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Q. A uniform thin bar of mass $6\, m$ and length $12 \,L$ is bent to make a regular hexagon. Its moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of the hexagon is:

System of Particles and Rotational Motion

Solution:

$x=2 L \cos 30^{\circ}=L \sqrt{3}$
$I=6\left(\frac{m(2 L)^{2}}{12}+m x^{2}\right)=20 \,m L^{2}$
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