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

Step 1: Understanding the Concept:
We use the Parallel Axis Theorem to find the moment of inertia of each side of the hexagon and then sum them up.
: Key Formula or Approach:
1. Moment of inertia of a rod about its center: \( I_{cm} = \frac{1}{12} M a^2 \).
2. Parallel axis theorem: \( I = I_{cm} + M r^2 \).
Step 2: Detailed Explanation:
- Total mass = \( 6m \), Total length = \( 12L \).
- Hexagon has 6 sides, so each side has mass \( M_s = m \) and length \( a = 2L \).
- The distance from the center of the regular hexagon to the center of each side is \( r = \frac{a}{2} \tan 60^\circ = L \sqrt{3} \).
- Moment of inertia of one side about the hexagon's center:
\[ I_{side} = \frac{1}{12} m (2L)^2 + m(L\sqrt{3})^2 \]
\[ I_{side} = \frac{4mL^2}{12} + 3mL^2 = \frac{1}{3} mL^2 + 3mL^2 = \frac{10}{3} mL^2 \]
- Total moment of inertia for all 6 sides:
\[ I_{total} = 6 \times I_{side} = 6 \times \frac{10}{3} mL^2 \]
\[ I_{total} = 20 mL^2 \]
Step 3: Final Answer:
The moment of inertia of the hexagon is 20 mL2.