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:
Show Hint
When a body is broken into parts and rearranged, use the parallel axis theorem to calculate the moment of inertia.
The moment of inertia of a uniform rod of mass \( m \) and length \( l \) about an axis through its centre and perpendicular to its length is \( I_{\text{rod}} = \frac{1}{12} m l^2 = \alpha \).
The rod is divided into two equal halves, each with mass \( \frac{m}{2} \) and length \( \frac{l}{2} \). These halves are positioned as depicted in the diagram.
For each half-rod:
- The moment of inertia about its centre of mass is \( I_{\text{half}} = \frac{1}{12} \left(\frac{m}{2}\right) \left(\frac{l}{2}\right)^2 = \frac{1}{48} m l^2 \).
The parallel axis theorem is applied to calculate the moment of inertia about the rod's centre. The distance from the centre of mass of each half-rod to this axis is \( \frac{l}{4} \).
Consequently, the moment of inertia for each half-rod about the specified axis is:
\[I_{\text{half, total}} = I_{\text{half}} + \left(\frac{m}{2}\right) \left(\frac{l}{4}\right)^2 = \frac{1}{48} m l^2 + \frac{1}{32} m l^2 = \frac{5}{96} m l^2\]
As 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 is equivalent to \( 2\alpha \), given \( \alpha = \frac{1}{12} m l^2 \).
Therefore, the moment of inertia about the axis in question is \( 2\alpha \).
