The moment of inertia of a rod of mass M and | vedclass

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(D) The moment of inertia of a rod about its center of mass is $I_{cm} = \frac{ML^2}{12}$.Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $d

Vedclass · Accepted Answer

$\frac{7ML^2}{48}$

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(D) The moment of inertia of a rod about its center of mass is $I_{cm} = \frac{ML^2}{12}$.Using the parallel axis theorem,$I = I_{cm} + Md^2$,where $d$ is the distance from the center of mass to the axis of rotation.The center of mass is at $L/2$ from the end. The axis is at $L/4$ from the end.Therefore,the distance $d = |L/2 - L/4| = L/4$.Substituting these values into the theorem:$I = \frac{ML^2}{12} + M(\frac{L}{4})^2$$I = \frac{ML^2}{12} + \frac{ML^2}{16}$Taking the least common multiple of $12$ and $16$,which is $48$:$I = \frac{4ML^2 + 3ML^2}{48} = \frac{7ML^2}{48}$.

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

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