Moment of inertia of rod, disc, ring, sphere — derivation table

Question

Derive the moment of inertia of (a) a thin uniform rod about its centre, (b) a uniform disc about its axis, (c) a ring about its axis, and (d) a solid sphere about its diameter. Present the results in a comparison table.

(JEE Main 2023 — MOI derivations are a perennial favourite)

Solution — Step by Step

Consider a rod of mass $M$ and length $L$ along the x-axis, centred at the origin. Take a small element $dm$ at distance $x$ from the centre, of length $dx$ .

Linear mass density: $\lambda = M/L$ , so $dm = \lambda\,dx = (M/L)\,dx$ .

I = \int_{-L/2}^{L/2} x^2\,dm = \frac{M}{L}\int_{-L/2}^{L/2} x^2\,dx = \frac{M}{L} \cdot \frac{x^3}{3}\Big|_{-L/2}^{L/2} = \frac{ML^2}{12}

Take a thin ring element at radius $r$ , width $dr$ . Area mass density: $\sigma = M/(\pi R^2)$ .

dm = \sigma \cdot 2\pi r\,dr = \frac{2M}{R^2}\,r\,dr

Every point on this ring is at distance $r$ from the axis:

I = \int_0^R r^2\,dm = \frac{2M}{R^2}\int_0^R r^3\,dr = \frac{2M}{R^2} \cdot \frac{R^4}{4} = \frac{MR^2}{2}

All mass sits at distance $R$ from the axis. No integration needed:

I = MR^2

This is the simplest case and serves as the building block for the disc derivation above.

Slice the sphere into thin discs perpendicular to the axis. A disc at height $y$ from the centre has radius $r = \sqrt{R^2 - y^2}$ and thickness $dy$ .

Volume density: $\rho = 3M/(4\pi R^3)$ .

Mass of disc: $dm = \rho \cdot \pi r^2\,dy = \frac{3M}{4R^3}(R^2 - y^2)\,dy$

MOI of each disc about the axis: $dI = \frac{1}{2}dm \cdot r^2 = \frac{3M}{8R^3}(R^2 - y^2)^2\,dy$

I = \frac{3M}{8R^3}\int_{-R}^{R}(R^2 - y^2)^2\,dy = \frac{3M}{8R^3} \cdot \frac{16R^5}{15} = \frac{2MR^2}{5}

Summary Table

Body	Axis	MOI	$k^2/R^2$
Thin rod (length $L$ )	Centre, perpendicular	$ML^2/12$	—
Thin rod (length $L$ )	End, perpendicular	$ML^2/3$	—
Ring (radius $R$ )	Through centre, perpendicular	$MR^2$	1
Disc (radius $R$ )	Through centre, perpendicular	$MR^2/2$	1/2
Solid sphere (radius $R$ )	Diameter	$2MR^2/5$	2/5
Hollow sphere (radius $R$ )	Diameter	$2MR^2/3$	2/3

Why This Works

The moment of inertia is the rotational analogue of mass — it measures how mass is distributed relative to the rotation axis. The farther the mass from the axis, the larger the MOI. That is why a ring ( $MR^2$ ) has a higher MOI than a disc ( $MR^2/2$ ) of the same mass and radius — the ring’s mass is all at the rim.

The $k^2/R^2$ ratio tells us how “spread out” the mass is. It directly determines rolling speed on inclines: smaller ratio means faster rolling. Solid sphere (2/5) beats disc (1/2) beats ring (1) every time.

For JEE, memorise the MOI values directly — do not derive them in the exam. But know the derivation well enough to reconstruct any one if you blank out. The disc derivation (integrating rings) is the most commonly asked derivation in boards.

Common Mistake

When using the parallel axis theorem ( $I = I_{CM} + Md^2$ ), students sometimes apply it to shift from one arbitrary axis to another. The theorem only works when one of the two axes passes through the centre of mass. To go from one non-CM axis to another, first shift to the CM axis, then shift out to the new axis.

Question

Solution — Step by Step

Summary Table

Why This Works

Common Mistake

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