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A disc of radius \(R_{1}\) having uniform surface density has a concentric hole of radius \(R_{2} \lt R_{1}\). If its mass is M, the principal moments of inertia are

  • A.\(\frac{M(R_{1}^{2} - R_{2}^{2})}{2}\), \(\frac{M(R_{1}^{2} - R_{2}^{2})}{4}\), \(\frac{M(R_{1}^{2} - R_{2}^{2})}{4}\)
  • B.\(\frac{M(R_{1}^{2} + R_{2}^{2})}{2}\), \(\frac{M(R_{1}^{2} + R_{2}^{2})}{4}\), \(\frac{M(R_{1}^{2} + R_{2}^{2})}{4}\)
  • C.\(\frac{M(R_{1}^{2} + R_{2}^{2})}{2}\), \(\frac{M(R_{1}^{2} + R_{2}^{2})}{4}\), \(\frac{M(R_{1}^{2} + R_{2}^{2})}{8}\)
  • D.\(\frac{M(R_{1}^{2} - R_{2}^{2})}{2}\), \(\frac{M(R_{1}^{2} - R_{2}^{2})}{4}\), \(\frac{M(R_{1}^{2} - R_{2}^{2})}{8}\)

Answer: B

The principal moments are \(\frac{M(R_{1}^{2} + R_{2}^{2})}{2}\), \(\frac{M(R_{1}^{2} + R_{2}^{2})}{4}\), \(\frac{M(R_{1}^{2} + R_{2}^{2})}{4}\).

About the axis through the centre perpendicular to the disc, an annulus of mass M has

\[ \begin{aligned} I_{z} &= \frac{M}{\pi(R_{1}^{2} - R_{2}^{2})}\int_{R_{2}}^{R_{1}} 2\pi r^{3}dr \\ &= \frac{M(R_{1}^{2} + R_{2}^{2})}{2} \end{aligned} \]

For a flat body the perpendicular axis theorem gives \(I_{x} + I_{y} = I_{z}\), and by symmetry \(I_{x} = I_{y} = \frac{M(R_{1}^{2} + R_{2}^{2})}{4}\).

The options with \(R_{1}^{2} - R_{2}^{2}\) come from subtracting moments without keeping the total mass fixed, and the options with 8 break the perpendicular axis theorem.

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