A thin disc of mass M and radius R has mass per unit area sigma(r) = kr^2, where r is the distance f

Q: Why do we take an elemental ring instead of a small square element?

Because the surface mass density $\sigma(r)$ depends only on the radial distance $r$. A ring of radius $r$ has a constant density throughout its area, making the integration much simpler.

Q: How did we get the area of the elemental ring as $2\pi r dr$?

If you cut the thin ring and unroll it, it forms a rectangle of length equal to its circumference ($2\pi r$) and width equal to its thickness ($dr$). Thus, the area is $2\pi r dr$.

Q: Why is the moment of inertia of the elemental ring $dm \cdot r^2$?

The elemental ring is essentially a collection of point masses all located at the same distance $r$ from the axis of rotation. The moment of inertia of a point mass is $mr^2$, so for the entire ring, it's $dm \cdot r^2$.

Q: What does a non-uniform density physically mean here?

It means the disc is heavier towards the edges. Since $\sigma(r) = kr^2$, the density is zero at the center and increases quadratically as you move outwards.

Q: Why is the final moment of inertia $\frac{2}{3}MR^2$ and not $\frac{1}{2}MR^2$ like a standard disc?

Because more mass is concentrated at the outer edges (larger $r$), those masses contribute more to the moment of inertia ($r^2$ dependence). This results in a higher overall moment of inertia compared to a uniform disc of the same total mass.

Question

A thin disc of mass $M$ and radius $R$ has mass per unit area $\sigma(r) = kr^2$, where $r$ is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is

Options:

(A) $\frac{MR^2}{2}$

(B) $\frac{MR^2}{6}$

(C) $\frac{MR^2}{3}$

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a thin disc of radius $R$. Unlike a standard uniform disc, this one has a trick up its sleeve: its mass is not distributed evenly!

The problem states that the surface mass density is given by $\sigma(r) = kr^2$. This means the density is zero at the exact center and increases quadratically as you move towards the outer edge.

To tackle this, we cannot use the standard formula directly. Instead, we must build the disc from scratch using calculus. We start by considering a tiny elemental ring of radius $r$ and an infinitesimally small thickness $dr$.

### The Mass of the Disc

Before we can find the moment of inertia, we need to understand the total mass $M$ of the disc.

First, let's find the area of our elemental ring. If you were to cut this ring and lay it flat, it would form a rectangle with a length equal to its circumference, $2\pi r$, and a width equal to its thickness, $dr$. Thus, the area is $dA = 2\pi r dr$.

The mass of this tiny ring, $dm$, is simply its area multiplied by the surface mass density at that radius:
$$dm = \sigma(r) dA = (kr^2)(2\pi r dr) = 2\pi k r^3 dr$$

To find the total mass $M$, we integrate this elemental mass from the center ($r=0$) to the outer edge ($r=R$):
$$M = \int_0^R 2\pi k r^3 dr = 2\pi k \left[ \frac{r^4}{4} \right]_0^R = \frac{\pi k R^4}{2}$$

### Calculating the Moment of Inertia

Now, let's focus on the moment of inertia.

For our elemental ring, all of its mass $dm$ is located at the exact same distance $r$ from the central axis. Therefore, its moment of inertia $dI$ is simply:
$$dI = dm \cdot r^2$$

Substituting our expression for $dm$ into this equation, we get:
$$dI = (2\pi k r^3 dr) \cdot r^2 = 2\pi k r^5 dr$$

Notice the $r^5$ term! This tells us that the outer rings contribute massively to the overall moment of inertia. To find the total moment of inertia $I$, we integrate $dI$ from $0$ to $R$:
$$I = \int_0^R 2\pi k r^5 dr = 2\pi k \left[ \frac{r^6}{6} \right]_0^R = \frac{\pi k R^6}{3}$$

### The Final Connection

We have the moment of inertia $I = \frac{\pi k R^6}{3}$, but the options are given in terms of the total mass $M$. We need to bridge this gap.

From our mass calculation, we know that $M = \frac{\pi k R^4}{2}$. We can rearrange this to isolate $\pi k$:
$$\pi k = \frac{2M}{R^4}$$

Now, substitute this back into our equation for $I$:
$$I = \frac{1}{3} \left( \frac{2M}{R^4} \right) R^6 = \frac{2}{3} MR^2$$

**And there we have it!** The moment of inertia of this non-uniform disc is **$\frac{2}{3} MR^2$**. Because the mass is pushed further out towards the edges compared to a uniform disc (which has $I = \frac{1}{2} MR^2$), it is harder to rotate, resulting in a larger moment of inertia.

Animated Solution for Physics - Rotational Motion: A thin disc of mass $M$ and radius $R$ has mass per unit area $σ (r) = k r^{2}$ , where $r$ is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is

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Visualized Solution

$Elemental Ring$

$Mass of Elemental Ring$

$Substituting Density$

$Total Mass of Disc$

$Moment of Inertia of Ring$

$Substituting Mass$

$Total Moment of Inertia$

$Final Expression$

The Sigma Insight: Moment of Inertia

Analyzing the Setup

The Mass of the Disc

Calculating the Moment of Inertia

The Final Connection

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Visualized Solution

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

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

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

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

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

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

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

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