A uniform thin bar of mass 6 ,kg and length 2 | vedclass

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(B) The total length of the bar is $L = 2.4 \,m$. Since it is bent into a hexagon with $6$ sides,the length of each side is $\ell = \frac{2.4}{6} = 0.

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$8$

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(B) The total length of the bar is $L = 2.4 \,m$. Since it is bent into a hexagon with $6$ sides,the length of each side is $\ell = \frac{2.4}{6} = 0.4 \,m$.The mass of the total bar is $M = 6 \,kg$,so the mass of each side is $m = \frac{6}{6} = 1 \,kg$.For an equilateral hexagon,the distance $r$ from the center to the midpoint of a side is given by $r = \ell \sin 60^{\circ} = \ell \frac{\sqrt{3}}{2}$.The moment of inertia of one side about an axis passing through its center and perpendicular to its length is $I_{cm} = \frac{m \ell^2}{12}$.Using the parallel axis theorem,the moment of inertia of one side about the center of the hexagon is $I_1 = I_{cm} + m r^2 = \frac{m \ell^2}{12} + m \left(\frac{\ell \sqrt{3}}{2}ight)^2 = \frac{m \ell^2}{12} + \frac{3 m \ell^2}{4} = \frac{m \ell^2 + 9 m \ell^2}{12} = \frac{10 m \ell^2}{12} = \frac{5}{6} m \ell^2$.Since there are $6$ such sides,the total moment of inertia is $I = 6 	imes I_1 = 6 	imes \frac{5}{6} m \ell^2 = 5 m \ell^2$.Substituting the values $m = 1 \,kg$ and $\ell = 0.4 \,m$:$I = 5 	imes 1 	imes (0.4)^2 = 5 	imes 0.16 = 0.8 \,kg \cdot m^2$.Thus,$I = 8 	imes 10^{-1} \,kg \cdot m^2$.

$A$ uniform thin bar of mass $6 \,kg$ and length $2.4 \,m$ 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 the hexagon is ...... $\times 10^{-1} \,kg \cdot m^2$.

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