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?

1 Answer
Feb 27, 2016

Answer:

#=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#