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 ______ × 10-1 kg m2.

Answer 1

8

Answer 2

The total mass of the bar is $M = 6$ kg and its total length is $L_{total} = 2.4$ m. When bent into a regular hexagon, each of the 6 sides has a mass $m = M/6 = 1$ kg and a length $s = L_{total}/6 = 0.4$ m.

The distance from the center of the hexagon to the midpoint of each side is $r = \frac{s}{2}\sqrt{3} = \frac{0.4}{2}\sqrt{3} = 0.2\sqrt{3}$ m.

The moment of inertia of a thin rod about its center of mass is $I_{rod, CM} = \frac{ms^2}{12}$. Using the parallel axis theorem, the moment of inertia of one side about the center of the hexagon is $I_{one\_side} = I_{rod, CM} + mr^2 = \frac{ms^2}{12} + mr^2$.

Substituting the values: $I_{one\_side} = \frac{(1 \text{ kg})(0.4 \text{ m})^2}{12} + (1 \text{ kg})(0.2\sqrt{3} \text{ m})^2 = \frac{0.16}{12} + (0.04 \times 3) = \frac{0.16}{12} + 0.12 = \frac{4}{300} + \frac{36}{300} = \frac{40}{300} = \frac{2}{15}$ kg m².

The total moment of inertia of the hexagon is $I_{total} = 6 \times I_{one\_side} = 6 \times \frac{2}{15} = \frac{12}{15} = \frac{4}{5} = 0.8$ kg m².

To express this in the required format: $0.8 = X \times 10^{-1}$, so $X = \frac{0.8}{10^{-1}} = 8$. The moment of inertia is $8 \times 10^{-1}$ kg m².