A72.0-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).