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

Answer 1

To solve this problem it is necessary to apply the concepts related to Torque and Inertia.

Torque is defined as the force applied to a given distance radius, that is

$\tau = F*r$

Replacing with our vales we have that

$\tau = 5*0.8$

$\tau = 4N.m$

On the other hand the definition of Inertia defines us that it is inversely proportional to angular acceleration and proportional to torque, that is to say

$I = \frac{\tau}{\alpha}$

$I = \frac{4}{2}$

$I = 2 kg.m^2$