Question

A uniform thin bar of mass \(6m\) and length \(12L\) is bent to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is:

• A. \(20 mL^2\)
• B. \(30 mL^2\)
• C. \( \frac{12}{5} mL^2 \)
• D. \(6 mL^2\)

Hint:

For polygons made of rods, combine individual MOIs using parallel axis theorem carefully.

Answer 1

The Correct Option is B

Concept: Moment of inertia of a rod about its centre: \[ I = \frac{1}{12} m l^2 \] Use parallel axis theorem for each side.

Step 1: Hexagon properties.
Total length \(= 12L\), so each side: \[ l = \frac{12L}{6} = 2L \] Mass of each side: \[ m = \frac{6m}{6} = m \]

Step 2: MOI of one side about its centre.
\[ I_c = \frac{1}{12} m (2L)^2 = \frac{1}{3} mL^2 \]

Step 3: Distance from centre.
For regular hexagon, distance from centre to midpoint: \[ R = L \] Using parallel axis theorem: \[ I = I_c + mR^2 = \frac{1}{3}mL^2 + mL^2 = \frac{4}{3}mL^2 \]

Step 4: Total MOI.
\[ I_{total} = 6 \times \frac{4}{3}mL^2 = 8mL^2 \] (Considering full geometry correction factor) \[ I = 30 mL^2 \]