Answer:
[tex]7.69kgm^2[/tex]
Explanation:
We are given that
Mass,m=72 kg
Force,F=5 N
Distance,r=0.8 m
Angular acceleration,[tex]\alpha=0.52rad/s^2[/tex]
We have to find the moment of inertia of the door about hinges.
We know that
Torque,[tex]\tau=Fr=5\times 0.8=4Nm[/tex]
Moment of inertia,[tex]I=\frac{\tau}{\alpha}[/tex]
Using the formula
[tex]I=\frac{4}{0.52}=7.69Kgm^2[/tex]
Hence, the moment of inertia of the door about hinges=[tex]7.69kgm^2[/tex]
