Parallel axis theorem — I = I_CM + Md² — proof

Derivation from first principles

HOW — choose smart coordinates. Let both axes be along $\hat z$ . Put the origin at the CM, so the CM-axis is the $z$ -axis through origin. Let the second axis pierce the $xy$ -plane at the point $(a, b)$ , where $d^2 = a^2 + b^2$ .

For a mass element at position $(x_i, y_i)$ (we ignore $z_i$ — distance to a $z$ -axis only uses $x,y$ ):

Distance² to CM-axis: $r_{i,CM}^2 = x_i^2 + y_i^2$
Distance² to the other axis: $r_i^2 = (x_i - a)^2 + (y_i - b)^2$

Now compute $I$ about the second axis: $I = \sum_i m_i\big[(x_i-a)^2 + (y_i-b)^2\big]$

Why this step? We just plugged the new axis location into the definition. Expand the squares: $I = \sum_i m_i\big[x_i^2 + y_i^2 - 2a x_i - 2b y_i + a^2 + b^2\big]$

Group the terms cleverly: $I = \underbrace{\sum_i m_i(x_i^2+y_i^2)}_{=\,I_{CM}} \;-\; 2a\underbrace{\sum_i m_i x_i}_{(\star)} \;-\; 2b\underbrace{\sum_i m_i y_i}_{(\star\star)} \;+\; (a^2+b^2)\underbrace{\sum_i m_i}_{=\,M}$

Why this step? The first sum is literally $I_{CM}$ . The last sum is total mass $M$ , and $a^2+b^2=d^2$ . The middle two are the secret.

Therefore: $\boxed{I = I_{CM} + Md^2}$

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

Common mistake Steel-manned mistakes

Mistake A: "I can connect any two parallel axes with $I_2 = I_1 + Md^2$ using their separation." Why it feels right: It looks symmetric and tidy. The fix: The $-2a\sum m_i x_i$ terms only vanish when the origin is the CM. So exactly one axis must pass through the CM. For two arbitrary axes, go through the CM (Example 3).

Mistake B: "Use $d$ = distance between the axes' starting points anywhere." Why it feels right: Any "distance" seems fine. The fix: $d$ is the perpendicular distance between the parallel axes, equivalently from the CM to the new axis. Measured along the line ⟂ to both.

Mistake C: "Parallel axis always increases $I$ , so I can also use it to decrease." You can — but only by moving toward the CM, and $I$ is minimum at the CM. The fix: Because $Md^2 \ge 0$ , the CM-axis gives the smallest possible $I$ among all parallel axes. There's no axis with smaller inertia parallel to it.

Recall Feynman: explain to a 12-year-old

Imagine spinning a hammer. Spinning it about the spot where it balances (its centre point) is easiest — least effort. Move your hand away from that balance point and spinning gets harder. How much harder? Pretend the whole hammer shrank into a tiny dot at the balance point, and you swing that dot in a circle of radius $d$ — that extra effort is exactly $Md^2$ . Add it to the "easy" balance-point effort and you get the effort for the new spot. That's it!

Flashcards

State the parallel axis theorem.

I = I_{CM} + Md^2

, with

d

= ⟂ distance between the parallel axes, one of which passes through the CM.

What single condition makes the theorem valid?

One axis must pass through the center of mass.

Why do the cross terms

-2a\sum m_i x_i

vanish in the proof?

Because the origin is the CM, so

\sum m_i x_i = \sum m_i y_i = 0

by definition of CM.

What is

d

in the theorem?

The perpendicular distance between the two parallel axes (= CM to the new axis).

Among all axes parallel to a given direction, which gives minimum I?

The one through the center of mass (since

Md^2 \ge 0

I

of a rod (mass M, length L) about its end?

\frac{1}{3}ML^2

, from

\frac{1}{12}ML^2 + M(L/2)^2

I

of a disc (mass M, radius R) about a perpendicular axis at its rim?

\frac{3}{2}MR^2

, from

\frac12 MR^2 + MR^2

How do you relate two non-CM parallel axes?

Compute each from the CM separately and subtract; never apply the theorem directly between them.

Connections

Moment of inertia — definition
Center of mass — definition and computation
Perpendicular axis theorem
Radius of gyration
Rotational kinetic energy
Physical pendulum

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, moment of inertia ye batata hai ki kisi cheez ko ghumana kitna mushkil hai — aur ye depend karta hai ki tum kaunse axis ke around ghuma rahe ho. Har axis ke liye alag se integration karna bahut tedious hai. Parallel axis theorem bolta hai: agar tumhe center of mass (CM) ke through wale axis ka $I_{CM}$ pata hai, toh kisi bhi parallel axis ka $I$ nikalna easy — bas $Md^2$ add kar do, jahan $d$ dono axes ke beech perpendicular distance hai.

Proof ka asli jaadu ye hai: jab tum origin ko CM par rakhte ho, toh $\sum m_i x_i$ aur $\sum m_i y_i$ dono zero ho jaate hain (kyunki CM ki definition hi yahi hai). Isi wajah se cross terms gayab ho jaate hain aur formula itna saaf-suthra aata hai: $I = I_{CM} + Md^2$ . Yaad rakho — ek axis ko zaroor CM se guzarna chahiye, warna ye formula kaam nahi karega.

Physical feel: $Md^2$ ka matlab hai poore mass ko ek chhote point ki tarah CM par rakh do aur use $d$ radius par ghumao. Wo extra effort hi $Md^2$ hai. Isliye CM-axis hamesha sabse sasta (minimum $I$ ) hota hai, kyunki $Md^2$ kabhi negative nahi ho sakta. Do alag-alag non-CM axes ke beech directly mat lagao — dono ko CM se nikaalo phir subtract karo, yahi sahi tareeka hai.

Parallel axis theorem — I = I_CM + Md² — proof

Derivation from first principles

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Rotational Mechanics

Connections