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Question. The collar C is moving downward with an acceleration of 1 m/s2. At the instant shown, it has a speed of 2 m/s which gives links CB and AB an angular velocity ωAB = ωCB = 10 rad/s. Find the angular accelerations of AB and CB at this instant.
Suggestions:(A) Calculate aAB and aCB(B) aB = aC + aB/C)t + aB/C)n(C) aB/C)t = raCB(D) aB/C)n = rωCB2
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