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(D) Let the three rods be $AB$ (left vertical),$CD$ (right vertical),and $EF$ (horizontal connecting rod). We want to find the moment of inertia about

Vedclass · Accepted Answer

$\frac{4Ml^2}{3}$

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(D) Let the three rods be $AB$ (left vertical),$CD$ (right vertical),and $EF$ (horizontal connecting rod). We want to find the moment of inertia about the axis passing through the rod $AB$.$1$. Moment of inertia of rod $AB$ about its own axis is $I_1 = 0$ (since the mass is distributed along the axis).$2$. Moment of inertia of rod $CD$ about the axis $AB$ is calculated using the parallel axis theorem: $I_2 = I_{cm} + Md^2 = \frac{Ml^2}{12} + M(l)^2 = \frac{13}{12}Ml^2$. Wait,the standard approach is: $I = I_{rod1} + I_{rod2} + I_{rod3}$.For rod $AB$: $I_{AB} = 0$.For rod $CD$: Since it is parallel to $AB$ at a distance $l$,$I_{CD} = M(l)^2 = Ml^2$.For rod $EF$: It is perpendicular to $AB$ and attached at the midpoint. Using the parallel axis theorem,$I_{EF} = I_{cm} + M(l/2)^2 = \frac{Ml^2}{12} + \frac{Ml^2}{4} = \frac{Ml^2}{3}$.Total Moment of Inertia $I = 0 + Ml^2 + \frac{Ml^2}{3} = \frac{4}{3}Ml^2$.

Three identical thin rods,each of length $l$ and mass $M$,are joined together to form the letter $H$. What is the moment of inertia of the system about one of the vertical sides of the $H$?

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