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BIC Select the elements from the list and add them to the canvas setting the appropriate attributes. Submit Previous Answers v Correct Part B At the instant shown, determine w nC, the magnitude of the angular velocity of link BC. Use the same values from Part A: 0 = 50.0 °, vc = 4.30 ft/s , l1 = 3.80 ft , and l2 = 1.00 ft Express your answer numerically in radians per second to three significant figures. • View Available Hint(s) Nνα ΑΣφ vec WBC = rad/s

BIC Select the elements from the list and add them to the canvas setting the appropriate attributes. Submit Previous Answers v Correct Part B At the instant shown, determine w nC, the magnitude of the angular velocity of link BC. Use the same values from Part A: 0 = 50.0 °, vc = 4.30 ft/s , l1 = 3.80 ft , and l2 = 1.00 ft Express your answer numerically in radians per second to three significant figures. • View Available Hint(s) Nνα ΑΣφ vec WBC = rad/s

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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Learning Goal: To describe the angular and linear velocity of systems under translational and rotational motion using the instantaneous center of zero velocity. The velocity of any point B in a rigid body can be described by VB = VA +w X rBA where w is the angular velocity, rBA is the relative position vector from point A to point B, and vĄ is the velocity of point A. Choosing point A to have zero velocity, VA = 0, simplifies the equation to VB = w X FBA The point chosen such that the velocity is zero is called the instantaneous center of zero velocity (IC). The IC lies along the instantaneous axis of zero velocity, which is perpendicular to the plane of motion. The intersection of the axis and plane of motion define the location of the IC. Because we chose point A as the IC, the equation becomes VB = wX FBIC Point B revolves around the IC. Because of this circular motion the direction of vg must always be perpendicular to 1BIC and vg's magnitude becomes VB = WTBIC Figure 1 of 1 VC
Transcribed Image Text:Learning Goal: To describe the angular and linear velocity of systems under translational and rotational motion using the instantaneous center of zero velocity. The velocity of any point B in a rigid body can be described by VB = VA +w X rBA where w is the angular velocity, rBA is the relative position vector from point A to point B, and vĄ is the velocity of point A. Choosing point A to have zero velocity, VA = 0, simplifies the equation to VB = w X FBA The point chosen such that the velocity is zero is called the instantaneous center of zero velocity (IC). The IC lies along the instantaneous axis of zero velocity, which is perpendicular to the plane of motion. The intersection of the axis and plane of motion define the location of the IC. Because we chose point A as the IC, the equation becomes VB = wX FBIC Point B revolves around the IC. Because of this circular motion the direction of vg must always be perpendicular to 1BIC and vg's magnitude becomes VB = WTBIC Figure 1 of 1 VC
ICIO VBIC Select the elements from the list and add them to the canvas setting the appropriate attributes. Submit Previous Answers v Correct Part B At the instant shown, determine wBC, the magnitude of the angular velocity of link BC. Use the same values from Part A: 0 = 50.0 ° , vc = 4.30 ft/s, l1 = 3.80 ft , and l2 = 1.00 ft . Express your answer numerically in radians per second to three significant figures. • View Available Hint(s) vec WBC = rad/s
Transcribed Image Text:ICIO VBIC Select the elements from the list and add them to the canvas setting the appropriate attributes. Submit Previous Answers v Correct Part B At the instant shown, determine wBC, the magnitude of the angular velocity of link BC. Use the same values from Part A: 0 = 50.0 ° , vc = 4.30 ft/s, l1 = 3.80 ft , and l2 = 1.00 ft . Express your answer numerically in radians per second to three significant figures. • View Available Hint(s) vec WBC = rad/s
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