The moment of inertia of a uniform thin rod o | vedclass

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Question

The distance $\mathrm{OC}=\frac{\mathrm{L}}{2}-\frac{\mathrm{L}}{3}=\frac{\mathrm{L}}{6}$

Applying the theorem of parallel axes.

$\mathrm{I}_{\m

Vedclass · Accepted Answer

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

Vedclass · Answer

The distance $\mathrm{OC}=\frac{\mathrm{L}}{2}-\frac{\mathrm{L}}{3}=\frac{\mathrm{L}}{6}$

Applying the theorem of parallel axes.

$\mathrm{I}_{\mathrm{C}}=1_{\mathrm{O}}+\mathrm{M}(\mathrm{OC})^{2}=\frac{\mathrm{ML}^{2}}{12}+\mathrm{M}\left(\frac{\mathrm{L}}{6}\right)^{2}=\frac{\mathrm{ML}^{2}}{9}$

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

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