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MHT CET · Physics · Rotational Motion

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. A \(\frac{4 \mathrm{ML}^2}{3}\)
  2. B \(\frac{\mathrm{ML}^2}{12}\)
  3. C \(\frac{\mathrm{ML}^2}{6}\)
  4. D \(\frac{2 \mathrm{ML}^2}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{4 \mathrm{ML}^2}{3}\)

Step-by-step Solution

Detailed explanation

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),
\(\therefore \quad \mathrm{I}_{\text {total }}=\frac{4 \mathrm{ML}^2}{3}\)