Answer:
[tex]I=0.54\ kg-m^2[/tex]
Explanation:
Given that,
Mass of a uniform rod, m = 12 kg
Length of the rod, l = 0.3 m
The moment of inertia rotating about an axis perpendicular to the rod and passing through the center of the rod is given by :
[tex]I=\dfrac{ml^2}{2}\\\\I=\dfrac{12\times (0.3)^2}{2}\\\\I=0.54\ kg-m^2[/tex]
So, the moment of inertia of the rod is [tex]0.54\ kg-m^2[/tex]
