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rewona [7]
4 years ago
14

72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is locat

ed 0.800 m from axis of the frictionless hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s2. What is the moment of inertia of the door about the hinges?
Physics
1 answer:
expeople1 [14]4 years ago
4 0

Answer:I=2 kg-m^2

Explanation:

Given

mass m=72 kg

Force F=5 N

door knob is located at a distance of r=0.8 m from axis

Angular acceleration of door \alpha =2 rad/s^2

Torque T=I\alpha =F\times r

where I=moment of inertia

5\times 0.8=I\times 2

I=2 kg-m^2

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