Question

A rod of length $l=10\pi \text{m}$ is rotating about a perpendicular bisector passing through its C.O.M and having moment of inertia, ${\pi }^2{\text{kg-m}}^2$. If same rod is bent into semi circular arc of radius $r$ as shown in figure and rotated about the same axis in same plane, then moment of inertia of arc will be

• A. ${\pi }^2{\text{kg-m}}^2$
• B. $\frac{{\pi }^2}{12}{\text{kg-m}}^2$
• C. $12{\text{kg-m}}^2$
• D. $\frac{12}{{\pi }^2}{\text{kg-m}}^2$

Answer 1

The correct option is C $12{\text{kg-m}}^2$

Given, length of rod $l=10\pi \text{m}$
Let $M$ be the mass of the rod.
Then, moment of inertia of rod about $x{x}^{\prime };(I{)}_{rod}=\frac{M{l}^2}{12}....(i)$

Moment of inertia of semi circular arc about axis $x{x}^{\prime };I_{arc}=M{r}^2.....\left(ii\right)$
From $\left(i\right)\text{and}\left(ii\right)$ $\frac{I_{rod}}{I_{arc}}=\frac{l^2}{12r^2}....\left(iii\right)$

When rod is bent into semicircular arc,
$l=\pi r\Rightarrow \frac{l}{r}=\pi ....\left(iv\right)$

From equation $\left(iii\right)$ and $\left(iv\right)$
$\frac{I_{rod}}{I_{arc}}=\frac{{\pi }^2}{12}\Rightarrow I_{arc}=\frac{12}{{\pi }^2}\times I_{rod}=\frac{12}{{\pi }^2}\times {\pi }^2=12{\text{kg-m}}^2$