The ends of the 0.38-m bar remain in contact with their respective support surfaces. End B has a velocity of 0.46 m/s and an acceleration of 0.59 m/s² in the directions shown. Determine the angular acceleration a (positive if counterclockwise, negative if clockwise) of the bar and the acceleration of end A (positive if up, negative if down). 24 0.38 m 101 Ban-0.59m Ug-0.46 ms Ancwers:

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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The ends of the 0.38-m bar remain in contact with their respective support surfaces. End B has a velocity of 0.46 m/s and an
acceleration of 0.59 m/s² in the directions shown. Determine the angular acceleration a (positive if counterclockwise, negative if
clockwise) of the bar and the acceleration of end A (positive if up, negative if down).
24
0.38 m
101
aB = 0.59 ms²
Ug -0.46 ms
B
Answers:
a =
i
! rad/s2
aA =
i
! m/s?
Transcribed Image Text:The ends of the 0.38-m bar remain in contact with their respective support surfaces. End B has a velocity of 0.46 m/s and an acceleration of 0.59 m/s² in the directions shown. Determine the angular acceleration a (positive if counterclockwise, negative if clockwise) of the bar and the acceleration of end A (positive if up, negative if down). 24 0.38 m 101 aB = 0.59 ms² Ug -0.46 ms B Answers: a = i ! rad/s2 aA = i ! m/s?
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