Question

The moment of inertia of a uniform semicircular wire of mass $m$ and radius r., about an axis passing through its centre of mass and perpendicular to its plane is

• A. $\frac{m{r}^2}{2}$
• B. $m{r}^2$
• C. $m{r}^2\left(1-\frac{4}{{\pi }^2}\right)$
• D. $m{r}^2\left(1+\frac{4}{{\pi }^2}\right)$

Answer 1

The correct option is C $m{r}^2\left(1-\frac{4}{{\pi }^2}\right)$

R.E.F image

using parallel axis theorem

$I_0=I_{cm}+m{d}^2$

$d=\frac{2R}{\pi }\Rightarrow$ distance of Com.

$I_{cm}=I_0-m(\frac{2R}{\pi }{)}^2$

$I_{cm}=M{R}^2[1-\frac{4}{{\pi }^2}]$