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.
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.
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Always shift MOI from rod centre to system centre using parallel axis theorem.
Updated On: Apr 23, 2026
- \(\frac{4}{3}Ml^2\)
- \(\frac{Ml^2}{3}\)
- \(\frac{Ml^2}{6}\)
- \(\frac{2}{3}Ml^2\)
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The Correct Option is A
Solution and Explanation
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
\[ I = 4 \times \frac{1}{3}Ml^2 = \frac{4}{3}Ml^2 \] Conclusion: \[ {(A)} \]
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
\[ I = 4 \times \frac{1}{3}Ml^2 = \frac{4}{3}Ml^2 \] Conclusion: \[ {(A)} \]
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