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The mechanism produces intermittent motion of link AB. If the sprocket S is turning with an angular velocity of Ws = 6 rad/s, determine the angular velocity of link AB at this instant. The sprocket S is mounted on a shaft which is separated from a collinear shaft attached to AB at A. The pin at C is attached to one of the chain links. Solution: Kinematic Diagram: Since link AB is rotating about the fixed point A, then VB is always directed perpendicular to link AB its magnitude is VB = ω AB rAB =0.2 ω AB . At the instant shown . Vs is directed at an angle 60° with the horizontal. Since point C is attached to chain , at the instant shown ,it moves vertically with a speed of Vc = WSrS = 6(0.175) = 1.05m/s Instantaneous center : The instantaneous center of zero velocity of link BC at the instant shown is located at the intersection point of extended lines drawn perpendicular from vB and vC . Using law of sines , we have rB/rC Sin105° = 0.15 sin 30° rB/rC = 0.2898 m rC/rC Sin45° = 0.15 sin 30° rC/rC = 0.2121 m The angular velocity of bar BC is given by ω BC = vC 1.05 = rC / rC 0.2121 = 4.950 rad/s Thus ,the angular velocity of link AB is give by vB = 0.2 ω BC rB/rC ω AB =4.950(0.2898) ω AB =7.17 rad/s