Question

A thin rod of length ' $4L$ ' and mass ' $4m$ ' is bent as shown. Find the moment of inertia about an axis through $O$ and perpendicular to the plane.

• A. $\frac{mL^2}{3}$
• B. $\frac{10mL^2}{3}$
• C. $\frac{mL^2}{12}$
• D. $\frac{mL^2}{24}$

Hint:

Shortcut: Divide bent rod into straight segments and apply parallel axis theorem carefully.

Answer 1

The Correct Option is B

Step 1: Understand geometry
The rod is bent into 4 equal segments, each of length $L$ and mass $m$. So total: \[ \text{Length} = 4L,\quad \text{Mass} = 4m \]

Step 2: Break into 4 rods
Each segment contributes moment of inertia about point $O$. We use: \[ I = I_{\text{cm}} + m d^2 \] (Parallel axis theorem)

Step 3: Contribution of each segment
Because of symmetry: - Two inner segments (meeting at $O$) - Two outer segments (farther from $O$) (i) Inner segments
Each has one end at $O$, so: \[ I = \frac{1}{3} m L^2 \] Two such rods: \[ I_{\text{inner}} = 2 \times \frac{1}{3} mL^2 = \frac{2mL^2}{3} \] (ii) Outer segments
Each outer rod is at distance $L$ from $O$ Using parallel axis theorem: \[ I = \frac{1}{3} mL^2 + mL^2 = \frac{4}{3} mL^2 \] Two such rods: \[ I_{\text{outer}} = 2 \times \frac{4}{3} mL^2 = \frac{8mL^2}{3} \]

Step 4: Total moment of inertia
\[ I = \frac{2mL^2}{3} + \frac{8mL^2}{3} = \frac{10mL^2}{3} \] Final Answer: Option (B)