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

Answer 1

`7.69kgm^2`

Step-by-step explanation:

We are given that

Mass,m=72 kg

Force,F=5 N

Distance,r=0.8 m

Angular acceleration,
`\alpha=0.52rad/s^2`

We have to find the moment of inertia of the door about hinges.

We know that

Torque,
`\tau=Fr=5* 0.8=4Nm`

Moment of inertia,
`I=(\tau)/(\alpha)`

Using the formula

`I=(4)/(0.52)=7.69Kgm^2`

Hence, the moment of inertia of the door about hinges=
`7.69kgm^2`