Three thin rods,each of mass 2M and length L, | vedclass

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Question

(A) The total moment of inertia of the system about the $x$-axis is the sum of the moments of inertia of the three individual rods about the $x$-axis:

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The total moment of inertia of the system about the $x$-axis is the sum of the moments of inertia of the three individual rods about the $x$-axis: $I_{	ext{total}} = I_x + I_y + I_z$.For a thin rod of mass $m$ and length $L$ rotating about an axis passing through one end and perpendicular to the rod,the moment of inertia is $I = \frac{mL^2}{3}$.$1$. For the rod along the $x$-axis: Since the rod lies on the $x$-axis,its distance from the $x$-axis is zero for all points. Thus,$I_x = 0$.$2$. For the rod along the $y$-axis: The axis of rotation ($x$-axis) is perpendicular to the rod and passes through one end (the origin). Here,$m = 2M$,so $I_y = \frac{(2M)L^2}{3} = \frac{2ML^2}{3}$.$3$. For the rod along the $z$-axis: Similarly,the $x$-axis is perpendicular to this rod and passes through one end. Here,$m = 2M$,so $I_z = \frac{(2M)L^2}{3} = \frac{2ML^2}{3}$.Summing these values: $I_{	ext{total}} = 0 + \frac{2ML^2}{3} + \frac{2ML^2}{3} = \frac{4ML^2}{3}$.

Three thin rods,each of mass $2M$ and length $L$,are placed along the $x, y,$ and $z$ axes,which are mutually perpendicular. One end of each rod is at the origin. The moment of inertia of the system about the $x$-axis is:

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