Question

Moment of inertia of a ring of radius ${}^{\prime }{r}^{\prime }\text{m}$ about an axis passing through the $COM$, perpendicular to its plane is $I{\text{kg-m}}^2$. The moment of inertia of the same ring about an axis passing through the diameter of the ring is $nI{\text{kg-m}}^2$. Find the value of $n$.

Answer 1

Given,
$I_{COM}=I{\text{kg-m}}^2$
Using perpendicular axis theorem,
$I_{dia}=\frac{I_{COM}}{2}=\frac{I}{2}=0.5I{\text{kg-m}}^2$
$\therefore n=0.5$