The moment of inertia of a rod of length l ab | vedclass

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(C) The moment of inertia of a rod of length $l$ and mass $m$ about an axis passing through its centre and perpendicular to its length is $I = \frac{m

Vedclass · Accepted Answer

$60$

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(C) The moment of inertia of a rod of length $l$ and mass $m$ about an axis passing through its centre and perpendicular to its length is $I = \frac{ml^2}{12}$.Thus,$ml^2 = 12I$.For a regular hexagon,the distance $x$ from the centre $O$ to the centre of each rod is the apothem of the hexagon. Since the hexagon is made of six equilateral triangles with side length $l$,the distance $x = l \sin(60^\circ) = \frac{\sqrt{3}}{2}l$.Using the parallel axis theorem,the moment of inertia of one rod about the axis passing through $O$ and perpendicular to the plane is $I_{	ext{rod}} = I_{	ext{cm}} + mx^2 = \frac{ml^2}{12} + m\left(\frac{\sqrt{3}}{2}light)^2 = \frac{ml^2}{12} + \frac{3ml^2}{4} = \frac{ml^2 + 9ml^2}{12} = \frac{10ml^2}{12} = \frac{5ml^2}{6}$.The total moment of inertia of the system of six rods is $I_{	ext{system}} = 6 	imes I_{	ext{rod}} = 6 	imes \frac{5ml^2}{6} = 5ml^2$.Substituting $ml^2 = 12I$,we get $I_{	ext{system}} = 5(12I) = 60I$.

The moment of inertia of a rod of length $l$ about an axis passing through its centre of mass and perpendicular to the rod is $I$. The moment of inertia of a hexagonal shape formed by six such rods,about an axis passing through its centre of mass and perpendicular to its plane,will be (in $I$)

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