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

Answer:

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

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 $\tau =F\times r=5\times 0.8=4N-m$

We know that torque is also given by

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

So $4=I\times 2$

$I=2kgm^2$