Parallel Axes Theorem — Moment of Inertia for JEE

You've memorised that a solid sphere has a moment of inertia of $\frac{2}{5}MR^2$ . Good. But the exam rarely asks about that axis. It asks about a tangent, an edge, a point off to the side. Do you re-derive an integral every time? No. You shift.

That shift is the parallel axes theorem, and it's one of the most useful tools in Class XI rotation. Let me show you how it works.

First, what is moment of inertia?

Mass tells you how hard it is to get a body moving. You feel it in Newton's second law:

$F = ma$

Now ask a different question. How hard is it to get a body spinning? That answer is its moment of inertia, $I$ . It shows up in the rotational twin of the equation above:

$\tau = I\alpha$

Torque on the left, angular acceleration on the right, and $I$ sitting in the middle — the rotational version of mass. To handle any rotation problem you need all three. The parallel axes theorem is how you get $I$ fast.

The theorem in one line

Here's the whole idea. If you know the moment of inertia about an axis through the centre of mass, you can get it about any parallel axis:

$I = I_{cm} + Md^2$

$I_{cm}$ — moment of inertia about the axis through the centre of mass.
$M$ — total mass of the body.
$d$ — the perpendicular distance between the two parallel axes.

Read it as a story: start at the centre of mass, slide the axis out by a distance $d$ , and the inertia grows by exactly $Md^2$ . Move the axis farther, spinning gets harder. That extra term is never negative, so the centre-of-mass axis always gives the smallest moment of inertia. Useful sanity check.

Two rules before you use it

It works for any rigid body — sphere, rod, disc, ring, you name it. But two conditions are non-negotiable:

The axes must be parallel.
One axis must pass through the centre of mass. Not "near" it. Through it.

That second rule trips people up, so look at the trap.

Take a rod. Axis 1 passes through a point a distance $a$ from the centre. Axis 2 passes through a point a distance $b$ from the centre. Both axes are perpendicular to the rod, neither passes through the middle.

Can you write $I_2 = I_1 + M(\text{something})^2$ directly? No. Neither axis is the centre-of-mass axis, so the theorem does not connect them in one step. Try it and you'll get a wrong answer with full confidence — the worst kind.

We'll fix that case in a minute. First, the clean one.

Worked example: sphere about a tangent

Find the moment of inertia of a solid sphere about a tangent line.

A tangent just grazes the surface. Before reaching for anything, ask the one question that matters:

Do I know $I$ about an axis parallel to the tangent that passes through the centre of mass?

Yes. An axis through the centre, parallel to any tangent, is just a diameter:

$I_{cm} = \frac{2}{5}MR^2$

The tangent sits a distance $d = R$ from that diameter (the radius). Perfect setup. Drop the numbers in:

$I = I_{cm} + Md^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$

No integral. One line. That's the whole point of the theorem.

Your turn. A uniform disc of mass $M$ and radius $R$ has $I_{cm} = \frac{1}{2}MR^2$ about its central axis (perpendicular to the disc). What is its moment of inertia about a parallel axis through a point on the rim?

Check: the rim is a distance $d = R$ from the centre, so $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$ .

The trickier case: go through the centre of mass

Now back to that rod with two off-centre axes. You can relate $I_1$ and $I_2$ — you just can't do it in one jump. Use the centre-of-mass axis as a stepping stone.

Write each one separately from the centre:

$I_1 = I_{cm} + Ma^2$ $I_2 = I_{cm} + Mb^2$

Now subtract. The $I_{cm}$ you never bothered to compute cancels clean:

$I_2 - I_1 = M(b^2 - a^2)$

A relation between the two axes, and you never needed the actual value at the centre. The centre of mass was just the bridge between them. As simple as that.

Recap

Moment of inertia $I$ is rotational mass: $\tau = I\alpha$ .
Parallel axes theorem: $I = I_{cm} + Md^2$ , where $d$ is the gap between the parallel axes.
One axis must pass through the centre of mass, and the axes must be parallel.
The centre-of-mass axis always gives the minimum $I$ — every shift only adds.
Two off-centre axes? Bridge them through the centre of mass and subtract.