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

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(A) The moment of inertia of a thin rod of mass $M$ and length $L$ about an axis passing through its centre and perpendicular to its length is given b

Vedclass · Accepted Answer

$\frac{I}{64}$

Vedclass · Answer

(A) The moment of inertia of a thin rod of mass $M$ and length $L$ about an axis passing through its centre and perpendicular to its length is given by $I = \frac{1}{12} M L^2$.When the rod is cut into four 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 part about an axis passing through its own centre and perpendicular to its length is given by:$I' = \frac{1}{12} M' (L')^2$Substituting the values of $M'$ and $L'$:$I' = \frac{1}{12} \left( \frac{M}{4} \right) \left( \frac{L}{4} \right)^2$$I' = \frac{1}{12} \left( \frac{M}{4} \right) \left( \frac{L^2}{16} \right)$$I' = \frac{1}{64} \left( \frac{1}{12} M L^2 \right)$Since $I = \frac{1}{12} M L^2$,we get:$I' = \frac{I}{64}$

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