A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density

Q: Why can we just subtract the moments of inertia?

Moment of inertia is a scalar quantity (more accurately, a tensor, but for a fixed axis it behaves like a scalar) that depends linearly on mass. Because mass is additive, the moment of inertia of a composite body is the sum of the moments of inertia of its parts. Removing a part is mathematically equivalent to adding a part with "negative mass".

Q: Why is the mass of the large disc 4 times the mass of the cavity?

For a uniform lamina, mass is proportional to the area. The area of a circle is $\pi r^2$. The large disc has a radius of $2R$, so its area is $\pi(2R)^2 = 4\pi R^2$. The cavity has a radius of $R$, so its area is $\pi R^2$. Therefore, the large disc has exactly 4 times the area, and thus 4 times the mass, of the cavity.

Q: How did we find the distance from the cavity center to point P?

We set up a coordinate system with the origin at $O(0,0)$. The cavity is centered at $C(0, R)$ and point $P$ is at $(2R, 0)$. Using the distance formula, the distance $r$ is $\sqrt{(2R - 0)^2 + (0 - R)^2} = \sqrt{4R^2 + R^2} = \sqrt{5}R$.

Q: Can we use the perpendicular axis theorem here?

The perpendicular axis theorem relates the moment of inertia about an axis perpendicular to the plane to the moments of inertia about two perpendicular axes in the plane. While true, it doesn't directly help us shift axes parallel to each other. The parallel axis theorem is the correct tool for shifting the axis of rotation from the center of mass to points $O$ and $P$.

Q: What if the cavity was not a circle but a square?

The principle of superposition would still apply perfectly! You would calculate the moment of inertia of the solid large disc, and then subtract the moment of inertia of the square cavity. You would just need to use the specific moment of inertia formula for a square plate and apply the parallel axis theorem to shift it to the required axes.

Question

A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is $I_O$ and $I_P$, respectively. Both these axes are perpendicular to the plane of the lamina. The ratio $\frac{I_P}{I_O}$ to the nearest integer is

Sigma Percentile · Accepted Answer

The problem of finding the moment of inertia of a body with a cavity is a classic in rotational mechanics. It might seem daunting at first, but it beautifully unravels when we apply the **Principle of Superposition**. Let's embark on this journey to find the ratio of the moments of inertia for our given lamina.

### Analyzing the Setup

Imagine a large, uniform circular metal plate of radius $2R$. From this plate, a smaller circular section of diameter $2R$ (and thus radius $R$) is removed. This removed section, which we'll call the cavity, is positioned such that its boundary passes through the center $O$ of the large disc.

We are tasked with finding the moment of inertia of the remaining shape (the lamina) about two specific axes: one passing through the center $O$, and another passing through a point $P$ located on the edge of the large disc, diametrically opposite to the cavity's position relative to the center.

### The Principle of Superposition

To tackle this, we use a powerful concept: the **Principle of Superposition**. Instead of trying to integrate over the complex shape of the lamina, we can treat it as a complete large disc with a smaller disc of "negative mass" superimposed on it.

Mathematically, the moment of inertia of the lamina about any axis is simply the moment of inertia of the complete large disc minus the moment of inertia of the removed cavity about that same axis:

$$
I_{	ext{lamina}} = I_{	ext{large}} - I_{	ext{cavity}}
$$

Before we calculate the moments of inertia, let's determine the masses. Assuming a uniform mass per unit area $\sigma$, the mass is proportional to the area.

The mass of the complete large disc is:

$$
m_{	ext{large}} = \sigma \pi (2R)^2 = 4\pi\sigma R^2
$$

The mass of the removed cavity is:

$$
m_{	ext{cavity}} = \sigma \pi (R)^2 = \pi\sigma R^2
$$

Notice that $m_{	ext{large}} = 4 m_{	ext{cavity}}$. This relationship will be handy!

### Conquering the Center: Moment of Inertia about O

Let's first calculate the moment of inertia about the axis passing through $O$, denoted as $I_O$.

For the complete large disc, the axis passes through its center of mass. Using the standard formula for a disc:

$$
I_{O, 	ext{large}} = \frac{1}{2} m_{	ext{large}} (2R)^2 = \frac{1}{2} (4\pi\sigma R^2) (4R^2) = 8\pi\sigma R^4
$$

Now, for the cavity. Its center $C$ is at a distance $R$ from $O$. We must use the **Parallel Axis Theorem** to find its moment of inertia about $O$:

$$
I_{O, 	ext{cavity}} = I_{C} + m_{	ext{cavity}} R^2 = \frac{1}{2} m_{	ext{cavity}} R^2 + m_{	ext{cavity}} R^2 = \frac{3}{2} m_{	ext{cavity}} R^2
$$

Substituting the mass of the cavity:

$$
I_{O, 	ext{cavity}} = \frac{3}{2} (\pi\sigma R^2) R^2 = 1.5\pi\sigma R^4
$$

Applying superposition, the net moment of inertia about $O$ is:

$$
I_O = 8\pi\sigma R^4 - 1.5\pi\sigma R^4 = 6.5\pi\sigma R^4
$$

### Shifting the Axis: Moment of Inertia about P

Next, we calculate the moment of inertia about the axis passing through $P$, denoted as $I_P$. Point $P$ is at a distance of $2R$ from the center $O$.

For the large disc, we again apply the Parallel Axis Theorem, shifting the axis from $O$ to $P$:

$$
I_{P, 	ext{large}} = I_{O, 	ext{large}} + m_{	ext{large}} (2R)^2 = 8\pi\sigma R^4 + (4\pi\sigma R^2)(4R^2) = 24\pi\sigma R^4
$$

Now comes the tricky part: the cavity. We need the exact distance $r$ from the cavity's center $C$ to point $P$. Let's set up a coordinate system with $O$ at the origin $(0,0)$. The cavity center $C$ is at $(0, R)$, and point $P$ is at $(2R, 0)$. Using the Pythagorean theorem:

$$
r = \sqrt{(2R - 0)^2 + (0 - R)^2} = \sqrt{4R^2 + R^2} = \sqrt{5}R
$$

Applying the Parallel Axis Theorem for the cavity about $P$:

$$
I_{P, 	ext{cavity}} = I_{C} + m_{	ext{cavity}} r^2 = \frac{1}{2} m_{	ext{cavity}} R^2 + m_{	ext{cavity}} (\sqrt{5}R)^2
$$

$$
I_{P, 	ext{cavity}} = 0.5\pi\sigma R^4 + 5\pi\sigma R^4 = 5.5\pi\sigma R^4
$$

Applying superposition once more, the net moment of inertia about $P$ is:

$$
I_P = 24\pi\sigma R^4 - 5.5\pi\sigma R^4 = 18.5\pi\sigma R^4
$$

### The Final Ratio

We have successfully calculated both $I_O$ and $I_P$. The final step is to find their ratio:

$$
\frac{I_P}{I_O} = \frac{18.5\pi\sigma R^4}{6.5\pi\sigma R^4} = \frac{18.5}{6.5} = \frac{37}{13}
$$

Calculating the decimal value:

$$
\frac{37}{13} \approx 2.846
$$

The question asks for the ratio to the nearest integer. Rounding $2.846$ to the nearest integer gives us our final answer: **3**.

Enter Numerical Value:

Visualized Solution

$I_{lamina}$

$I_{lamina} = I_{large} - I_{cavity}$

$m = σ A$

$I_{O, large}$

$I_{O, cavity}$

$I_{O}$

$I_{P, large}$

$r_{C P}$

$I_{P, cavity}$

$I_{P}$

$\frac{I _{P}}{I _{O}}$

The Sigma Insight: Moment of Inertia

Analyzing the Setup

The Principle of Superposition

Conquering the Center

Shifting the Axis

The Final Ratio

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ilamina​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ilamina​=Ilarge​−Icavity​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=σA

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IO,large​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IO,cavity​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IO​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IP,large​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } rCP​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IP,cavity​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IP​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } IO​IP​​