Question

A 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/s 2 . What is the moment of inertia of the door about the hinges?

Answer 1

Moment of inertia will be
`I=2kgm^2`

Step-by-step explanation:

We have given mass of the person m = 72 kg

Radius r = 0.8 m

Force is given F = 5 N

Angular acceleration
`\alpha =2rad/sec^2`

Torque is given by
`\tau =F* r=5* 0.8=4N-m`

We know that torque is also given by

`\tau =I\alpha`, here I is moment of inertia and
`\alpha` is angular acceleration

So
`4=I* 2`

`I=2kgm^2`