Question

The moment of inertia of a uniform rod of mass \( m \) and length \( l \) is \( \alpha \) when rotated about an axis passing through the centre and perpendicular to the length. If the rod is broken into equal halves and arranged as shown, then the moment of inertia about the given axis is:

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

Hint:

When a body is broken into parts and rearranged, use the parallel axis theorem to calculate the moment of inertia.

Answer 1

The Correct Option is A

The moment of inertia of a uniform rod of mass \( m \) and length \( l \) about an axis passing through the centre and perpendicular to its length is given by: \[ I_{\text{rod}} = \frac{1}{12} m l^2 = \alpha \] Now, the rod is broken into two equal halves, each of length \( \frac{l}{2} \) and mass \( \frac{m}{2} \). These two halves are arranged as shown in the diagram. For each half-rod: - The moment of inertia about the centre of mass is \( I_{\text{half}} = \frac{1}{12} \left(\frac{m}{2}\right) \left(\frac{l}{2}\right)^2 = \frac{1}{48} m l^2 \). Using the parallel axis theorem, we need to account for the moment of inertia about the axis passing through the centre of the rod. The distance from the centre of mass of each half-rod to the axis is \( \frac{l}{4} \). Thus, the moment of inertia for each half-rod about the axis is: \[ I_{\text{half, total}} = I_{\text{half}} + \left(\frac{m}{2}\right) \left(\frac{l}{4}\right)^2 = \frac{1}{48} m l^2 + \frac{1}{32} m l^2 = \frac{5}{96} m l^2 \] Since there are two such halves, the total moment of inertia is: \[ I_{\text{total}} = 2 \times \frac{5}{96} m l^2 = \frac{5}{48} m l^2 \] This simplifies to \( 2\alpha \), where \( \alpha = \frac{1}{12} m l^2 \). Therefore, the moment of inertia about the given axis is \( 2\alpha \).