Question

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?

Answer 1

`=40.5kgm^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, "rod"} = int int_Q int rho x^2 dV`
`= int_{-l//2}^{l//2} rho x^2 s dx`
`= |rho s {x^3}/{3}|_{-l//2}^{l//2}`
`= {rho s}/{3} (l^3/{8} + l^3/8)`
We know that mass of the rod `m=rho s l`

`:. I_{C, "rod"}= {ml^2}/12`

In the given question
`m=6kg and l=9m`. Plugging the values in the expression
`I_C=(6xx9^2)/12=40.5kgm^2`