A uniform rod of mass M and length L has a mo | vedclass

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(D) The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to the rod is given b

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$4I$

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(D) The moment of inertia of a uniform rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to the rod is given by: $I = \frac{ML^2}{12}$ According to the parallel axis theorem,the moment of inertia $I'$ about an axis passing through one of its ends and parallel to the central axis is: $I' = I + Md^2$ Here,the distance between the two axes is $d = L/2$. Substituting the values: $I' = \frac{ML^2}{12} + M(L/2)^2$ $I' = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2 + 3ML^2}{12} = \frac{4ML^2}{12} = \frac{ML^2}{3}$ Since $I = \frac{ML^2}{12}$,we have $ML^2 = 12I$. Substituting this into the expression for $I'$: $I' = \frac{12I}{3} = 4I$

$A$ uniform rod of mass $M$ and length $L$ has a moment of inertia $I$ about an axis passing through its center and perpendicular to the rod. What will be its moment of inertia about an axis passing through one of its ends and perpendicular to the rod?

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