One quarter section is cut from a uniform circular disc of radius R. This section has a mass M. It i

Q: Why can't we just use the formula $\frac{1}{2} M R^2$ directly since it's a part of a disc?

You actually can! The formula $I = \frac{1}{2} M R^2$ is valid for any sector of a uniform disc of mass $M$ and radius $R$. The "full disc" method is just a clever way to prove it without using integration.

Q: Does the moment of inertia depend on the angle of the sector?

If the mass $M$ is kept constant, the moment of inertia $\frac{1}{2} M R^2$ does not depend on the angle. However, if the density is constant, a larger angle means more mass, which increases the moment of inertia.

Q: What if the axis of rotation was along one of the straight edges of the quarter disc?

Then the moment of inertia would be different. By the perpendicular axis theorem, it would be $\frac{1}{4} M R^2$ about a straight edge.

Q: Why is the mass of the full disc $4M$ and not $M$?

The problem states that the *quarter section* itself has a mass of $M$. If you assemble four such sections to make a full disc, the total mass becomes $M + M + M + M = 4M$.

Q: Is moment of inertia a vector? Why can we just add them?

Moment of inertia is a tensor, but for a fixed axis of rotation, it behaves like a scalar. Because it is defined as an integral over the mass distribution ($I = \int r^2 dm$), you can simply add the moments of inertia of different parts of a body about the same axis.

Question

One quarter section is cut from a uniform circular disc of radius $R$. This section has a mass $M$. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

Options:

(A) $\frac{1}{2} M R^2$

(B) $\frac{1}{4} M R^2$

(C) $\frac{1}{8} M R^2$

(D) $\sqrt{2} M R^2$

Sigma Percentile · Accepted Answer

The problem of finding the moment of inertia of a quarter disc might seem like it requires complex calculus at first glance. However, physics is often about finding elegant symmetries that simplify our lives. Let's embark on a journey to solve this problem using a beautiful superposition trick!

### The Geometry of the Setup

Imagine you are holding a slice of pizza, but this is a perfectly uniform quarter section of a circular disc. We are given that the mass of this specific quarter section is **$M$**, and its radius is **$R$**.

The axis of rotation is a line passing through the center of the original disc (the tip of the pizza slice) and perpendicular to its plane. Our goal is to find the moment of inertia of this quarter disc about this central axis.

### The Superposition Trick

Instead of diving into integration, let's use our imagination. What if we take three more identical quarter discs and arrange them around the center to complete the full circle?

By doing this, we have constructed a full uniform disc. Now, let's think about the mass of this new, complete disc. Since one quarter section has a mass of $M$, the total mass of the full disc must be four times that amount.

Therefore, the mass of the full disc is **$4M$**.

### The Master Equation

We already know the standard formula for the moment of inertia of a full uniform disc about its central axis. For a disc of mass $m$ and radius $R$, the moment of inertia is given by:

$$I = \frac{1}{2} m R^2$$

Let's apply this formula to our newly constructed full disc. We substitute the total mass $m = 4M$ into the equation:

$$I_{full} = \frac{1}{2} (4M) R^2 = 2MR^2$$

This is the moment of inertia of the entire disc.

### Final Calculation

Now, we use a fundamental property of the moment of inertia: it is an additive scalar quantity. This means the total moment of inertia of a body is simply the sum of the moments of inertia of its individual parts.

Since our full disc is composed of four identical quarter sections, and the setup is perfectly symmetric, each quarter section must contribute exactly one-fourth to the total moment of inertia.

So, to find the moment of inertia of our single quarter section, we divide the total moment of inertia by four:

$$I_{quarter} = \frac{I_{full}}{4} = \frac{2MR^2}{4} = \frac{1}{2}MR^2$$

And there we have it! The moment of inertia of the quarter disc is **$\frac{1}{2}MR^2$**.

### A Deeper Insight: The Integration Perspective

You might wonder, is it a coincidence that the formula for the quarter disc ($\frac{1}{2}MR^2$) looks exactly like the formula for a full disc? Not at all!

The moment of inertia is defined by the integral $I = \int r^2 dm$. This integral depends entirely on how the mass is distributed radially from the axis of rotation. If you take a full disc and cut out a sector of any angle, the *radial* distribution of mass remains identical.

Therefore, for *any* sector of a uniform disc with mass $M$ and radius $R$, the moment of inertia about the central axis will always be **$\frac{1}{2}MR^2$**. The superposition trick we used is just a clever, intuitive way to prove this universal truth without writing a single integral!

Select Answer:

Visualized Solution

Visualizing the Quarter Disc

The Superposition Trick

Mass of the Full Disc

Moment of Inertia of Full Disc

Calculating $I_{f u l l}$

Symmetry and Additivity

Final Calculation

The Integration Perspective

The Sigma Insight: Moment of Inertia

The Geometry of the Setup

The Superposition Trick

The Master Equation

Final Calculation

A Deeper Insight

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Select Answer:

Visualized Solution

Visualizing the Quarter Disc

The Superposition Trick

Mass of the Full Disc

Moment of Inertia of Full Disc

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

Symmetry and Additivity

Final Calculation

The Integration Perspective

The Sigma Insight: Moment of Inertia

The Geometry of the Setup

The Superposition Trick

The Master Equation

Final Calculation

A Deeper Insight

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The moment of inertia of uniform semi-circular disc of mass M and radius r about a line perpendicular to the plane of the disc through the centre is

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