Question

Four thin rods of same mass \(M\) and length \(l\) form a square. Find moment of inertia about axis through centre and perpendicular to plane.

• A. \(\frac{4}{3}Ml^2\)
• B. \(\frac{Ml^2}{3}\)
• C. \(\frac{Ml^2}{6}\)
• D. \(\frac{2}{3}Ml^2\)

Hint:

Always shift MOI from rod centre to system centre using parallel axis theorem.

Answer 1

The Correct Option is A

Understanding the Concept: Use parallel axis theorem.

Step 1: MOI of one rod about its centre
\[ I = \frac{1}{12}Ml^2 \]

Step 2: Shift to square centre
Distance = \(\frac{l}{2}\) \[ I = \frac{1}{12}Ml^2 + M\left(\frac{l}{2}\right)^2 = \frac{1}{12}Ml^2 + \frac{1}{4}Ml^2 = \frac{1}{3}Ml^2 \]

Step 3: Total for 4 rods
\[ I = 4 \times \frac{1}{3}Ml^2 = \frac{4}{3}Ml^2 \] Final Answer: \[ {(A)} \]