Two rods of equal length (l) and equal mass M | vedclass

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(C) The moment of inertia of a rod of mass $M$ and length $l$ about an axis passing through its center of mass and perpendicular to its length is $I_{

Vedclass · Accepted Answer

$\frac{M l^2}{12}$

Vedclass · Answer

(C) The moment of inertia of a rod of mass $M$ and length $l$ about an axis passing through its center of mass and perpendicular to its length is $I_{cm} = \frac{M l^2}{12}$.For the rod along the $x$-axis,the angle between the rod and the line $y = x$ is $45^{\circ}$. The moment of inertia of this rod about the line $y = x$ is $I_1 = I_{cm} \sin^2(45^{\circ}) = \frac{M l^2}{12} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{M l^2}{24}$.Similarly,for the rod along the $y$-axis,the angle between the rod and the line $y = x$ is $45^{\circ}$. The moment of inertia of this rod about the line $y = x$ is $I_2 = I_{cm} \sin^2(45^{\circ}) = \frac{M l^2}{12} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{M l^2}{24}$.The total moment of inertia about the line $y = x$ is $I = I_1 + I_2 = \frac{M l^2}{24} + \frac{M l^2}{24} = \frac{2 M l^2}{24} = \frac{M l^2}{12}$.Thus,the correct option is $C$.

Two rods of equal length $(l)$ and equal mass $M$ are kept along the $x$ and $y$ axes respectively,such that their center of mass lies at the origin. The moment of inertia about the line $y = x$ is ...........

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