The moment of inertia of a rod of mass M and | vedclass

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Question

(B) The moment of inertia of the original rod of mass $M$ and length $L$ about an axis through its center is given by:$\alpha = \frac{ML^2}{12} \quad

Vedclass · Accepted Answer

$\alpha / 4$

Vedclass · Answer

(B) The moment of inertia of the original rod of mass $M$ and length $L$ about an axis through its center is given by:$\alpha = \frac{ML^2}{12} \quad \dots (i)$When the rod is cut into two equal parts,each part has mass $M' = M/2$ and length $L' = L/2$.For each part,the moment of inertia about an axis passing through its center and perpendicular to its length is:$I_{part} = \frac{M'(L')^2}{12} = \frac{(M/2)(L/2)^2}{12} = \frac{ML^2}{96}$In the cross shape,each rod is placed such that its center coincides with the center of the cross. The axis of rotation is perpendicular to the plane of the cross. For each rod,this axis passes through its center and is perpendicular to its length.Thus,the total moment of inertia of the cross is the sum of the moments of inertia of the two rods:$\alpha' = I_{part} + I_{part} = 2 	imes \frac{ML^2}{96} = \frac{ML^2}{48}$Comparing this with equation $(i)$:$\alpha' = \frac{1}{4} \left( \frac{ML^2}{12} ight) = \frac{\alpha}{4}$Therefore,the correct option is $B$.

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