Question

Four identical thin rods each of mass M and length $l$, form a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is

• A. $\frac{2}{3}Ml^2$
• B. $\frac{13}{3}Ml^2$
• C. $\frac{1}{3}Ml^2$
• D. $\frac{4}{3}Ml^2$

Answer 1

The Correct Option is D

To find the moment of inertia of a square frame made of four identical thin rods, each of mass \( M \) and length \( l \), about an axis through the center and perpendicular to its plane, we proceed as follows:

1. Firstly, consider the geometry of the problem. In a square, each rod is parallel or perpendicular to the axes when the square is aligned with the coordinate axes. The axis about which we need to calculate the moment of inertia passes through the center of the square and is perpendicular to its plane.

2. The square's diagonal can be calculated using Pythagoras' theorem as \( \sqrt{l^2 + l^2} = \sqrt{2}l \). The center of the square will be at half of the diagonal distance from any corner.

3. The moment of inertia of each rod about an axis through its center and along its length is given by \( \frac{1}{12} Ml^2 \).

4. Use the parallel axis theorem to calculate the moment of inertia for each rod. According to the parallel axis theorem, if we know the moment of inertia of a body about an axis passing through its center of mass, we can find the moment of inertia about any parallel axis using the formula:

   I = I_{cm} + Md^2

   where \( d \) is the distance between the two axes.

5. For a rod aligned along one side of the square, the distance \( d \) from the center of the square to the center of a rod (which is positioned at the edge of the square) can be calculated as \( \frac{l}{2\sqrt{2}} \) where the diagonal was \( \sqrt{2}l \). This gives:

   d = \frac{l}{\sqrt{2}} = \frac{l}{2}

6. Thus, the moment of inertia for each rod about the center of the square:

   I_{\text{rod}} = \frac{1}{12} M l^2 + M \left(\frac{l}{2}\right)^2 = \frac{1}{12} M l^2 + \frac{M l^2}{4} = \frac{1}{12} M l^2 + \frac{3}{12} M l^2 = \frac{4}{12} M l^2 = \frac{1}{3} M l^2

7. The total moment of inertia for the frame which has four rods is:

   I_{\text{total}} = 4 \times \frac{1}{3} M l^2 = \frac{4}{3} M l^2

This calculation confirms that the correct answer is \frac{4}{3} Ml^2.