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(B) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its center of mass and perpendicular to its l

Vedclass · Accepted Answer

$7ML^2/48$

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(B) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its center of mass and perpendicular to its length is $I_{cm} = ML^2/12$.Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance between the center of mass and the new axis.The center of mass is at $L/2$ from one end. The new axis is at $L/4$ from the same end.Therefore,the distance $d = |L/2 - L/4| = L/4$.Substituting these values into the parallel axis theorem:$I = ML^2/12 + M(L/4)^2$$I = ML^2/12 + ML^2/16$Taking the least common multiple of $12$ and $16$,which is $48$:$I = (4ML^2 + 3ML^2) / 48 = 7ML^2/48$.

The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through a point at a distance $L/4$ from one of its ends and perpendicular to the length of the rod is

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