Question

A thin uniform rod of mass $M$ and length $L$ has its moment of inertia $I_{rod}$ about its perpendicular bisector. This rod is bend in the form of a circular shape of radius $r$. Now, circle has its moment of inertia $I_{\text{circle}}$ about its centre and perpendicular to its plane. Find the ratio of $I_{rod}:I_{\text{circle}}$.

• A. $\frac{{\pi }^2}{3}$
• B. $\frac{{\pi }^2}{12}$
• C. $\frac{{\pi }^2}{4}$
• D. $1$

Answer 1

The correct option is A $\frac{{\pi }^2}{3}$


Given
moment of inertia of rod about its perpendicular bisector = $I_{rod}$
$I_{rod}=\frac{M{L}^2}{12}----\left(1\right)$

Moment of inertia of circular ring of radius ${}^{\prime }{r}^{\prime }$ and having mass ${}^{\prime }{M}^{\prime }$ about the axis passing through its centre and perpendicular to its plane $I_{\text{circle}}$
$I_{\text{circle}}=M{r}^2------\left(2\right)$

but $L=2\pi r(\because$Rod bend into circular ring.

From equation (1) and (2)
$I_{rod}=\frac{M(2\pi r{)}^2}{12}=\frac{4{\pi }^2}{12}M{r}^2=\frac{{\pi }^3}{3}M{r}^2$

$I_{rod}=\frac{{\pi }^2}{3}\left(I_{\text{circle}}\right)$
So, $\frac{I_{\text{rod}}}{I_{\text{circle}}}=\frac{{\pi }^2}{3}$