Mass per unit area of a circular disc of radius a depends on the distance r from its centre as sigma

Q: Why can't we just find the total mass and use $I = \frac{1}{2}Ma^2$?

The formula $I = \frac{1}{2}Ma^2$ is strictly derived for a disc with a *uniform* mass distribution. When density varies with radius, the mass is distributed differently (here, more mass is towards the edge), which fundamentally changes the moment of inertia.

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

Imagine cutting the thin ring and unrolling it. It forms a rectangle with a length equal to the circumference ($2\pi r$) and a width equal to the thickness ($dr$). The area is simply length $\times$ width.

Q: Why is the moment of inertia of the elemental ring $dm \cdot r^2$ and not $\frac{1}{2} dm \cdot r^2$?

The elemental shape we chose is a *thin ring*, not a solid disc. For a thin ring where all the mass $dm$ is at the exact same distance $r$ from the axis, the moment of inertia is exactly $dm \cdot r^2$.

Q: What if the density varied with the angle $\theta$ instead of the radius $r$?

If density varied with angle, our elemental ring approach wouldn't work because the density would change along the ring itself. We would instead have to choose a thin sector (a wedge) of angle $d\theta$ as our element and integrate over the angle from $0$ to $2\pi$.

Q: How do we know the final units are correct for moment of inertia?

Let's check! $A$ has units of $kg/m^2$ and $B$ has units of $kg/m^3$. The term $A/4$ is $kg/m^2$, and $aB/5$ is $m \cdot (kg/m^3) = kg/m^2$. Multiplying this by $a^4$ ($m^4$) gives $kg \cdot m^2$, which is the exact standard unit for moment of inertia.

Question

Mass per unit area of a circular disc of radius $a$ depends on the distance $r$ from its centre as $\sigma(r) = A + Br$. The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is

Options:

(A) $2\pi a^4 \left( \frac{A}{4} + \frac{B}{5} \right)$

(B) $2\pi a^4 \left( \frac{aA}{4} + \frac{B}{5} \right)$

(C) $\pi a^4 \left( \frac{A}{4} + \frac{aB}{5} \right)$

(D) $2\pi a^4 \left( \frac{A}{4} + \frac{aB}{5} \right)$

Sigma Percentile · Accepted Answer

### The Challenge of Variable Density

Imagine you are holding a circular disc. Usually, we assume the mass is spread evenly across the entire surface. But what if it isn't?

In this fascinating problem, the mass per unit area, or surface mass density $\sigma(r)$, is not a constant. It depends on the distance $r$ from the center, given by the equation $\sigma(r) = A + Br$.

Because the density changes as we move outward, we cannot simply use the standard formula $I = \frac{1}{2}Ma^2$. We need to break the disc down into tiny, manageable pieces and use the power of calculus!

### Slicing the Disc into Rings

To tackle this, let's visualize slicing the disc into infinitely many thin, concentric rings.

Consider one such elemental ring at a distance $r$ from the center, with a microscopic thickness $dr$.

If we were to cut this thin ring and unroll it, it would look like a very long, thin rectangle. The length of this rectangle would be the circumference of the ring, $2\pi r$, and its width would be $dr$.

Therefore, the area of this elemental ring is:
$$dA = 2\pi r dr$$

Now, how much mass does this tiny ring hold? We simply multiply its area by the density at that specific distance $r$.

$$dm = \sigma(r) dA = (A + Br) 2\pi r dr$$

### The Moment of Inertia of the Element

Every piece of mass in our elemental ring is at the exact same distance $r$ from the central axis.

For a perfect ring, the moment of inertia is just its mass multiplied by the square of its radius.

So, the tiny moment of inertia $dI$ contributed by our elemental ring is:
$$dI = dm \cdot r^2$$

Let's substitute our expression for $dm$ into this equation:
$$dI = (A + Br) 2\pi r dr \cdot r^2$$

Expanding this out, we get a beautiful polynomial expression:
$$dI = 2\pi (A r^3 + B r^4) dr$$

### The Grand Summation

We have the moment of inertia for just one microscopic ring. To find the total moment of inertia $I$ for the entire disc, we must add up the contributions from all these rings, starting from the center ($r = 0$) all the way to the outer edge ($r = a$).

This continuous addition is exactly what integration is for!

$$I = \int_0^a dI = \int_0^a 2\pi (A r^3 + B r^4) dr$$

Since $2\pi$ is a constant, we can pull it outside the integral. Now, we integrate the polynomial terms using the power rule:

$$I = 2\pi \left[ A \frac{r^4}{4} + B \frac{r^5}{5} \right]_0^a$$

Finally, we apply the limits of integration. Substituting the upper limit $a$ and subtracting the lower limit $0$, we get:

$$I = 2\pi \left( A \frac{a^4}{4} + B \frac{a^5}{5} \right)$$

To make it match our options, let's factor out $a^4$:

$$I = 2\pi a^4 \left( \frac{A}{4} + \frac{aB}{5} \right)$$

And there we have it! By breaking a complex, non-uniform object into simple rings, we've successfully calculated its total moment of inertia.

Animated Solution for Physics - Rotational Motion: Mass per unit area of a circular disc of radius $a$ depends on the distance $r$ from its centre as $σ (r) = A + B r$ . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is

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Visualized Solution

$Elemental Ring Setup$

$Mass of the Element$

$Moment of Inertia of Element$

$Substitution$

$Setting up the Integral$

$Integration$

$Final Answer$

The Sigma Insight: Moment of Inertia

The Challenge of Variable Density

Slicing the Disc into Rings

The Moment of Inertia of the Element

The Grand Summation

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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 Setup

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

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

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

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

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

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