# What is the moment of inertia of a rod with a mass of 6 kg and length of 9 m that is spinning around its center?

Feb 27, 2016

$= 40.5 k g {m}^{2}$

#### Explanation:

Let $l$ be the length, $m$ mass, $s$ area of cross section and $\rho$ the density of a thin rod rotating about its center.

Its moment of inertia about a perpendicular axis through its center of mass is determined by the following volume integral.

If the rod is placed along the $x$-axis and the center of rotation be the origin, the volume integral reduces to length integral,

${I}_{C , \text{rod}} = \int {\int}_{Q} \int \rho {x}^{2} \mathrm{dV}$
$= {\int}_{- l / 2}^{l / 2} \rho {x}^{2} s \mathrm{dx}$
$= | \rho s \frac{{x}^{3}}{3} {|}_{- l / 2}^{l / 2}$
$= \frac{\rho s}{3} \left({l}^{3} / \left\{8\right\} + {l}^{3} / 8\right)$
We know that mass of the rod $m = \rho s l$

$\therefore {I}_{C , \text{rod}} = \frac{m {l}^{2}}{12}$

In the given question
$m = 6 k g \mathmr{and} l = 9 m$. Plugging the values in the expression
${I}_{C} = \frac{6 \times {9}^{2}}{12} = 40.5 k g {m}^{2}$