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:$I=2 kg-m^2$

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

$5\times 0.8=I\times 2$

$I=2 kg-m^2$