Parallel axis theorem — I = I_CM + Md² — proof: Visual walkthrough

Step 0 — What does "moment of inertia" even mean?

WHY squared and not just $r_{i}$ ? Because spinning something far out sweeps a bigger circle and moves faster for the same turn-rate — the two effects multiply, giving one factor of $r$ each. That is the same $r^{2}$ you meet again in Rotational kinetic energy. Hold onto it.

Figure — Parallel axis theorem — I = I_CM + Md² — proof

Look at the figure: the axis is the vertical teal line, one grain $m_{i}$ sits off to the side, and the burnt-orange arrow is its perpendicular distance $r_{i}$ — the shortest rope from grain to axis. That arrow, squared and mass-weighted, is one term of the sum.

Step 1 — Two parallel axes: the whole problem in one picture

WHAT we want: $I$ about Axis 2, given we already know $I_{C M}$ (the $I$ about Axis 1).

WHY bother: computing $I$ from scratch for a new axis means redoing the whole sum. The theorem lets us reuse $I_{C M}$ and just pay a correction. That reuse is the entire payoff.

Both teal dots are where the axes pierce the page (the axes stick straight up out of the paper toward you). The plum segment between them is $d$ . Notice one axis sits exactly on the balance point — that condition is doing secret work we'll cash in at Step 5.

Step 2 — Choose clever coordinates (origin at the CM)

Let the second axis pierce the page at the point $(a, b)$ :

$a$ — how far the new axis sits to the right of the CM,
$b$ — how far it sits up.

Since $d$ is the straight-line distance from CM $(0, 0)$ to $(a, b)$ , Pythagoras gives $d^{2} = a^{2} + b^{2}$

$a^{2} + b^{2}$ — horizontal-shift² plus vertical-shift²,
$= d^{2}$ — the perpendicular gap, squared. This is just the hypotenuse rule on the little plum triangle.

The orange grid is centered on the CM (that's why the origin cross sits on it). The new axis lands at $(a, b)$ ; the dotted right triangle shows $a$ , $b$ , and the plum hypotenuse $d$ .

Step 3 — Write the two distances for one grain

WHY subtract $a$ and $b$ ? Distance is always "grain minus the thing you're measuring from." To the origin you subtract $(0, 0)$ (so nothing changes); to the shifted axis you subtract $(a, b)$ . Each bracket is one leg of a right triangle from grain to that axis, and squaring-and-adding is Pythagoras again.

Same grain, two ropes: the orange rope $r_{i, C M}$ goes to the CM-axis, the teal rope $r_{i}$ goes to the shifted axis. The little legs $(x_{i} - a)$ and $(y_{i} - b)$ are drawn dotted — those are the brackets inside the formula.

Step 4 — Plug into the sum and expand

WHAT: put the new-axis distance into $I = \sum_{i} m_{i} r_{i}^{2}$ : $I = \sum_{i} m_{i} [(x_{i} - a)^{2} + (y_{i} - b)^{2}]$

WHY expand? The body is buried inside the brackets. Multiplying out separates "grain-only" pieces from "shift-only" pieces so we can recognise old friends: $I = \sum_{i} m_{i} [grain’s spread x_{i}^{2} + y_{i}^{2} cross terms - 2 a x_{i} - 2 b y_{i} + = d^{2} a^{2} + b^{2}]$

Now split the single sum into four separate sums (allowed — adding is adding, in any order): $I = (A) i \sum m_{i} (x_{i}^{2} + y_{i}^{2}) - 2 a (B) i \sum m_{i} x_{i} - 2 b (C) i \sum m_{i} y_{i} + (a^{2} + b^{2}) (D) i \sum m_{i}$

Term by term:

(A) $\sum m_{i} (x_{i}^{2} + y_{i}^{2})$ — every grain's distance² to the CM-axis, mass-weighted. That is literally $I_{C M}$ .
(B), (C) — the mysterious cross terms; $a$ and $b$ pulled out front because they're the same for every grain.
(D) $\sum m_{i} = M$ — just the total mass; and $(a^{2} + b^{2}) = d^{2}$ .

The figure sorts the four colored buckets: (A) in orange = $I_{C M}$ , (D) in teal = $M d^{2}$ , and (B),(C) in plum = the "?" we must resolve next.

Step 5 — The cross terms vanish (the heart of the proof)

WHAT this means physically: " $\sum m_{i} x_{i} = 0$ " says the mass to the right of the CM perfectly balances the mass to the left (that's what a balance point is). So the cross terms cancel not by luck but by the meaning of "center of mass."

Each grain contributes $m_{i} x_{i}$ : orange bars point right (positive $x$ ), teal bars point left (negative $x$ ). Stack them and they cancel to a flat zero line — the visual death of the cross terms.

Step 6 — Collect the survivors

With (B) and (C) gone, only (A) and (D) remain: $I = (A): real spread about CM I_{C M} + (D): point mass M at distance d M d^{2}$ $I = I_{C M} + M d^{2}$

$I_{C M}$ — the honest inertia of the body about its balance line,
$M d^{2}$ — the inertia the body would have if it shrank to a single dot of mass $M$ sitting $d$ away. (See it in Radius of gyration, where $I = M k^{2}$ makes $k$ an "effective distance.")

WHY it's beautiful: the correction depends only on how far you moved, never on the body's shape. Move any body distance $d$ off its CM and you pay the same $M d^{2}$ .

Step 7 — The edge and degenerate cases (never leave a gap)

The plum curve is $I (d) = I_{C M} + M d^{2}$ — a parabola, bottoming out at the CM ( $d = 0$ ). Two sample axes A and B sit on the curve; their heights subtract, their $d$ 's do not.

The one-picture summary

Read it left to right: start with the raw sum, plant the origin on the CM, expand into four buckets, watch the two plum cross-term buckets collapse to zero, and what survives is $I_{C M} + M d^{2}$ . That's the entire derivation in one frame.

Recall Feynman retelling — the whole walkthrough in plain words

Spinning something is hard in proportion to how far its mass sits from the spin line, counted squared. We asked: if I already know how hard it is to spin about the balance point, how hard is it about some parallel line a distance $d$ away? We laid a grid centred on the balance point and wrote each grain's distance to the new line. When we multiplied it out, four kinds of terms appeared: (1) the grains' spread about the balance point — that's the "easy" number we already knew; (2) two mixed terms; (3) the total mass times the gap squared. The mixed terms are just "how lopsided is the mass around the balance point" — and around a balance point, by definition, the lopsidedness is zero. Left and right cancel exactly. So they vanish. What's left: the easy number, plus mass-times-gap-squared — the effort of swinging the whole thing as one dot at distance $d$ . Home base is cheapest; every step out costs $M d^{2}$ .

Connections

1.5.06 Parallel axis theorem — I = I_CM + Md² — proof (Hinglish)
Moment of inertia — definition
Center of mass — definition and computation
Perpendicular axis theorem
Radius of gyration
Rotational kinetic energy
Physical pendulum