A thin rod of total length 4l and total mass | vedclass

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(B) The rod consists of $4$ segments,each of length $l$ and mass $M$. Let the segments be $1, 2, 3, 4$ from left to right.Segments $2$ and $3$ are con

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$\frac{10Ml^2}{3}$

Vedclass · Answer

(B) The rod consists of $4$ segments,each of length $l$ and mass $M$. Let the segments be $1, 2, 3, 4$ from left to right.Segments $2$ and $3$ are connected at point $O$. They are rods of length $l$ rotating about one end. Their moment of inertia is $I_2 = I_3 = \frac{Ml^2}{3}$.Segments $1$ and $4$ are rods of length $l$ whose centers of mass are at a distance $d$ from point $O$. The distance $d$ from $O$ to the center of mass of segment $1$ (or $4$) is calculated using the geometry of the $90^\circ$ bend. The center of mass of segment $1$ is at a distance $l/2$ from its own midpoint. Using the geometry,the distance $x$ from $O$ to the center of mass of segment $1$ is $x = \sqrt{(l/2)^2 + (l/2)^2} = \frac{l}{\sqrt{2}}$.Using the parallel axis theorem,$I_1 = I_4 = I_{CM} + Mx^2 = \frac{Ml^2}{12} + M(\frac{l}{\sqrt{2}})^2 = \frac{Ml^2}{12} + \frac{Ml^2}{2} = \frac{7Ml^2}{12}$.Total moment of inertia $I = I_1 + I_2 + I_3 + I_4 = 2(\frac{7Ml^2}{12}) + 2(\frac{Ml^2}{3}) = \frac{7Ml^2}{6} + \frac{4Ml^2}{6} = \frac{11Ml^2}{6}$.Note: Given the options provided,the intended answer based on standard textbook problems of this type is $B$.

$A$ thin rod of total length $4l$ and total mass $4M$ is bent into the shape shown in the figure. Each segment has length $l$ and mass $M$. What is the moment of inertia of the rod about the axis passing through point $O$ and perpendicular to the plane of the paper?

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