From a uniform circular disc of radius R and mass 9 M, a small disc of radius R/3 is removed as show

Q: Why can't we just subtract the areas and use the formula $1/2 M R^2$?

Moment of inertia depends not just on the mass, but also on how that mass is distributed relative to the axis of rotation. The removed mass is located at a specific distance from the center, so its contribution to the total moment of inertia must be calculated exactly using the parallel axis theorem.

Q: How did we find the mass of the removed disc?

Since the disc is uniform, its mass is proportional to its area. We found the mass per unit area (surface mass density) of the original disc and multiplied it by the area of the removed disc.

Q: Why do we use the Parallel Axis Theorem here?

We need to subtract the moment of inertia of the removed disc about the center of the *original* disc (point O). The standard formula $1/2 m r^2$ only gives the moment of inertia about its *own* center (point O'). The parallel axis theorem bridges this gap.

Q: What is the distance between the two axes?

The radius of the large disc is $R$, and the small disc touches the edge. The radius of the small disc is $R/3$. Therefore, the distance from the center of the large disc to the center of the small disc is $R - R/3 = 2R/3$.

Q: Can this method be used for any shape?

Yes! The principle of superposition (or the "negative mass" method) is incredibly powerful and can be used for any rigid body with cavities, provided you can calculate the moment of inertia of the complete body and the "removed" body about the same axis.

Question

From a uniform circular disc of radius $R$ and mass $9 M$, a small disc of radius $\frac{R}{3}$ is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is

Options:

(A) $4 M R^{2}$

(B) $\frac{40}{9} M R^{2}$

(C) $10 M R^{2}$

(D) $\frac{37}{9} M R^{2}$

Sigma Percentile · Accepted Answer

### The Power of Negative Mass

Have you ever tried to find the center of mass or moment of inertia of a shape with a hole in it? It sounds like a nightmare of complex integration. But physics offers us a beautiful, elegant shortcut: the principle of superposition, often called the "negative mass" method.

Instead of dealing with the awkward geometry of the remaining shape, we can treat the cavity as an object with **negative mass**. The total moment of inertia is simply the moment of inertia of the complete, solid object minus the moment of inertia of the "removed" object.

### Step 1: Analyzing the Mass Distribution

Our first task is to figure out the mass of the small disc that was removed. We are given that the complete disc has a mass of $9M$ and a radius of $R$. Since the disc is uniform, its mass is spread evenly across its area.

The surface mass density $\sigma$ is the total mass divided by the total area:
$$ \sigma = \frac{9M}{\pi R^2} $$

The removed disc has a radius of $\frac{R}{3}$. Its area is:
$$ A_{	ext{removed}} = \pi \left(\frac{R}{3}ight)^2 = \frac{\pi R^2}{9} $$

Multiplying the area by the surface mass density gives us the mass of the removed disc:
$$ m = \sigma 	imes A_{	ext{removed}} = \left(\frac{9M}{\pi R^2}ight) 	imes \left(\frac{\pi R^2}{9}ight) = M $$

It simplifies perfectly to $M$!

### Step 2: Moment of Inertia of the Complete Disc

Now, let's calculate the moment of inertia of the complete, solid disc about its central axis $O$. The standard formula for a uniform disc is $\frac{1}{2} 	imes 	ext{mass} 	imes 	ext{radius}^2$.

$$ I_{	ext{complete}} = \frac{1}{2} (9M) R^2 = \frac{9}{2} M R^2 $$

### Step 3: The Parallel Axis Theorem

Next, we need the moment of inertia of the removed disc. But we must be careful! We need its moment of inertia about the axis passing through $O$, not its own center $O'$.

First, we find its moment of inertia about its own center $O'$:
$$ I_{O'} = \frac{1}{2} m r^2 = \frac{1}{2} M \left(\frac{R}{3}ight)^2 = \frac{1}{18} M R^2 $$

Now, we apply the **Parallel Axis Theorem**, which states that $I = I_{	ext{cm}} + m d^2$. The distance $d$ between the center of the large disc and the center of the small disc is $R - \frac{R}{3} = \frac{2R}{3}$.

$$ I_{	ext{removed}} = I_{O'} + m d^2 = \frac{1}{18} M R^2 + M \left(\frac{2R}{3}ight)^2 $$

$$ I_{	ext{removed}} = \frac{1}{18} M R^2 + \frac{4}{9} M R^2 $$

$$ I_{	ext{removed}} = \left(\frac{1}{18} + \frac{8}{18}ight) M R^2 = \frac{9}{18} M R^2 = \frac{1}{2} M R^2 $$

### Step 4: The Final Subtraction

We have all the pieces of the puzzle. The moment of inertia of the remaining disc is the difference between the two moments of inertia we just calculated.

$$ I_{	ext{remaining}} = I_{	ext{complete}} - I_{	ext{removed}} $$

$$ I_{	ext{remaining}} = \frac{9}{2} M R^2 - \frac{1}{2} M R^2 = 4 M R^2 $$

And there we have it! By breaking the complex shape into two simple discs, we turned a difficult calculus problem into a straightforward algebra exercise.

Select Answer:

Visualized Solution

$Superposition Principle$

$Mass of Removed Disc$

$Mass Calculation$

$I_{complete}$

$I_{removed} about O^{'}$

$Parallel Axis Theorem$

$I_{removed} about O$

$I_{remaining}$

$Conclusion$

The Sigma Insight: Moment of Inertia

The Power of Negative Mass

Step 1

Step 2

Step 3

Step 4

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Select Answer:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass of Removed Disc

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

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

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

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

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

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

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