In a four bar chain ABCD, AD is fixed and is 150 mm long. The crank AB is 40 mm long and rotates at 120 r.p.m. clockwise, while the link CD = 80 mm oscillates about D. BC and AD are of equal length. Find the angular acceleration of link CD when angle BAD = 60°
Given
"N_{BA}=120 r.p.m \\implies \\omega_{BA}=2 \\pi *\\frac{120}{60}=12.568 rad\/s"
Since the length of the crank AB = 40 mm = 0.04 m , therefore velocity of B with respect to A or even velocity of B is given by
"v_{BA}=v_B = \\omega _{BA}*AB=12.568 * 0.04 = 0.503 m\/s"
By the method of measurement , we find that
"v_{CD}= v_C= 0.385 m\/s"
Angular velocity of link CD , "\\omega _{CD}=\\frac{v_{CD}}{CD}=\\frac{0.385}{0.08}=4.8 rad\/s" . Clockwise about point D
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