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The 0.2-lb 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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Answer #1

K Sin -0.27m *3in = 0.076277 CM B kabel N/m mo moc m=0.2 lb 0.0907kg cm Τη ΔΑΑΒ΄ ; B tar 36BB AB A cm AB to 2 20.1057m • 2) B

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