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.
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.
Solution
To solve this problem, we need to understand the properties of an equilateral hexagon and the formula for the moment of inertia.
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An equilateral hexagon can be divided into 6 equilateral triangles. Since the total length of the bar is 2.4 meters, each side of the hexagon (and each side of the equilateral triangles) is 2.4/6 = 0.4 meters.
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The formula for the moment of inertia (I) of a thin rod rotating about an axis through its center of mass and perpendicular to the length of the rod is I = (m*l^2)/12, where m is the mass of the rod and l is the length of the rod.
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However, in this case, we have 6 rods (the sides of the hexagon), so we need to multiply the moment of inertia of one rod by 6.
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The mass of each rod is the total mass divided by 6, or 6/6 = 1 kg.
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Substituting these values into the formula gives I = 6 * (1*(0.4^2))/12 = 0.08 kg*m^2.
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To express this in the form requested, we need to multiply by 10 to the power of 1, which gives 0.8 × 10–1 kg*m^2.
So, the moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the hexagon is 0.8 × 10–1 kg*m^2.
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