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 $m{r}^2\left(1-\frac{k}{{\pi }^2}\right)$ then find the value of $k$.

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (C)

Moment of inertia about $z$-axis,
$I_z=m{r}^2$ (about centre of mass)
Applying parallel axes theorem,
$I_z=I_{cm}+m{k}^2$
$I_{cm}=I_z-m{\left(\frac{2}{\pi }r\right)}^2$
$=m{r}^2-\frac{m4r^2}{{\pi }^2}=m{r}^2\left(1-\frac{4}{{\pi }^2}\right)$
i.e., $k=4$