + ax (ax + x) + 20 x Vel A + ad From the solution to Sample Problem 5/17, we make use of the values o = 2k rad/s, xcA = -4k rad/s, and Vrel A = -450 (i) mm/s and write arA = xCAXCA + acZA x acXA + (aCX x rCA) Vel A= (225/V2)i - 4k -( 4k x 225/V2 {( I r are x j ar = 0 since o = constant Vel A Substitution into the relative-acceleration equation yields (225cA + 3600i = ] -(225cA + 3600i) = -900/V2i + 7 i Ii mm/ss Equating separately the i and j terms gives astream (225cA + 3600)/V2= -900/pi + 7 t rem and PP )N -(225cA + 3600)/V2= -1800/jJ Solving for the two unknowns gives aiC Ax = eiCA = 32 rad/s e and itc If desired, the acceleration of A may also be written as arA = -(225)((5/2)(1 - j) + (3600)/V2(i + j) + 764Oi - 2550] mm/s We make use here of the geometric representation of the relative-acceleration equation to further clarify the problem. The geometric approach may be used as an alternate solution. Again, we introduce point P on OD coincident with A. The equivalent scalar terms are (ar)/(aC x | ac x ac | = acA, normal to CA, sense unknown = |acA x (acA x ac=| = ac^ca | from A to C (probably acA - |ac x x r= 0 sinice o = constant Ac.freq) Psi ]20 x Val| | 20x Val, directed as shown 20t Along Along OD,s ense unknown Helpful Hints If the slot had been curved with a radius of curvature a, the term ar would have had a component ar'' , normal to the slot and directed toward the center of curvature in addition to its component along the slot. 5/160 The disk rotates about a fixed axis through O with angular velocity co = 5 rad/sec and angular acceleration a = 3 rad/sec^2 in the directions shown at a certain instant. The small sphere A moves in the circular slot, and at the same instant, B = 30°, B = 2 rad/sec, and B = -4 rad/sec^2. Determine the absolute velocity and acceleration of A at this instant. Problem 5/160 5/163 An experimental vehicle A travels with constant speed v relative to the earth along a north-south track. Determine the Coriolis acceleration aCO as a function of the latitude 0. Assume an earth-fixed rotating frame Bxyz and a spherical earth. If the vehicle speed is v = 500 km/h, determine the magnitude of the Coriolis acceleration at (a) the equator and (b) the north pole. 5/182 The crank OA revolves clockwise with a constant angular velocity of 10 rad/s within a limited arc of its motion. For the position θ = 30° determine the angular velocity of the slotted link CB and the acceleration of A as measured relative to the slot in CB. 5/183 The Geneva wheel of Prob. 5/56 is shown again here. Determine the angular acceleration α2 of wheel C for the instant when θ = 20°. Wheel A has a constant clockwise angular velocity of 2 rad/s. 5/191 The pin A in the bell crank AOD is guided by the flanges of the collar B, which slides with a constant velocity vB of 3 ft/sec along the fixed shaft for an interval of motion. For the position θ = 30° determine the acceleration of the plunger CE, whose upper end is positioned by the radial slot in the bell crank.