Two identical rods each of mass M and length | vedclass

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(B) Let the two rods lie in the $xy$-plane,with one rod along the $x$-axis and the other along the $y$-axis. The moment of inertia of each rod about a

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

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(B) Let the two rods lie in the $xy$-plane,with one rod along the $x$-axis and the other along the $y$-axis. The moment of inertia of each rod about an axis passing through its center and perpendicular to its length is $\frac{Ml^2}{12}$.For the system,the moment of inertia about the $z$-axis (perpendicular to the plane of the rods and passing through the intersection point) is the sum of the moments of inertia of the two rods about this axis:$I_z = I_x + I_y = \frac{Ml^2}{12} + \frac{Ml^2}{12} = \frac{Ml^2}{6}$.By the perpendicular axes theorem,$I_z = I_{B_1} + I_{B_2}$,where $B_1$ and $B_2$ are the bisector axes. Due to the symmetry of the system,$I_{B_1} = I_{B_2}$.Therefore,$2I_{B_1} = \frac{Ml^2}{6}$,which gives $I_{B_1} = I_{B_2} = \frac{Ml^2}{12}$.

Two identical rods each of mass $M$ and length $l$ are joined in a crossed position as shown in the figure. The moment of inertia of this system about a bisector ($B_1$ or $B_2$) is:

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