Question

The moment of inertia of a rod about an axis through its centre and perpendicular to it is $\frac{1}{12}M{L}^2$ (where M is the mass and L the length of the rod). The rod is bent from the middle so that the two halves make an angle of ${60}^o$. The moment of inertia of the bent rod about the same axis would be :

• A. $\frac{M{L}^2}{48}$
• B. $\frac{M{L}^2}{12}$
• C. $\frac{M{L}^2}{24}$
• D. $\frac{M{L}^2}{8\sqrt{3}}$

Answer 1

The correct option is B $\frac{M{L}^2}{12}$

The total length of the rod is L. Since, the rod is bent at the middle, so each part of it will have same length $\frac{L}{2}$ and mass $\frac{M}{2}$.

Moment of inertia of each part about an axis passing through its one end

$=\frac{1}{3}\left(\begin{array}{c}\frac{M}{2}\end{array}\right){\left(\begin{array}{c}\frac{L}{2}\end{array}\right)}^2$

Hence, net moment of inertia about an axis Passing through its Middle point O is

$l=\frac{1}{3}\left[\begin{array}{c}\frac{M}{2}\end{array}\right]{\left[\begin{array}{c}\frac{L}{2}\end{array}\right]}^2+\frac{1}{3}\left[\begin{array}{c}\frac{M}{2}\end{array}\right]{\left[\begin{array}{c}\frac{L}{2}\end{array}\right]}^2$

$=\frac{1}{3}\left[\begin{array}{c}\frac{M{L}^2}{8}+\frac{M{L}^2}{8}\end{array}\right]=\frac{M{L}^2}{12}$