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The ends of the 0.63-m bar remain in contact with their respective support surfaces. End B has a velocity of 0.40 m/s and an acceleration of 0.47 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). A 28 105 Answers: a = aA = i i 0.63 m B 0.274 0.196 -ag=0.47m/s² UB=0.40 m/s rad/s² m/s²

The ends of the 0.63-m bar remain in contact with their respective support surfaces. End B has a velocity of 0.40 m/s and an acceleration of 0.47 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). A 28 105 Answers: a = aA = i i 0.63 m B 0.274 0.196 -ag=0.47m/s² UB=0.40 m/s rad/s² m/s²

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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### Problem Statement The ends of the 0.63-m bar remain in contact with their respective support surfaces. End \( B \) has a velocity of 0.40 m/s and an acceleration of 0.47 m/s\(^2\) in the directions shown. Determine the angular acceleration \( \alpha \) (positive if counterclockwise, negative if clockwise) of the bar and the acceleration of end \( A \) (positive if up, negative if down). ### Diagram Explanation The diagram illustrates a bar \( AB \) of length 0.63 meters. The bar is inclined between two support surfaces at points \( A \) and \( B \) with angles of \( 28^\circ \) and \( 105^\circ \) to the ground, respectively. Point \( B \) has: - A velocity (\( v_B \)) of 0.40 m/s in a downward and slightly rightward direction from the horizontal line. - An acceleration (\( a_B \)) of 0.47 m/s\(^2\) in the same direction as the velocity (\( v_B \)). ### Given Data - Length of the bar, \( L = 0.63 \text{ m} \) - Velocity at end \( B \), \( v_B = 0.40 \text{ m/s} \) - Acceleration at end \( B \), \( a_B = 0.47 \text{ m/s}^2 \) ### Required 1. Angular acceleration, \( \alpha \), of the bar. 2. Acceleration of end \( A \), \( a_A \). ### Answers - Angular acceleration, \( \alpha = 0.274 \text{ rad/s}^2 \) - Acceleration of end \( A \), \( a_A = 0.196 \text{ m/s}^2 \)
Transcribed Image Text:### Problem Statement The ends of the 0.63-m bar remain in contact with their respective support surfaces. End \( B \) has a velocity of 0.40 m/s and an acceleration of 0.47 m/s\(^2\) in the directions shown. Determine the angular acceleration \( \alpha \) (positive if counterclockwise, negative if clockwise) of the bar and the acceleration of end \( A \) (positive if up, negative if down). ### Diagram Explanation The diagram illustrates a bar \( AB \) of length 0.63 meters. The bar is inclined between two support surfaces at points \( A \) and \( B \) with angles of \( 28^\circ \) and \( 105^\circ \) to the ground, respectively. Point \( B \) has: - A velocity (\( v_B \)) of 0.40 m/s in a downward and slightly rightward direction from the horizontal line. - An acceleration (\( a_B \)) of 0.47 m/s\(^2\) in the same direction as the velocity (\( v_B \)). ### Given Data - Length of the bar, \( L = 0.63 \text{ m} \) - Velocity at end \( B \), \( v_B = 0.40 \text{ m/s} \) - Acceleration at end \( B \), \( a_B = 0.47 \text{ m/s}^2 \) ### Required 1. Angular acceleration, \( \alpha \), of the bar. 2. Acceleration of end \( A \), \( a_A \). ### Answers - Angular acceleration, \( \alpha = 0.274 \text{ rad/s}^2 \) - Acceleration of end \( A \), \( a_A = 0.196 \text{ m/s}^2 \)
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