Question

Name: The flywheel rotates with angular velocity ? = 3 rad/s and angular acceleration ? : 5 rad/s. Determine (a) the motion of the flywheel, rod AB, and rod BC. Then determine the angular acceleration of (b) link AB and (c) link BC at the instant shown. Use: a = 0.4 m, b = 0.5 m, r = 0.3 m, d= 4, and e-3. (Hint: Use Translating Frames analysis.) 4. Motion: 1. Flywheel-> 2. Rod AB-> 3. Rod BC-

Answer 1

part A: Velocity of points B VB = wcB times CB = WCB times 0.4J = 0.4WcD(- i) Velocity of point A VA = WKtimesoA = 2ktimes0.3j = 0.6(- i) Velocity of A with respect to B VA/B = WABtimesBA = WABktimes(- AB cos thetai + AB sintheatj) = - AB costheta WABj + AB sintheta WAB(- i) VA/B = - AB sintheta WABi - AB cos WBAj VA = VB + VA/B - 0.6i = - 0.4 WcDi - AB sintheta WABi - AB costhetaWAB j = > WAB = 0 i = - 0.6 = - 0.4WCD + 0 WCD = 1.5 rad/s partB: The angualar acceleration of link AB is aB = aA + alphaABtimesrB/A - W^2rB/A - (aB) ti - (0.6) ^2j = - 6(0.3) i - (2) ^2(0.3) j + (aAB) r times (0.4i - 0.3j) - 0 - (aB) i - 0.9j = - 1.8j + 0.4 alphaABj + 0.3alphaAB compare the coefficient of i in above eqn - (aB) t = - 1.8 + 0.3 alphaAB compare the coefficient of j in above eqn - 0.9 = - 1.2 + 0.4alphaAB alphaAB = 0.9 + 1.2/0.4 alpha AB = 0.75 rad/s^2 part C: - (abeta) t = - 1.8 + 0.3(0.75) = - 1.5 (aB) t = 1.575m/s^2 angular acceleration, alphaBC = 1.575/0.4 = 3.94 rad/s^2