Question

The moment of inertia of slim rod about a transverse axis through one
end is M(L^2)/3, where M is the mass of the rod and L its length, and L^2 is L
squared. What is the moment of inertia of the same rod if we choose a new
parallel axis at a distance of L/3 from the original axis?

Answer 1

From the Huygens–Steiner theorem:

$J=J_C+Md^2,$

$J_C=\frac{ML^2}{12},$

$d=\frac L2 -\frac L3=\frac L6,$

so $J=\frac{ML^2}{12}+\frac{ML^2}{36}=\frac{ML^2}{9}.$