I_edge = I_cm + Md² for a disc about its edge.

Parallel Axis Theorem — I About Edge of Disc

Question

A uniform disc of mass $M$ and radius $R$ rotates about an axis passing through its edge and perpendicular to its plane. Find its moment of inertia.

This is a direct application of the Parallel Axis Theorem — one of the most reliable 4-mark setups in JEE Main.

Solution — Step by Step

The moment of inertia of a disc about its own central axis (perpendicular to the plane) is:

I_{cm} = \frac{MR^2}{2}

This is a standard result you must know cold. No derivation needed in the exam — just cite it.

The new axis passes through the edge of the disc, parallel to the central axis. The distance between these two parallel axes is exactly $d = R$ .

Draw it out: one axis at the center, one at the rim — the gap between them is one radius.

The theorem states:

I = I_{cm} + Md^2

This holds only when the reference axis passes through the centre of mass. Substitute:

I_{edge} = \frac{MR^2}{2} + MR^2

I_{edge} = \frac{MR^2}{2} + \frac{2MR^2}{2} = \frac{3MR^2}{2}

Final answer: $I_{edge} = \dfrac{3MR^2}{2}$

Why This Works

The Parallel Axis Theorem accounts for two contributions to rotational inertia when we shift the axis away from the centre of mass. The first term, $I_{cm}$ , captures the “spinning about own centre” part. The second term, $Md^2$ , adds the extra inertia due to the entire mass $M$ being displaced a distance $d$ from the new axis.

Think of it this way: if you had all the mass concentrated at the centre of mass, it would contribute $Md^2$ to the new axis. The $I_{cm}$ term corrects for the fact that mass is actually spread out, not concentrated at a point.

The theorem only works from a CM axis to a parallel axis — never between two arbitrary parallel axes. This direction matters.

Alternative Method — Integration (Verify the Formula)

If you want to derive $I_{edge}$ from scratch, set up coordinates with the edge axis at the origin. For a point at position $(x, y)$ on the disc, the perpendicular distance from the edge axis is $\sqrt{(x-R)^2 + y^2}$ if we place the centre at $(R, 0)$ .

Integrating $\int r^2 \, dm$ over the disc in polar coordinates (centred at the disc’s own centre) and expanding:

I_{edge} = \int (r^2 - 2Rx\cos\theta + R^2) \, dm

The cross term $\int 2Rx\cos\theta \, dm$ vanishes by symmetry (it’s the first moment about the CM, which is zero by definition). What remains gives exactly $I_{cm} + MR^2$ — confirming the Parallel Axis Theorem algebraically.

This route is 10× longer. Use it only if asked to derive rather than apply.

Common Mistake

The most frequent error: using $I_{cm} = \frac{MR^2}{4}$ instead of $\frac{MR^2}{2}$ .

The value $\frac{MR^2}{4}$ is the moment of inertia about a diameter — an axis lying in the plane of the disc. Here, our axis is perpendicular to the plane. Wrong $I_{cm}$ → wrong final answer, and you lose all marks since the method step is trivial.

When you see “axis through edge, perpendicular to plane” — the central axis result is $\frac{MR^2}{2}$ . Always check the axis orientation before picking your $I_{cm}$ .

For JEE, memorise the “edge of disc” result directly: $I_{edge} = \frac{3MR^2}{2}$ . It appears often enough that recognising it on sight saves 30 seconds — enough to attempt one more question.

Question

Solution — Step by Step

Why This Works

Alternative Method — Integration (Verify the Formula)

Common Mistake

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