Parallel Axis Theorem

In short

The moment of inertia of a rigid body about any axis equals its moment of inertia about a parallel axis through the centre of mass plus Md^2, where M is the total mass and d is the perpendicular distance between the two axes: I = I_{\text{cm}} + Md^2. The centre-of-mass axis always gives the smallest moment of inertia among all parallel axes.

Pick up a cricket bat and try to spin it. Grip it near the centre of the blade and twist — the bat rotates without much effort. Now hold the very end of the handle and try to twirl the whole bat around that point. The same bat suddenly feels much heavier, much harder to set spinning. Nothing about the bat changed — same mass, same shape, same wood. What changed is where you placed the axis of rotation.

This is not a vague feeling. There is an exact formula that tells you how much harder the rotation becomes when you shift the axis away from the centre of mass. It is called the parallel axis theorem, and it says: every centimetre you move the axis away from the centre of mass costs you an extra Md^2 in moment of inertia. The cost grows as the square of the distance — shift the axis twice as far, and the penalty quadruples.

Why a shifted axis changes the moment of inertia

The moment of inertia I measures how hard it is to start or stop a body's rotation about a given axis. It depends on two things: how much mass the body has, and how far that mass sits from the axis. A mass element far from the axis contributes much more to I than one sitting close to it — because I = \int r^2 \, dm, and the r^2 factor makes distant mass count disproportionately.

When the axis passes through the centre of mass, the mass is distributed as "close on average" to the axis as it can be. Some mass elements are to the left of the axis, some to the right, some above, some below — but their average position is the axis, because that is what "centre of mass" means.

Now shift the axis to one side. Every mass element on the far side of the body is now farther from the axis than before. Some elements on the near side are closer, true — but the r^2 weighting means the far-side elements pay a much bigger penalty than the near-side elements gain. The net effect is always an increase in I. You can never reduce the moment of inertia by moving the axis away from the centre of mass.

Seeing it with a dumbbell

Take two identical cricket balls — each of mass m = 0.16 kg — and tape them to the ends of a light stick of length 2a = 0.40 m, one ball at each end. The centre of mass is at the midpoint.

Axis through the centre (CM axis): each ball is at distance a = 0.20 m.

I_{\text{cm}} = m \cdot a^2 + m \cdot a^2 = 2ma^2 = 2 \times 0.16 \times 0.04 = 0.0128 \text{ kg·m}^2

Axis through one end: the near ball sits on the axis (distance 0), the far ball is at distance 2a = 0.40 m.

I_{\text{end}} = m \cdot 0^2 + m \cdot (2a)^2 = 4ma^2 = 4 \times 0.16 \times 0.04 = 0.0256 \text{ kg·m}^2

The end-axis moment of inertia is exactly double the centre-axis value. The far ball, now at twice its original distance, contributes four times as much as before. The near ball, now at distance zero, contributes nothing. The net effect: I doubled.

Check this against I_{\text{cm}} + Md^2: total mass M = 2m = 0.32 kg, displacement d = a = 0.20 m.

I_{\text{cm}} + Md^2 = 0.0128 + 0.32 \times 0.04 = 0.0128 + 0.0128 = 0.0256 \text{ kg·m}^2 \quad \checkmark

The theorem gives the right answer. The Md^2 term captures exactly how much extra inertia the shift creates — it is the moment of inertia that the total mass M, if concentrated entirely at the centre of mass, would have about the new axis.

The parallel axis theorem — statement and proof

Parallel Axis Theorem

For a rigid body of total mass M with moment of inertia I_{\text{cm}} about an axis through its centre of mass, the moment of inertia about any parallel axis at perpendicular distance d from the centre-of-mass axis is:

I = I_{\text{cm}} + Md^2

Reading the formula. I_{\text{cm}} is the moment of inertia you would look up in a table — the one about the "natural" axis through the centre of mass. Md^2 is the penalty for shifting the axis: the entire mass M treated as if it were concentrated at the centre of mass, sitting at distance d from the new axis. The two terms add because the shift always makes rotation harder, never easier.

The proof

The proof is short — one integral, one algebraic expansion, and the definition of the centre of mass doing the heavy lifting.

Take the cross-section of the body perpendicular to both axes. Place the origin at the centre of mass O, with the x-axis pointing from O toward the new axis point P, which sits at distance d from O. Both rotation axes are perpendicular to this cross-section — they point into the page.

Cross-section of a rigid body. $O$ marks the centre of mass, $P$ marks the new axis at distance $d$. Both axes point into the page. The mass element $dm$ sits at position $(x, y)$ from $O$.

A mass element dm at position (x, y) relative to O has:

Distance from the CM axis (through O): \sqrt{x^2 + y^2}
Distance from the new axis (through P): \sqrt{(x - d)^2 + y^2}

Now write the moment of inertia about the new axis and expand it step by step.

Step 1. Write the definition of I_P.

I_P = \int \left[(x - d)^2 + y^2\right] dm

Why: moment of inertia is \int r_\perp^2 \, dm. The perpendicular distance from the axis through P to the element at (x, y) is \sqrt{(x-d)^2 + y^2}, so r_\perp^2 = (x-d)^2 + y^2.

Step 2. Expand the square (x - d)^2 = x^2 - 2xd + d^2.

I_P = \int \left[x^2 - 2xd + d^2 + y^2\right] dm

Why: expanding the bracket lets you separate the integral into three recognisable terms.

Step 3. Split into three separate integrals.

I_P = \int (x^2 + y^2) \, dm \;-\; 2d \int x \, dm \;+\; d^2 \int dm

Why: the displacement d is a constant (it does not depend on which mass element you are looking at), so it comes out of the integral. Addition distributes over integration, giving three independent terms.

Step 4. Identify each integral.

\displaystyle\int (x^2 + y^2) \, dm = I_{\text{cm}} — the moment of inertia about the CM axis, by definition.
\displaystyle\int x \, dm = M\bar{x} — this is M times the x-coordinate of the centre of mass. But the origin is at the centre of mass, so \bar{x} = 0. This entire term vanishes.
\displaystyle\int dm = M — the total mass.

Why: the middle term is the key to the entire proof. It vanishes because the first axis passes through the centre of mass. If you tried to apply this derivation between two axes that both miss the CM, the cross-term would not be zero and the neat I_{\text{cm}} + Md^2 formula would not hold. This is why one of the two axes must always pass through the CM.

Step 5. Substitute the three results.

I_P = I_{\text{cm}} + 0 + Md^2

\boxed{I = I_{\text{cm}} + Md^2}

Why: the cross-term killed itself, leaving the cleanest possible result. The moment of inertia about any parallel axis is the CM value plus a pure penalty term Md^2 for shifting the axis. No complicated geometry, no shape-dependent correction — just mass times distance squared.

That is the entire proof. Five lines of algebra, one physical fact (the definition of centre of mass), and a theorem you will use hundreds of times.

Explore the theorem

The formula I = I_{\text{cm}} + Md^2 is a parabola in d: the moment of inertia increases quadratically as you move the axis away from the centre of mass. The interactive figure below plots this for a uniform rod of mass 1 kg and length 1 m, where I_{\text{cm}} = \frac{ML^2}{12} \approx 0.083 kg·m².

Drag the red point to shift the rotation axis along a 1 kg, 1 m uniform rod. At $d = 0$ (centre of mass), $I$ is at its minimum: $\frac{1}{12}$ kg·m². At $d = \pm 0.5$ m (either end), $I$ reaches $\frac{1}{3}$ kg·m² — four times larger. The parabolic shape shows the $d^2$ penalty at work.

Notice the parabola's shape: even a small shift away from the centre of mass produces a noticeable jump in I, and the penalty accelerates as d grows. At the ends of the rod (d = \pm 0.5 m), I is four times the centre value. The symmetry of the parabola confirms that it does not matter which direction you shift the axis — the moment of inertia depends only on |d|, not on where along the body the axis ends up.

This is why a cricket bat feels dramatically different depending on where you grip it. Gripping near the centre of mass gives the lowest moment of inertia — the bat is easiest to swing quickly. Gripping at the end of the handle maximises d and makes the bat feel heavy. Top-order batsmen who need quick wrist work grip the bat high up the handle; lower-order sloggers who want power can afford a longer lever and a bigger I.

Worked examples

Example 1: A cricket stump about its end

A cricket stump is approximately a uniform rod of length L = 0.71 m and mass M = 0.20 kg. From the standard results, a uniform rod about an axis through its centre perpendicular to its length has I_{\text{cm}} = \frac{ML^2}{12}. Find the moment of inertia about a parallel axis through one end of the stump.

The cricket stump modelled as a uniform rod. The CM axis (red, dashed) passes through the centre; the end axis (dark, dashed) is at distance $d = L/2$ from it.

Step 1. Identify the knowns.

M = 0.20 kg, L = 0.71 m, I_{\text{cm}} = \frac{ML^2}{12}, d = \frac{L}{2} = 0.355 m.

Why: the end of the stump is exactly L/2 from the centre — that is the perpendicular distance between the two parallel axes.

Step 2. Compute I_{\text{cm}}.

I_{\text{cm}} = \frac{ML^2}{12} = \frac{0.20 \times 0.71^2}{12} = \frac{0.20 \times 0.504}{12} = \frac{0.1008}{12} = 0.0084 \text{ kg·m}^2

Why: this is the tabulated result for a uniform rod about its centre. You derive it by integrating \int_{-L/2}^{L/2} x^2 \,\frac{M}{L}\, dx — the details are in Moment of Inertia of Standard Bodies.

Step 3. Apply the parallel axis theorem.

I_{\text{end}} = I_{\text{cm}} + Md^2 = 0.0084 + 0.20 \times 0.355^2

= 0.0084 + 0.20 \times 0.126 = 0.0084 + 0.0252 = 0.0336 \text{ kg·m}^2

Why: the Md^2 penalty (0.0252 kg·m²) is three times larger than I_{\text{cm}} itself (0.0084 kg·m²). Shifting the axis from the centre to the end tripled the penalty portion of the moment of inertia.

Step 4. Verify against the known direct formula.

I_{\text{end}} = \frac{ML^2}{3} = \frac{0.20 \times 0.504}{3} = \frac{0.1008}{3} = 0.0336 \text{ kg·m}^2 \quad \checkmark

Why: ML^2/3 is the independently derived result for a rod about one end. Getting the same answer from the parallel axis theorem confirms the theorem — and shows you do not need to memorise ML^2/3 separately. If you know ML^2/12 and the theorem, one line gives you the end-axis value.

Result: I_{\text{end}} = 0.0336 kg·m² = \frac{ML^2}{3}, exactly four times I_{\text{cm}}.

What this shows: The parallel axis theorem converts I_{\text{cm}} = ML^2/12 into I_{\text{end}} = ML^2/3 in a single step. The factor of 4 comes from the d^2 scaling: shifting the axis by half the rod's length quadrupled the moment of inertia. The Md^2 term alone contributed three-quarters of the total.

Example 2: A metal disc about its rim

A flat circular steel plate — like a weight plate in a gym — has mass M = 2.0 kg and radius R = 0.15 m. For a uniform disc, I_{\text{cm}} = \frac{MR^2}{2} about the central axis perpendicular to the disc. The plate is mounted on a nail at its rim, so it can swing about an axis through the rim point, parallel to the original central axis. Find the moment of inertia about this rim axis.

The disc with its centre (CM axis, red) and a point on the rim (rim axis, dark). The distance between the two parallel axes is $d = R$.

Step 1. Identify the knowns.

M = 2.0 kg, R = 0.15 m, I_{\text{cm}} = \frac{MR^2}{2}, d = R = 0.15 m.

Why: the rim is exactly one radius from the centre — the farthest any axis can be from the CM while still touching the disc.

Step 2. Compute I_{\text{cm}}.

I_{\text{cm}} = \frac{MR^2}{2} = \frac{2.0 \times 0.15^2}{2} = \frac{2.0 \times 0.0225}{2} = 0.0225 \text{ kg·m}^2

Why: the standard result for a uniform disc about its central axis, from Moment of Inertia of Standard Bodies.

Step 3. Apply the parallel axis theorem.

I_{\text{rim}} = I_{\text{cm}} + Md^2 = 0.0225 + 2.0 \times 0.15^2 = 0.0225 + 0.045 = 0.0675 \text{ kg·m}^2

Why: with d = R, the penalty Md^2 = MR^2 is twice the original I_{\text{cm}} = MR^2/2. The shift from centre to rim tripled the total moment of inertia.

Step 4. Express in closed form.

I_{\text{rim}} = \frac{MR^2}{2} + MR^2 = \frac{3MR^2}{2}

Why: this compact form is worth remembering for quick calculations. For any uniform disc, the rim axis always has exactly \frac{3}{2} times the moment of inertia of the central axis.

Result: I_{\text{rim}} = 0.0675 kg·m² = \frac{3MR^2}{2}, exactly three times I_{\text{cm}}.

What this shows: Shifting the axis from the centre to the rim tripled the moment of inertia. The penalty Md^2 = 0.045 kg·m² was twice the original I_{\text{cm}} = 0.0225 kg·m². If this plate were used as a pendulum, swinging about the rim nail would be noticeably slower than swinging about the centre, because the larger I means a larger time period.

The minimum moment of inertia is always about the centre of mass

The parallel axis theorem contains a quiet but powerful consequence: among all axes in a given direction, the one through the centre of mass gives the smallest moment of inertia.

The proof is immediate. Since I = I_{\text{cm}} + Md^2 and both M > 0 and d^2 \geq 0:

I \geq I_{\text{cm}}

Equality holds only when d = 0 — when the axis passes through the centre of mass.

This is not just a mathematical curiosity. It has a direct physical consequence: if a body is free to choose its own axis of rotation (no external constraints forcing a particular axis), it rotates about its centre of mass. A cricket bat thrown tumbling through the air after a batsman loses grip — it rotates about its centre of mass, never about its end. A spinning top on ice, free of friction, naturally rotates about its CM axis. The minimum-I axis requires the least "effort" to maintain a given angular velocity, and nature always takes the path of least resistance.

This minimum property also gives you a quick sanity check: if you compute I about some axis and get a value smaller than the tabulated I_{\text{cm}}, you have made an error. The centre-of-mass value is the floor — no parallel axis can go below it.

Common confusions

"The theorem works between any two parallel axes." Not directly. One of the two axes must pass through the centre of mass. If you know I about an axis at distance d_1 from the CM and want I about another parallel axis at distance d_2 from the CM, you must first go back to the CM: I_{\text{cm}} = I_1 - Md_1^2, then go out to the new axis: I_2 = I_{\text{cm}} + Md_2^2. The going-deeper section below shows the combined formula.
"d is any distance between the axes." Be precise: d is the perpendicular distance between the two parallel axes — the shortest distance between two parallel lines in space. In most textbook problems this is straightforward, but in three-dimensional problems with tilted bodies, measure d carefully.
"The theorem applies to any pair of axes." No — the two axes must be parallel. If they point in different directions, their moments of inertia are not related by this theorem. For axes that are perpendicular to each other in a flat body, you need the perpendicular axis theorem instead.
"I_{\text{cm}} means the moment of inertia about any axis through the CM." Not quite. A body has different moments of inertia about different axes through the CM. A rod has ML^2/12 about an axis perpendicular to its length through the centre, but zero about the axis along its length. In the parallel axis theorem, I_{\text{cm}} refers to the moment of inertia about the specific CM axis that is parallel to the axis you are interested in.
"The Md^2 term is the total moment of inertia about the new axis." No — Md^2 is the moment of inertia that a point mass M located at the centre of mass would have about the new axis. The actual moment of inertia also includes I_{\text{cm}}, which accounts for how the mass is spread around the centre. A thin ring and a solid disc of the same mass and radius have different I_{\text{cm}} values, so shifting their axes by the same d gives different results even though Md^2 is identical.

If you are comfortable with the parallel axis theorem and can apply it to rods, discs, and rings, you can stop here — you have the tool and know how to use it. What follows is for readers preparing for JEE Advanced or anyone who wants to see where the theorem leads.

Jumping between two non-CM parallel axes

Suppose you know the moment of inertia I_A about an axis A at distance d_A from the centre of mass, and you want I_B about a parallel axis B at distance d_B from the CM. Apply the theorem twice — once to go back to the CM, once to go out to B:

I_A = I_{\text{cm}} + Md_A^2 \quad \Longrightarrow \quad I_{\text{cm}} = I_A - Md_A^2

I_B = I_{\text{cm}} + Md_B^2 = I_A - Md_A^2 + Md_B^2

\boxed{I_B = I_A + M(d_B^2 - d_A^2)}

The route always passes through the centre of mass as an intermediate step, even though I_{\text{cm}} cancels from the final expression. This two-step technique is standard in JEE problems: given I about one edge of a body, find I about the opposite edge by computing d_A and d_B from the CM.

Radius of gyration form

The moment of inertia is sometimes written as I = Mk^2, where k is the radius of gyration — the distance at which you would need to concentrate all the mass in a single point to get the same I. In terms of k, the parallel axis theorem becomes:

Mk^2 = Mk_{\text{cm}}^2 + Md^2

\boxed{k^2 = k_{\text{cm}}^2 + d^2}

The squared radii of gyration add like the two shorter sides of a right triangle. The radius of gyration about any displaced axis is always larger than k_{\text{cm}} — a restatement of the minimum-I property in the language of k.

The compound pendulum

A compound pendulum (or physical pendulum) is any rigid body swinging under gravity about a fixed horizontal axis that does not pass through the centre of mass. Unlike a simple pendulum, the mass is distributed throughout the body, not concentrated at a single point.

If the pivot is at distance l from the centre of mass, the moment of inertia about the pivot is I = I_{\text{cm}} + Ml^2 = M(k_{\text{cm}}^2 + l^2). For small oscillations, the restoring torque is \tau = -Mgl\sin\theta \approx -Mgl\theta, and Newton's second law for rotation gives:

M(k_{\text{cm}}^2 + l^2)\,\ddot{\theta} = -Mgl\,\theta

This is simple harmonic motion with time period:

T = 2\pi\sqrt{\frac{k_{\text{cm}}^2 + l^2}{gl}}

Two limiting cases reveal the physics. When l is very large (pivot far from the CM), T \approx 2\pi\sqrt{l/g} — the familiar simple-pendulum formula, because the mass distribution barely matters when the body is far from the pivot. When l is very small (pivot near the CM), T \to \infty — the pendulum barely oscillates because the gravitational torque arm l is tiny.

Between these extremes, there is a distance l that minimises T. Minimising (k_{\text{cm}}^2 + l^2)/l = k_{\text{cm}}^2/l + l gives l = k_{\text{cm}} — the pendulum oscillates fastest when the pivot is at a distance equal to the radius of gyration from the centre of mass. The minimum time period is:

T_{\min} = 2\pi\sqrt{\frac{2k_{\text{cm}}}{g}}

This result appears regularly in JEE Advanced problems. For a uniform rod (k_{\text{cm}} = L/\sqrt{12}), the optimal pivot is at L/(2\sqrt{3}) \approx 0.29\,L from the centre — not at the end (where T is larger because I is too big), and not at the centre (where T blows up because the torque arm vanishes).

A disc with a hole — the subtraction trick

A classic JEE problem: a uniform disc of mass M and radius R has a circular hole of radius R/2 punched out. The hole is centred at a distance R/2 from the disc's centre. Find I of the remaining body about the original disc's central axis.

The trick is subtraction: the remaining body = full disc − cut-out disc. Moments of inertia about the same axis are additive, so I_{\text{remaining}} = I_{\text{full}} - I_{\text{cutout, about original centre}}.

The cut-out disc has area \pi(R/2)^2 = \pi R^2/4, so its mass is M_c = M/4. Its radius is R/2, so its moment of inertia about its own centre is:

I_c = \frac{M_c (R/2)^2}{2} = \frac{(M/4)(R^2/4)}{2} = \frac{MR^2}{32}

Now shift this to the original disc's centre using the parallel axis theorem. The centre of the cut-out is at distance R/2 from the original centre:

I_{c,\text{shifted}} = \frac{MR^2}{32} + \frac{M}{4}\left(\frac{R}{2}\right)^2 = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{3MR^2}{32}

The full disc has I_{\text{full}} = MR^2/2 = 16MR^2/32. Therefore:

I_{\text{remaining}} = \frac{16MR^2}{32} - \frac{3MR^2}{32} = \frac{13MR^2}{32}

Without the parallel axis theorem, you could not shift the cut-out disc's I to the required axis, and this problem would require a difficult direct integration. The theorem turns a hard geometry problem into clean arithmetic.

Where this leads next

Perpendicular Axis Theorem — for flat bodies, the moment of inertia about an axis perpendicular to the plane equals the sum of moments about two in-plane axes.
Torque — the rotational analogue of force, and the reason moment of inertia matters for dynamics.
Newton's Second Law for Rotation — \tau = I\alpha, where I is the very quantity the parallel axis theorem computes.
Angular Momentum — L = I\omega, connecting moment of inertia to the conserved quantity in rotation.
Conservation of Angular Momentum — what happens when a spinning body changes its moment of inertia by redistributing its mass.