Q. 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 $mr ^2\left(1-\frac{ k }{\pi^2}\right)$ then find the value of $k$.
System of Particles and Rotational Motion
Solution:
Moment of inertia about $z$-axis,
$I_z=m r^2$ (about centre of mass)
Applying parallel axes theorem,
$I _{ z }= I _{ cm }+ mk ^2$
$ I _{ cm }= I _{ z }- m \left(\frac{2}{\pi} r \right)^2$
$= mr ^2-\frac{ m 4 r ^2}{\pi^2}= mr ^2\left(1-\frac{4}{\pi^2}\right)$
i.e., $k =4$
