Question

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

• A. $\dfrac{M(R_1^2 - R_2^2)}{2},\; \dfrac{M(R_1^2 - R_2^2)}{4},\; \dfrac{M(R_1^2 - R_2^2)}{4}$
• B. $\dfrac{M(R_1^2 + R_2^2)}{2},\; \dfrac{M(R_1^2 + R_2^2)}{4},\; \dfrac{M(R_1^2 + R_2^2)}{4}$
• C. $\dfrac{M(R_1^2 + R_2^2)}{2},\; \dfrac{M(R_1^2 + R_2^2)}{4},\; \dfrac{M(R_1^2 + R_2^2)}{8}$
• D. $\dfrac{M(R_1^2 - R_2^2)}{2},\; \dfrac{M(R_1^2 - R_2^2)}{4},\; \dfrac{M(R_1^2 - R_2^2)}{8}$

Hint:

Moments of inertia for bodies with holes are evaluated by subtracting the contribution of the missing region.

Answer 1

The Correct Option is B

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).