A thin circular plate of mass M and radius R has its density varying as rho(r) = rho_0 r with rho_0

Q: Why can't we just use the standard formula $I = \frac{3}{2}MR^2$ for a disc about its edge?

That standard formula only applies to a **uniform** disc where the density is constant everywhere. In this problem, the density $\rho(r) = \rho_0 r$ varies with the radius, meaning the disc is heavier towards the edges. You must derive it from scratch using integration.

Q: Why do we apply the parallel axis theorem to the elementary ring instead of the whole disc?

You can actually do either! You can find the total moment of inertia about the center first ($I_{cm} = \int dI_c$) and then apply the theorem to the whole disc ($I_{edge} = I_{cm} + MR^2$). Both methods are mathematically identical and will give you the exact same result of $\frac{8}{5}MR^2$.

Q: What does $\rho(r) = \rho_0 r$ physically mean?

It means the mass density is directly proportional to the distance from the center. At the exact center ($r=0$), the density is zero. As you move outwards, the material gets progressively denser and heavier.

Q: How did we get the area of the elementary ring as $2\pi x dx$?

Imagine taking a very thin ring and cutting it open to lay it flat. It forms a long, thin rectangle. The length of this rectangle is the circumference of the ring ($2\pi x$), and its width is the thickness ($dx$). Therefore, the area is length $\times$ width = $2\pi x dx$.

Q: Why is calculating the total mass $M$ important if it's not given as a specific number?

In physics, final answers are almost always required in terms of the macroscopic, observable properties of the system (like total mass $M$ and radius $R$), rather than internal arbitrary constants like $\rho_0$. Calculating $M$ allows us to substitute $\rho_0$ out of our final equation.

Question

A thin circular plate of mass $M$ and radius $R$ has its density varying as $\rho(r) = \rho_0 r$ with $\rho_0$ as constant and $r$ is the distance from its centre. The moment of inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is $I = aMR^2$. The value of the coefficient $a$ is

Options:

(A) $\frac{1}{2}$

(B) $\frac{3}{5}$

(C) $\frac{8}{5}$

(D) $\frac{3}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a circular plate where the mass isn't spread out evenly. Instead, it gets denser as you move from the center to the edge. Mathematically, this is given by the density function $\rho(r) = \rho_0 r$. Our goal is to find the moment of inertia of this plate about an axis passing through its edge and perpendicular to its plane.

To tackle this, we can't just use the standard formula for a uniform disc. We need to build the disc from scratch using calculus. Let's consider a thin elementary ring of radius $x$ and thickness $dx$.

### The Mass of the Disc
First, let's find the total mass $M$ of the disc. The mass of our elementary ring, $dm$, is its density multiplied by its area. If you unroll the ring, it forms a rectangle of length $2\pi x$ and width $dx$. Therefore, the area is $dA = 2\pi x dx$.

$$dm = \rho(x) dA = (\rho_0 x) (2\pi x dx) = 2\pi \rho_0 x^2 dx$$

To find the total mass $M$, we integrate $dm$ from the center ($x=0$) to the edge ($x=R$):

$$M = \int_0^R 2\pi \rho_0 x^2 dx = 2\pi \rho_0 \left[ \frac{x^3}{3} \right]_0^R = \frac{2\pi \rho_0 R^3}{3}$$

Keep this expression safe; we will need it later to substitute back into our final answer!

### Moment of Inertia via Parallel Axis Theorem
Now, we need the moment of inertia about the edge. We can find the moment of inertia of our elementary ring about the central axis ($dI_c$) and then use the **Parallel Axis Theorem** to shift it to the edge.

The moment of inertia of the ring about the center is simply its mass times its radius squared:

$$dI_c = dm \cdot x^2 = (2\pi \rho_0 x^2 dx) \cdot x^2 = 2\pi \rho_0 x^4 dx$$

According to the Parallel Axis Theorem, the moment of inertia of this ring about the edge axis is:

$$dI = dI_c + dm \cdot R^2$$

Substituting our expressions for $dI_c$ and $dm$:

$$dI = 2\pi \rho_0 x^4 dx + (2\pi \rho_0 x^2 dx) R^2$$

### Final Calculation
To find the total moment of inertia $I$, we integrate $dI$ from $0$ to $R$:

$$I = \int_0^R \left( 2\pi \rho_0 x^4 + 2\pi \rho_0 R^2 x^2 \right) dx$$

$$I = 2\pi \rho_0 \left[ \frac{x^5}{5} + R^2 \frac{x^3}{3} \right]_0^R$$

$$I = 2\pi \rho_0 R^5 \left( \frac{1}{5} + \frac{1}{3} \right) = 2\pi \rho_0 R^5 \left( \frac{8}{15} \right) = \frac{16\pi \rho_0 R^5}{15}$$

Finally, we must express this result in terms of the total mass $M$. Let's cleverly factor out the expression for $M$ that we found earlier:

$$I = \frac{8}{5} \left( \frac{2\pi \rho_0 R^3}{3} \right) R^2$$

Since the term in the parentheses is exactly $M$, we get:

$$I = \frac{8}{5} M R^2$$

Comparing this with the given equation $I = aMR^2$, we find that the coefficient $a$ is **$\frac{8}{5}$**.

Select Answer:

Visualized Solution

\text{Elementary Ring Setup}

\text{Mass of Elementary Ring}

\text{Total Mass of the Disc}

\text{Moment of Inertia about Center}

\text{Parallel Axis Theorem}

\text{Moment of Inertia of Ring about Edge}

\text{Total Moment of Inertia}

\text{Relating } I \text{ to } M

\text{Finding the Coefficient } a

The Sigma Insight: Moment of Inertia

Analyzing the Setup Imagine a circular plate where the mass isn't spread out evenly

The Mass of the Disc First, let's find the total mass $M$ of the disc

Moment of Inertia via Parallel Axis Theorem Now, we need the moment of inertia about the edge

Final Calculation

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