A thin metal rod of mass M and length L is cu | vedclass

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

Vedclass · Accepted Answer

$\frac{I}{64}$

Vedclass · Answer

(D) The moment of inertia of a rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to its length is given by $I = \frac{ML^2}{12}$.When the rod is cut into $4$ equal parts,each part has a mass $M' = \frac{M}{4}$ and a length $L' = \frac{L}{4}$.The moment of inertia $I'$ of each small part about an axis passing through its own center and perpendicular to its length is $I' = \frac{M'(L')^2}{12}$.Substituting the values of $M'$ and $L'$: $I' = \frac{(\frac{M}{4}) \cdot (\frac{L}{4})^2}{12} = \frac{M \cdot \frac{L^2}{16}}{4 \cdot 12} = \frac{ML^2}{12 \cdot 64}$.Since $I = \frac{ML^2}{12}$,we get $I' = \frac{I}{64}$.

$A$ thin metal rod of mass $M$ and length $L$ is cut into $4$ equal parts by cutting it perpendicular to its length. If the moment of inertia of the rod about an axis passing through its centre and perpendicular to its axis is $I$,then what is the moment of inertia of each part about a similar axis?

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