Question

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 located 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?

Answer 1

Answer:
`I=2 kg-m^2`

Step-by-step 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* r`

where I=moment of inertia

`5* 0.8=I* 2`

`I=2 kg-m^2`