# What is the moment of inertia of a rod with a mass of 3 kg and length of 4 m that is spinning around its center?

Moment of inertia of given rod is $4 k g {m}^{2}$.
As the rod is spinning on its own centre , we are considering the moment of inertia of rod with axis perpendicular to rod at midpoint i.e., $I = \left(m {r}^{2}\right) / 12$
This can be done by calculating the moment of inertia of a system $I = \Sigma {m}_{i} {\left({r}_{i}\right)}^{2}$.
$I = \left[3 \times {\left(4\right)}^{2}\right] / 12 = 4 k g {m}^{2}$.