Question

A thin rod of mass $M$ and length $L$ is rotating about an axis perpendicular to the length of rod and passing through a point of rod at a distance $\frac{L}{6}$ from centre of rod. Its moment of inertia is:

• A. $\frac{M{L}^2}{24}$
• B. $\frac{M{L}^2}{18}$
• C. $\frac{M{L}^2}{12}$
• D. $\frac{M{L}^2}{9}$

Answer 1

The correct option is D $\frac{M{L}^2}{9}$

Moment of inertia about the center of rod is $I_{cm}=\frac{M{L}^2}{12}$

Now using parallel axis theorem $I=I_{cm}+M{d}^2$ here $d=\frac{L}{6}$

So $I=\frac{M{L}^2}{12}+\frac{M{L}^2}{36}=\frac{M{L}^2}{9}$