A uniform thin bar of mass 6kg and length 24 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 hexagon is…………..* 10^-1 kg m².

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Vishal Baghel | Contributor-Level 10

5 months ago

Let m_AB = m = 1 kg

AB = 0.4 m = $l$

d = OM = $\sqrt[]{0.16 - 0.04} = \sqrt[]{0.12}$

$d = 2 \sqrt[]{3} * 1 0^{- 1}$

$I_{A B a b o u t O} = \frac{m l^{2}}{12} + m d^{2}$

$∴ l_{h e x a g o n, a b o u t O} = 6 [\frac{m l^{2}}{12} + m d^{2}] = \frac{m l^{2}}{2} + 6 m d^{2}$

$= \frac{1 * {(0.4)}^{2}}{2} + 6 * 1 * {(2 \sqrt[]{3} * 1 0^{- 1})}^{2} = \frac{0.16}{2} + 6 * 4 * 3 * 1 0^{- 2}$

= 0.08 + 0.72 = 0.8 kgm² = 8 * 10^-¹ kgm²

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According to question, we can write

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A uniform thin bar of mass 6kg and length 24 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 hexagon is…………..* 10-1 kg m2.

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A uniform thin bar of mass 6kg and length 24 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 hexagon is…………..* 10^-1 kg m².