Question

A thin rod of length $L$ and mass $M$ is bent at the middle point O at an angle of ${60}^{\circ }$. The moment of inertia of the rod about an axis passing through O and perpendicular to the plane of the rod will be:

• A. $\frac{M{L}^2}{6}$
• B. $\frac{M{L}^2}{12}$
• C. $\frac{M{L}^2}{24}$
• D. $\frac{M{L}^2}{3}$

Answer 1

The correct option is B

$\frac{M{L}^2}{12}$

Step 1: Given

Length of the rod = $L$

Length of each part = $\raisebox{1ex}{$L$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Mass of the rod = $M$

Mass of each part = $\raisebox{1ex}{$M$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Step 2: Formula used and calculation

Moment of inertia of rod = $\frac{M{L}^2}{3}$

Moment of inertia of complete structure about the point O is given by

$$\begin{gathered} I=\frac{1}{3}\frac{M}{2}{\left(\frac{L}{2}\right)}^2+\frac{1}{3}\frac{M}{2}{\left(\frac{L}{2}\right)}^2 \\ \Rightarrow I=\frac{1}{3}(\frac{M{L}^2}{8}+\frac{M{L}^2}{8}) \\ \Rightarrow I=\frac{1}{3}\frac{M{L}^2}{4} \\ \Rightarrow I=\frac{M{L}^2}{12} \end{gathered}$$

Hence, option (b) is correct.