Question

The moment of inertia of a thin rod of mass $M$ and length $L$ about an axis passing through the point at a distance $\frac{L}{4}$ from one of its end and perpendicular to the rod is

• A. $\frac{7M{L}^2}{48}$
• B. $\frac{M{L}^2}{12}$
• C. $\frac{M{L}^2}{9}$
• D. $\frac{M{L}^2}{3}$

Answer 1

The correct option is A $\frac{7M{L}^2}{48}$

Using $I=I_{cm}+M{d}^2$ (parallel axis theorem) where, $d=\frac{L}{2}-\frac{L}{4}=\frac{L}{4}\&I_{cm}=\frac{M{L}^2}{12}$
Therefore, $I=\frac{M{L}^2}{12}+\frac{M{L}^2}{16}=\frac{7M{L}^2}{48}$