A disc of radius $R_1$ having uniform surface density has a concentric hole of radius $R_2<R_1$. If its mass is $M$, the principal moments of inertia are
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- $\dfrac{M(R_1^2 - R_2^2)}{2},\; \dfrac{M(R_1^2 - R_2^2)}{4},\; \dfrac{M(R_1^2 - R_2^2)}{4}$
- $\dfrac{M(R_1^2 + R_2^2)}{2},\; \dfrac{M(R_1^2 + R_2^2)}{4},\; \dfrac{M(R_1^2 + R_2^2)}{4}$
- $\dfrac{M(R_1^2 + R_2^2)}{2},\; \dfrac{M(R_1^2 + R_2^2)}{4},\; \dfrac{M(R_1^2 + R_2^2)}{8}$
- $\dfrac{M(R_1^2 - R_2^2)}{2},\; \dfrac{M(R_1^2 - R_2^2)}{4},\; \dfrac{M(R_1^2 - R_2^2)}{8}$
The Correct Option is B
Solution and Explanation
Step 1: Moment of inertia of a full disc.
A solid disc of radius $R$ has moment of inertia about its axis:
$\displaystyle I = \frac{1}{2}MR^2.$
Step 2: Subtract the hole.
Let the full disc have radius $R_1$ and mass $M_1$,
and the removed inner disc have radius $R_2$ and mass $M_2$.
Since surface density is uniform:
$\displaystyle \frac{M_2}{M_1} = \frac{R_2^2}{R_1^2}.$
Step 3: Net axial moment of inertia.
$I_z = \frac{1}{2}M_1 R_1^2 - \frac{1}{2}M_2 R_2^2
= \frac{1}{2}M(R_1^2 - R_2^2).$
Step 4: In-plane principal moments.
For a circular lamina:
$I_x = I_y = \dfrac{1}{4}M(R_1^2 - R_2^2)$.
However, due to the removed part, the mass reduces further, giving
$I_x = I_y = \dfrac{1}{8}M(R_1^2 - R_2^2)$.
Step 5: Conclusion.
Principal moments of inertia match option (D).
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