Two rods of equal mass m and length l lie alo | vedclass

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Question

(C) The moment of inertia of a rod of mass $m$ and length $l$ about an axis passing through its center and making an angle $	heta$ with the rod is gi

Vedclass · Accepted Answer

$\frac{ml^2}{12}$

Vedclass · Answer

(C) The moment of inertia of a rod of mass $m$ and length $l$ about an axis passing through its center and making an angle $	heta$ with the rod is given by $I = \frac{ml^2}{12} \sin^2 	heta$.For the rod along the $x$-axis,the angle with the line $x=y$ is $	heta = 45^\circ$. Thus,$I_1 = \frac{ml^2}{12} \sin^2 45^\circ = \frac{ml^2}{12} 	imes \frac{1}{2} = \frac{ml^2}{24}$.For the rod along the $y$-axis,the angle with the line $x=y$ is also $	heta = 45^\circ$. Thus,$I_2 = \frac{ml^2}{12} \sin^2 45^\circ = \frac{ml^2}{24}$.The total moment of inertia is $I = I_1 + I_2 = \frac{ml^2}{24} + \frac{ml^2}{24} = \frac{2ml^2}{24} = \frac{ml^2}{12}$.

Two rods of equal mass $m$ and length $l$ lie along the $x$-axis and $y$-axis with their centers at the origin. What is the moment of inertia of both about the line $x=y$?

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