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AUBURN MECH 6710 - Velocity and Acceleration Analysis

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4 Velocity and Acceleration Analysis 14 Homework 6:Velocity and Acceleration AnalysisProblem 4.5The mechanism in Fig. P4.5 has the dimensions: AB=200 mm, AC=600 mm,BD=1000 mm, La=150 mm, and Lb=250 mm. The driver link 1 rotates witha constant angular speed of n=60 rpm. Find the velocities and the accel-erations of the mechanism for φ=φ1=120◦. For φ=120◦the position of themechanism is given by: xB=-0.1 m, yB=0.173205 m, xC=-0.6 m, yC=0 m,xD=-1.04491 m, yD=-0.154122 m, φ2=0.333473 rad, φ5=0.940376 rad.0 yφAx1 LaLbDCBE3 2 4 5 0 0 nFigure P4.5: Mechanism P4.54 Velocity and Acceleration Analysis 2Results˙xB1=-1.08828 m/s, ˙yB1=-0.628319 m/s, ¨xB1=3.94784 m/s2, ¨yB1=-6.83786 m/s2,vC2C3=-1.23399 m/s, ω2=-0.448799 rad/s, ˙xD2=-1.23518 m/s, ˙yD2=-0.204243 m/s,acorC2C3x=-0.362557 m/s2, acorC2C3y=1.04661 m/s2, aC2C3=1.59873 m/s2, α3=-16.7458 rad/s2, ¨xD2=-1.34318 m/s2, ¨yD2=9.05135 m/s2, vD5D4=0.893105 m/s,ω5=-1.75371 rad/s, acorD5D4x=2.53037 m/s2, acorD5D4y=-1.84656 m/s2, aD5D4=-4.98108m/s2, α5=-6.57248 rad/s2.rA = [0, 0, 0] (m)rB = [ -0.1, 0.173, 0] (m)rC = [ -0.6, 0, 0] (m)rD = [ -1.04, -0.154, 0] (m)rE = [ -0.75, 0.25, 0] (m)phi2 = 120 (degrees)phi2 = phi3 = 19.1 (degrees)phi4 = phi5 = 53.9 (degrees)omega1 = [0, 0, 6.28] (rad/s)alpha1 = [0, 0, 0] (rad/s^2)vB = vB1 = vB2 = [ -1.09, -0.628, 0] (m/s)aB = aB1 = aB2 = [ 3.95, -6.84, 0] (m/s^2)omega2z = -0.449 (rad/s)vC23 = -1.23 (m/s)omega3 = omega2 = [0, 0, -0.449] (rad/s)vC2C3 = [ -1.17, -0.404, 0] (m/s)alpha2z = -16.7 (rad/s)aC23 = 1.6 (m/s^2)alpha2 = alpha3 = [0, 0, -16.7] (rad/s^2)aC2C3 = [ 1.51, 0.523, 0] (m/s^2)vD2 = vD4 = [ -1.24, -0.204, 0] (m/s)4 Velocity and Acceleration Analysis 3omega5z = -1.75 (rad/s)vD54 = 0.893 (m/s)omega4 = omega5 = [0, 0, -1.75] (rad/s)vD5D4 = [ 0.526, 0.721, 0] (m/s)aD54cor = [ 2.53, -1.85, 0 ] (m/s^2)alpha5z = -6.57 (rad/s^2)aD54 = -4.98 (m/s^2)alpha4 = alpha5 = [0, 0, -6.57] (rad/s^2)aD5D4 = [ -2.94, -4.02, -0] (m/s^2)4 Velocity and Acceleration Analysis 4Problem 4.6The mechanism in Fig. P4.6 has the dimensions: AB=150 mm, AC=350 mm,BD=530 mm, DE=300 mm, EF =200 mm, La=55 mm, and Lb=125 mm.The constant angular speed of the driver link 1 is n=30 rpm. Find the veloc-ities and the accelerations of the mechanism for φ=φ1=120◦. For φ=120◦theposition of the mechanism is given by: xB=-0.075 m, yB=0.129904 m, xD=-0.554223 m, yD=-0.0964704 m, xE=-0.482421 m, yE=0.19481 m, φ2=0.441306 rad,φ4=1.32911 rad, φ5=-0.356559 rad.0 yφAx1 LaLbDCBEF2 3 4 5 0 0 nFigure P4.6: Mechanism P4.64 Velocity and Acceleration Analysis 5Results˙xB=-0.408105 m/s, ˙yB=-0.235619 m/s, ¨xB=0.74022 m/s2, ¨yB=-1.2821 m/s2,vC3C2x=0.42465 m/s, vC3C2y=0.200595 m/s, ω2=-0.127362 rad/s, ω4=-1.16943 rad/s,ω5=1.37953 rad/s, ˙xE=-0.0963053 m/s, ˙yE=-0.258552 m/s, acorC3C2x=0.0510963 m/s2,acorC3C2y=-0.108168 m/s2, α2=-5.24453 rad/s2, aC3C2x=-0.114494 m/s2, aC3C2y=-0.0540842 m/s2, ¨xD=-0.439231 m/s2, ¨yD=1.23487 m/s2, α4=-3.94679 rad/s2,α5=-3.66019 rad/s2, ¨xE=0.612199 m/s2, ¨yE=0.553139 m/s2.rA = [ 0.000, 0.000, 0] (m)rD = [-0.554, -0.096, 0] (m)rB = [-0.075, 0.130, 0] (m)rC = [-0.350, 0.000, 0] (m)rE = [-0.482, 0.195, 0] (m)rF = [-0.295, 0.125, 0] (m)phi2 = 25.285 (degrees)phi3 = 25.285 (degrees)phi4 = 40.509 (degrees)phi5 = -20.429 (degrees)Velocity and acceleration analysisomega1=[0, 0, 3.142] (rad/s)alpha1=[0, 0, 0] (rad/s^2)vB=[-0.408, -0.236, 0] (m/s)aB=[ 0.740, -1.282, 0] (m/s^2)vC2 = vB + omega2 x rBC = vC3 + vC23 =>x-axis: 0.13*omega2z - 0.904*v23 - 0.408 = 0y-axis: - 0.275*omega2z - 0.427*v23 - 0.236 = 0=>omega2z = -0.127 (rad/s)vC23 = -0.470 (m/s)omega3=omega2=[0, 0, -0.127](rad/s)vC2C3=[-0.425, -0.201, -0](m/s)4 Velocity and Acceleration Analysis 6aC23cor=[-0.051, 0.108, 0](m/s^2)aC2=aB+omega2xrBC-(omega2.omega2)rBC=aC3+aC23+aC23cor =>x-axis: 0.13*alpha2z - 0.904*a23 + 0.796 = 0y-axis: - 0.427*a23 - 0.275*alpha2z - 1.39 = 0=>alpha2z = -5.245 (rad/s)aC23 = 0.127 (m/s)alpha3=alpha2=[0, 0, -5.245](rad/s^2)aC3C2=[ 0.114, 0.054, 0](m/s^2)vD=[-0.437, -0.175, 0](m/s)aD=[-0.439, 1.235, 0](m/s^2)vE = vD + omega4 x rDE = vF + omega5 x rFE =>x-axis: 0.0698*omega5z - 0.291*omega4z - 0.437 = 0y-axis: 0.0718*omega4z + 0.187*omega5z - 0.175 = 0=>omega4z= -1.169 (rad/s)omega5z= 1.380 (rad/s)omega4=[0, 0, -1.169](rad/s)omega5=[0, 0, 1.380](rad/s)vE=[-0.096, -0.259, 0](m/s)aE = aD + alpha4 x rDE - (omega4.omega4)rDEaE = aF + alpha5 x rFE - (omega5.omega5)rFEx-axis: 0.0698*alpha5z - 0.291*alpha4z - 0.894 = 0y-axis: 0.0718*alpha4z + 0.187*alpha5z + 0.969 = 0=>alpha4z= -3.947 (rad/s^2)alpha5z= -3.660 (rad/s^2)alpha4=[0, 0, -3.947] (rad/s^2)alpha5=[0, 0, -3.660] (rad/s^2)4 Velocity and Acceleration Analysis 7aE=[ 0.612, 0.553, 0]


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