The figure shows a uniform rod of length L an | vedclass

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(B) Let the rod have mass $M$ and length $L$. The moment of inertia of the rod about an axis passing through the origin $O$ and perpendicular to the r

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a circle

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(B) Let the rod have mass $M$ and length $L$. The moment of inertia of the rod about an axis passing through the origin $O$ and perpendicular to the rod (the $z$-axis) is $I_O = \frac{1}{3}ML^2$.Consider a point $P(x, y)$ in the $xy$-plane. The moment of inertia of the rod about an axis passing through $P$ and perpendicular to the $xy$-plane is given by the parallel axis theorem: $I_P = I_{CM} + Md^2$,where $d$ is the distance from the center of mass of the rod to point $P$.The center of mass of the rod is at $(L/2, 0)$. The distance $d$ from $(L/2, 0)$ to $(x, y)$ is $d^2 = (x - L/2)^2 + y^2$.The moment of inertia about the center of mass is $I_{CM} = \frac{1}{12}ML^2$.Thus,$I_P = \frac{1}{12}ML^2 + M((x - L/2)^2 + y^2)$.Setting $I_P = I_O$,we get: $\frac{1}{12}ML^2 + M((x - L/2)^2 + y^2) = \frac{1}{3}ML^2$.$(x - L/2)^2 + y^2 = \frac{1}{3}L^2 - \frac{1}{12}L^2 = \frac{1}{4}L^2$.This is the equation of a circle with center at $(L/2, 0)$ and radius $L/2$.

The figure shows a uniform rod of length $L$ and mass $M$ lying along the $x$-axis with one end at the origin $O$. The locus of all points $(x, y)$ in the $xy$-plane,about which the moment of inertia of the rod is the same as that about $O$,is:

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