Three thin rods, each mass ' 2 M ' and length ' $L$ ' are placed along $x, y$ and $z$ axis which are…

Three thin rods, each mass ' 2 M ' and length ' $L$ ' are placed along $x, y$ and $z$ axis which are mutually perpendicular. One end of each rod is at origin. Moment of inertia of the system about x -axis is
  1. $\frac{4 \mathrm{ML}^2}{3}$
  2. $\frac{\mathrm{ML}^2}{12}$
  3. $\frac{\mathrm{ML}^2}{6}$
  4. $\frac{2 \mathrm{ML}^2}{3}$

Solution

Total moment of inertia, $\mathrm{I}_{\text {total }}=\mathrm{I}_{\mathrm{x}}+\mathrm{I}_{\mathrm{y}}+\mathrm{I}_{\mathrm{z}}...(i)$
Moment of inertia of thin rod, when axis is passing through one end and perpendicular to the rod, $\mathrm{I}=\frac{\mathrm{ML}^2}{3}$...(ii) $\mathrm{I}_{\mathrm{y}}=\frac{\mathrm{M}_{\mathrm{y}} \mathrm{L}^2}{3}=\frac{2 \mathrm{ML}^2}{3}$ ...[From(ii)] ... (given, $\mathrm{M}_{\mathrm{y}}=2 \mathrm{M}$ ) Similarly, $\cdot \mathrm{I}_{\mathrm{z}}=\frac{\mathrm{M}_{\mathrm{2}} \mathrm{L}^2}{3}=\frac{2 \mathrm{ML}^2}{3}$ ...[From(ii)] $\ldots\left(\right.$ given, $\left.\mathrm{M}_{\mathrm{z}}=2 \mathrm{M}\right)$ $\mathrm{I}_{\mathrm{x}}=0$ $\ldots(\because$ Rod lies along X -axis $)$ Substituting in (i), $\therefore \quad \mathrm{I}_{\text {total }}=\frac{4 \mathrm{ML}^2}{3}$

Asked in: MHT CET 2024 (09 May Shift 2)

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