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A 18-kg rear hatch of a vehicle opens as shown in the figure and can be modeled as a uniform 0.6-m long slender rod. Knowing that the tailgate is released from rest in the position shown in the figure, determine the angular velocity of the tailgate as it impacts the car body. I have attached Solution Too, You can Refer It( There Mass=16 kg) In our Problem Mass=18 Kg That is Only difference.

A 18-kg rear hatch of a vehicle opens as shown in the figure and can be modeled as a uniform 0.6-m long slender rod. Knowing that the tailgate is released from rest in the position shown in the figure, determine the angular velocity of the tailgate as it impacts the car body. I have attached Solution Too, You can Refer It( There Mass=16 kg) In our Problem Mass=18 Kg That is Only difference.

International Edition---engineering Mechanics: Statics, 4th Edition
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter9: Moments And Products Of Inertia Of Areas
Section: Chapter Questions
Problem 9.46P
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A 18-kg rear hatch of a vehicle opens as shown in the figure and can be modeled as a uniform 0.6-m long slender rod. Knowing that the tailgate is released from rest in the position shown in the figure, determine the angular velocity of the tailgate as it impacts the car body. I have attached Solution Too, You can Refer It( There Mass=16 kg) In our Problem Mass=18 Kg That is Only difference.
55" T15⁰
Transcribed Image Text:55" T15⁰
Given m = 16 kg AB = 0.6m B Here angle = 90°-15° = 75° B = 98-55-15° = 20° G is the centroid of rear hatch. 'B' AG = AB = 0.3 m Mass moment of inertia of a rod of length AB=0.6 and mass m kg about one end A. is 2 2 m (AB) ² 16 (0-6) ² = = = 1.92 kg. m² 3 3 height AD = AG-AG Cos 20 = AG(1-COS20) = 0·3(1-Cos20) = 0.0146 m AC = AG-AG Cos 75° = AG (1-(0575) 0.3(1-C0575) = 0.1852 m thus h = AD-AC = 0.1706 m Now by Conservation of energy principle IA Now A K BG D d h 15° G mgh = = = IAW²; w is angular velocity of rear hatch before impact 16x 9.81 x 0-1706 = 1/2x1.92 0² = 5.28 rad/s
Transcribed Image Text:Given m = 16 kg AB = 0.6m B Here angle = 90°-15° = 75° B = 98-55-15° = 20° G is the centroid of rear hatch. 'B' AG = AB = 0.3 m Mass moment of inertia of a rod of length AB=0.6 and mass m kg about one end A. is 2 2 m (AB) ² 16 (0-6) ² = = = 1.92 kg. m² 3 3 height AD = AG-AG Cos 20 = AG(1-COS20) = 0·3(1-Cos20) = 0.0146 m AC = AG-AG Cos 75° = AG (1-(0575) 0.3(1-C0575) = 0.1852 m thus h = AD-AC = 0.1706 m Now by Conservation of energy principle IA Now A K BG D d h 15° G mgh = = = IAW²; w is angular velocity of rear hatch before impact 16x 9.81 x 0-1706 = 1/2x1.92 0² = 5.28 rad/s
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