Question

A uniform thin bar of mass 6 kg and length 2.4 meter is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of hexagon is __________ x 10\^{-1} kg m².

Hint:

For any regular polygon made of wire, $I = \frac{Ma^2}{12} (1 + 3\cot^2(\pi/n))$, where $n$ is number of sides and $M$ is total mass.

Answer 1

Correct Answer: 8

Step 1: Total length $L = 2.4 \text{ m}$. Side of hexagon $a = 2.4/6 = 0.4 \text{ m}$. Mass of each side $m = 6/6 = 1 \text{ kg}$.
Step 2: For one side, $I$ about center of hexagon using parallel axis theorem: $I_{side} = \frac{ma^2}{12} + mh^2$, where $h = \frac{\sqrt{3}}{2}a$ (distance from center to side).
Step 3: $I_{side} = \frac{ma^2}{12} + m(\frac{3a^2}{4}) = \frac{ma^2 + 9ma^2}{12} = \frac{10}{12}ma^2 = \frac{5}{6}ma^2$.
Step 4: Total $I = 6 \times \frac{5}{6}ma^2 = 5ma^2 = 5(1)(0.4)^2 = 5 \times 0.16 = 0.8 \text{ kg m}^2$. In scientific notation: $8 \times 10^{-1} \text{ kg m}^2$.