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

Vedclass · Accepted Answer

$\frac{ML^2}{9}$

Vedclass · Answer

(B) The moment of inertia of a uniform thin rod of length $L$ and mass $M$ about an axis passing through its center of mass and perpendicular to the rod is $I_{cm} = \frac{ML^2}{12}$.We need to find the moment of inertia about an axis at a distance $d = \frac{L}{3}$ from one end.The distance of this axis from the center of mass is $a = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$.Using the parallel axis theorem,$I = I_{cm} + Ma^2$.Substituting the values,$I = \frac{ML^2}{12} + M\left(\frac{L}{6}\right)^2$.$I = \frac{ML^2}{12} + \frac{ML^2}{36} = \frac{3ML^2 + ML^2}{36} = \frac{4ML^2}{36} = \frac{ML^2}{9}$.

The moment of inertia of a uniform thin rod of length $L$ and mass $M$ about an axis passing through a point at a distance of $\frac{L}{3}$ from one of its ends and perpendicular to the rod is

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