What is the moment of inertia of a rod with a mass of 6 kg6kg and length of 9 m9m that is spinning around its center?

1 Answer
A08
Feb 27, 2016

=40.5kgm^2=40.5kgm2

Explanation:

Let ll be the length, mm mass, ss area of cross section and rhoρ 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 xx-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 dVIC,rod=∫∫Q∫ρx2dV
= int_{-l//2}^{l//2} rho x^2 s dx =∫l/2−l/2ρx2sdx
= |rho s {x^3}/{3}|_{-l//2}^{l//2}=∣∣∣ρsx33∣∣∣l/2−l/2
= {rho s}/{3} (l^3/{8} + l^3/8)=ρs3(l38+l38)
We know that mass of the rod m=rho s lm=ρsl

:. 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

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