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

The moment of inertia of a rod about an axis through its center and perpendicular to its length is \( \alpha = \frac{ML^2}{12} \), where \( M \) is mass and \( L \) is length. When the rod is divided into two equal segments, each segment has mass \( \frac{M}{2} \) and length \( \frac{L}{2} \). The moment of inertia for each segment is denoted by \( \alpha' \). For a cross shape, the total moment of inertia is the sum of the moments of inertia of the two segments, accounting for their distances from the rod's center. Applying the parallel axis theorem yields: \[ \alpha' = 2 \times \frac{M L^2}{48} = \frac{\alpha}{4} \] Consequently, the correct moment of inertia is \( \frac{\alpha}{4} \).