Question

Four thin rods of same mass M and same length L form a square as shown in figure. Moment of inertia of this system about an axis passing through point O and perpendicular to its plane is

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

Hint:

Use the rod about its centre perpendicular axis and shift it by the parallel axis theorem.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
O is the centre of the square. Each rod has its centre at distance \(\dfrac{L}{2}\) from O.

Step 2: One rod
About an axis through O, \(I=\dfrac{ML^{2}}{12}+M\left(\dfrac{L}{2}\right)^{2}=\dfrac{ML^{2}}{3}\).

Step 3: Four rods
\(I=4\times\dfrac{ML^{2}}{3}=\dfrac{4}{3}ML^{2}\).

Final Answer:
The moment of inertia is \(\dfrac{4}{3}ML^{2}\). \[ \boxed{\text{(A) }\dfrac{4}{3}\mathrm{ML}^{2}} \]