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 KgThat is Only difference. 55"T15⁰ Given m = 16 kgAB = 0.6mBHere angle = 90°-15° = 75°B = 98-55-15° = 20°G is the centroid of rear hatch.'B'AG = AB = 0.3 mMass moment of inertia of a rod of length AB=0.6andmass m kg about one end A. is22m (AB) ² 16 (0-6) ²=== 1.92 kg. m²33heightAD = AG-AG Cos 20 = AG(1-COS20)= 0·3(1-Cos20) = 0.0146 mAC = AG-AG Cos 75° = AG (1-(0575)0.3(1-C0575)= 0.1852 mthush = AD-AC = 0.1706 mNow by Conservation of energy principleIANowAKBGDdh15°Gmgh = = = IAW²; w is angular velocity of rearhatch before impact16x 9.81 x 0-1706 = 1/2x1.92 0²= 5.28 rad/s

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Answered: A 18-kg rear hatch of a vehicle opens…

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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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.

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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