Classical-Mechanics · Unit 20 · Video 7 · Interactive Practice

Proving the Parallel Axis Theorem

IKey Formulas

Formula	Name	What you need
$\vec{r}_{S,dm} = \vec{r}_{S,\text{cm}} + \vec{r}_{dm}$	Vector triangle	Three position vectors, no distances yet
$\vec{r}_{S,\perp,dm} = \vec{r}_{S,\perp,\text{cm}} + \vec{r}_{\perp,dm}$ , $\ \left\|\vec{r}_{S,\perp,\text{cm}}\right\| = d_{S,\text{cm}}$	Perpendicular pieces	Both axes parallel, so one split serves both
$I_S = \displaystyle\int_{\text{body}} dm\, d_{S,\text{cm}}^2 + \int_{\text{body}} dm\,(r_{\perp,dm})^2 + 2\,\vec{r}_{S,\perp,\text{cm}}\cdot\!\int_{\text{body}} dm\,\vec{r}_{\perp,dm}$	The three integrals	Expand $\vec{r}_{S,\perp,dm}\cdot\vec{r}_{S,\perp,dm}$
$I_S = I_{\text{cm}} + m\,d_{S,\text{cm}}^2$	Parallel axis theorem	$I_{\text{cm}}$ , the total mass and the axis offset

Key Insight: The body's shape enters the proof in exactly one place — the cross term $2\,\vec{r}_{S,\perp,\text{cm}}\cdot\int dm\,\vec{r}_{\perp,dm}$ — and that integral is $\vec{0}$ because the element vectors $\vec{r}_{dm}$ are measured from the centre of mass. Shift the second axis off the centre of mass and the cross term survives, which is why $m\,d^2$ is the whole correction only for a centre-of-mass axis.

IIThe Vector Triangle and Its Perpendicular Shadow

The three position vectors close a triangle; so do their pieces perpendicular to the two axes.

r⊥,dm 1.40 r∥,dm 1.60

💡 The split is taken against whatever direction the axes point, not against $z$ — which is why the same proof still covers a turning bicycle, whose wheel axes are not parallel, and a precessing top, whose axis sweeps a cone.

IIIWhy the Cross Term Dies

The mass-weighted chain of element vectors closes on itself from one point only.

Body Vectors measured from

💡 Two displaced axes can be related only through the centre of mass: apply the theorem to each and subtract to get $I_{S'} = I_S + m\left(d_{S',\text{cm}}^2 - d_{S,\text{cm}}^2\right)$ , valid only while both axes stay parallel to the centre-of-mass axis.

IVThree Sums, One Body

For this body the three terms are $m\,d_{S,\text{cm}}^2$ , $I_{\text{cm}}$ , and a cross term that is exactly zero.

Step 1

Step 1 — Sum the definition directly

I_S = \sum_i m_i\,(r_{S,\perp,i})^2

= 3(2) + 2(20) + 1(34) + 2(20)

= 6 + 40 + 34 + 40 = 120\ \mathrm{kg\,m^2}

Step 2 — Split every element vector

\vec{r}_{S,\perp,i} = \vec{r}_{S,\perp,\text{cm}} + \vec{r}_{\perp,i}

\vec{r}_{S,\perp,\text{cm}} = (3, 0)\ \mathrm{m}

is the same for every element; only

\vec{r}_{\perp,i}

changes.

Step 3 — Square and expand

(r_{S,\perp,i})^2 = d_{S,\text{cm}}^2 + (r_{\perp,i})^2 + 2\,\vec{r}_{S,\perp,\text{cm}}\cdot\vec{r}_{\perp,i}

I_S = \sum_i m_i d_{S,\text{cm}}^2 + \sum_i m_i (r_{\perp,i})^2 + 2\,\vec{r}_{S,\perp,\text{cm}}\cdot\sum_i m_i \vec{r}_{\perp,i}

Step 4 — Evaluate the three sums

\sum_i m_i d_{S,\text{cm}}^2 = m\,d_{S,\text{cm}}^2 = 8(3)^2 = 72

\sum_i m_i (r_{\perp,i})^2 = I_{\text{cm}} = 15 + 10 + 13 + 10 = 48

2\,\vec{r}_{S,\perp,\text{cm}}\cdot\sum_i m_i \vec{r}_{\perp,i} = 2\,(3,0)\cdot(0,0) = 0

Step 5 — Assemble

I_S = I_{\text{cm}} + m\,d_{S,\text{cm}}^2 = 48 + 72 = 120\ \mathrm{kg\,m^2}\ \checkmark

The same number step 1 reached by summing

m_i (r_{S,\perp,i})^2

element by element.

VQuiz Questions

Problem 1 · Applying the Theorem

Given: a body of mass $m = 6\ \mathrm{kg}$ with $I_{\text{cm}} = 40\ \mathrm{kg\,m^2}$ — find $I_S$ about a parallel axis a distance $d_{S,\text{cm}} = 2\ \mathrm{m}$ from the centre-of-mass axis.

✅ Correct!

40 + (6)(2)^2 = 40 + 24 = 64\ \mathrm{kg\,m^2}

; moving off the centre of mass can only increase

I

❌ The mass is missing. That is

I_{\text{cm}} + d^2

; the first integral is

d^2\int dm = m\,d^2

, so the total mass multiplies the square.

❌ Close — the offset is squared.

m\,d

12

, but the term comes from

d^2\int dm

, not

d\int dm

❌ Wrong sign. The correction

m\,d^2

is a sum of squares and is added;

I_{\text{cm}}

is the smallest moment of inertia among all parallel axes.

❌ Not quite. Use

I_S = I_{\text{cm}} + m\,d_{S,\text{cm}}^2

with

m = 6

and

d = 2

Show solution

The theorem needs only three numbers, since the shape has already been absorbed into $I_{\text{cm}}$ :

I_S = I_{\text{cm}} + m\,d_{S,\text{cm}}^2 = 40 + (6)(2)^2 = 40 + 24 = 64\ \mathrm{kg\,m^2}

The correction $m\,d^2$ came from the integral $\int dm\, d_{S,\text{cm}}^2$ , in which $d_{S,\text{cm}}$ is the same constant for every element and factors out, leaving $\int dm = m$ .

Problem 2 · Two Displaced Axes

Given: the centre-of-mass axis and two parallel axes $A$ and $B$ all lie in one plane, with $A$ and $B$ on the same side of the centre of mass at $d_A = 3\ \mathrm{m}$ and $d_B = 5\ \mathrm{m}$ — so $A$ and $B$ are $2\ \mathrm{m}$ apart. The body has $m = 8\ \mathrm{kg}$ and $I_A = 120\ \mathrm{kg\,m^2}$ — find $I_B$ .

✅ Correct!

I_{\text{cm}} = 120 - 8(3)^2 = 48

, then

I_B = 48 + 8(5)^2 = 248\ \mathrm{kg\,m^2}

— equivalently

I_B = I_A + m\left(d_B^2 - d_A^2\right) = 120 + 8(16)

❌ That applies the theorem between two non-cm axes.

I_A + m(2)^2

would need the cross term for the expansion about

A

to vanish, and it does not: neither

A

nor

B

passes through the centre of mass.

❌ The old correction is still in there.

I_A

already carries

m\,d_A^2

; strip it off before adding

m\,d_B^2

❌ That is $I_{\text{cm}} + m\left(d_B^2 - d_A^2\right)$ . The difference of squares must be added to

I_A

, not to

I_{\text{cm}}

❌ Not quite. Route through the centre-of-mass axis: first recover

I_{\text{cm}}

from

I_A

, then step out to

B

Show solution

Step 1 — Come back to the centre of mass. The theorem applies to $A$ because $A$ 's partner axis is the centre-of-mass axis:

I_{\text{cm}} = I_A - m\,d_A^2 = 120 - 8(3)^2 = 120 - 72 = 48\ \mathrm{kg\,m^2}

Step 2 — Step out to $B$ .

I_B = I_{\text{cm}} + m\,d_B^2 = 48 + 8(5)^2 = 48 + 200 = 248\ \mathrm{kg\,m^2}

Subtracting the two applications gives the general rule from the video,

I_B = I_A + m\left(d_B^2 - d_A^2\right) = 120 + 8(25 - 9) = 248\ \mathrm{kg\,m^2}

Note what this rule is not: $I_A + m\,(2\ \mathrm{m})^2 = 152\ \mathrm{kg\,m^2}$ is wrong. Expanding about $A$ instead of the centre of mass leaves a cross term $2\,\vec{r}_{B,\perp,A}\cdot\int dm\,\vec{r}_{\perp,A,dm}$ that does not vanish, because $\int dm\,\vec{r}$ is zero only when measured from the centre of mass.

Problem 3 · Three Particles, One Axis

Given: three particles in the plane perpendicular to a family of parallel axes: $m_1 = 1\ \mathrm{kg}$ at $(-4, 2)\ \mathrm{m}$ , $m_2 = 2\ \mathrm{kg}$ at $(0, -2)\ \mathrm{m}$ , $m_3 = 1\ \mathrm{kg}$ at $(4, 2)\ \mathrm{m}$ .

Where is the centre of mass?

What is $I_S$ about the parallel axis through $(0, 3)\ \mathrm{m}$ ?

✅ Correct!

I_{\text{cm}} = 48\ \mathrm{kg\,m^2}

and

d = 3\ \mathrm{m}

, so

I_S = 48 + 4(9) = 84\ \mathrm{kg\,m^2}

— the same value the direct sum

17 + 50 + 17

gives.

❌ That is the plain average of the three positions. Each position must be weighted by its own mass: the

2\ \mathrm{kg}

particle counts twice.

❌ Mixed weights. That divides

\sum_i \vec{r}_i

m

; the numerator must be

\sum_i m_i\vec{r}_i

❌ The heavy particle does pull downward, but not that far. Its

2\ \mathrm{kg}

y = -2

is balanced exactly by two

1\ \mathrm{kg}

particles at

y = +2

❌ Check the centre of mass. Compute

\sum_i m_i \vec{r}_i

first, then divide by

m = 4\ \mathrm{kg}

❌ Check $I_S$ . Get

I_{\text{cm}}

from the three particles, then add

m\,d_{S,\text{cm}}^2

with

d

measured between the two axes.

Show solution

Step 1 — Centre of mass. With $m = 1 + 2 + 1 = 4\ \mathrm{kg}$ :

\sum_i m_i\vec{r}_i = 1(-4, 2) + 2(0, -2) + 1(4, 2) = (0, 0) \ \Rightarrow\ \vec{r}_{\text{cm}} = (0,0)\ \mathrm{m}

Step 2 — Moment of inertia about the centre-of-mass axis.

I_{\text{cm}} = 1(16 + 4) + 2(0 + 4) + 1(16 + 4) = 20 + 8 + 20 = 48\ \mathrm{kg\,m^2}

Step 3 — Move the axis. The axis through $(0,3)$ is $d_{S,\text{cm}} = 3\ \mathrm{m}$ from the centre-of-mass axis:

I_S = I_{\text{cm}} + m\,d_{S,\text{cm}}^2 = 48 + 4(3)^2 = 84\ \mathrm{kg\,m^2}

Check directly. The perpendicular vectors from $S$ are $(-4,-1)$ , $(0,-5)$ and $(4,-1)$ :

I_S = 1(17) + 2(25) + 1(17) = 17 + 50 + 17 = 84\ \mathrm{kg\,m^2}\ \checkmark

Problem 4 · The Cross Term That Survives

Given: the same three particles. A student expands about the point $P = (0, 2)\ \mathrm{m}$ instead of the centre of mass, writing $\vec{r}_{S,i} = \vec{r}_{S,P} + \vec{r}_{P,i}$ with $S = (0, 3)\ \mathrm{m}$ , so that $\vec{r}_{S,P} = (0,-1)\ \mathrm{m}$ .

What is $\sum_i m_i \vec{r}_{P,i}$ ?

So the cross term $2\,\vec{r}_{S,P}\cdot\sum_i m_i \vec{r}_{P,i}$ equals

✅ Exactly.

\sum_i m_i\vec{r}_{P,i} = m\left(\vec{r}_{\text{cm}} - \vec{r}_P\right) = (0,-8)

and

2(0,-1)\cdot(0,-8) = 16\ \mathrm{kg\,m^2}

— the term the centre of mass would have killed.

❌ That integral is zero only from the centre of mass.

\sum_i m_i\vec{r}_{P,i} = m\left(\vec{r}_{\text{cm}} - \vec{r}_P\right)

, which vanishes precisely when

P

is the centre of mass.

❌ Sign reversed.

\vec{r}_{P,i}

runs from

P

to the particle, so it is

\vec{r}_i - \vec{r}_P

, and the centre of mass lies below

P

❌ The mass factor is missing.

\vec{r}_{\text{cm}} - \vec{r}_P = (0,-2)\ \mathrm{m}

still has to be multiplied by

m = 4\ \mathrm{kg}

❌ Check the sum. Add the three vectors

m_i\left(\vec{r}_i - \vec{r}_P\right)

, or use

m\left(\vec{r}_{\text{cm}} - \vec{r}_P\right)

directly.

❌ Check the dot product. Both vectors point in

-\hat{y}

, so their dot product is positive:

2\,(0)(0) + 2(-1)(-8)

Show solution

Step 1 — The sum from $P$ . Since $\sum_i m_i\vec{r}_i = m\,\vec{r}_{\text{cm}}$ for any origin,

\sum_i m_i \vec{r}_{P,i} = \sum_i m_i\left(\vec{r}_i - \vec{r}_P\right) = m\left(\vec{r}_{\text{cm}} - \vec{r}_P\right) = 4\left[(0,0) - (0,2)\right] = (0,-8)\ \mathrm{kg\,m}

Step 2 — The cross term. With $\vec{r}_{S,P} = (0,-1)\ \mathrm{m}$ :

2\,\vec{r}_{S,P}\cdot\sum_i m_i\vec{r}_{P,i} = 2\left[(0)(0) + (-1)(-8)\right] = 16\ \mathrm{kg\,m^2}

Step 3 — The expansion still balances, but only with that term. About $P$ , $I_P = 1(16) + 2(16) + 1(16) = 64\ \mathrm{kg\,m^2}$ and $m\left|\vec{r}_{S,P}\right|^2 = 4(1) = 4\ \mathrm{kg\,m^2}$ , so

I_S = I_P + m\left|\vec{r}_{S,P}\right|^2 + 16 = 64 + 4 + 16 = 84\ \mathrm{kg\,m^2}\ \checkmark

which agrees with Problem 3. Dropping the cross term would have given $68\ \mathrm{kg\,m^2}$ : the clean statement $I_S = I_{\text{cm}} + m\,d^2$ is a property of the centre-of-mass axis, not of any pair of parallel axes.

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