Question

The ends of the 0.35-m bar remain in contact with their respective support surfaces. End B has a velocity of 0.45 m/s and an acceleration of 0.51 m/s2 in the directions shown. Determine the angular acceleration α (positive if counterclockwise, negative if clockwise) of the bar and the acceleration of end A (positive if up, negative if down).

28 0.35 m 106 B -Ug = 0.51 m/s2 Ug = 0.45 ms Answers: a = 0.74 Orad/s² X ад 0.2490 Um/s²

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Acceleration of end a is upward (aa) CA 28° + 106° + 0 = 180° ансова 180°-106 - 28° = 46° :28 5106 Ao casugo From con onstrai

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