A 72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. T?

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/s2. What is the moment of inertia of the door about the hinges?

- - The torque he applies to the door is

(force) x (distance from the hinge)

= (5 N) x (0.8 m) = 4 N-m.

- - Torque = (moment of inertia) x (angular acceleration)

4 N-m = (moment of inertia) x (2 / s²)

Moment of inertia = (4 N-m) / (2 / s²)

= (4 kg-m²/s²) / (2 / s²)

= 2 kg-m²

Note that the pusher's mass makes no difference, only the force exerted

perpendicular to the door. It could be an elephant pushing lightly, or even

a mosquito if he could get up the required 5 N (about 18 ounces of force).

Torque = F*r = 5*0.8

Inertia = Torque/Angular acceleration = 5*0.8/2