Question

The moment of inertia of a rod about an axis passing through its centre and perpendicular to its length is \[ I=\frac{1}{12}ML^2, \] where \(M\) is the mass and \(L\) is the length of the rod. The rod is bent in the middle so that the two halves make an angle of \(60^\circ\). The moment of inertia of the bent rod about the same axis would be

• A. \[ \frac{1}{12}ML^2 \]
• B. \[ \frac{1}{8\sqrt3}ML^2 \]
• C. \[ \frac{1}{24}ML^2 \]
• D. \[ \frac{1}{48}ML^2 \]

Hint:

For an axis perpendicular to the plane through the bend point, the moment of inertia depends only on the distances of mass elements from the axis. Bending the rod changes direction but not these distances, so the moment of inertia remains unchanged.

Answer 1

The Correct Option is A

Concept: The axis passes through the midpoint (bend point) and remains perpendicular to the plane of the bent rod. The moment of inertia depends only on the distance of each mass element from the axis. Bending the rod does not change these distances.

Step 1: Divide the rod into two equal halves. Each half has \[ \text{Length}=\frac{L}{2}, \qquad \text{Mass}=\frac{M}{2}. \] The axis passes through the common end of both halves.

Step 2: Find the moment of inertia of one half about the bend point. For a rod of length \(l\) about an axis through one end and perpendicular to its length, \[ I=\frac13 ml^2. \] Here, \[ m=\frac{M}{2}, \qquad l=\frac{L}{2}. \] Therefore, \[ I_1 = \frac13 \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2. \] \[ I_1 = \frac{ML^2}{24}. \]

Step 3: Add the moments of inertia of the two halves. Since both halves have the same moment of inertia about the same axis, \[ I = 2I_1. \] \[ I = 2\left(\frac{ML^2}{24}\right). \] \[ I = \frac{ML^2}{12}. \] Notice that the angle between the two halves does not appear in the calculation because the distance of each mass element from the axis remains unchanged. Therefore, \[ \boxed{ I=\frac{1}{12}ML^2 } \] \[ \boxed{\text{Answer = (A)}} \]