Question

A uniform rod of mass $m$ is bent into the form of a semicircle of radius $R$. The moment of inertia of the rod about an axis passing through A and perpendicular to the plane of the paper is

• A. $\frac{2}{3}m{R}^2$
• B. $m{R}^2$
• C. $\frac{5}{\pi }m{R}^2$
• D. 2$m{R}^2$

Answer 1

Answer: (D)

2$m{R}^2$

$\begin{array}{c}\text{Given-}\ast \text{A semicircular ring of mass}=m\\ \ast \text{Radius of ring}=R\end{array}$

$\begin{array}{c}\text{Let us consider two such semi-circular rings}\\ \text{Therefore the moment of Inertia of full ring}\\ I=2m\cdot R^2\\ \text{I semicircular ring}=m{R}^2\end{array}$

$\begin{array}{c}\text{By parallel axis theorem -}\\ \text{Moment of Inertia about point}A-\\ \begin{array}{cc}{I}_A& =I_{\text{Semicircular ring}}+m{R}^2\\ & =m{R}^2+m{R}^2\\ \Rightarrow I_A& =2m{R}^2\end{array}\\ \text{Hence, option (D) is correct.}\end{array}$