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The 0.2-16 bar is released from rest in the position shown. Find the angular velocity of the bar when it reaches a position 3

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3 in 5 in Given data, K = 4.5 lb/ft M = 0.2 lb 230 2 f G B consider O ABB!! (G is centre BB! AB tan 30 F 5 tan30 = 2.8867 infrom pythagoras theorem, Bc Final length of spring B C² + BB12 J3² + 2.88672 = 4.1632 in Given undeformed length of spring-Now, wi=0 and Iz MK = (0-2)(3) = 1-6667 lb in? O — Ź I wg + mg (1.4433) + 1 k ( 8.) ² & 1 Iuf + mg hit 1 ks 2 mag.: (0-2 X38

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