Question

Moment of inertia of a rod of mass \( M \) and length \( L \) about an axis passing through its center and normal to its length is \( \alpha \). Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to the plane containing cross is:

• A. \( \alpha \)
• B. \( \frac{\alpha}{4} \)
• C. \( \frac{\alpha}{8} \)
• D. \( \frac{\alpha}{2} \)

Hint:

To calculate the moment of inertia of composite shapes, use the parallel axis theorem and add the individual moments of inertia.

Answer 1

The Correct Option is B

Let the moment of inertia of the rod about the axis passing through its center and normal to its length be \( \alpha = \frac{ML^2}{12} \), where \( M \) is the mass and \( L \) is the length.  
Now, the rod is cut into two equal parts, each having mass \( \frac{M}{2} \) and length \( \frac{L}{2} \). 
Each part has a moment of inertia \( \alpha' \). 
For the cross shape, the total moment of inertia will be the sum of the moments of inertia of the two parts, considering the distance from the center of the rod. 
After using the parallel axis theorem, we get: \[ \alpha' = 2 \times \frac{M L^2}{48} = \frac{\alpha}{4} \] Thus, the correct option is \( \frac{\alpha}{4} \).