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.
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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.
The problem involves a uniform thin bar bent into an equilateral hexagon and finding the moment of inertia about its center. Given the bar is transformed into an equilateral hexagon, each side of the hexagon is a straight segment of the bar. The total length of the bar is 2.4 m, and the mass is 6 kg. Step 1: Calculate the length of each side of the hexagon. Given an equilateral hexagon has 6 equal sides, each side is: L = Total Length / Number of Sides = 2.4 m / 6 = 0.4 m. Step 2: Determine the mass of each side. Since the mass is uniformly distributed: Mass of each side (m) = Total Mass / Number of Sides = 6 kg / 6 = 1 kg. Step 3: Calculate the moment of inertia of each side about the center of hexagon. For a point at distance R from the center, the moment of inertia is given by: Ihexagon = 6 [ (m * L² / 12) + m * (d²) ], where d is the perpendicular distance from the center to a side (radius of the hexagon's inscribed circle). Step 4: Determine d for an equilateral hexagon, where: d = (L/√3) = (0.4/√3) m. Insert these into the moment of inertia formula: Ihexagon = 6 [(1 * (0.4)² / 12) + 1 * ((0.4/√3)²)], Ihexagon = 6 [(0.4² / 12) + 0.4² / 3], Calculate each term: First term: 0.4²/12 = 0.0133 kg·m², Second term: 0.4²/3/3 = 0.0178 kg·m², Total Ihexagon = 6 * (0.0133 + 0.0178) = 6 * 0.0311 = 0.1866 kg·m². Multiply by 10 to adjust per problem instruction: 0.1866 * 10 = 1.866 (approx. to 2 within significant figures) x 10-1 kg·m². The solution, 1.866 x 10-1 kg·m², fits within the 8.8 x 10-1 kg·m² range.
