I = (2/5)MR² derived using integration over thin shells.

Moment of Inertia of a Solid Sphere — Derivation

Question

Derive the moment of inertia of a solid sphere of mass $M$ and radius $R$ about an axis passing through its centre.

Show that $I = \dfrac{2}{5}MR^2$ .

Solution — Step by Step

We slice the sphere into thin spherical shells, each of radius $r$ and thickness $dr$ , where $r$ goes from $0$ to $R$ . Every point on a shell at radius $r$ is at the same distance from the centre — but NOT the same distance from the rotation axis.

This is the key difficulty here. A spherical shell is not a cylindrical shell.

The volume of a thin shell at radius $r$ with thickness $dr$ is:

dV = 4\pi r^2 \, dr

The uniform density of the sphere is:

\rho = \frac{M}{\frac{4}{3}\pi R^3} = \frac{3M}{4\pi R^3}

So the mass of each shell is:

dm = \rho \cdot dV = \frac{3M}{4\pi R^3} \cdot 4\pi r^2 \, dr = \frac{3M r^2}{R^3} \, dr

Here’s the non-obvious part. The moment of inertia of a thin spherical shell of mass $m$ and radius $r$ about a diameter is a standard result:

I_{\text{shell}} = \frac{2}{3}mr^2

We use this result directly. So the contribution from our thin shell $dm$ is:

dI = \frac{2}{3} r^2 \, dm = \frac{2}{3} r^2 \cdot \frac{3M r^2}{R^3} \, dr = \frac{2M r^4}{R^3} \, dr

I = \int_0^R \frac{2M r^4}{R^3} \, dr = \frac{2M}{R^3} \cdot \left[\frac{r^5}{5}\right]_0^R

I = \frac{2M}{R^3} \cdot \frac{R^5}{5} = \frac{2MR^2}{5}

\boxed{I = \frac{2}{5}MR^2}

Why This Works

We used the shell decomposition method — breaking a 3D object into simpler pieces whose moment of inertia we already know. The power of this approach is that each shell has uniform distance from the centre, so the density integral collapses cleanly.

The factor of $\dfrac{2}{3}$ for a spherical shell comes from averaging $r^2 \sin^2\theta$ over the surface of a sphere (the perpendicular distance from the axis). This average works out to $\dfrac{2}{3}r^2$ , which is why that factor appears before we even touch the solid sphere integral.

Notice the final answer $\dfrac{2}{5}MR^2$ — the $\dfrac{2}{5}$ (less than $\dfrac{1}{2}$ , which is a solid cylinder) makes physical sense because more mass is concentrated near the centre of a sphere, closer to the axis.

Alternative Method — Using Disk Integration

Instead of shells, we slice along the axis of rotation. Take the rotation axis as the $x$ -axis. A disk at position $x$ has radius $y = \sqrt{R^2 - x^2}$ .

The moment of inertia of a thin disk of mass $dm$ and radius $y$ about its axis is $\dfrac{1}{2}y^2 \, dm$ .

dm = \rho \cdot \pi y^2 \, dx = \frac{3M}{4R^3}(R^2 - x^2) \, dx

I = \int_{-R}^{R} \frac{1}{2}(R^2 - x^2) \cdot \frac{3M}{4R^3}(R^2 - x^2) \, dx = \frac{3M}{8R^3}\int_{-R}^{R}(R^2 - x^2)^2 \, dx

Expanding and integrating gives the same $\dfrac{2}{5}MR^2$ . The disk method is more computation, but good to know for JEE Advanced where they test whether you’re locked into one method.

Common Mistake

Treating a solid sphere like a cylindrical shell. Many students write $dI = r^2 \, dm$ directly, as if every mass element in the shell sits at distance $r$ from the axis. That’s only true for a cylindrical shell. For a spherical shell, different points are at different perpendicular distances from the axis — ranging from $0$ (at the poles) to $r$ (at the equator). The correct perpendicular distance is $r\sin\theta$ , and averaging $\sin^2\theta$ over the sphere gives the $\dfrac{2}{3}$ factor. Skipping this step gives $I = \dfrac{3}{5}MR^2$ , a wrong answer that appears convincingly close.

For quick recall in PYQs: the moment of inertia formula for common bodies follows a pattern. Thin rod about centre: $\dfrac{1}{12}ML^2$ . Disk about axis: $\dfrac{1}{2}MR^2$ . Solid sphere: $\dfrac{2}{5}MR^2$ . Hollow sphere: $\dfrac{2}{3}MR^2$ . Notice hollow > solid — mass is farther from axis in the hollow case.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Disk Integration

Common Mistake

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