Ratio of Moments of Inertia for Lamina with Circular Cutout

To find the ratio of the moments of inertia $\frac{I_P}{I_O}$, we can use the principle of superposition by treating the cut-out lamina as a complete uniform disc of radius $2R$ minus a smaller circular disc of diameter $2R$ (radius $R$).

1. Mass Distribution

Let $\sigma$ be the uniform mass density per unit area of the disc.

• The mass of the complete large disc of radius $R_1 = 2R$ is: $M = \sigma \cdot \pi (2R)^2 = 4\pi \sigma R^2$

• The mass of the removed small disc of radius $r = R$ is: $m = \sigma \cdot \pi R^2 = \frac{M}{4}$

Set up a Cartesian coordinate system with the origin at the center $O(0, 0)$ of the original disc:

• The center of the small removed disc is at $C(0, R)$.
• Point $P$ lies on the edge of the large disc along the x-axis, at coordinates $P(2R, 0)$.

---

2. Moment of Inertia about Axis passing through $O$ ($I_O$)

1. For the complete large disc about $O$: $I_{O, \text{large}} = \frac{1}{2} M R_1^2 = \frac{1}{2} M (2R)^2 = 2 M R^2$

2. For the removed small disc about $O$:

   • Moment of inertia about its own center $C(0, R)$: $I_{C, \text{small}} = \frac{1}{2} m r^2 = \frac{1}{2} \left(\frac{M}{4}\right) R^2 = \frac{1}{8} M R^2$
   • Using the parallel axis theorem to shift the axis to $O$, where the distance $OC = R$: $I_{O, \text{small}} = I_{C, \text{small}} + m (OC)^2 = \frac{1}{8} M R^2 + \left(\frac{M}{4}\right) R^2 = \frac{3}{8} M R^2$

3. For the remaining lamina about $O$: $I_O = I_{O, \text{large}} - I_{O, \text{small}} = 2 M R^2 - \frac{3}{8} M R^2 = \frac{13}{8} M R^2$

---

3. Moment of Inertia about Axis passing through $P$ ($I_P$)

1. For the complete large disc about $P$:

   • Using the parallel axis theorem, where the distance $OP = 2R$: $I_{P, \text{large}} = I_{O, \text{large}} + M (OP)^2 = 2 M R^2 + M (2R)^2 = 6 M R^2$

2. For the removed small disc about $P$:

   • The square of the distance between the center of the small disc $C(0, R)$ and point $P(2R, 0)$ is: $CP^2 = (2R - 0)^2 + (0 - R)^2 = 4R^2 + R^2 = 5R^2$
   • Using the parallel axis theorem: $I_{P, \text{small}} = I_{C, \text{small}} + m (CP)^2 = \frac{1}{8} M R^2 + \left(\frac{M}{4}\right) (5R^2) = \frac{1}{8} M R^2 + \frac{5}{4} M R^2 = \frac{11}{8} M R^2$

3. For the remaining lamina about $P$: $I_P = I_{P, \text{large}} - I_{P, \text{small}} = 6 M R^2 - \frac{11}{8} M R^2 = \frac{37}{8} M R^2$

---

4. Ratio $\frac{I_P}{I_O}$

Taking the ratio of $I_P$ to $I_O$: $\frac{I_P}{I_O} = \frac{\frac{37}{8} M R^2}{\frac{13}{8} M R^2} = \frac{37}{13} \approx 2.846$

Rounding to the nearest integer: $\frac{I_P}{I_O} \approx 3$