question_answer

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

Correct Answer: C

Solution :

[c] Moment of inertia about z-axis, \[{{I}_{z}}=m{{r}^{2}}\] (about centre of mass) Applying parallel axes theorem,   \[{{I}_{z}}={{I}_{cm}}+m{{k}^{2}}\] \[{{I}_{cm}}={{I}_{z}}-m{{\left( \frac{2}{\pi }r \right)}^{2}}\] \[=m{{r}^{2}}-\frac{m4{{r}^{2}}}{{{\pi }^{2}}}=m{{r}^{2}}\left( 1-\frac{4}{{{\pi }^{2}}} \right)\]i.e., k=4