Two uniform thin identical rods AB and CD eac | vedclass

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(C) Let the intersection point of the rods be the origin $(0,0)$.For rod $AB$ lying along the $x$-axis,the moment of inertia about the $x$-axis is $I_

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the intersection point of the rods be the origin $(0,0)$.For rod $AB$ lying along the $x$-axis,the moment of inertia about the $x$-axis is $I_x = 0$ and about the $y$-axis is $I_y = \frac{ML^2}{12}$.For rod $CD$ lying along the $y$-axis,the moment of inertia about the $x$-axis is $I_x = \frac{ML^2}{12}$ and about the $y$-axis is $I_y = 0$.The total moment of inertia of the cross about the $x$-axis is $I_x = 0 + \frac{ML^2}{12} = \frac{ML^2}{12}$.The total moment of inertia of the cross about the $y$-axis is $I_y = \frac{ML^2}{12} + 0 = \frac{ML^2}{12}$.The line $EF$ is a bisector in the $xy$-plane,making an angle of $45^{\circ}$ with both axes.The moment of inertia about an axis in the plane of the object is given by $I = I_x \sin^2 	heta + I_y \cos^2 	heta - 2 I_{xy} \sin 	heta \cos 	heta$.Since the rods are symmetric about the axes,the product of inertia $I_{xy} = 0$.Thus,$I_{EF} = I_x \sin^2 45^{\circ} + I_y \cos^2 45^{\circ} = \frac{ML^2}{12} \left( \frac{1}{2} ight) + \frac{ML^2}{12} \left( \frac{1}{2} ight) = \frac{ML^2}{12}$.

Two uniform thin identical rods $AB$ and $CD$ each of mass $M$ and length $L$ are joined so as to form a cross as shown. The moment of inertia of the cross about a bisector line $EF$ is (Line $EF$ is in the plane of the cross and bisects the angle between the rods).

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